ALCANI
Rezolvare: \(C_{n}H_{2n + 2}\) este formula generală a unui alcan și are \(M = 12n + 2n + 2 = 14n + 2\)
Atunci: \(14n + 2 = 44 \rightarrow 14n = 42 \rightarrow n = 3\)
\[\rightarrow C_{3}H_{8}\ este\ formula\ moleculară\ a\ alcanului,\ propan\]
Rezolvare: \(C_{n}H_{2n + 2} + {\frac{3n + 1}{2}O}_{2} \rightarrow n{CO}_{2} + (n + 1)H_{2}O\)
Atunci: la (14n+2) g alcan……………….44n g bioxid de carbon
88 g alcan………………………..264 g bioxid de carbon
264\(\bullet\)(14n+2)=44\(n \bullet\) 88\(\rightarrow\) 84n+12= 88n\(\rightarrow\)12 = 4n\(\rightarrow\)n =3,
deci \(C_{3}H_{8}\ este\ formula\ moleculară\ a\ alcanului,\ propan\)
Rezolvare: \(C_{n}H_{2n + 2} \rightarrow C_{n}H_{2n} + nH_{2}\)
\((14n + 2)g\ \ alcan\ldots\ldots\ldots\ldots\ldots\ldots\ldots.22,4 \bullet n\ \ L\ \ hidrogen\)
\[90\ g\ \ alcan\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.67,2\ L\ \ hidrogen\]
\[\rightarrow 14n + 2 = 30 \rightarrow n = 2 \rightarrow C_{2}H_{6\ }este\ formula\ moleculară\ a\ alcanului,\ etan\]
Rezolvare: \({CH}_{4} + O_{2} \rightarrow C + 2H_{2}O\)
\[22,4\ L\ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 12\ g\ C\]
\[112\ L\ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C\ \rightarrow x = 60\ g\ \ C\]
Rezolvare: \({CH}_{4} + \frac{1}{2}O_{2} \rightarrow CO + 2H_{2}\)
\[16g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(1 + 2) \bullet 22,4\ L\ gaz\]
\[160g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ L\ gaz \rightarrow x = 672\ L\ gaz\]
Rezolvare: \({CH}_{4} + H_{2}O \rightarrow CO + 3H_{2}\)
\[16g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(1 + 3) \bullet 22,4\ L\ gaz\]
\[80g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ L\ gaz \rightarrow x = 448\ L\ gaz\]
Rezolvare: \({CH}_{4} + {NH}_{3} + \frac{3}{2}O_{2} \rightarrow HCN + 3H_{2}O\)
\[17\ g\ {NH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.27\ g\ HCN\]
\[85\ g\ {NH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ g\ HCN \rightarrow x = 135\ g\ acid\ cianhidric\ HCN\]
Rezolvare: \(2{CH}_{4} \rightarrow C_{2}H_{2} + {3H}_{2}\)
\[16g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 22,4\ L\ acetilenă\]
\[80g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ \ \ L\ acetilenă \rightarrow x = 112\ L\ acetilenă\]
Rezolvare: \({CH}_{4} + \frac{1}{2}O_{2} \rightarrow {CH}_{3}OH\)
\[16g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 32\ g\ alcool\ metilic\]
\[240g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ \ g\ alcool\ metilic\]
\[\rightarrow x = 480\ g\ alcool\ metilic\ {CH}_{3}OH\]
Rezolvare: \({CH}_{4} + O_{2} \rightarrow {O = CH}_{2} + H_{2}O\)
\[16g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 3O\ g\ aldehidă\ formică\ {CH}_{2}O\]
\[240g\ \ \ {CH}_{4}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ \ aldehidă\ formică\ {CH}_{2}Og\ \]
\[\rightarrow x = 450\ g\ aldehidă\ formică\ {CH}_{2}O\]
Rezolvare:
\[\left\{ \begin{array}{r}
C\ \frac{82,76}{12} = 6,89 \rightarrow \frac{6,89}{6,89} = 1 \rightarrow 2 \\
H\ \frac{17,24}{1} = 17,24 \rightarrow \frac{17,24}{6,89} = 2,5 \rightarrow 5
\end{array} \rightarrow C_{2}H_{5} \right.\ \ formula\ brută\]
\[n = \frac{M_{fm}}{M_{fb}} = \frac{58}{29} = 2 \rightarrow C_{4}H_{10},\ butan\ este\ formula\ moleculară\]
ALCHENE
12. Care este alchena cu masa molară egală cu 42 g ?
Rezolvare:\(C_{n}H_{2n}\ cu\ M = 14n\ \ și\ 14n = 42 \rightarrow n = 3\ \rightarrow C_{3}H_{6}\ ,\)
\[\ sau\ {CH}_{2} = CH - {CH}_{3},\ propenă\]
13. Ce cantitate de etenă se obține prin cracarea a 220 g propan?
Rezolvare: \({CH}_{3} - {CH}_{2} - {CH}_{3} \rightarrow {CH}_{2} = {CH}_{2} + {CH}_{4}\)
\[44\ g\ C_{3}H_{8}\ldots\ldots\ldots\ldots\ldots 28\ g\ C_{2}H_{4}\]
\[220\ g\ C_{3}H_{8}\ldots\ldots\ldots\ldots\ldots x\ g\ C_{2}H_{4} \rightarrow x = 140\ g\ C_{2}H_{4}\ etenă\]
14. Ce cantitate de etenă se obține din 138 g alcool etilic?
Rezolvare: \({CH}_{3} - {CH}_{2} - OH \rightarrow {CH}_{2} = {CH}_{2} + H_{2}O\)
\[46\ g\ {CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..28\ g\ {CH}_{2} = {CH}_{2}\]
\[138\ g\ {CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ g\ {CH}_{2} = {CH}_{2} \rightarrow x = 84\ g\ {CH}_{2} = {CH}_{2}(etenă)\]
15. Ce cantitate de propenă se obține prin reacția de eliminare a acidului clorhidric din 785 g 1 – clor – propan?
Rezolvare: \(Cl - {CH}_{2} - {CH}_{2} - {CH}_{3} \rightarrow {CH}_{2} = CH - {CH}_{3} + HCl\)
\[78,5\ g\ Cl - {CH}_{2} - {CH}_{2} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots 44\ g\ {CH}_{2} = CH - {CH}_{3}\]
\[785\ g\ Cl - {CH}_{2} - {CH}_{2} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{2} = CH - {CH}_{3}\]
\[\rightarrow x = 440\ g\ {CH}_{2} = CH - {CH}_{3}\ (propenă)\ \]
16. Ce cantitate de etan se obține prin hidrogenarea a 84 g etenă?
Rezolvare:\({CH}_{2} = {CH}_{2} + H_{2} \rightarrow {CH}_{3} - {CH}_{3}\)
\[{28\ g\ CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots 30g\ {CH}_{3} - {CH}_{3}\]
\[{84\ g\ CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots x\ g\ {CH}_{3} - {CH}_{3} \rightarrow x = 90\ g\ {CH}_{3} - {CH}_{3},\ etan\ \]
Rezolvare:\({CH}_{2} = {CH}_{2} + {Br}_{2} \rightarrow {CH}_{2}Br - {CH}_{2}Br\)
\[{160\ gBr}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 188\ g\ {CH}_{2}Br - {CH}_{2}Br\]
\[{320\ gBr}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{2}Br - {CH}_{2}Br\]
\[\rightarrow x = 376\ g\ {CH}_{2}Br - {CH}_{2}Br,\ 1,2 - dibrometan\]
18. Ce cantitate de 2 – clorpropan se obține prin adiția acidului clorhidric la 176 g propenă?
Rezolvare:\({CH}_{2} = CH - {CH}_{3} + HCl \rightarrow {CH}_{3} - CHCl - {CH}_{3}\)
\[{44g\ CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots.80,5\ g{CH}_{3} - CHCl - {CH}_{3}\]
\[{176g\ CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots x\ g{CH}_{3} - CHCl - {CH}_{3}\]
\[\rightarrow x = 322\ g\ {CH}_{3} - CHCl - {CH}_{3},\ 2 - clorpropan\]
19. Ce cantitate alcool etilic se obține prin adiția apei la 140 g etenă?
Rezolvare:\({CH}_{2} = {CH}_{2} + HOH \rightarrow {CH}_{3} - {CH}_{2} - OH\)
\[28\ g\ {CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots\ldots..46\ g\ {CH}_{3} - {CH}_{2} - OH\]
\[140\ g\ {CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots\ldots.x\ \ g\ {CH}_{3} - {CH}_{2} - OH\]
\[\rightarrow x = 230\ g\ {CH}_{3} - {CH}_{2} - OH,\ alcool\ etilic\]
20. Ce cantitate de 1,2 – propandiol se obține prin oxidarea cu permanganat de potasiu în soluție bazică, a 210 g propenă?
Rezolvare: \({CH}_{2} = CH - {CH}_{3} + \lbrack O\rbrack + \ H_{2}O\overset{ {KMnO}_{4},\ {Na}_{2}{CO}_{3}}{\rightarrow}\ {CH}_{2}OH - CHOH - {CH}_{3}\)
\[42\ g{CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 76\ g\ {CH}_{2}OH - CHOH - {CH}_{3}\]
\[210\ g{CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{2}OH - CHOH - {CH}_{3}\]
\[\rightarrow x = 310\ g\ {CH}_{2}OH - CHOH - {CH}_{3},\ 1,2 - propandiol\ \]
21. Ce cantitate de 1,2 – etandiol (glicol) se obține din 168 g etenă, prin oxidare cu oxigen molecular, urmată de hidroliză ?
Rezolvare: \({CH}_{2} = {CH}_{2} + \frac{1}{2}O_{2} \rightarrow oxid\ de\ etenă\overset{HOH}{\rightarrow}{CH}_{2}OH - {CH}_{2}OH\)
\[28\ g{CH}_{2} = {CH}_{2}\ldots\ldots\ldots 62\ g{CH}_{2}OH - {CH}_{2}OH\]
\[168\ g{CH}_{2} = {CH}_{2}\ldots\ldots x\ g{CH}_{2}OH - {CH}_{2}OH \rightarrow x = 372\ g{CH}_{2}OH - {CH}_{2}OH,\ 1,2 - etandiol\]
22. Ce cantitate de acid acetic se obține prin oxidarea a 210 g propenă, cu permanganat de potasiu în mediu de acid sulfuric?
Rezolvare: \({CH}_{2} = CH - {CH}_{3} + 5\lbrack O\rbrack\overset{KMnO_{4},H_{2}{SO}_{4}}{\rightarrow}{CO}_{2} + H_{2}O + HOOC - {CH}_{3}\)
\[42g\ {CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.60\ g{CH}_{3} - COOH\]
\[210g\ {CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots x\ g{CH}_{3} - COOH \rightarrow x = 300g\ {CH}_{3} - COOH,\ acid\ acetic\]
23. Ce cantitate de acid acetic se obține prin oxidarea a 280 g 2 – butenă , cu permanganat de potasiu în mediu de acid sulfuric?
Rezolvare: \({CH}_{3} - CH = CH - {CH}_{3} + 4\lbrack O\rbrack\overset{KMnO_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - COOH + HOOC - {CH}_{3}\)
\[56\ g{CH}_{3} - CH = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 120\ g{CH}_{3} - COOH\]
\[280\ g{CH}_{3} - CH = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots x\ g{CH}_{3} - COOH\]
\[\rightarrow x = 600g{\ CH}_{3} - COOH,\ acid\ acetic\]
24. Ce cantitate de acetonă se obține prin oxidarea a 280 g 2metil,2 – butenă , cu permanganat de potasiu în mediu de acid sulfuric?
Rezolvare: \({CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3} + 3\lbrack O\rbrack\overset{KMnO_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - CO - {CH}_{3} + HOOC - {CH}_{3}\)
\[{7O\ gCH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..58\ g{CH}_{3} - CO - {CH}_{3}\]
\[{280\ gCH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ g{CH}_{3} - CO - {CH}_{3}\]
\[\rightarrow x = 232\ g\ {CH}_{3} - CO - {CH}_{3},\ acetonă\]
25. Care este randamentul reacției de polimerizare a etenei, dacă din 900 g etenă se obțin 540 g polietenă?
Rezolvare: \({nCH}_{2} = {CH}_{2} \rightarrow - ({ {CH}_{2} - {CH}_{2})}_{n} -\)
\(28\ n\ g\ etenă\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.28\ n\ g\ polietenă\)
\[900\ g\ etenă\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x_{teoretic}\ g\ polietenă \rightarrow x_{t} = 900\ g\ polietenă\]
\[ƞ = \frac{x_{real}}{x_{teoretic}} = \frac{540}{900} = 0,6 \rightarrow randament\ 60\%\]
ALCHINE
Rezolvare:\(C_{n}H_{2n - 2}\ cu\ M = 14n - 2 = 40 \rightarrow n = 3 \rightarrow C_{3}H_{4}\ \ sau\ CH \equiv C - {CH}_{3},\ propină\)
27. Ce volum (c.n.) de acetilenă se obține prin reacția a 100 g carbid cu apa?
Rezolvare:\({CaC}_{2} + 2H_{2}O \rightarrow CH \equiv CH + {Ca(OH)}_{2}\)
\[{64\ g\ CaC}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 22,4\ L\ CH \equiv CH\]
\[{100\ g\ CaC}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ L\ CH \equiv CH \rightarrow x = \ 35L\ CH \equiv CH,\ acetilenă\]
28. Ce cantitate de etenă se obține prin adiția hidrogenului la 260 g acetilenă?
Rezolvare: \(CH \equiv CH + H_{2} \rightarrow {CH}_{2} = {CH}_{2}\)
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots 28g\ {CH}_{2} = {CH}_{2}\]
\[260\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{2} = {CH}_{2} \rightarrow x = 280\ g\ {CH}_{2} = {CH}_{2},\ etenă\]
29. Ce cantitate de etan se obține prin adiția hidrogenului la 520 g acetilenă?
Rezolvare: \(CH \equiv CH + {2H}_{2} \rightarrow {CH}_{3} - {CH}_{3}\)
\[26\ g\ CH \equiv CH + {2H}_{2} \rightarrow {30\ g\ CH}_{3} - {CH}_{3}\]
\[520\ g\ CH \equiv CH + {2H}_{2} \rightarrow {x\ g\ CH}_{3} - {CH}_{3} \rightarrow x = 600\ g\ {CH}_{3} - {CH}_{3},\ etan\]
30. Ce cantitate de acetilenă este necesară pentru a adiționa 640 g brom?
Rezolvare: \(CH \equiv CH + {2Br}_{2} \rightarrow {CH\ Br}_{2} - {CHBr}_{2}\)
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots.320\ g{Br}_{2}\]
\[x\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 640\ g{Br}_{2} \rightarrow x = 52\ g\ CH \equiv CH,\ acetilenă\]
31. Ce cantitate de tetracloretan se obține prin adiția clorului la 156 g acetilenă ?
Rezolvare: \(CH \equiv CH + {2Cl}_{2} \rightarrow {CH\ Cl}_{2} - {CHCl}_{2}\)
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..168\ g\ {CH\ Cl}_{2} - {CHCl}_{2}\]
\[156\ g\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH\ Cl}_{2} - {CHCl}_{2}\]
\[\rightarrow x = 1008\ g\ {CH\ Cl}_{2} - {CHCl}_{2},\ tetracloretan\]
