Bài tập về hiệu suất phản ứng este hóa

Đặt \(n_{CH_3COOH}=x\) và \(n_{C_2H_5OH}=y\)  mol

+ m _hh = 60x + 46y = 25,8 (1)

+ Đốt cháy hỗn hợp: n_H2O = \(\dfrac{{23,4}}{{18}}=1,3\) (mol)

CH₃COOH → 2H₂O

         x       →      2x

C₂H₅OH  →  3H₂O

      y →              3y

n_H2O = 2x + 3y = 1,3 (2)

Giải hệ (1) và (2) được x = 0,2 và y = 0,3

=> %m_CH3COOH = \(\dfrac{{0,2\times 60}}{{25,8}}\times100\)% = 46,5%

 %m_C2H5OH = 53,5%

Gọi hiệu suất của phản ứng este hóa là h.

PTHH:\(C{H_3}{\text{COO}}H + {C_2}{H_5}OH\overset {{t^0},{H_2}S{O_{4{\text{đ}}}}} \leftrightarrows C{H_3}{\text{CO}}O{C_2}{H_5} + {H_{_2}}O\)

BĐ:     0,2                0,3

(0,2/1 < 0,3/1 nên hiệu suất tính theo axit)

Pư:      0,2h    →      0,2h                 →          0,2h

Sau:  0,2 - 0,2h       0,3-0,2h                       0,2h

=> m _este = m_CH3COOC2H5 = 14,08 gam

=> 0,2h.88 = 14,08 => h = 0,8 = 80%

Gọi:\(n_{CH_3COOH}=x\)

\(n_{C_2H_5OH}=y\)

\(n_{CH_3COOH}=0,5\) mol

\(n_{C_2H_5OH}=2\) mol

\(C{H_3}{\text{COO}}H + {C_2}{H_5}OH\overset {{t^0},{H_2}S{O_{4{\text{đ}}}}} \leftrightarrows C{H_3}{\text{CO}}O{C_2}{H_5} + {H_{_2}}O\)

=> Ancol dư , axit hết

=> n_este = n_CH3COOH

H% = 0,5.60% = 0,3 mol

=> m_CH3COOC2H5 = 26,4g

Vì 10 gam hỗn hợp 2 axit trung hòa vừa đủ 0,19 mol NaOH

$ \to $ n_axit = n_NaOH = 0,19 mol

${n_{ancol}} = \dfrac{{9,2}}{{46}} = 0,2\,\,mol$

Vì n_axit < n_ancol $ \to $Phản ứng tính theo số mol axit

H = 90% $ \to $ n_{ancol phản ứng} = 0,19.90% = 0,171 mol và ${m_{{H_2}O}} = 0,19.0,9 = 0,171\,\,mol$

Bảo toàn khối lượng: m_{este phản ứng} + m_{ancol phản ứng} = m_este + m_nước

$ \to 0,9.10 + 0,171.46 = {m_{este}} + 0,171.18$

$ \to {m_{este}} = 13,788\,\,gam$

\({C_2}{H_5}OH{\rm{ }} + {\rm{ }}C{H_3}COOH\; \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} \;C{H_3}COO{C_2}{H_5} + {\rm{ }}{H_2}O\)

=> n_{C2H5OH phản ứng} = n_{este sinh ra} = 0,125 mol

=> hiệu suất phản ứng =    \({{0,125} \over {0,3}}.100\% = 41,67\% \)

Chọn D

\(\eqalign{
& {n_{C{H_3}{\rm{COO}}H}} = 0,15\,\,mol;\,{n_{{C_2}{H_5}OH}} = 0,1\,\,mol \cr
& {C_2}{H_5}OH{\rm{ }} + {\rm{ }}C{H_3}COOH\; \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} \;C{H_3}COO{C_2}{H_5} + {\rm{ }}{H_2}O \cr}\)

Vì  \({n_{C{H_3}{\rm{COO}}H}} > {n_{{C_2}{H_5}OH}}\)=> hiệu suất phản ứng tính theo C₂H₅OH

n_este = 0,075 mol => n_{ancol phản ứng} = 0,075 mol

\(H\% = {{0,075} \over {0,1}}.100 = 75\% \)

