Đáp án + Giải thích các bước giải:

$a\big)$

Đặt $n_{CH_3COOH}=x(mol);n_{C_2H_5OH}=y(mol)$

$\to 60x+46y=25,8(1)$

$n_{H_2}=\frac{5,6}{22,4}=0,25(mol)$

$2C_2H_5OH+2K\to 2C_2H_5OK+H_2$

$2CH_3COOH+2K\to 2CH_3COOK+H_2$

Theo PT: $x+y=2n_{H_2}=0,5(2)$

Từ $(1)(2)\to x=0,2;y=0,3$

$\to \%m_{CH_3COOH}=\frac{0,2.60}{25,8}.100\%\approx 46,51\%$

$\to \%m_{C_2H_5OH}=100-46,51=53,49\%$

$b\big)$

$n_{CH_3COOC_2H_5(tt)}=\frac{13,2}{88}=0,15(mol)$

$CH_3COOH+C_2H_5OH\buildrel{{t^o,H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

Vì $n_{CH_3COOH}<n_{C_2H_5OH}\to H$ tính theo $CH_3COOH$

Theo PT: $n_{CH_3COOH(tt)}=n_{CH_3COOC_2H_5(tt)}=0,15(mol)$

$\to H=\frac{0,15}{0,2}.100\%=75\%$

Em tham khảo nha :

\(\begin{array}{l}
2C{H_3}COOH + 2K \to 2C{H_3}COOK + {H_2}\\
2{C_2}{H_5}OH + 2K \to 2{C_2}{H_5}OK + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:C{H_3}COOH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
60a + 46b = 25,8\\
\dfrac{a}{2} + \dfrac{a}{2} = 0,25
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,3\\
{m_{C{H_3}COOH}} = 0,2 \times 60 = 12g\\
\% C{H_3}COOH = \dfrac{{12}}{{25,8}} \times 100\%  = 46,5\% \\
\% {C_2}{H_5}OH = 100 - 46,5 = 53,5\% \\
b)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5}OH + {H_2}O\\
\dfrac{{0,2}}{1} < \dfrac{{0,3}}{1} \Rightarrow {C_2}{H_5}OH\text{ dư}\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,2 \times 88 = 17,6g\\
H = \dfrac{{13,2}}{{17,6}} \times 100\%  = 75\% 
\end{array}\)