32. Ce cantitate de clorură de vinil se obține prin adiția acidului clorhidric la 104 g acetilenă?
Rezolvare: \(CH \equiv CH + HCl \rightarrow CHCl = {CH}_{2}\)
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots 62,5\ gCHCl = {CH}_{2}\]
\[104\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ CHCl = {CH}_{2} \rightarrow x = \ 250\ g\ CHCl = {CH}_{2},\ clorură\ de\ vinil\]
33. Ce cantitate de acetat de vinil se obține prin adiția acidului acetic la 182 g acetilenă?
Rezolvare: \(CH \equiv CH + {CH}_{3}COOH \rightarrow CH_{2} = CH - O - CO - {CH}_{3}\)
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.86\ g\ CH_{2} = CH - O - CO - {CH}_{3}\]
\[182\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ CH_{2} = CH - O - CO - {CH}_{3}\]
\[\rightarrow x = 602\ g\ CH_{2} = CH - O - CO - {CH}_{3},\ acetat\ de\ vinil\]
34. Ce cantitate de acrilonitril se obține prin adiția acidului cianhidric la 208 g acetilenă?
Rezolvare: \(CH \equiv CH + HCN \rightarrow CH_{2} = CH - CN\)
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots 53\ g\ CH_{2} = CH - CN\]
\[208\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots x\ g\ CH_{2} = CH - CN \rightarrow x = 424\ g\ CH_{2} = CH - CN,\ acrilonitril\]
35. Ce cantitate de acetaldehidă se obține prin adiția apei la 130 g acetilenă?
Rezolvare:
\[CH \equiv CH + HOH \rightarrow {CH}_{2} = CH - OH \leftrightarrow {CH}_{3} - HC = O\]
\[26\ g\ CH \equiv CH + HOH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 44\ g\ {CH}_{3} - HC = O\]
\[130\ g\ CH \equiv CH + HOH\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{3} - HC = O\]
\[\rightarrow x = 220\ g{CH}_{3} - HC = O,\ acetaldehidă\ \]
36. Ce cantitate de acid oxalic se obține prin oxidarea cu permanganat de potasiu a 208 g acetilenă?
Rezolvare:
\[CH \equiv CH + 4(\ O) \rightarrow HOOC - COOH\]
\[26\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots..90\ g\ HOOC - COOH\]
\[208\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots..x\ g\ HOOC - COOH \rightarrow x = 720g\ HOOC - COOH,\ acid\ oxalic\]
37. Ce cantitate de bioxid de carbon se obține prin arderea a 104 g acetilenă?
Rezolvare:
\[2CH \equiv CH + 5O_{2} \rightarrow {4CO}_{2} + 2H_{2}O\]
\[52\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots{176\ g\ CO}_{2}\]
\[104\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots x{\ g\ CO}_{2} \rightarrow x = 352\ g\ {CO}_{2},\ bioxid\ de\ carbon\]
38. Ce cantitate de sodiu reacționează cu acetilena pentru a obține 140 g acetilură disodică?
Rezolvare: \(CH \equiv CH + 2\ Na\ \rightarrow H_{2} + NaC \equiv CNa\)
\[46\ g\ Na\ldots\ldots\ldots\ldots\ldots\ldots..70g\ NaC \equiv CNa\]
\[x\ g\ Na\ldots\ldots\ldots\ldots\ldots\ldots..140g\ NaC \equiv CNa \rightarrow x = 92\ g\ NaC \equiv CNa,\ acetilură\ disodică\]
39. Ce cantitate de clorură diamino cupru (I) reacționează cu acetilena pentru a obține 456 g de acetilură de cupru (I)?
Rezolvare: \(CH \equiv CH + 2\left\lbrack Cu{({NH}_{3})}_{2} \right\rbrack Cl\ \rightarrow CuC \equiv CCu + 2{NH}_{4}Cl + 2{NH}_{3\ }\)
\[267\ g\ \left\lbrack Cu{({NH}_{3})}_{2} \right\rbrack Cl\ \ldots\ldots\ldots\ldots\ldots\ldots..152\ g\ CuC \equiv CCu\]
\[x\ g\ \left\lbrack Cu\left( {NH}_{3} \right)_{2} \right\rbrack Cl\ \ldots\ldots\ldots\ldots\ldots\ldots..456\ g\ CuC \equiv CCu\]
\[\rightarrow x = \ 801\ g\left\lbrack Cu{({NH}_{3})}_{2} \right\rbrack Cl,\ clorură\ diamino\ cupru(I)\ \]
40. Ce cantitate de acetilenă reacționează cu hidroxidul diamino argint(I), știind că s – au obținut ca produs secundar, 170 g amoniac?
Rezolvare: \(CH \equiv CH + 2\left\lbrack Ag\left( {NH}_{3} \right)_{2} \right\rbrack OH\ \rightarrow AgC \equiv CAg + 2{NH}_{4}Cl + 4{NH}_{3\ } + 2H_{2}O\)
\[26\ gCH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 68\ g\ {NH}_{3\ }\]
\[x\ gCH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 170\ g\ {NH}_{3\ } \rightarrow x = 65\ g\ CH \equiv CH,\ acetilenă\]
Rezolvare:\(\left\{ \begin{array}{r}
C,\frac{92,3}{12} = 7,69 \rightarrow 1 \\
H,\frac{7,69}{1} = 7,69 \rightarrow 1
\end{array} \right.\ \rightarrow CH\ este\ formula\ brută \rightarrow n = \frac{M_{fm}}{M_{fb}} = \frac{26}{13} = 2\)
\[\rightarrow C_{2}H_{2}\ este\ formula\ moleculară\]
42. Ce cantitate de benzen se obține prin trimerizarea a 300 g acetilenă, dacă randamentul reacției de trimerizare este 30 %?
Rezolvare: \(3CH \equiv CH \rightarrow C_{6}H_{6}\)
\[78\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots..23,4\ gC_{6}H_{6}\]
\[300\ g\ CH \equiv CH\ldots\ldots\ldots\ldots\ldots\ldots..x\ gC_{6}H_{6} \rightarrow x = 90\ g\ C_{6}H_{6},\ benzen\]
ARENE
43. Ce cantitate de acid azotic reacționează cu benzenul pentru a obține 984 g nitrobenzen?
Rezolvare:\(C_{6}H_{6} + HO{NO}_{2} \rightarrow C_{6}H_{5}{NO}_{2} + H_{2}O\)
\[63\ g\ HO{NO}_{2}\ldots\ldots\ldots.123\ g\ C_{6}H_{5}{NO}_{2}\]
\[x\ g\ HO{NO}_{2}\ldots\ldots\ldots.984\ g\ C_{6}H_{5}{NO}_{2} \rightarrow x = 504\ g\ HO{NO}_{2}\ ,\ acid\ azotic\ \]
44. Ce cantitate de acid benzen – sulfonic se obține prin sulfonarea a 234 g benzen?
Rezolvare: \(C_{6}H_{6} + HOSO_{3}H \rightarrow C_{6}H_{5} - {SO}_{3}H + H_{2}O\)
\[{78\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots\ldots.{158\ g\ C}_{6}H_{5} - {SO}_{3}H\]
\[{234\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots\ldots.x{\ g\ C}_{6}H_{5} - {SO}_{3}H\]
\[\rightarrow x = 474\ g\ {\ \ C}_{6}H_{5} - {SO}_{3}H,\ acid\ benzen - sulfonic\]
45. Ce cantitate de etilbenzen se obține prin alchilarea a 390 g benzen cu etenă?
Rezolvare: \(C_{6}H_{6} + {CH}_{2} = {CH}_{2}\overset{ {AlCl}_{3}}{\rightarrow}C_{6}H_{5} - {CH}_{2} - {CH}_{3}\)
\[{\ \ 78\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots\ldots 106\ g\ C_{6}H_{5} - {CH}_{2} - {CH}_{3}\]
\[{\ \ 390\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots x\ g\ C_{6}H_{5} - {CH}_{2} - {CH}_{3} \rightarrow x = 530\ g\ C_{6}H_{5} - {CH}_{2} - {CH}_{3},\ etilbenzen\]
46. Ce cantitate de izopropilbenzen se obține prin alchilarea a 468 g benzen cu propenă?
Rezolvare: 
\(C_{6}H_{6} + {CH}_{2} = CH - {CH}_{3}\overset{ {AlCl}_{3}}{\rightarrow}C_{6}H_{5} - CH{({CH}_{3})}_{2}\)
\[{78\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..120\ g\ C_{6}H_{5} - CH{({CH}_{3})}_{2}\]
\[{468\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..x\ g\ C_{6}H_{5} - CH\left( {CH}_{3} \right)_{2}\]
\[\rightarrow x = 720\ gC_{6}H_{5} - CH\left( {CH}_{3} \right)_{2},\ izopropilbenzen\ \]
47. Ce cantitate de fenil, metil – cetonă ( acetofenonă ) se obține prin acilarea benzenului cu 157 g clorură de acetil?
Rezolvare:

\[C_{6}H_{6} + {CH}_{3} - CO - Cl\overset{ {AlCl}_{3}}{\rightarrow}C_{6}H_{5} - CO - {CH}_{3}\]
\[78,5\ g\ {CH}_{3} - CO - Cl\ldots\ldots\ldots\ldots.{120\ g\ C}_{6}H_{5} - CO - {CH}_{3}\]
\[157\ g\ {CH}_{3} - CO - Cl\ldots\ldots\ldots\ldots.{x\ g\ C}_{6}H_{5} - CO - {CH}_{3}\]
\[\rightarrow x = 240\ g{120\ g\ C}_{6}H_{5} - CO - {CH}_{3},\ acetofenonă\]
48. Ce cantitate de ciclohexan se obține prin hidrogenarea a 234 g benzen?
Rezolvare: \(C_{6}H_{6} + {3H}_{2}\overset{Pt,\ 250℃}{\rightarrow}C_{6}H_{12}\)
\[{78\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..84\ gC_{6}H_{12}\]
\[234{g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..x\ g\ C_{6}H_{12} \rightarrow x = 252\ gC_{6}H_{12},\ ciclohexan\]
49. Ce cantitate de hexaclorciclohexan se obține prin clorurarea a 156 g benzen?
Rezolvare: \(C_{6}H_{6} + {3Cl}_{2}\overset{lumină}{\rightarrow}C_{6}H_{6}{Cl}_{6}\)
\[{78\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..291\ gC_{6}H_{6}{Cl}_{6}\]
\[{156\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..x\ g\ C_{6}H_{6}{Cl}_{6} \rightarrow x = 582\ gC_{6}H_{6}{Cl}_{6},\ hexaclorciclohexan\]
50. Ce cantitate de acid benzoic se obține prin oxidarea la catena laterală a 276 g toluen?
Rezolvare: \(C_{6}H_{5} - {CH}_{3} + 3\lbrack O\rbrack\overset{ {KMnO}_{4,\ \ }\ \ \ H_{2}{SO}_{4}}{\rightarrow}C_{6}H_{5} - COOH + H_{2}O\)
\[92g\ C_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots.122\ g\ C_{6}H_{5} - COOH\]
\[276g\ C_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C_{6}H_{5} - COOH \rightarrow x = 366\ g\ C_{6}H_{5} - COOH,\ acid\ benzoic\]
51. Ce cantitate de benzen se obține prin dehidrogenarea a 168 g ciclohexan?
Rezolvare:

\[\ C_{6}H_{12} \rightarrow C_{6}H_{6} + {3H}_{2}\]
\[84\ g\ C_{6}H_{12}\ldots\ldots\ldots\ldots 78\ g\ C_{6}H_{6}\]
\[168\ g\ C_{6}H_{12}\ldots\ldots\ldots x\ g\ C_{6}H_{6} \rightarrow 156\ g\ C_{6}H_{6},\ benzen\]
52. Ce cantitate de toluen se obține prin dehidrogenarea a 294 g metilciclohexan?
\[Rezolvare:\]

\[\ C_{6}H_{11} - {CH}_{3} \rightarrow C_{6}H_{5} - {CH}_{3} + \ 3H_{2}\]
\[{98\ g\ C}_{6}H_{11} - {CH}_{3}\ldots\ldots\ldots\ldots..{92\ g\ C}_{6}H_{5} - {CH}_{3}\]
\[{294\ g\ C}_{6}H_{11} - {CH}_{3}\ldots\ldots\ldots\ldots..x{\ g\ C}_{6}H_{5} - {CH}_{3} \rightarrow x = 276\ g\ {\ \ C}_{6}H_{5} - {CH}_{3},\ toluen\]
53. Ce cantitate de orto – xilen se obține prin dehidrogenarea a 224 g de
1,2 – dimetilciclohexan?
\[Rezolvare:\ C_{6}H_{10}{({CH}_{3})}_{2} \rightarrow C_{6}H_{4}{({CH}_{3})}_{2} + {3H}_{2}\]
\[112\ g\ C_{6}H_{10}{({CH}_{3})}_{2}\ldots\ldots\ldots\ldots 106\ g\ C_{6}H_{4}{({CH}_{3})}_{2}\]
\[224\ g\ C_{6}H_{10}{({CH}_{3})}_{2}\ldots\ldots\ldots\ldots x\ g\ C_{6}H_{4}\left( {CH}_{3} \right)_{2} \rightarrow x = 212\ g\ C_{6}H_{4}\left( {CH}_{3} \right)_{2},\ xilen\]
54. Ce cantitate de toluen se obține prin dehidrogenarea a 100 g normal – heptan ?
Rezolvare: \(C_{7}H_{16} \rightarrow C_{6}H_{5} - {CH}_{3} + {4H}_{2}\)
\[100\ gC_{7}H_{16}\ldots\ldots\ldots\ldots\ldots\ldots\ldots 92\ gC_{6}H_{5} - {CH}_{3} \rightarrow 92\ g\ toluen\]
55. Ce cantitate de toluen este necesară pentru a obține prin hidrogenare 156 g benzen?
Rezolvare:\(C_{6}H_{5} - CH_{3} + H_{2} \rightarrow C_{6}H_{6} + {CH}_{4}\)
\[{92\ g\ C}_{6}H_{5} - CH_{3}\ldots\ldots\ldots\ldots..78\ g\ C_{6}H_{6}\]
\[{x\ g\ C}_{6}H_{5} - CH_{3}\ldots\ldots\ldots\ldots..156\ g\ C_{6}H_{6} \rightarrow x = 184\ g{\ C}_{6}H_{5} - CH_{3},\ toluen\]