\(\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HCOOH{\rm{ }} + {\rm{ }}C{H_3}OH\;\; \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {\rm{ }}HCOOC{H_3}\;{\rm{ }} + {\rm{ }}{H_2}O\;\;\;\;{\rm{ }}{K_{CB}}{\rm{ }} = {\rm{ 9}}\;\;\;\; \cr
& bd(mol)\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr
& pu(mol)\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \cr
& cb(mol)\,\,\,\,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \cr
& {K_{Cb}} = {{\left[ {C{H_3}COO{C_2}{H_5}} \right].\left[ {{H_2}O} \right]} \over {\left[ {C{H_3}COOH} \right]{\rm{. }}\left[ {{C_2}{H_5}OH} \right]}} = {{{x^2}} \over {{{\left( {1 - x} \right)}^2}}} = 9 < = > x = \left[ \matrix{
x = 0,75(nhan) \hfill \cr
x = 1,5 > 1(loai) \hfill \cr} \right. \cr
& H\% = {{n\,{\,_{phan\,\,ung}}} \over {n\,{\,_{ban\,\,dau}}}}.100 = {{0,75} \over 1}.100 = 75\% \cr} \)

Tóm tắt quá trình:\(4,44\;(g)\;\left\{ \matrix{
C{H_3}COOH \hfill \cr
C{H_3}OH \hfill \cr} \right.\left\langle \matrix{
\underbrace {C{H_3}COOC{H_3}}_{0,06\;mol} + {H_2}O \hfill \cr
\buildrel { + \;{O_2}} \over
\longrightarrow C{O_2}:{n_{C{O_2}}} = 0,4mol \hfill \cr} \right.\)

- Gọi \(\left\{ \matrix{
C{H_3}OH:x \hfill \cr
C{H_3}COOH:y \hfill \cr} \right. \to \left\{ \matrix{
32x + 60y = 12,2 \hfill \cr
x + 2y = 0,4 \hfill \cr} \right. \to \left\{ \matrix{
x = 0,1 \hfill \cr
y = 0,15 \hfill \cr} \right. \Rightarrow H = {{0,06} \over {0,1}}.100\% = 60\% \)

\(\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HCOOH + R'OH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HCOOR' + {H_2}O \cr
& bd(mol)\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15 \cr
& pu(mol)\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = {{3,7} \over {{M_{R'}} + 45}} \cr
& cb(mol)\,\,0,1 - x\,\,\,\,\,\,\,\,\,\,\,0,15 - x\,\,\,\,\,\,\,\,\,\,\,x \cr
& {n_{ancol}} + {n_{axit\,\,du}} = 2{n_{H2}} = 0,15\,\,mol< = > (0,1 - x) + (0,15 - x) = 0,15 = > x = 0,05 \cr
& Do\,\,{n_{ancol}} > {n_{{\rm{ax}}it}} = > H\% = {{{n_{axit(pu)}}} \over {{n_{axit(bd)}}}}.100 = 50\% \cr
& = > {M_{R'}} = 29: - {C_2}{H_5} \cr
& = > {\rm{Ancol}}:{C_2}{H_5}OH \cr}\)

\(\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}COO{C_2}{H_5} + NaOH  \to C{H_3}COONa + {C_2}{H_5}OH \cr
& (mol)\,\,\,\,\,0,24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0,24 \cr
& H\% = {{m\,{\,_{thuc.te}}} \over {m\,{\,_{ly\,thuyet}}}}.100 = > m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} = 0,24.82.{{70} \over {100}} = 13,776gam \cr}\)

\(\eqalign{
& \underbrace {\left\{ \matrix{
{C_2}{H_5}OH:2a\,mol \hfill \cr
C{H_3}OH:amol \hfill \cr} \right.}_{6,2(gam)}\buildrel {C{H_3}COOH:0,2mol} \over
\longrightarrow \left\{ \matrix{
C{H_3}COO{C_2}{H_5}:2a \hfill \cr
C{H_3}COOC{H_3}:a \hfill \cr} \right. \cr
& = > a = 0,05mol \cr}\)

Nx: n_{hh axit}< n _ancol => Este tính theo axit.

=> m _{lý thuyết} = 0,05. 74+ 0,1. 88 =12,5gam.