56. Ce cantitate de acid ftalic se obține prin oxidarea la catena laterală a 184 g xilen?
Rezolvare: \(106\ g\ { {CH}_{3} - C}_{6}H_{4} - {CH}_{3} + 6\lbrack O\rbrack\overset{ {KMnO}_{4,\ \ }\ \ \ H_{2}{SO}_{4}}{\rightarrow}{HOOC - C}_{6}H_{4} - COOH + {2H}_{2}O\)
\[92g\ C_{8}H_{10}\ldots\ldots\ldots\ldots\ldots\ldots.106\ g\ C_{8}H_{6}O_{4}\]
\[184g\ C_{8}H_{10}\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C_{8}H_{6}O_{4} \rightarrow x = 212\ g\ C_{8}H_{6}O_{4},\ acid\ ftalic\ \]
HIDROCARBURI AROMATICE POLICICLICE
Rezolvare:
\[\left\{ \begin{array}{r}
C\frac{93,75}{12} = 7,81 \rightarrow \frac{7,81}{6,25} = 1,25 \rightarrow 5 \\
H\frac{6,25}{1} = 6,25 \rightarrow \frac{6,25}{6,25}\ = 1 \rightarrow 4
\end{array} \right.\ \rightarrow C_{5}H_{4}\ este\ formula\ brută\]
\[\rightarrow n = \frac{M_{fm}}{M_{fb}} = \frac{128}{64} = 2 \rightarrow C_{10}H_{8}\ este\ formula\ moleculară\]
58. Ce cantitate de \(\mathbf{\alpha - nitro - naftalină}\) se obține prin nitrarea a 640 g naftalină?
Rezolvare:
\[C_{10}H_{8} + {HONO}_{2} \rightarrow C_{10}H_{7} - {NO}_{2} + H_{2}O\]
\[128\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.173\ g\ C_{10}H_{7}{NO}_{2}\]
\[640\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ g\ C_{10}H_{7}{NO}_{2}\]
\(\rightarrow x = 865\ g\ C_{10}H_{7}{NO}_{2},\ \alpha - nitro - naftalină\)
59. Ce cantitate de \(\mathbf{\ \ \ \ acid\ \alpha - naftalin - sulfonic}\) se obține din 256 g naftalină?
Rezolvare:
\[C_{10}H_{8} + {HOSO}_{3}H\overset{100℃}{\rightarrow}C_{10}H_{7} - {SO}_{3}H + H_{2}O\]
\[128\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots.208\ g\ C_{10}H_{7} - {SO}_{3}H\]
\[256\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots.x\ g\ C_{10}H_{7} - {SO}_{3}H\]
\(\rightarrow x = 416\ g\ C_{10}H_{7} - {SO}_{3}H,\ acid\ \alpha - naftalin - sulfonic\)
60. Ce cantitate de \(\mathbf{\ \ \ \ acid\ \beta - naftalin - sulfonic}\) se obține din 64 g naftalină?
Rezolvare:
\[C_{10}H_{8} + {HOSO}_{3}H\overset{180℃}{\rightarrow}C_{10}H_{7} - {SO}_{3}H + H_{2}O\]
\[128\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots.208\ g\ C_{10}H_{7} - {SO}_{3}H\]
\[64\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots.x\ g\ C_{10}H_{7} - {SO}_{3}H\]
\(\rightarrow x = 104\ g\ C_{10}H_{7} - {SO}_{3}H,\ acid\ \beta - naftalin - sulfonic\)
61. Ce cantitate de decalină se obține prin hidrogenarea a 64 g naftalină?
Rezolvare:
\[C_{10}H_{8} + {5H}_{2}\overset{Ni}{\rightarrow}C_{10}H_{18}\]
\[128\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots 138\ g\ C_{10}H_{18}\]
\[64\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C_{10}H_{18} \rightarrow x = 69\ g\ C_{10}H_{18},\ decalină\]
62. Ce cantitate de acid ftalic se obține prin oxidarea a 384 g naftalină?
Rezolvare:
\[C_{10}H_{8} + \frac{9}{2}O_{2}\overset{V_{2}O_{5},\ 350℃}{\rightarrow}HOOC - C_{6}H_{4} - COOH + 2{CO}_{2} + H_{2}O\]
\[128\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots.166\ g\ HOOC - C_{6}H_{4} - COOH\]
\[384\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots.x\ \ g\ HOOC - C_{6}H_{4} - COOH\]
\[\rightarrow x = 498\ g\ HOOC - C_{6}H_{4} - COOH,\ acid\ ftalic\]
63. Ce cantitate de antracen este necesară pentru a obține prin oxidare cu dicromat de potasiu
în mediu de acid acetic, 104 g de antrachinonă ?
Rezolvare:
\[C_{14}H_{10} + 3\lbrack O\rbrack\overset{K_{2}{Cr}_{2}O_{7},{CH}_{3}COOH}{\rightarrow}C_{10}H_{8}O_{2} + H_{2}O\]
\[{178\ gC}_{14}H_{10}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 208\ g\ C_{14}H_{8}O_{2} + H_{2}O\]
\[{x\ gC}_{14}H_{10}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 104\ g\ C_{14}H_{8}O_{2} \rightarrow x = 89\ g\ C_{14}H_{10},\ antracen\]
DERIVAȚI HALOGENAȚI
și conține 24,74%C și 2,06 %H și 73,19% Cl?
Rezolvare:\(\left\{ \begin{array}{r}
C\frac{24,74}{12} = 2,06 \\
H\frac{2,06}{1} = 2,06 \\
Cl\frac{73,19}{35,5} = 2,06
\end{array} \rightarrow CHCl\ este\ formula\ brută \right.\ \)
\[n = \frac{M_{fm}}{M_{fb}} = \frac{291}{48,5} = 6 \rightarrow C_{6}H_{6}{Cl}_{6}\ este\ formula\ moleculară\]
65. Ce cantitate de etanol se obține prin hidroliza în mediu bazic a 129 g cloretan?
Rezolvare:
\[{CH}_{3} - {CH}_{2} - Cl\overset{NaOH,\ HOH}{\rightarrow}{CH}_{3} - {CH}_{2} - OH + NaCl\]
\[\ 64,5\ g\ {CH}_{3} - {CH}_{2} - Cl\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 46\ g\ {CH}_{3} - {CH}_{2} - OH\]
\[129\ g\ {CH}_{3} - {CH}_{2} - Cl\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{3} - {CH}_{2} - OH\]
\[\rightarrow x = \ 92\ g\ {CH}_{3} - {CH}_{2} - OH\ ,\ etanol\]
3. Ce cantitate de aldehidă benzoică se obține prin hidroliza a 483 g clorură de benziliden ?
Rezolvare:
\[C_{6}H_{5} - {CHCl}_{2}\overset{2NaOH,\ HOH}{\rightarrow}C_{6}H_{5} - CH = O + 2NaCl\]
\[161\ g\ C_{6}H_{5} - {CHCl}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots 106\ g\ C_{6}H_{5} - CH = O\]
\[483\ g\ C_{6}H_{5} - {CHCl}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C_{6}H_{5} - CH = O\]
\[\rightarrow x = 318\ gC_{6}H_{5} - CH = O,\ benzaldehidă\ \ \]
66. Ce cantitate de acid acetic se obține prin hidroliza a 267 g gem. tricloretan ?
Rezolvare:
\[{CH}_{3} - {CCl}_{3}\overset{3NaOH,\ HOH}{\rightarrow}{CH}_{3} - COOH + 3NaCl\]
\[{133,5\ g\ CH}_{3} - {CCl}_{3}\ldots\ldots\ldots\ldots\ldots\ldots..{60\ g\ CH}_{3} - COOH\]
\[{267\ g\ CH}_{3} - {CCl}_{3}\ldots\ldots\ldots\ldots\ldots\ldots..{x\ g\ CH}_{3} - COOH\]
\[\rightarrow x = 120\ g\ {\ CH}_{3} - COOH\]
Rezolvare:
\[{CH}_{3} - I + {NH}_{3} \rightarrow {CH}_{3} - {NH}_{2} + HI\]
\[{142g\ CH}_{3} - I\ldots\ldots\ldots\ldots\ldots\ldots..17g\ {NH}_{3}\]
\[{284g\ CH}_{3} - I\ldots\ldots\ldots\ldots\ldots\ldots..x\ g\ {NH}_{3} \rightarrow x = 34\ g\ {NH}_{3},\ amoniac\]
68. Se obține toluen din benzen și clorură de metil. Ce cantitate de clorură de metil reacționează cu 156 g benzen?
Rezolvare: \(C_{6}H_{6} + {CH}_{3}Cl\overset{ {AlCl}_{3}}{\rightarrow}C_{6}H_{5} - {CH}_{3} + HCl\)
\[78\ g\ C_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..50,5\ g\ {CH}_{3}Cl\]
\[156\ g\ C_{6}H_{6}\ldots\ldots\ldots\ldots\ldots.x\ g\ {CH}_{3}Cl \rightarrow x = 101\ g\ {CH}_{3}Cl,\ clorură\ de\ metil\]
69. Se obține propenă prin eliminare de hidracid din clorpropan. Ce cantitate de propenă se obține din 157 g clorpropan?
Rezolvare: \({CH}_{3} - {CH}_{2} - {CH}_{2}Cl\overset{KOH,\ \ alcool}{\rightarrow}{CH}_{3} - CH = {CH}_{2} + HCl\)
\[78,\ 5\ g{CH}_{3} - {CH}_{2} - {CH}_{2}Cl\ldots\ldots\ldots\ldots\ldots\ldots 42\ g{CH}_{3} - CH = {CH}_{2}\]
\[157\ g{CH}_{3} - {CH}_{2} - {CH}_{2}Cl\ldots\ldots\ldots\ldots\ldots\ldots x\ g{CH}_{3} - CH = {CH}_{2}\]
\[\rightarrow x = 84\ g\ {CH}_{3} - CH = {CH}_{2},\ propenă\ \]
70. Ce cantitate de benzen reacționează cu bromul pentru a obține 314 g brombenzen?
Rezolvare:\(C_{6}H_{6} + {Br}_{2} \rightarrow C_{6}H_{5}Br + HBr\)
\[{78\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..157\ g\ C_{6}H_{5}Br\]
\[{x\ g\ C}_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..314\ g\ C_{6}H_{5}Br \rightarrow x = 156\ g\ C_{6}H_{6},\ brombenzen\]
71. Ce cantitate de monoclorbenzen se obține prin clorurarea a 156 g benzen?
Rezolvare:\(C_{6}H_{6} + {Cl}_{2}\overset{ {FeCl}_{3}}{\rightarrow}C_{6}H_{5}Cl + HCl\)
\[78\ g\ C_{6}H_{6}\ldots\ldots\ldots.112,5\ g\ C_{6}H_{5}Cl\]
\[156\ g\ C_{6}H_{6}\ldots\ldots\ldots.x\ g\ C_{6}H_{5}Cl \rightarrow x = 225\ g\ \ C_{6}H_{5}Cl,\ monoclorbenzen\]
72. Ce cantitate de clorură de benzil se obține prin clorurarea la catena laterală a 184 g toluen?
Rezolvare: \(C_{6}H_{5} - {CH}_{3} + {Cl}_{2}\overset{lumină}{\rightarrow}C_{6}H_{5} - {CH}_{2}Cl + HCl\)
\[{92\ g\ C}_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots.126,5\ g\ C_{6}H_{5} - {CH}_{2}Cl\]
\({184\ g\ C}_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots x\ g\ C_{6}H_{5} - {CH}_{2}Cl \rightarrow x = 253\ g\ C_{6}H_{5} - {CH}_{2}Cl,\ clorură\ de\ benzil\)
73. Ce cantitate de clorură de benziliden se obține prin clorurarea la catena laterală a 184 g toluen?
Rezolvare: \(C_{6}H_{5} - {CH}_{3} + 2{Cl}_{2}\overset{lumină}{\rightarrow}C_{6}H_{5} - CH{Cl}_{2} + 2HCl\)
\[{92\ g\ C}_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots.161\ g\ C_{6}H_{5} - CH{Cl}_{2}\]
\[{184\ g\ C}_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots x\ g\ C_{6}H_{5} - CH{Cl}_{2}\]
\[\rightarrow x = 322\ gC_{6}H_{5} - CH{Cl}_{2},\ clorură\ de\ benziliden\]
\[Rezolvare:\ C_{6}H_{5} - {CH}_{3} + 3{Cl}_{2}\overset{lumină}{\rightarrow}C_{6}H_{5} - C{Cl}_{3} + 3HCl\]
\[{92\ g\ C}_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots.195,5\ g\ C_{6}H_{5} - {CCl}_{3}\]
\[{368\ g\ C}_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots.x\ g\ C_{6}H_{5} - {CCl}_{3} \rightarrow x = 782\ g\ \ C_{6}H_{5} - {CCl}_{3}\]
75. Monobrombenzenul reacționează cu magneziul în mediu de eter anhidru.
Să se calculeze cantitatea de bromură de fenil magneziu care se obține în urma reacției dintre 471 g monobrombenzen și magneziu.
\[Rezolvare:\ C_{6}H_{5} - Br + Mg\ \rightarrow \ C_{6}H_{5} - MgBr\]
\[{157\ g\ C}_{6}H_{5} - Br\ldots\ldots\ldots\ldots..{181\ g\ C}_{6}H_{5} - MgBr\]
\[{471\ g\ C}_{6}H_{5} - Br\ldots\ldots\ldots\ldots..{x\ g\ C}_{6}H_{5} - MgBr\]
\[\rightarrow x = 543\ g\ {\ C}_{6}H_{5} - MgBr,\ bromură\ de\ fenil\ magneziu\]
Rezolvare: \({CH}_{3}I + KCN \rightarrow {CH}_{3} - CN + KI\)
\[{142\ g\ CH}_{3}I\ldots\ldots\ldots..65\ g\ KCN\]
\[{x\ g\ CH}_{3}I\ldots\ldots\ldots..195\ g\ KCN \rightarrow x = 426\ g\ {CH}_{3}I,\ iodmetan\]
77. Ce cantitate de acid ftalic se obține prin oxidarea la catena laterală a 212 g xilen?
Rezolvare: \(\ { {CH}_{3} - C}_{6}H_{4} - {CH}_{3} + 6\lbrack O\rbrack\overset{ {KMnO}_{4,\ \ }\ \ \ H_{2}{SO}_{4}}{\rightarrow}{HOOC - C}_{6}H_{4} - COOH + {2H}_{2}O\)
\[106g\ C_{8}H_{10}\ldots\ldots\ldots\ldots\ldots\ldots.166\ g\ C_{8}H_{6}O_{4}\]
\[212g\ C_{8}H_{10}\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C_{8}H_{6}O_{4} \rightarrow x = 332\ g\ C_{8}H_{6}O_{4},\ acid\ ftalic\ \]
Rezolvare:
\[{CH}_{3}I + {NH}_{3} \rightarrow {CH}_{3} - {NH}_{2} + HI\]
\[{142\ g\ CH}_{3}I\ldots\ldots\ldots\ldots\ldots\ldots\ldots 17g\ {NH}_{3}\]
\[{284\ g\ CH}_{3}I\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {NH}_{3} \rightarrow x = 34g\ {NH}_{3},\ amoniac\ \]
ALCOOLI
Rezolvare:\({CH}_{3}Cl + H_{2}O\overset{NaOH}{\rightarrow}{CH}_{3} - OH\)
\(\ \ 50,5\ g\ {CH}_{3}Cl\ldots\ldots\ldots\ldots.32g\ {CH}_{3} - OH\)
\[101\ g\ {CH}_{3}Cl\ldots\ldots\ldots\ldots.x\ g\ {CH}_{3} - OH \rightarrow x = 64g\ {CH}_{3} - OH,\ metanol\]
80. Cât etanol se obține prin adiția apei în prezența acidului sulfuric la 84 g etenă?
Rezolvare:\({CH}_{2} = {CH}_{2} + HOH\overset{H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - {CH}_{2} - OH\)
\[28\ g\ {CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots..46\ g{CH}_{3} - {CH}_{2} - OH\]
\[84\ g\ {CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots..x\ g{CH}_{3} - {CH}_{2} - OH \rightarrow x = 138g\ {CH}_{3} - {CH}_{2} - OH,\ etanol\ \]
81. Cât 2 – propanol se obține prin adiția apei în prezența acidului sulfuric la 126 g propenă?
Rezolvare:\({CH}_{2} = CH - {CH}_{3} + HOH\overset{H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - CH(OH) - {CH}_{3}\)
\[{42\ g\ CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots.60\ g\ {CH}_{3} - CH(OH) - {CH}_{3}\]
\[{126\ g\ CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots x\ g\ {CH}_{3} - CH(OH) - {CH}_{3}\]
\[\rightarrow x = 180\ g\ {CH}_{3} - CH(OH) - {CH}_{3},\ 2 - propanol\]
82. Cât etanol se obține prin hidrogenarea a 88 g etanal?
Rezolvare:\(\ {CH}_{3} - HC = O + H - H\overset{Ni}{\rightarrow}{CH}_{3} - {CH}_{2} - OH\)
\[44g\ {CH}_{3} - HC = O\ldots\ldots\ldots\ldots\ldots\ldots 46\ g\ {CH}_{3} - {CH}_{2} - OH\]
\[88g\ {CH}_{3} - HC = O\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{3} - {CH}_{2} - OH\]
\[\rightarrow x = 92\ g{CH}_{3} - {CH}_{2} - OH,\ etanol\ \]
83. Cât 2 – propanol se obține prin hidrogenarea a 174 g propanonă?
Rezolvare:\(\ {({CH}_{3})}_{2}C = O + H - H\overset{Ni}{\rightarrow}\left( {CH}_{3} \right)_{2}CH - OH\)
\[{58\ g\ ({CH}_{3})}_{2}C = O\ldots\ldots\ldots\ldots\ldots..60\ g\ \left( {CH}_{3} \right)_{2}CH - OH\]
\[{174\ g\ \left( {CH}_{3} \right)}_{2}C = O\ldots\ldots\ldots\ldots\ldots..x\ \ g\ \left( {CH}_{3} \right)_{2}CH - OH\]
\[\rightarrow x = 180\ g\ \left( {CH}_{3} \right)_{2}CH - OH,\ 2 - propanol\]
84. Cât etoxid de sodiu se obține prin reacția a 69 g sodiu cu etanolul?
Rezolvare:\({CH}_{3} - {CH}_{2} - OH + Na \rightarrow {CH}_{3} - {CH}_{2} - ONa + \frac{1}{2}H_{2}\)
\[23\ g\ Na\ldots\ldots\ldots\ldots\ldots 68\ g\ {CH}_{3} - {CH}_{2} - ONa\]
\[69\ g\ Na\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 204\ g\ {CH}_{3} - {CH}_{2} - ONa\ (\ etoxid\ de\ sodiu)\]
85. Câtă propenă se obține prin eliminarea intramoleculară a apei, din 120 g 2 – propanol?
Rezolvare:\({CH}_{3} - CH(OH) - {CH}_{3}\overset{H_{2}{SO}_{4}}{\rightarrow}{CH}_{2} = CH - {CH}_{3}\)
\[{60\ g\ CH}_{3} - CH(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots..42\ g{CH}_{2} = CH - {CH}_{3}\]
\[{120\ g\ CH}_{3} - CH(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots..x \rightarrow x = 84\ g\ {CH}_{2} = CH - {CH}_{3}\ (propenă)\]
86. Cât dietileter se obține prin eliminarea intermoleculară a apei, din 92 g etanol?
Rezolvare:\({CH}_{3} - {CH}_{2} - OH + HO - \ {CH}_{2} - {CH}_{3}\overset{ {\ puțin\ H}_{2}{SO}_{4}}{\rightarrow}C_{2}H_{5} - O - C_{2}H_{5}\)
\(92\ g{CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots..74\ gC_{2}H_{5} - O - C_{2}H_{5}(dietileter)\)
Rezolvare:\({CH}_{3} - OH + HOSO_{3}H \rightarrow {CH}_{3} - OSO_{3}H + H_{2}O\)
\[{32\ g\ CH}_{3} - OH\ldots\ldots\ldots\ldots 112\ g{CH}_{3} - OSO_{3}H\]
\[{96\ g\ CH}_{3} - OH\ldots\ldots\ldots\ldots x \rightarrow x = 336\ g{CH}_{3} - OSO_{3}H\]
88. Cât acetat de etil se obține prin reacția cu acidul acetic a 460 g etanol?
Rezolvare:\({CH}_{3} - COOH + C_{2}H_{5} - OH \rightarrow {CH}_{3} - COOC_{2}H_{5} + H_{2}O\)
\[46\ gC_{2}H_{5} - OH\ldots\ldots\ldots\ldots\ldots.88\ g\ {CH}_{3} - COOC_{2}H_{5}\]
\[460gC_{2}H_{5} - OH\ldots\ldots x \rightarrow x = 880\ g\ {CH}_{3} - COOC_{2}H_{5}\ (acetat\ de\ etil)\ \]
89. Cât propanal se obține prin oxidarea blândă cu bicromat de potasiu în mediu de acid sulfuric a 120 g propanol?
Rezolvare:\({CH}_{3} - {C\ H}_{2} - {CH}_{2} - OH\overset{K_{2}{Cr}_{2}O_{7},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - {C\ H}_{2} - HC = O\)
\[60\ g\ {CH}_{3} - {C\ H}_{2} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..58\ g{CH}_{3} - {C\ H}_{2} - HC = O\]
\[120\ g\ {CH}_{3} - {C\ H}_{2} - {CH}_{2} - OH\ldots x \rightarrow x = 116\ g{CH}_{3} - {C\ H}_{2} - HC = O(\ propanal\ )\]
90. Cât acid propionic se obține prin oxidarea energică cu permanganat de potasiu în mediu de acid sulfuric a 240 g propanol?
Rezolvare:\({CH}_{3} - {C\ H}_{2} - {CH}_{2} - OH\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - {C\ H}_{2} - COOH\)
\[60\ g\ {CH}_{3} - {C\ H}_{2} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots 74\ g{CH}_{3} - {C\ H}_{2} - COOH\]
\[240\ g\ {CH}_{3} - {C\ H}_{2} - {CH}_{2} - OH\ldots x \rightarrow x = 296\ g{CH}_{3} - {C\ H}_{2} - COOH\ (acid\ propionic)\]
91. Câtă acetonă se obține prin oxidarea blândă cu bicromat de potasiu în mediu de acid sulfuric a 180 g 2 – propanol?
Rezolvare:\({CH}_{3} - CH(OH) - {CH}_{3}\overset{K_{2}{Cr}_{2}O_{7},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - CO - {CH}_{3}\)
\[60\ g{CH}_{3} - CH(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{58\ gCH}_{3} - CO - {CH}_{3}\]
\[180\ g{CH}_{3} - CH(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots{x \rightarrow x = 174\ gCH}_{3} - CO - {CH}_{3}\ (acetonă)\]
Rezolvare:\(CO + 2H_{2}\overset{ZnO,\ {Cr}_{2}O_{3},350℃,\ 250atm}{\Rightarrow}{CH}_{3} - OH\)
\[32\ g\ gaz\ de\ sinteză\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 32\ g{CH}_{3} - OH\]
\[x\ g\ gaz\ de\ sinteză\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 64\ g{CH}_{3} - OH \rightarrow x = 64\ g\ gaz\ de\ sinteză\]
93. Cât etanol se obține prin fermentația alcoolică a 540 g glucoză?
Rezolvare: \(C_{6}H_{12}O_{6}\overset{Saccharomyces\ cerevisiae}{\rightarrow}{2CO}_{2} + 2C_{2}H_{5} - OH\)
\[180\ g\ C_{6}H_{12}O_{6}\ldots\ldots\ldots\ldots\ldots 92\ gC_{2}H_{5} - OH\]
\[540\ g\ C_{6}H_{12}O_{6}\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 276\ gC_{2}H_{5} - OH\ (etanol)\]
94. Cât glicol se obține prin oxidarea etenei la oxid de etenă, urmată de hidroliză, dacă s – au folosit 140 g etenă?
Rezolvare:
\[{CH}_{2} = {CH}_{2} + \frac{1}{2}O_{2}\overset{Ag,\ 250{^\circ}C}{\rightarrow}C_{2}H_{4}O(oxid\ de\ etenă)\overset{H_{2}O}{\rightarrow}HO - {CH}_{2} - {CH}_{2} - OH\]
\[28\ g{CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.62\ gHO - {CH}_{2} - {CH}_{2} - OH\]
\[140\ g{CH}_{2} = {CH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ gHO - {CH}_{2} - {CH}_{2} - OH\]
\[\rightarrow x = 310gHO - {CH}_{2} - {CH}_{2} - OH,\ glicol\]
\[\mathbf{CH}_{\mathbf{2}}\mathbf{= CH -}\mathbf{CH}_{\mathbf{3}}\overset{\mathbf{+}\mathbf{Cl}_{\mathbf{2}}\mathbf{,\ 500{^\circ}C}}{\rightarrow}\overset{\rightarrow}{\mathbf{- HCl}}\mathbf{CH}_{\mathbf{2}}\mathbf{= CH -}\mathbf{CH}_{\mathbf{2}}\mathbf{Cl}\overset{\mathbf{NaOH,\ }\mathbf{H}_{\mathbf{2}}\mathbf{O}}{\rightarrow}\overset{\rightarrow}{\mathbf{- NaCl}}\mathbf{CH}_{\mathbf{2}}\mathbf{= CH -}\mathbf{CH}_{\mathbf{2}}\mathbf{- OH\ }\]
\[\mathbf{CH}_{\mathbf{2}}\mathbf{= CH -}\mathbf{CH}_{\mathbf{2}}\mathbf{- OH}\overset{\mathbf{Cl}_{\mathbf{2}}}{\rightarrow}\mathbf{Cl - CH}_{\mathbf{2}}\mathbf{- (Cl)CH -}\mathbf{CH}_{\mathbf{2}}\mathbf{- OH}\overset{\mathbf{+ 2}\mathbf{NaOH,}\mathbf{H}_{\mathbf{2}}\mathbf{O}}{\rightarrow}\overset{\rightarrow}{\mathbf{- 2}\mathbf{NaCl}}\]
\(\mathbf{\rightarrow}\mathbf{HO - CH}_{\mathbf{2}}\mathbf{- (HO)CH -}\mathbf{CH}_{\mathbf{2}}\mathbf{- OH}\)
Rezolvare: \({CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{HO - CH}_{2} - (HO)CH - {CH}_{2} - OH\)
\(42\ g\ {CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.92g\ {HO - CH}_{2} - (HO)CH - {CH}_{2} - OH\)
\(210\ g\ {CH}_{2} = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {HO - CH}_{2} - (HO)CH - {CH}_{2} - OH\)
\(\rightarrow x = 460\ g\ {HO - CH}_{2} - (HO)CH - {CH}_{2} - OH,\ glicerină\)
96. Câtă acroleină se obține prin deshidratarea a 230 g glicerină?
Rezolvare:
\[{HO - CH}_{2} - (HO)CH - {CH}_{2} - OH\overset{H_{2}{SO}_{4}}{\rightarrow}\overset{\rightarrow}{- {2H}_{2}O}{CH}_{2} = CH - CHO\]
\[{92g\ HO - CH}_{2} - (HO)CH - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.56\ g\ {CH}_{2} = CH - CHO\]
\[{230g\ HO - CH}_{2} - (HO)CH - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CH}_{2} = CH - CHO\]
\[\rightarrow x = 140\ g\ {CH}_{2} = CH - CHO,\ acroleină\]
97. Cât trinitrat de glicerină se obține prin reacția de esterificare a 460 g glicerină cu acidul azotic?
Rezolvare:
\[{CH}_{2}(OH) - CH(OH) - {CH}_{2}(OH) + 3HO{NO}_{2} \rightarrow {CH}_{2}\left( O - {NO}_{2} \right) - CH\left( O - {NO}_{2} \right) - {CH}_{2}({O - NO}_{2})\ \]
\[92\ g{CH}_{2}(OH) - CH(OH) - {CH}_{2}(OH)\ldots\ldots\ldots 227\ g\ {CH}_{2}\left( O - {NO}_{2} \right) - CH\left( O - {NO}_{2} \right) - {CH}_{2}({O - NO}_{2})\ \]
\[460\ g{CH}_{2}(OH) - CH(OH) - {CH}_{2}(OH)\ldots\ldots\ldots x\ g\ {CH}_{2}\left( O - {NO}_{2} \right) - CH\left( O - {NO}_{2} \right) - {CH}_{2}({O - NO}_{2})\]
\[x = 1135\ g\ {CH}_{2}\left( O - {NO}_{2} \right) - CH\left( O - {NO}_{2} \right) - {CH}_{2}\left( {O - NO}_{2} \right),\ trinitrat\ de\ glicerină\]
98. Ce cantitate de bioxid de carbon se obține prin descompunerea a 454 g trinitrat de glicerină?
Rezolvare:
\[{4C}_{3}H_{5}{({ONO}_{2})}_{3} \rightarrow 12{CO}_{2} + 10H_{2}O + 6N_{2} + O_{2}\]
\[{908\ g\ C}_{3}H_{5}{({ONO}_{2})}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 528\ g\ {CO}_{2}\]
\[{454\ g\ C}_{3}H_{5}{({ONO}_{2})}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ {CO}_{2} \rightarrow x = 264\ g\ {CO}_{2},\ bioxid\ de\ carbon\]
FENOLI
Rezolvare:
\[\left\{ \begin{array}{r}
C\ \frac{65,45}{12} = 5,45 \rightarrow \frac{5,45}{1,81} = 3 \\
H\frac{5,45}{1} = 5,45 \rightarrow \frac{5,45}{1,81} = 3 \\
O\frac{29}{16} = 1,81 \rightarrow \frac{1,81}{1,81} = 1
\end{array} \right.\ \rightarrow C_{3}H_{3}O\ este\ formula\ brută\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{110}{55} = 2\]
\[\rightarrow {(C_{3}H_{3}O)}_{2}\ sau\ C_{6}H_{6}O_{2}\ este\ formula\ moleculară,\ poate\ fi\ hidrochinonă\]
100. Cât fenol se obține din 90 g benzensulfonat de sodiu prin topire alcalină?
Rezolvare:
\[C_{6}H_{5} - {SO}_{3}Na + NaOH\overset{300℃}{\rightarrow}C_{6}H_{5} - OH + {Na}_{2}{SO}_{3}\]
\[180\ g\ C_{6}H_{5} - {SO}_{3}Na\ldots\ldots\ldots\ldots\ldots\ldots\ldots..94\ gC_{6}H_{5} - OH\]
\[90\ g\ C_{6}H_{5} - {SO}_{3}Na\ldots\ldots\ldots\ldots\ldots\ldots\ldots..x\ gC_{6}H_{5} - OH \rightarrow x = 47\ g\ C_{6}H_{5} - OH,\ fenol\]
101. Cât \(\mathbf{\propto -}\)naftol rezultă folosind ca materii prime 32 g naftalină, acid sulfuric și hidroxid de sodiu?