\(m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} = 12,5.{{60} \over {100}} = 7,5gam\)

\(\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HOOH + C{H_3}OH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HCOOC{H_3} + {H_2}O \cr
& bd(mol)\,\,\,0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2 \cr
& pu(mol)\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\leftarrow 0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0,2 \cr
& cb(mol)\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2 \cr
& m{\,_{ly\,thuyet}} = 0,2.60 = 12gam \cr
& m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} = 12.{{80} \over {100}} = 9,6gam \cr} \)

\(\left. \matrix{
{C_n}{H_{2n + 2}}O:x \hfill \cr
{C_n}{H_{2n}}{O_2}:y \hfill \cr} \right\} = > \,TQ:{C_n}{H_y}{O_z}\)

\(\eqalign{
& {C_x}{H_y}{O_z}\buildrel { + {O_2},{t^0}} \over
\longrightarrow {n_{C{O_2}}} \cr
& {{0,5} \over n} \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5 \cr} \)

=> M _{trung bình} = 27,2n

=> (14n + 18) < 27,2n < 14n + 32

=> 1,36 < n < 2,4=> n = 2

=> C₂H₆O (0,1 mol), C₂H₄O₂ (0,15 mol)

=> n _este = n_C2H6O . h% = 0,06mol

=> m _CH3COOC2H5= 5,28g

\(\eqalign{
& {{\rm{n}}_{{\rm{ax}}it}} = {{150} \over {60}} = 2,5mol \cr
& {n_{ancol}} = {{176} \over {88}} = 2mol \cr
& {\left( {C{H_3}} \right)_2}CHC{H_2}C{H_2}OH + C{H_3}COOH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COOC{H_2}C{H_2}CH{(C{H_3})_2} + {H_2}O \cr
& Nx:2 < 2,5 = > {n_{{\rm{es}}te}} = 2mol \cr
& m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} \cr
& = > {{\rm{m}}_{{\rm{es}}te}} = 2.130.{{55} \over {100}} = 143g \cr} \)

\(\eqalign{
& {m_{Ca{C_2}}} = 50.64\% = 32kg \cr
& m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} \cr
& H = {H_1}.{H_2} \cr
& Ca{C_2}\buildrel {60\% } \over
\longrightarrow C{H_3}{\rm{COO}}H\buildrel {92\% } \over
\longrightarrow C{H_3}{\rm{COO}}{C_2}{H_5} \cr} \)

64 đvc                    60 đvc                      88đvc

32kg                                                   →  44kg

=> Khối lượng este thực tế thu được là  = 44.92%.60%=24,288kg

CH₂=C(CH₃)COOH + CH₃OH → CH₂=C(CH₃)COOCH₃ →  Trùng hợp

=> n_{mắt xích} = n_este = 2 kmol

=>  n_axit = 2 . 100/40 . 100/50 = 10 kmol

=> m_axit = 860 kg

\(\eqalign{
& {\rm{Ax}}it\,\,RCOOH:a(mol)\buildrel { + CaCO3} \over
\longrightarrow C{O_2}:0,15mol \cr
& 2{\rm{ }}R - COOH{\rm{ }} + {\rm{ }}CaC{O_3}{\rm{ }} \to {\rm{ }}{\left( {R - COO} \right)_2} - Ca{\rm{ }} + {\rm{ }}C{O_2}{\rm{ }} + {\rm{ }}{H_2}O \cr
& {n_{RCOOH}} = 2{n_{CO2}} = 0,3mol \cr
& Ancol\,\,\overline R OH:b(mol)\buildrel { + Na} \over
\longrightarrow {H_2}:0,1mol \cr
& \overline R OH + Na \to \overline R ONa + {1 \over 2}{H_2} \cr
& {n_{\overline R OH}} = 2{n_{H2}} = 0,16mol \cr
& \underbrace {A:RCOOH}_{a(gam):0,3mol} + \underbrace {B:\overline R OH}_{11,7(gam):0,3mol}\buildrel {h\% } \over
\longrightarrow \underbrace {RCOO\overline R }_{m(gam)} + {H_2}O \cr
& BTKL:(a + 11,7).h\% = m + 18.0,3.h\% = > m = (a + 6,3)h\% \cr}\)