Rezolvare:
\[C_{10}H_{8} + HO{SO}_{3}H\overset{- H_{2}O}{\rightarrow}C_{10}H_{7} - {SO}_{3}H\overset{+ NaOH}{\rightarrow}\overset{\rightarrow}{- H_{2}O}C_{10}H_{7} - {SO}_{3}Na\overset{+ NaOH}{\rightarrow}C_{10}H_{7} - OH + {Na}_{2}{SO}_{3}\]
\[128\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..144\ gC_{10}H_{7} - OH\]
\[32\ g\ C_{10}H_{8}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..x\ gC_{10}H_{7} - OH \rightarrow x = 36\ gC_{10}H_{7} - OH, \propto - naftol\ \]
102. Cât fenol se obține prin oxidarea a 240 g izopropilbenzen ?
Rezolvare: \(C_{6}H_{5} - CH\left( {CH}_{3} \right)_{2} + O_{2} \rightarrow \overset{H_{2}{SO}_{4}}{\rightarrow}C_{6}H_{5} - OH + CO{({CH}_{3})}_{2}\)
\[{120\ gC}_{6}H_{5} - CH\left( {CH}_{3} \right)_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 94\ g\ C_{6}H_{5} - OH\]
\[{240\ gC}_{6}H_{5} - CH\left( {CH}_{3} \right)_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ g\ C_{6}H_{5} - OH\]
\[\rightarrow x = 188\ g\ C_{6}H_{5} - OH,\ fenol\]
103. Cât fenol rezultă din reacția a 464 g fenoxid de sodiu cu acidul carbonic?
Rezolvare:
\[C_{6}H_{5} - ONa + H_{2}{CO}_{3} \rightarrow C_{6}H_{5} - OH + {NaHCO}_{3}\]
\[116\ g\ C_{6}H_{5} - ONa\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{94\ g\ C}_{6}H_{5} - OH\]
\[464\ g\ C_{6}H_{5} - ONa\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{x\ g\ C}_{6}H_{5} - OH\]
\[\rightarrow x = \ {376\ g\ C}_{6}H_{5} - OH,\ fenol\]
104. Cât fenol reacționează cu 120 g hidroxid de sodiu ?
Rezolvare:
\[C_{6}H_{5} - OH + NaOH \rightarrow C_{6}H_{5} - ONa + H_{2}O\]
\[94\ g\ C_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 40\ g\ NaOH\]
\[x\ g\ C_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 120\ g\ NaOH\]
\[\rightarrow x = 282g\ C_{6}H_{5} - OH,\ fenol\ \]
105. Cât monoclorbenzen rezultă din reacția a 188 g fenol cu acidul clorhidric?
Rezolvare:
\(\ C_{6}H_{5} - OH + HCl \rightarrow C_{6}H_{5} - Cl + H_{2}O\ \)
\[{94\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots..112,5\ g\ C_{6}H_{5} - Cl\]
\[{188\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ g\ C_{6}H_{5} - Cl\]
\[\rightarrow x = 225\ gC_{6}H_{5} - Cl,\ monoclorbenzen\]
106. Cât tribromfenol rezultă prin bromurarea a 282 g fenol?
Rezolvare:
\(\ C_{6}H_{5} - OH + 3{Br}_{2} \rightarrow C_{6}H_{2}{Br}_{3}OH + 3HBr\ \)
\[{94\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots 331\ g\ C_{6}H_{2}{Br}_{3}OH\]
\[{282\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots x\ g\ C_{6}H_{2}{Br}_{3}OH\]
\[\rightarrow x = 993\ g\ C_{6}H_{2}{Br}_{3}OH,\ tribromfenol\]
107. Cât ciclohexanol rezultă prin hidrogenarea a 188 g fenol?
Rezolvare:
\[C_{6}H_{5} - OH + 3H_{2}\overset{presiune,\ Ni,\ 200℃}{\rightarrow} \rightarrow C_{6}H_{11}OH\]
\({94\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.100\ gC_{6}H_{11}OH\)
\[{188\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ gC_{6}H_{11}OH\]
\[\rightarrow x = \ 200\ gC_{6}H_{11}OH,\ ciclohexanol\]
108. Cât acid picric se obține prin reacția a 282 g fenol cu amestec sulfonitric, la încălzire?
Rezolvare:
{width=”7.036231408573928in” height=”2.1805522747156605in”}
\({94\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.229\ gC_{6}H_{3}N_{3}O_{7}\)
\[{282\ g\ C}_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x\ gC_{6}H_{3}N_{3}O_{7}\]
\[\rightarrow x = 687gC_{6}H_{3}N_{3}O_{7}\ acid\ picric\ sau\ 2,4,6 - trinitrofenol\]
AMINE
Rezolvare:
\[\left\{ \begin{array}{r}
C\ \frac{78,5}{12} = 6,54 \rightarrow \frac{6,54}{0,93} = 7 \\
H\frac{8,41}{1} = 8,41 \rightarrow \frac{8,41}{0,93} = 9 \\
N\frac{13,08}{14} = 0,93 \rightarrow \frac{0,93}{0,93} = 1
\end{array} \right.\ \rightarrow C_{7}H_{9}N\ este\ formula\ brută\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{107}{107} = 1\]
\[\rightarrow C_{7}H_{9}N\ este\ formula\ moleculară\ sau\ {CH}_{3} - C_{6}H_{4} - {NH}_{2}\]
110. Ce structură poate avea o amină care are M = 45 și care conține 53,33%C, 15,55% H și 31,11%N ?
Rezolvare:
\[\left\{ \begin{array}{r}
C\ \frac{53,33}{12} = 4,44 \rightarrow \frac{4,44}{2,22} = 2 \\
H\frac{15,55}{1} = 15,55 \rightarrow \frac{15,55}{2,22} = 7 \\
N\frac{31,11}{14} = 2,22 \rightarrow \frac{2,22}{2,22} = 1
\end{array} \right.\ \rightarrow C_{2}H_{7}N\ este\ formula\ brută\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{45}{45} = 1\]
\[\rightarrow C_{2}H_{7}N\ este\ și\ formula\ moleculară\]
\[\rightarrow \mathbf{CH}_{\mathbf{3}}\mathbf{-}\mathbf{CH}_{\mathbf{2}}\mathbf{-}\mathbf{NH}_{\mathbf{2}}\mathbf{,}\ etilamina,\ o\ amină\ primară\ \]
\(sau\ \mathbf{CH}_{\mathbf{3}}\mathbf{- NH -}\mathbf{CH}_{\mathbf{3}}\), dimetilamina, o amină secundară
111. Ce structură poate avea o amină care are M = 59 și care conține 61,01%C, 15,25% H și 23,72%N ?
Rezolvare:
\[\left\{ \begin{array}{r}
C\ \frac{61,01}{12} = 5,08 \rightarrow \frac{5,08}{1,69} = 3 \\
H\frac{15,25}{1} = 15,25 \rightarrow \frac{15,25}{1,69} = 9 \\
N\frac{23,72}{14} = 1,69 \rightarrow \frac{1,69}{1,69} = 1
\end{array} \right.\ \rightarrow C_{3}H_{9}N\ este\ formula\ brută\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{59}{59} = 1\]
\[\rightarrow C_{3}H_{9}N\ este\ și\ formula\ moleculară\]
\[{CH}_{3} - {CH}_{2} - {CH}_{2} - {NH}_{2},\ propil - amina,\ amină\ primară\]
\[{CH}_{3} - CH\left( {NH}_{2} \right) - {CH}_{3},\ izopropil - amina,\ amină\ primară\]
\[{CH}_{3} - {CH}_{2} - NH - {CH}_{3},\ etil,metil - amina,\ amină\ secundară\]
\[{CH}_{3} - {N(CH}_{3}) - {CH}_{3},\ trimetil - amina,\ amină\ terțiară\]
112. Ce structură poate avea o amină cu M = 60 și care conține 40%C, 46,66%N, 13,33%H ?
Rezolvare:
\[\left\{ \begin{array}{r}
C\ \frac{40}{12} = 3,33 \rightarrow \frac{3,33}{3,33} = 1 \\
H\frac{13,33}{1} = 13,33 \rightarrow \frac{13,33}{3,33} = 4 \\
N\frac{46,66}{14} = 3,33 \rightarrow \frac{3,33}{3,33} = 1
\end{array} \right.\ \rightarrow CH_{4}N\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{60}{30} = 2\]
\[\rightarrow C_{2}H_{8}N_{2}\ este\ și\ formula\ moleculară\ \]
\[și\ poate\ fi\ H_{2}N - {CH}_{2} - {CH}_{2} - {NH}_{2},\ etilendiamină\]
113. Să se calculeze ce cantitate de anilină se obține prin reducerea catalitică a 369 g nitrobenzen.
Rezolvare:
\[C_{6}H_{5} - {NO}_{2}\overset{+ {6H}^{+}}{\rightarrow}\overset{\rightarrow}{+ 6e^{-}}C_{6}H_{5} - {NH}_{2} + 2H_{2}O\]
\[123\ gC_{6}H_{5} - {NO}_{2}\ldots\ldots\ldots\ldots\ldots..93\ gC_{6}H_{5} - {NH}_{2}\]
\[369\ gC_{6}H_{5} - {NO}_{2}\ldots\ldots\ldots\ldots\ldots..x \rightarrow x = 279gC_{6}H_{5} - {NH}_{2},\ anilină\]
114. Să se calculeze ce cantitate de \(\mathbf{\alpha - naftilamină}\) se obține prin reducerea catalitică a 346g \(\mathbf{\alpha - nitonaftalină}\) .
Rezolvare:
\[C_{10}H_{7} - {NO}_{2}\overset{+ {6H}^{+}}{\rightarrow}\overset{\rightarrow}{+ 6e^{-}}C_{10}H_{7} - {NH}_{2} + 2H_{2}O\]
\[{173\ gC}_{10}H_{7} - {NO}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{143\ gC}_{10}H_{7} - {NH}_{2}\]
\({346\ gC}_{10}H_{7} - {NO}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = {286\ gC}_{10}H_{7} - {NH}_{2},\alpha - naftilamină\)
115. Să se calculeze cantitatea de etilamină care se obține prin reducerea a 164 g acetonitril.
Rezolvare:
\[{CH}_{3} - CN\ \overset{+ 2H_{2}}{\rightarrow}{CH}_{3} - {CH}_{2} - {NH}_{2}\]
\[41\ g\ \ {CH}_{3} - CN\ \ldots\ldots\ldots\ldots..45\ g\ {CH}_{3} - {CH}_{2} - {NH}_{2}\]
\[164\ g\ \ {CH}_{3} - CN\ \ldots\ldots\ldots\ldots..x \rightarrow x = \ 180\ g\ {CH}_{3} - {CH}_{2} - {NH}_{2},\ etilamină\]
116. Să se calculeze cantitatea de acetanilidă care se obține prin acilarea cu acid acetic a 276 g anilină.
Rezolvare:
\[C_{6}H_{5} - {NH}_{2} + {CH}_{3} - COOH \rightarrow C_{6}H_{5} - NH - CO - {CH}_{3} + H_{2}O\]
\[92\ gC_{6}H_{5} - {NH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 135\ g\ C_{6}H_{5} - NH - CO - {CH}_{3}\]
\[276g\ C_{6}H_{5} - {NH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow = \mathbf{405\ g\ }\mathbf{C}_{\mathbf{6}}\mathbf{H}_{\mathbf{5}}\mathbf{- NH - CO -}\mathbf{CH}_{\mathbf{3}}\mathbf{,\ acetanilidă}\]
117. Să se calculeze cantitatea de benzanilidă care se obține prin acilarea cu clorură de benzoil a 184 g anilină.
Rezolvare:
\[C_{6}H_{5} - {NH}_{2} + C_{6}H_{5} - COCl \rightarrow C_{6}H_{5} - NH - CO - C_{6}H_{5} + HCl\]
\[92\ g\ C_{6}H_{5} - {NH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 197\ g\ C_{6}H_{5} - NH - CO - C_{6}H_{5}\]
\[184\ g\ C_{6}H_{5} - {NH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 394\ g\ C_{6}H_{5} - NH - CO - C_{6}H_{5},\ benzanilidă\]
118. Să se calculeze cantitatea de acetil p – toluidină care se obține prin acilarea cu clorură de acetil a 214 g p – toluidină.
Rezolvare:
\[{CH}_{3} - C_{6}H_{4} - {NH}_{2} + {CH}_{3} - COCl \rightarrow {CH}_{3} - C_{6}H_{4} - NH - CO - {CH}_{3} + HCl\]
\[{107\ g\ CH}_{3} - C_{6}H_{4} - {NH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..149\ g\ {CH}_{3} - C_{6}H_{4} - NH - CO - {CH}_{3}\]
\[{214\ g\ CH}_{3} - C_{6}H_{4} - {NH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\]
\[\rightarrow x = 298\ g\ {CH}_{3} - C_{6}H_{4} - NH - CO - {CH}_{3},\ acetil\ p - toluidină\]
ALDEHIDE
Rezolvare:
\[{CH}_{4} + O_{2} \rightarrow {CH}_{2}O + H_{2}O\]
\[{16\ gCH}_{4}\ldots\ldots\ldots\ldots..30g{CH}_{2}O\]
\[{160\ gCH}_{4}\ldots\ldots\ldots\ldots.x \rightarrow x = 300g{CH}_{2}O\ aldehidă\ formică\]
Rezolvare:
\[{CH}_{3}OH\overset{Cu}{\rightarrow}\overset{\rightarrow}{250℃}{CH}_{2} = O + H_{2}\]
\[32\ g{CH}_{3}OH\ldots\ldots\ldots\ldots 30g{CH}_{2} = O\]
\[64\ g{CH}_{3}OH\ldots\ldots\ldots\ldots x \rightarrow x = 60g{CH}_{2} = O,\ aldehidă\ formică\]
121. Ce cantitate de aldehidă acetică se obține prin oxidarea în prezența dicromatului de potasiu și a acidui sulfuric, a 92 g alcool etilic?
Rezolvare:
\[{CH}_{3} - {CH}_{2} - OH + \left\lceil O \right\rceil \rightarrow {CH}_{3} - CH = O\]
\[46\ g{CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots..44\ g{CH}_{3} - CH = O\]
\[92\ g{CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = 88\ g{CH}_{3} - CH = O,\ aldehidă\ acetică\]
122. Ce cantitate de benzaldehidă se obține prin hidroliza a 483 g clorură de benziliden, în mediu slab bazic?
Rezolvare:
\[C_{6}H_{5} - {CHCl}_{2}\overset{hidroliză}{\rightarrow}C_{6}H_{5} - HC = O\]
\[{161\ g\ \ C}_{6}H_{5} - {CHCl}_{2}\ldots\ldots\ldots\ldots\ldots\ldots..106\ gC_{6}H_{5} - HC = O\]
\[{483\ g\ \ C}_{6}H_{5} - {CHCl}_{2}\ldots\ldots\ldots\ldots.x \rightarrow x = \ 318\ gC_{6}H_{5} - HC = O,\ benzaldehidă\]
Rezolvare:
\[{CH}_{2} = O + H_{2} \rightarrow {CH}_{3}OH\]
\[{30\ gCH}_{2} = O\ldots\ldots\ldots\ldots\ldots.32\ g{CH}_{3}OH\]
\[{90\ gCH}_{2} = O\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = 96\ g{CH}_{3}OH,\ metanol\]
124. Ce cantitate de etanol se obține prin adiția hidrogenului la 88 g de acetaldehidă?
Rezolvare:
\[{CH}_{3} - CH = O + H_{2} \rightarrow {CH}_{3} - {CH}_{2} - OH\]
\[44g{CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots\ldots 46g{CH}_{3} - {CH}_{2} - OH\]
\[88g{CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 92g{CH}_{3} - {CH}_{2} - OH,\ etanol\]
125. Ce cantitate de metilcianhidrină se obține prin adiția acidului cianhidric la 132 g acetaldehidă?
Rezolvare:
\({CH}_{3} - CH = O + HCN \rightarrow {CH}_{3} - HC(OH) - CN\)
\(44\ g\ {CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots\ldots..71\ g{CH}_{3} - HC(OH) - CN\)
\(132\ g\ {CH}_{3} - CH = O\ldots\ldots\ldots x \rightarrow x = \ 213\ g{CH}_{3} - HC(OH) - CN,\ metilcianhidrină\)
126. Ce cantitate de aldol se obține prin condensarea aldolică a 176 g acetaldehidă?
Rezolvare:
\[\ \ \ \ \ \ \ \ \ \ \ \ {CH}_{3} - CH = O + {CH}_{3} - CH = O \rightarrow {CH}_{3} - HC(OH) - {CH}_{2} - CH = O\]
\(88\ g{CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 88\ g{CH}_{3} - HC(OH) - {CH}_{2} - CH = O\)
\(176\ g{CH}_{3} - CH = O\ldots\ldots\ldots\ldots x \rightarrow x = 176\ g{CH}_{3} - HC(OH) - {CH}_{2} - CH = O,\ aldol\)
127. Ce cantitate de aldehidă crotonică se obține prin condensarea crotonică a 264 g acetaldehidă?
Rezolvare:
\({CH}_{3} - CH = O + {CH}_{3} - CH = O \rightarrow {CH}_{3} - CH = CH - CH = O + H_{2}O\)
\({88\ gCH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots 70\ g{CH}_{3} - CH = CH - CH = O\)
\[{264\ gCH}_{3} - CH = O\ldots x \rightarrow x = 210\ g{CH}_{3} - CH = CH - CH = O,\ aldehidă\ crotonică\]
Rezolvare:
\[C_{6}H_{5} - OH + {CH}_{2} = O + C_{6}H_{5} - OH\overset{m.\ acid}{\rightarrow}HO - C_{6}H_{4} - {CH}_{2} - C_{6}H_{4} - OH + H_{2}O\]
\[188\ gC_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 200\ g\ HO - C_{6}H_{4} - {CH}_{2} - C_{6}H_{4} - OH\]
\[94\ gC_{6}H_{5} - OH\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 100\ g\ HO - C_{6}H_{4} - {CH}_{2} - C_{6}H_{4} - OH\]
Rezolvare:
\(C_{6}H_{5} - OH + {CH}_{2} = O\overset{mediu\ bazic}{\rightarrow}HO - C_{6}H_{4} - {CH}_{2}OH\)
\[{30\ g\ CH}_{2} = O\ldots\ldots\ldots\ldots\ldots\ldots.124\ gHO - C_{6}H_{4} - {CH}_{2}OH\]
\[{270\ g\ CH}_{2} = O\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 1116gHO - C_{6}H_{4} - {CH}_{2}OH,\ alcool\ hidroxibenzilic\]
130. Ce cantitate de acid acetic se obține prin oxidarea cu permanganat de potasiu sau cu dicromat de potasiu a 220 g aldehidă acetică?
Rezolvare:
\[{CH}_{3} - CH = O + \lbrack O\rbrack \rightarrow {CH}_{3} - COOH\]
\[44g\ {CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots 60g\ {CH}_{3} - COOH\]
\[220g\ {CH}_{3} - CH = O\ldots\ldots\ldots\ldots x \rightarrow x = 300g\ {CH}_{3} - COOH\ (acid\ acetic)\]
Rezolvare:
\[{CH}_{2} = O + 2\left\lbrack Ag({NH}_{3})_{2} \right\rbrack OH \rightarrow H - COOH + {4NH}_{3} + H_{2}O + 2Ag\]
\(30\ g\ {CH}_{2} = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.216\ g\ Ag\)
\(90\ g\ {CH}_{2} = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x\ \rightarrow x = \ 648\ g\ Ag\)
CETONE
132. Ce cantitate de acetonă se obține prin oxidarea cu permanganat de potasiu în mediu de acid sulfuric a 140 g de 2 – metil – 2 – butenă ?
Rezolvare:
\[{CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3} + 3\lbrack O\rbrack\overset{ {KMnO}_{4}}{\rightarrow}\overset{\rightarrow}{H_{2}{SO}_{4}}{CH}_{3} - CO - {CH}_{3} + HOOC - {CH}_{3}\]
\[70\ g\ {CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots..58g\ {CH}_{3} - CO - {CH}_{3}\]
\[140\ g\ {CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3}\ldots\ldots\ldots\ldots x \rightarrow x = 116g\ {CH}_{3} - CO - {CH}_{3},\ acetonă\]
133. Ce cantitate de acetonă se obține prin oxidarea cu permanganat de potasiu în mediu de acid sulfuric a 168 g de 2,3 – dimetil – 2 – butenă ?
Rezolvare:
\[{CH}_{3} - C\left( {CH}_{3} \right) = C\left( {CH}_{3} \right) - {CH}_{3} + 2\lbrack O\rbrack\overset{ {KMnO}_{4}}{\rightarrow}\overset{\rightarrow}{H_{2}{SO}_{4}}{CH}_{3} - CO - {CH}_{3} + {CH}_{3} - CO - {CH}_{3}\]
\[\ \ \ \ \ \ \ \ \ 84\ g\ {CH}_{3} - C\left( {CH}_{3} \right) = C\left( {CH}_{3} \right) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 116\ g\ {CH}_{3} - CO - {CH}_{3}\]
\(168\ g\ {CH}_{3} - C\left( {CH}_{3} \right) = C\left( {CH}_{3} \right) - {CH}_{3}\ldots\ldots x \rightarrow x = 232\ \ g\ {CH}_{3} - CO - {CH}_{3},\ acetonă\)
134. Ce cantitate de acetonă se obține prin oxidarea a 180 g izopropanol cu agenți oxidanți moderați?
Rezolvare:
\({CH}_{3} - HC(OH) - {CH}_{3} + \lbrack O\rbrack \rightarrow {CH}_{3} - CO - {CH}_{3} + H_{2}O\)
\({\ \ \ 60\ g\ CH}_{3} - HC(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots 58\ g\ {CH}_{3} - CO - {CH}_{3} + H_{2}O\)
\({\ 180\ g\ CH}_{3} - HC(OH) - {CH}_{3}\ldots\ldots\ldots\ldots x \rightarrow x = 174\ g{CH}_{3} - CO - {CH}_{3},\ acetonă\)
135. Ce cantitate de acetonă se obține prin adiția apei la 80 g propină?
Rezolvare:
\[{CH}_{3} - C \equiv CH + HOH \rightarrow {CH}_{3} - CO - {CH}_{3}\]
\[40\ g\ {CH}_{3} - C \equiv CH\ldots\ldots\ldots\ldots..58\ g\ {CH}_{3} - CO - {CH}_{3}\]
\[80\ g\ {CH}_{3} - C \equiv CH\ldots\ldots\ldots\ldots x \rightarrow x = \ 116\ g\ {CH}_{3} - CO - {CH}_{3},\ acetonă\]
136. Ce cantitate de 2 – propanol se obține prin adiția hidrogenului la 174 g acetonă ?
Rezolvare:
\[{CH}_{3} - CO - {CH}_{3} + H_{2} \rightarrow {CH}_{3} - HC(OH) - {CH}_{3}\]
\[\ 58\ g{CH}_{3} - CO - {CH}_{3}\ldots\ldots\ldots\ldots 60\ g{CH}_{3} - HC(OH) - {CH}_{3}\]
\[174\ g{CH}_{3} - CO - {CH}_{3}\ldots\ldots\ldots x \rightarrow x = 180\ g{CH}_{3} - HC(OH) - {CH}_{3},\ 2 - propanol\]
137. Ce cantitate de dimetilcianhidrină se obține prin adiția acidului cianhidric la 232 g acetonă?
Rezolvare:
\({CH}_{3} - CO - {CH}_{3} + HCN \rightarrow ({CH}_{3})_{2}C(OH) - CN\)
58 g \({CH}_{3} - CO - {CH}_{3}\ldots\ldots\ldots\ldots\ldots 85\ g({CH}_{3})_{2}C(OH) - CN\)
232 g \({CH}_{3} - CO - {CH}_{3}\ldots\ldots x\ \rightarrow x = 340\ g({CH}_{3})_{2}C(OH) - CN,\ dimetilcianhidrină\)
Rezolvare:
\[{({CH}_{3})}_{2}C = O + {CH}_{3} - CO - {CH}_{3} \rightarrow \left( {CH}_{3} \right)_{2}C(OH) - {CH}_{2} - CO - {CH}_{3}\]
139. Să se calculeze ce cantitate de ciclohexanonă se obține prin dehidrogenarea a 300 g ciclohexanol.
Rezolvare:
\[C_{6}H_{11} - OH\overset{dehidrogenare}{\rightarrow}\ C_{6}H_{10}O\]
100 g \(C_{6}H_{11} - OH\ldots\ldots\ldots\ldots\ldots..\ 90g\ C_{6}H_{10}O\)
300 g \(C_{6}H_{11} - OH\ldots\ldots\ldots\ldots\ldots..\ x \rightarrow x = 270g\ C_{6}H_{10}O,\ ciclohexanonă\)
140. Să se calculeze ce cantitate de ciclohexanonă se obține prin oxidarea catalitică, cu aer, a 42 g ciclohexan.
Rezolvare:
\[C_{6}H_{12}\overset{oxidare\ catalitică}{\rightarrow}\ C_{6}H_{10}O\]
\[84\ gC_{6}H_{12}\ldots\ldots\ldots\ldots\ldots\ldots\ldots{90\ gC}_{6}H_{10}O\]
42 g \(C_{6}H_{12}\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 45\ C_{6}H_{10}O,\ ciclohexanonă\)
ACIZI CARBOXILICI
Rezolvare:
\[\left\{ \begin{array}{r}
C\frac{26}{12} = 2,16\ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow \frac{2,16}{2,16} = 1 \rightarrow 1 \\
H\frac{4,34}{1} = 4,34\ \ \rightarrow \frac{4,34}{2,16} = 2\ \ \rightarrow 2 \\
O\frac{69,56}{16} = 4,34 \rightarrow \frac{4,34}{2,16} = 2 \rightarrow 2
\end{array} \rightarrow {CH}_{2}O_{2}\ \ \ este\ formula\ brută \right.\ \ \]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{46}{46} = 1\]
\[\rightarrow {CH}_{2}O_{2}\ este\ și\ formulă\ moleculară \rightarrow HCOOH,\ acid\ formic\]
Rezolvare:
\[\left\{ \begin{array}{r}
C\frac{40}{12} = 3,33\ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow 1 \\
H\frac{6,66}{1} = 6,66\ \ \ \ \rightarrow 2 \\
O\frac{53,33}{16} = 3,33 \rightarrow 1
\end{array} \rightarrow {CH}_{2}O\ este\ formula\ brută \right.\ \ \]
\[\rightarrow C_{2}H_{4}O_{2}\ este\ formula\ moleculară\ pt.\ că\ gruparea\ carboxil\ are\ 2\ atomi\ de\ oxigen\]
\[\rightarrow {CH}_{3} - COOH,\ acid\ acetic\]
Rezolvare:
\[\left\{ \begin{array}{r}
C\frac{26,66}{12} = 2,22\ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow \frac{2,22}{2,22} = 1 \rightarrow 1 \\
H\frac{2,22}{1} = 2,22\ \ \rightarrow \frac{2,22}{2,22} = 1\ \ \rightarrow 1 \\
O\frac{71,11}{16} = 4,44 \rightarrow \frac{4,44}{2,22} = 2 \rightarrow 2
\end{array} \rightarrow CHO_{2}\ \ \ este\ formula\ brută \right.\ \ \]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{90}{45} = 2\]
\[\rightarrow {C_{2}H}_{2}O_{4}\ este\ \ formulă\ moleculară \rightarrow HOOC - COOH,\ acid\ oxalic\]
Rezolvare:
\[\begin{array}{r}
C\frac{57,83}{12} = 4,81\ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow \frac{4,81}{2,4} = 2 \rightarrow 4 \\
H\frac{3,61}{1} = 3,61\ \ \rightarrow \frac{3,61}{2,4} = 1,5\ \ \rightarrow 3 \\
O\frac{38,55}{16} = 2,40 \rightarrow \frac{2,4}{2,4} = 1 \rightarrow 2
\end{array} \rightarrow C_{4}H_{3}O_{2}\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{166}{82} = 2\]
\[\rightarrow {C_{8}H}_{6}O_{4}\ este\ \ formulă\ moleculară \rightarrow HOOC - C_{6}H_{4} - COOH,\ acid\ ftalic\]
Rezolvare:
C \(\frac{49,31}{12} = 4,10 \rightarrow \frac{4,10}{2,74} = 1,5\)
H \(\frac{6,84}{1} = 6,84 \rightarrow \frac{6,84}{2,74} = 2,5\)
O \(\frac{43,83}{16} = 2,74 \rightarrow \frac{2,74}{2,74} = 1\)
\[\rightarrow C_{3}H_{5}O_{2}\ este\ formula\ brută\]
\[n = \frac{M_{formula\ moleculară}}{M_{formula\ brută}} = \frac{146}{73} = 2\]
\[\rightarrow {C_{6}H}_{10}O_{4}\ este\ \ formulă\ moleculară\]
\[\rightarrow ar\ putea\ fi\ HOOC - {CH}_{2} - {CH}_{2} - {CH}_{2} - {CH}_{2} - COOH,\ acid\ adipic\]
146. Ce cantitate de acid stearic se obține prin oxidarea cu oxigen molecular din aer, la 100 \(\mathbf{℃}\) și în prezență de catalizatori a 127 g octadecan ?
Rezolvare:
\[{CH}_{3} - ({CH}_{2})_{16} - {CH}_{3}\overset{oxidare}{\rightarrow}{CH}_{3} - ({CH}_{2})_{16} - COOH + H_{2}O\]
\[{254\ g\ CH}_{3} - ({CH}_{2})_{16} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.284\ g{CH}_{3} - ({CH}_{2})_{16} - COOH\]
\[{127\ g\ CH}_{3} - ({CH}_{2})_{16} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots x\]
\[\rightarrow x = 142\ g{CH}_{3} - ({CH}_{2})_{16} - COOH,\ acid\ stearic\]
147. Ce cantitate de acid acetic se obține prin oxidarea cu permanganat de potasiu în mediu de acid sulfuric a 84 g propenă?