\(\eqalign{
& {n_{CO2}} = 0,45mol \cr
& {n_{H2O}} = 0,4mol \cr
& {n_{CO2}} > {n_{H2O}} = > {\rm{Este ko}}\,{\rm{no}} \cr
& M\left\{ \matrix{
X:ROH:{C_n}{H_{2n + 2}}O:a(mol) \hfill \cr
Y:R'C{\rm{OO}}H:{C_n}{H_{2n - 2}}{O_2}:b(mol;b > a) \hfill \cr} \right. \cr
& \left\{ \matrix{
{C_n}{H_{2n + 2}}O:a(mol) \hfill \cr
{C_n}{H_{2n - 2}}{O_2}:b(mol) \hfill \cr} \right.\buildrel { + O2} \over
\longrightarrow \left\{ \matrix{
C{O_2}:0,45\,\,mol \hfill \cr
{H_2}O:0,4\,\,mol \hfill \cr} \right. \cr
& = > \left\{ \matrix{
a + b = 0,15 \hfill \cr
\buildrel {BTNT:C} \over
\longrightarrow an + bn = 0,45 \hfill \cr
\buildrel {BTNT:H} \over
\longrightarrow a(n + 1) + b(n - 1) = 0,4 \hfill \cr} \right. = > \left\{ \matrix{
a = 0,05 \hfill \cr
b = 0,1 \hfill \cr
n = 3 \hfill \cr} \right. \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_3}{H_7}OH + {C_2}{H_3}COOH \to {C_2}{H_3}COO{C_3}{H_7} + {H_2}O \cr
& bd(mol)\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1 \cr
& pu(mol)\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0,05 \cr
& m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} \cr
& {{\rm{m}}_{{\rm{es}}te}} = 0,05.114.{{60} \over {100}} = 3,42gam \cr} \)

Gọi số mol ancol và axit lần lượt là x và y

Vì 2 chất đều no đơn chức nên khi đốt cháy : n_H2O = n_CO2 + n_ancol = 2,31 + x

Bảo toàn O : n_O2.2 + n_ancol + 2n_axit = 2n_CO2 + n_H2O => n_O2= (3,465 – y) mol

Bảo toàn khối lượng : m_X + m_O2 = m_CO2 + m_H2O

=> 9x + 16y = 9,45 => 9(x+y) < 9,45 < 16(x+y)

=> 0,59 < (x + y) < 1,05

Gọi số C trong mỗi chất là n => n_CO2 = 2,31 = n.(x + y) => (x + y) = 2,31/n

=> 2,2 < n < 3,9 =>n = 3

=> x = 0,41 ; y = 0,36 mol

C₂H₅COOH + C₃H₇OH → C₂H₅COOC₃H₇ + H₂O

0,36                  0,41  →   0,36

=> Tính theo chất phản ứng thiếu (axit) => n_este = n_axit = 0,36 mol

=>\(m\,{\,_{thuc.te}} = m\,{\,_{ly\,thuyet}}.{{H\% } \over {100}} = 25,056gam\)

\(\begin{array}{*{20}{l}}{{n_{C{H_3}COOH}} = 0,2mol}\\{{n_{C{H_3}COO{C_2}{H_5}}} = 0,125mol}\end{array}\)

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}COOH + {C_2}{H_5}OH\overset {H2SO4} \leftrightarrows C{H_3}COO{C_2}{H_5} + {H_2}O$

\(\begin{array}{l}bd:(mol)\,\,\,0,2\\pu:(mol)\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to x = 0,125\\cb:(mol)\,\,\,\,\,0,2 - x\\H\%  = \dfrac{{n\,{\,_{phan\,\,ung}}}}{{n\,{\,_{ban\,\,dau}}}}.100 = \dfrac{{0,125}}{{0,2}}.100 = 62,5\% \end{array}\)

n _CH3COOH  = 0,1 mol

n _C2H5OH = 0,2 mol

\(\begin{align}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}OH~+\text{ }C{{H}_{3}}COOH~~\rightleftarrows ~C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+\text{ }{{H}_{2}}O \\ & bd:(mol)\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1 \\ & pu:(mol)\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & cb:(mol)\,\,\,\,\,\,\,\,\,0,2-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\dfrac{5,5}{88}=0,0625 \\ \end{align}\)

Hiệu suất tính theo axit.

Hiệu suất:

\(H\%  = \dfrac{{n\,{\,_{phan\,\,ung}}}}{{n\,{\,_{ban\,\,dau}}}}.100 = \dfrac{{0,0625}}{{0,1}}.100 = 62,5\% \)