Rezolvare:
\[\ {CH}_{3} - CH = {CH}_{2} + 5\lbrack O\rbrack\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - COOH + {CO}_{2} + H_{2}O\]
\[42\ g\ {CH}_{3} - CH = {CH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{60\ g\ CH}_{3} - COOH\]
\[84\ g\ {CH}_{3} - CH = {CH}_{2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = {120\ g\ CH}_{3} - COOH,\ acid\ acetic\]
148. Ce cantitate de acid acetic se obține prin oxidarea cu permanganat de potasiu în mediu de acid sulfuric a 210 g 2 – pentenă ?
Rezolvare:
\[{CH}_{3} - CH = CH - {CH}_{2} - {CH}_{3} + 4\lbrack O\rbrack\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - COOH + HOOC - {CH}_{2} - {CH}_{3}\]
\[{70\ g\ CH}_{3} - CH = CH - {CH}_{2} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots 60\ g{CH}_{3} - COOH\]
\[\ \ \ \ {210\ g\ CH}_{3} - CH = CH - {CH}_{2} - {CH}_{3}\ldots\ldots\ldots\ldots.x \rightarrow x = \ 180\ g{CH}_{3} - COOH,\ acid\ acetic\]
149. Ce cantitate de acid acetic se obține prin oxidarea cu permanganat de potasiu în mediu de acid sulfuric a 140 g 2 – metil – 2 – butenă ?
Rezolvare:
\[{CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3} + 3\lbrack O\rbrack\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - CO - {CH}_{3} + HOOC - {CH}_{3}\]
\[70\ g\ {CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 60\ g\ {CH}_{3} - COOH\]
\[140\ g\ {CH}_{3} - C\left( {CH}_{3} \right) = CH - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 120\ g\ {CH}_{3} - COOH,\ acid\ acetic\]
150. Ce cantitate de acid tereftalic se obține prin oxidarea cu aer, în prezență de catalizatori a 212 g paraxilen?
Rezolvare:
\[{CH}_{3} - C_{6}H_{4} - {CH}_{3} + {3O}_{2}\overset{oxidare}{\rightarrow}\ HOOC - C_{6}H_{4} - COOH + 2H_{2}O\]
\[106\ g{CH}_{3} - C_{6}H_{4} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots.166\ gHOOC - C_{6}H_{4} - COOH\]
\[212\ g{CH}_{3} - C_{6}H_{4} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = 332\ gHOOC - C_{6}H_{4} - COOH,\ acid\ tereftalic\]
151. Ce cantitate de acid maleic se obține prin oxidarea, la încălzire și în prezența \(\mathbf{V}_{\mathbf{2}}\mathbf{O}_{\mathbf{5}}\) a 39 g benzen?
Rezolvare:
\[C_{6}H_{6}\overset{oxidare}{\rightarrow}HOOC - CH = CH - COOH + 2{CO}_{2} + H_{2}O\]
\[78\ g\ C_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..116\ g\ HOOC - CH = CH - COOH\]
\[39\ g\ C_{6}H_{6}\ldots\ldots\ldots\ldots\ldots..x \rightarrow x = 58\ g\ HOOC - CH = CH - COOH,\ acid\ maleic\]
152. Ce cantitate de acid acetic se obține prin oxidarea energică a 23 g etanol?
Rezolvare:
\[{CH}_{3} - {CH}_{2} - OH\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - COOH\]
\[46\ g{CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots..60\ g{CH}_{3} - COOH\]
\[23\ g{CH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = 30\ g{CH}_{3} - COOH,\ acid\ acetic\]
Rezolvare:
\[{CH}_{3} - CH(OH) - {CH}_{3}\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - COOH + HCOOH\]
\[{60\ g\ CH}_{3} - CH(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.46\ gHCOOH\]
\[{180\ g\ CH}_{3} - CH(OH) - {CH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 138\ gHCOOH,\ \ acid\ formic\]
154. Ce cantitate de acid propionic se obține prin oxidarea cu permanganat de potasiu sau cu dicromat de potasiu a 116 g propanal?
Rezolvare:
\[{CH}_{3} - {CH}_{2} - CH = O\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}{CH}_{3} - {CH}_{2} - COOH\]
\[58\ g{CH}_{3} - {CH}_{2} - CH = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots.74\ g{CH}_{3} - {CH}_{2} - COOH\]
\[116\ g{CH}_{3} - {CH}_{2} - CH = O\ldots\ldots\ldots\ldots x \rightarrow x = \ 148\ g{CH}_{3} - {CH}_{2} - COOH,\ acid\ propionic\]
Rezolvare:
\[{CH}_{2} = O + 2\left\lbrack Ag({NH}_{3})_{2} \right\rbrack OH \rightarrow H - COOH + {4NH}_{3} + H_{2}O + 2Ag\]
\[30\ g\ {CH}_{2} = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots..46\ gH - COOH\]
\[90\ g\ {CH}_{2} = O\ldots\ldots\ldots\ldots\ldots\ldots\ldots..x \rightarrow x = \ 138\ gH - COOH\]
156. Ce cantitate de acid acetic se obține prin oxidarea cu agenți oxidanți ( permanganat de potasiu sau dicromat de potasiu ) a 88 g acetaldehidă?
Rezolvare:
\[{CH}_{3} - CH = O\overset{oxidare}{\rightarrow}{CH}_{3} - COOH\]
\[{44g\ CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots.60\ g{CH}_{3} - COOH\]
\[{88g\ CH}_{3} - CH = O\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 120\ g{CH}_{3} - COOH,\ acid\ acetic\]
157. Ce cantitate de acid acetic se obține prin hidroliza catalizată de acizi sau baze a 264 g acetat de etil?
Rezolvare:
\[{CH}_{3} - CO - O - {CH}_{2} - {CH}_{3} + H_{2}O \leftrightarrow \ {CH}_{3} - COOH + HO - {CH}_{2} - {CH}_{3}\]
\[\ \ \ \ \ \ \ \ \ \ \ 88g\ \ {CH}_{3} - CO - O - {CH}_{2} - {CH}_{3}\ldots\ldots\ldots\ldots.60\ g\ {CH}_{3} - COOH\]
\(264g\ \ {CH}_{3} - CO - O - {CH}_{2} - {CH}_{3}\ldots\ldots\ldots\ldots x \rightarrow x = 180\ g\ {CH}_{3} - COOH,\ acid\ acetic\)
158. Ce cantitate de acid acetic se obține prin hidroliza la fierbere, catalizată de acizi sau baze, a 82 g acetonitil?
Rezolvare:
\({CH}_{3} - CN + 2H_{2}O \rightarrow \ {CH}_{3} - COOH + \ {NH}_{3}\)
\(41\ g\ {CH}_{3} - CN\ldots\ldots\ldots\ldots\ldots\ldots.60\ g\ {CH}_{3} - COOH\)
\(82\ g\ {CH}_{3} - CN\ldots\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = \ 120\ g\ {CH}_{3} - COOH,\ acid\ acetic\)
159. Ce cantitate de acid benzoic se obține prin hidroliza a 363 g benzamidă?
Rezolvare:
\[C_{6}H_{5} - CO - {NH}_{2} + H_{2}O \rightarrow C_{6}H_{5} - COOH + {NH}_{3}\]
\[{121\ g\ C}_{6}H_{5} - CO - {NH}_{2}\ldots\ldots\ldots\ldots\ldots 122gC_{6}H_{5} - COOH\]
\[{363\ g\ C}_{6}H_{5} - CO - {NH}_{2}\ldots\ldots\ldots\ldots x \rightarrow x = 366gC_{6}H_{5} - COOH,\ acid\ benzoic\]
160. Ce cantitate de acetat de sodiu se obține prin reacția a 30 g acid acetic cu carbonat acid de sodiu?
Rezolvare:
\[{CH}_{3} - COOH + NaH{CO}_{3} \rightarrow {CH}_{3} - COONa + H_{2}O + {CO}_{2}\]
\[60g{CH}_{3} - COOH\ldots\ldots\ldots\ldots\ldots\ldots 82\ g{CH}_{3} - COONa\]
\[30g{CH}_{3} - COOH\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 41\ g{CH}_{3} - COONa,\ acetat\ de\ sodiu\]
161. Ce cantitate de acetat de sodiu se obține prin reacția a 120 g acid acetic cu cianură de sodiu?
Rezolvare:
\[{CH}_{3} - COOH + NaCN \rightarrow {CH}_{3} - COONa + HCN\]
\[\ \ \ \ \ \ 60g{CH}_{3} - COOH\ldots\ldots\ldots\ldots\ldots\ldots 82\ g{CH}_{3} - COONa\]
\[120g{CH}_{3} - COOH\ldots\ldots\ldots\ldots x \rightarrow x = 164\ g{CH}_{3} - COONa,\ acetat\ de\ sodiu\]
Rezolvare:
\[HCOOH + \ C_{6}H_{5} - ONa \rightarrow HCOONa + C_{6}H_{5} - OH\]
**
\[46\ g\ HCOOH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.68\ g\ HCOONa\]
\[23\ g\ HCOOH\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = 34\ g\ HCOONa,\ formiat\ de\ sodiu\]
163. Ce cantitate de acetat de sodiu se obține prin reacția a 90 g acid acetic cu sodiu?
Rezolvare:
\[{CH}_{3} - COOH + Na \rightarrow {CH}_{3} - COONa + \frac{1}{2}H_{2}\]
60 g \({CH}_{3} - COOH\ldots\ldots\ldots\ldots.82{g\ CH}_{3} - COONa\)
90 g \({CH}_{3} - COOH\ldots\ldots\ldots\ldots.x \rightarrow x = 123\ g{CH}_{3} - COONa,\ acetat\ de\ sodiu\)
164. Ce cantitate de acetat de calciu se obține prin reacția a 60 g acid acetic cu oxidul de calciu?
Rezolvare:
\[{2CH}_{3} - COOH + CaO \rightarrow ({CH}_{3} - COO)_{2}Ca + H_{2}O\]
\[{120\ g\ CH}_{3} - COOH\ldots\ldots\ldots\ldots\ldots..158\ g({CH}_{3} - COO)_{2}Ca\]
\[{60\ g\ CH}_{3} - COOH\ldots\ldots\ldots\ldots\ldots..x\ \rightarrow x = 79\ g({CH}_{3} - COO)_{2}Ca,\ acetat\ de\ calciu\]
165. Ce cantitate de acetat de sodiu se obține prin reacția a 180 g acid acetic cu hidroxidul de sodiu?
Rezolvare:
\[{CH}_{3} - COOH + NaOH \rightarrow {CH}_{3} - COONa + H_{2}O\]
60 g \({CH}_{3} - COOH\ldots\ldots\ldots\ldots.82\ g\ {CH}_{3} - COONa\)
180 g \({CH}_{3} - COOH\ldots\ldots\ldots\ldots.x \rightarrow x = 246\ g\ {CH}_{3} - COONa,\ acetat\ de\ sodiu\)
166. Ce cantitate de ester se obține prin reacția a 240 g acid acetic cu alcoolul etilic?
Rezolvare:
\[{CH}_{3} - COOH + {CH}_{3} - {CH}_{2} - OH \leftrightarrow {CH}_{3} - CO - O - {CH}_{2} - {CH}_{3} + H_{2}O\]
\[{60\ gCH}_{3} - COOH\ldots\ldots\ldots\ldots\ldots\ldots.88\ g{CH}_{3} - CO - O - {CH}_{2} - {CH}_{3}\]
\[{240\ gCH}_{3} - COOH\ldots\ldots x \rightarrow x = 352\ g{CH}_{3} - CO - O - {CH}_{2} - {CH}_{3},\ acetat\ de\ etil\]
167. Ce cantitate de acetamidă se obține în urma reacției cu amoniacul, a 120 g acid acetic?
Rezolvare:
\[{CH}_{3} - COOH + {NH}_{3} \rightarrow {CH}_{3} - CO - {NH}_{2} + H_{2}O\]
60g \({CH}_{3} - COOH\ldots\ldots\ldots\ldots.59\ g\ {CH}_{3} - CO - {NH}_{2}\)
120g \({CH}_{3} - COOH\ldots\ldots\ldots\ldots x \rightarrow x = \ 118\ g\ {CH}_{3} - CO - {NH}_{2},\ acetamidă\)
Rezolvare:
\[CO + NaOH \rightarrow H - COONa\]
\[40\ g\ NaOH\ldots\ldots\ldots\ldots\ldots 68\ g\ H - COONa\]
\[20\ g\ NaOH\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 34\ g\ H - COONa,\ formiat\ de\ sodiu\]
Rezolvare:
\[H - COONa + H_{2}{SO}_{4} \rightarrow H - COOH + NaH{SO}_{4}\]
\[68\ g\ H - COONa\ldots\ldots\ldots\ldots.46g\ H - COOH\]
\[102\ g\ H - COONa\ldots\ldots\ldots\ldots x \rightarrow x = \ 69g\ H - COOH,\ acid\ formic\]
Rezolvare:
\[H - COOH\overset{+ O}{\rightarrow}\ H_{2}O + {CO}_{2}\]
\[46g\ H - COOH\ldots\ldots\ldots\ldots 44\ g\ {CO}_{2}\]
\[23g\ H - COOH\ldots\ldots\ldots\ldots x\ \rightarrow x = 22\ g\ {CO}_{2},\ bioxid\ de\ carbon\]
Rezolvare:
\(H - COOH \rightarrow H_{2} + {CO}_{2}\)
\[46g\ H - COOH\ldots\ldots\ldots\ldots 44\ g\ {CO}_{2}\]
\[92g\ H - COOH\ldots\ldots\ldots\ldots x \rightarrow x = 88\ g\ {CO}_{2},\ bioxid\ de\ carbon\]
Rezolvare:
\[H - COOH \rightarrow CO + H_{2}O\]
\[46\ g\ H - COOH\ldots\ldots\ldots..28\ g\ CO\]
\[138\ g\ H - COOH\ldots\ldots\ldots x\ \rightarrow x = 84\ g\ CO,\ monoxid\ de\ carbon\]
173. Ce cantitate de acid acetic se obține prin fermentația acetică a 92 g etanol?
Rezolvare:
\[{CH}_{3} - {CH}_{2} - OH + O_{2}\overset{fermentație\ acetică}{\rightarrow}{CH}_{3} - COOH + H_{2}O\]
\[{\ 46\ gCH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots.60\ g\ {CH}_{3} - COOH\]
\[{\ 92\ gCH}_{3} - {CH}_{2} - OH\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 120\ g\ {CH}_{3} - COOH,\ acid\ acetic\]
Rezolvare:
\[HOOC - COOH + \lbrack O\rbrack \rightarrow 2{CO}_{2} + H_{2}O\]
\[90\ gHOOC - COOH\ldots\ldots\ldots\ldots 88\ g\ {CO}_{2}\]
\[180\ gHOOC - COOH\ldots\ldots\ldots x \rightarrow x = 176\ g\ {CO}_{2},\ bioxid\ de\ carbon\]
Rezolvare:
\[HOOC - COOH \rightarrow CO + {CO}_{2} + H_{2}O\]
\[90\ gHOOC - COOH\ldots\ldots\ldots\ldots\ldots.44\ g{CO}_{2}\]
\[180\ gHOOC - COOH\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 88\ g{CO}_{2},\ bioxid\ de\ carbon\]
176. Ce cantitate de acid benzoic se obține prin oxidarea a 194 g toluen?
Rezolvare:
\[C_{6}H_{5} - {CH}_{3} + 3\lbrack O\rbrack\overset{ {KMnO}_{4},H_{2}{SO}_{4}}{\rightarrow}C_{6}H_{5} - COOH + H_{2}O\]
\[92\ g\ C_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots{122\ gC}_{6}H_{5} - COOH\]
\[184\ g\ C_{6}H_{5} - {CH}_{3}\ldots\ldots\ldots\ldots\ldots x = 244{\ gC}_{6}H_{5} - COOH,\ acid\ benzoic\]
Rezolvare:
\[2H - COOH + Mg \rightarrow (H - COO)_{2}Mg + H_{2}\]
\[24g\ Mg\ldots\ldots\ldots..2\ g\ H_{2}\]
\[48g\ Mg\ldots\ldots\ldots..x \rightarrow x = 4\ g\ H_{2},\ hidrogen\]
Rezolvare:
\[C_{n}H_{2n + 1} - COOH \rightarrow (n + 1){CO}_{2} + (n + 1)H_{2}\]
\[\rightarrow 2 = 2n + 2 \rightarrow n = 0\]
\[\rightarrow HCOOH \rightarrow {CO}_{2} + H_{2}\]
AMIDE
Rezolvare:
\[\left\{ \begin{array}{r}
\left\{ \begin{array}{r}
C\frac{26,55}{12} = 2,21 \rightarrow 1 \\
H\frac{6,66}{1} = 6,66 \rightarrow 3
\end{array} \right.\ \\
\left\{ \begin{array}{r}
N\frac{31,11}{14} = 2,22 \rightarrow 1 \\
O\frac{35,55}{16} = 2,22 \rightarrow 1
\end{array} \right.\
\end{array} \right.\ \rightarrow {CH}_{3}NO\ este\ formula\ brută\]
\[n = \frac{45}{45} = 1 \rightarrow {CH}_{3}NO\ este\ și\ formulă\ moleculară \rightarrow HC - CO - {NH}_{2},\ formamidă\]
180. Ce cantitate de acetamidă se obține din acid acetic, prin tratare cu 170 g amoniac?
Rezolvare:
\[{CH}_{3} - COOH + {NH}_{3} \rightarrow {CH}_{3} - CO - {NH}_{2} + H_{2}O\]
\[17g\ {NH}_{3}\ldots\ldots\ldots\ldots.59g{CH}_{3} - CO - {NH}_{2}\]
\[170g\ {NH}_{3}\ldots\ldots\ldots x \rightarrow x = 590g{CH}_{3} - CO - {NH}_{2},\ acetamidă\]
181. Ce cantitate de benzamidă se obține din clorură de benzoil, prin tratare cu 340 g amoniac?
Rezolvare:
\[C_{6}H_{5} - CO - Cl + {NH}_{3} \rightarrow C_{6}H_{5} - CO - {NH}_{2} + HCl\]
\[{17\ gNH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots 121\ gC_{6}H_{5} - CO - {NH}_{2}\]
\[{340\ gNH}_{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 2420\ gC_{6}H_{5} - CO - {NH}_{2},\ benzamidă\]
Rezolvare:
\[H - CO - O - {CH}_{2} - {CH}_{3} + {NH}_{3} \rightarrow H - CO - {NH}_{2} + HO - {CH}_{2} - {CH}_{3}\]
\[{17\ gNH}_{3}\ldots\ldots\ldots\ldots\ldots 45\ gH - CO - {NH}_{2}\]
\[{170\ gNH}_{3}\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 450\ gH - CO - {NH}_{2},\ formamidă\]
183. Ce cantitate de acetamidă se obține prin hidroliza a 82 g acetonitril?
Rezolvare:
\[{CH}_{3} - CN + HOH \rightarrow {CH}_{3} - CO - {NH}_{2}\]
\[41\ g{CH}_{3} - CN\ldots\ldots\ldots\ldots\ldots\ldots 59\ g{CH}_{3} - CO - {NH}_{2}\]
\[82\ g{CH}_{3} - CN\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 118\ g{CH}_{3} - CO - {NH}_{2},\ acetamida\]
184. Ce cantitate de uree se obține prin încălzirea cu bioxid de carbon, la presiune a 170 g amoniac?
Rezolvare:
\[{CO}_{2} + 2{NH}_{3}\overset{150℃}{\rightarrow}H_{2}N - CO - {NH}_{2} + H_{2}O\]
\[34\ g\ {NH}_{3}\ldots\ldots\ldots\ldots 60\ g\ H_{2}N - CO - {NH}_{2}\]
\[170\ g\ {NH}_{3}\ldots\ldots\ldots x \rightarrow x = 300\ g\ H_{2}N - CO - {NH}_{2},\ uree\]
185. Ce cantitate de acid acetic se obține prin hidroliza a 177 g acetamidă?
Rezolvare:
\[{CH}_{3} - CO - {NH}_{2} + HOH\overset{H^{+}}{\rightarrow}{CH}_{3} - COOH + {NH}_{4}^{+}\]
\[{59\ g\ CH}_{3} - CO - {NH}_{2}\ldots\ldots\ldots\ldots\ldots..60\ g{CH}_{3} - COOH\]
\[{177\ g\ CH}_{3} - CO - {NH}_{2}\ldots\ldots\ldots\ldots\ldots.x \rightarrow x = 180\ g{CH}_{3} - COOH,\ acid\ acetic\]
186. Ce cantitate de etilamină se obține prin hidroliza a 118 g acetamidă?
Rezolvare:
\[{CH}_{3} - CO - {NH}_{2} + 2H_{2} \rightarrow {CH}_{3} - {CH}_{2} - {NH}_{2} + H_{2}O\]
\[{59\ g\ CH}_{3} - CO - {NH}_{2}\ldots\ldots\ldots\ldots\ldots 45\ g\ {CH}_{3} - {CH}_{2} - {NH}_{2}\]
\[{118\ g\ CH}_{3} - CO - {NH}_{2}\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 90\ g\ {CH}_{3} - {CH}_{2} - {NH}_{2},\ etilamină\]
187. Ce cantitate de acetonitril se obține prin deshidratarea a 236 g acetamidă?
Rezolvare:
\[{CH}_{3} - CO - {NH}_{2}\overset{P_{2}O_{5}}{\rightarrow}{CH}_{3} - CN + H_{2}O\]
\[{59\ g\ CH}_{3} - CO - {NH}_{2}\ldots\ldots\ldots\ldots.{41\ g\ CH}_{3} - CN\]
\[{236\ g\ CH}_{3} - CO - {NH}_{2}\ldots\ldots\ldots\ldots.x \rightarrow x = {164\ g\ CH}_{3} - CN,\ acetonitril\]
AMINOACIZI
188. Ce cantitate de glicil – glicină se obține din 300 g glicină ?
Rezolvare:
\[{NH}_{2} - {CH}_{2} - COOH + {NH}_{2} - {CH}_{2} - COOH\overset{- H_{2}O}{\rightarrow}{NH}_{2} - {CH}_{2} - CO - NH - {CH}_{2} - COOH\]
\[150\ g\ glicină\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 132\ g\ glicil - glicină\]
\[300\ g\ glicină\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = \ 264\ g\ glicil - glicină\]
189. Care este masa tripeptidului glicil – glicil – alanină?
Rezolvare:
\[{NH}_{2} - {CH}_{2} - COOH + {NH}_{2} - {CH}_{2} - COOH + {NH}_{2} - CH\left( {CH}_{3} \right) - COOH\ \overset{- {2H}_{2}O}{\rightarrow}\]
\[\rightarrow {NH}_{2} - {CH}_{2} - CO - NH - {CH}_{2} - CO - NH - CH\left( {CH}_{3} \right) - COOH\ \]
\[M = (75 + 75 + 89 - 18 - 18)g = 203\ g\]
Rezolvare:
\[\left\{ \begin{array}{r}
\left\{ \begin{array}{r}
C\frac{40,44}{12} = 3,37 \rightarrow 3 \\
H\frac{7,86}{1} = 7,86 \rightarrow 7
\end{array} \right.\ \\
\left\{ \begin{array}{r}
O\frac{35,95}{16} = 2,24 \rightarrow 2 \\
N\frac{15,73}{14} = 1,09 \rightarrow 1
\end{array} \right.\
\end{array} \right.\ \rightarrow C_{3}H_{7}O_{2}N\ care\ are\ M = 89 \rightarrow C_{3}H_{7}O_{2}N\ \ este\ și\ f.moleculară\]
\[{NH}_{2} - CH\left( {CH}_{3} \right) - COOH\ care\ este\ formula\ aminoacidului\ alanină\]
191. Ce cantitate de sare de sodiu se obține prin reacția a 600 g glicină cu hidroxidul de sodiu ?
Rezolvare:
\[{NH}_{2} - {CH}_{2} - COOH + NaOH \rightarrow {NH}_{2} - {CH}_{2} - COONa + H_{2}O\]
\[{75\ g\ NH}_{2} - {CH}_{2} - COOH\ldots\ldots\ldots\ldots..97\ g\ {NH}_{2} - {CH}_{2} - COONa\]
\({600\ g\ NH}_{2} - {CH}_{2} - COOH\ldots\ldots\ldots x \rightarrow x = 776\ g\ {NH}_{2} - {CH}_{2} - COONa,\ sare\) de sodiu
192. Ce cantitate de ester de aminoacid se obține prin reacția 178 g alanină cu alcoolul etilic ?
Rezolvare:
\[{NH}_{2} - CH\left( {CH}_{3} \right) - COOH + HO - {CH}_{2} - {CH}_{3} \rightarrow {NH}_{2} - CH\left( {CH}_{3} \right) - CO - O - {CH}_{2} - {CH}_{3} + H_{2}O\]
\[89\ g\ {NH}_{2} - CH\left( {CH}_{3} \right) - COOH\ldots\ldots\ldots 117\ g\ {NH}_{2} - CH\left( {CH}_{3} \right) - CO - O - {CH}_{2} - {CH}_{3}\]
\[178\ g\ {NH}_{2} - CH\left( {CH}_{3} \right) - COOH\ldots\ldots x\]
\[\rightarrow x = 234\ g\ {NH}_{2} - CH\left( {CH}_{3} \right) - CO - O - {CH}_{2} - {CH}_{3},\ ester\ de\ aminoacid\]
193. Ce cantitate de clorură acidă se obține prin reacția a 150 g glicină cu pentaclorura de fosfor?
Rezolvare:
\[{NH}_{2} - {CH}_{2} - COOH + \ {PCl}_{5} \rightarrow {NH}_{2} - {CH}_{2} - CO - Cl + HCl + {POCl}_{3}\]
\[75\ g\ {NH}_{2} - {CH}_{2} - COOH\ldots\ldots\ldots\ldots\ldots\ldots\ldots.93,5\ g{NH}_{2} - {CH}_{2} - CO - Cl\]
\[150\ g\ {NH}_{2} - {CH}_{2} - COOH\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 187\ g{NH}_{2} - {CH}_{2} - CO - Cl,\ clorură\ acidă\]
194. Ce cantitate de produs rezultă prin reacția a 300 g glicină cu acidul clorhidric?
Rezolvare:
\[{NH}_{2} - {CH}_{2} - COOH + HCl \rightarrow {NH}_{3}^{+} - {CH}_{2} - COOH\ {Cl}^{-}\]
\[75\ g\ {NH}_{2} - {CH}_{2} - COOH\ldots\ldots\ldots\ldots.111,5\ g\ {NH}_{3}^{+} - {CH}_{2} - COOH\ {Cl}^{-}\]
\[300\ g\ {NH}_{2} - {CH}_{2} - COOH\ldots\ldots\ldots\ldots.446\ g\ {NH}_{3}^{+} - {CH}_{2} - COOH\ {Cl}^{-},\ sare\ de\ amoniu\]
195. Ce cantitate de aminoacid acilat rezultă din reacția a 267 g alanină cu clorura de acetil?
Rezolvare:
\[{NH}_{2} - CH\left( {CH}_{3} \right) - COOH + {CH}_{3} - COCl \rightarrow {CH}_{3} - CO - NH - CH\left( {CH}_{3} \right) - COOH + HCl\]
\[{89\ g\ NH}_{2} - CH\left( {CH}_{3} \right) - COOH\ldots\ldots\ldots 131\ g{CH}_{3} - CO - NH - CH\left( {CH}_{3} \right) - COOH\]
\[{267\ g\ NH}_{2} - CH\left( {CH}_{3} \right) - COOH\ldots\ldots x\]
\[\rightarrow x = 393\ g{CH}_{3} - CO - NH - CH\left( {CH}_{3} \right) - COOH,\ aminoacid\ acilat\]
MONOZAHARIDE
196. Ce cantitate de glucoză se obține prin hidroliza acidă a 81 g amidon?
Rezolvare:
\[{(C_{6}H_{10}O_{5})}_{n} + nH_{2}O\overset{HCl}{\rightarrow}nC_{6}H_{12}O_{6}\]
\[162\ g\ amidon\ldots\ldots\ldots\ldots\ldots\ldots 180\ g\ glucoză\]
\[81\ g\ amidon\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 90\ g\ glucoză\]
197. Ce cantitate de acid gluconic rezultă prin oxidarea a 45 g glucoză cu reactiv Fehling ?
Rezolvare:
\(C_{6}H_{12}O_{6}(glucoză) + {Cu(OH)}_{2} \rightarrow C_{6}H_{12}O_{7}(acid\ gluconic) + {Cu}_{2}O + 2H_{2}O\)
\[180\ g\ glucoză\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 196\ g\ acid\ gluconic\]
\[45\ g\ glucoză\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x\ = \ 49\ g\ acid\ gluconic\]
198. Ce cantitate de argint se depune prin reacția a 45 g glucoză cu reactiv Tollens?
Rezolvare:
\[C_{6}H_{12}O_{6}(glucoză) + 2Ag{({NH}_{3})}_{2}OH \rightarrow C_{6}H_{12}O_{7}(a.\ gluconic) + 4{NH}_{3} + H_{2}O + 2Ag\]
\[180\ g\ C_{6}H_{12}O_{6}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots 216\ g\ Ag\]
\[45\ g\ C_{6}H_{12}O_{6}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots x \rightarrow x = 54\ g\ Ag\]
199. Ce cantitate de pentaacetilglucoză se obține prin esterificarea a 270 g glucoză cu clorură de acetil?
Rezolvare:
\(C_{6}H_{12}O_{6}(glucoză) + 5{CH}_{3} - COCl \rightarrow\)
\[O = CH - \left( CH - O - CO - {CH}_{3} \right)_{4} - {CH}_{2} - O - CO - {CH}_{3} + 5HCl\]
\[180\ g\ glucoză\ldots\ldots\ldots\ldots.390\ g\ pentaacetilglucoză\]
\[270\ g\ glucoză\ldots\ldots\ldots\ldots.x \rightarrow x = 585\ g\ pentaacetilglucoză\]
200. Ce cantitate de alcool etilic se obține din 100 g glucoză prin fermentarea a 90 g glucoză?
Rezolvare:
\(C_{6}H_{12}O_{6}(glucoză) \rightarrow 2C_{2}H_{5}OH + 2{CO}_{2}\)
\({180\ g\ C}_{6}H_{12}O_{6}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..92\ g\ C_{2}H_{5}OH\)
\({90\ g\ C}_{6}H_{12}O_{6}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..x \rightarrow x = 46\ g\ C_{2}H_{5}OH\)