Đáp án + Giải thích các bước giải:

$n_{CH_3COOH}=\frac{9}{60}=0,15(mol)$

$n_{C_3H_7OH}=n_{C_2H_5OH}=\frac{10,6}{46+60}=0,1(mol)$

$\begin{cases} C_3H_7OH\\ C_2H_5OH \end{cases}+CH_3COOH\xrightarrow{H_2SO_{4(đ,n)}}\begin{cases} CH_3COOC_3H_7\\ CH_3COOC_2H_5 \end{cases}$

Vì $n_{C_3H_7OH}=n_{C_2H_5OH}<n_{CH_3COOH}$

$\to$ Hiệu suất tính theo hỗn hợp ancol

Theo PT: $n_{CH_3COOC_3H_7(lt)}=n_{CH_3COOC_2H_5(lt)}=0,1(mol)$

$\to \begin{cases}m_{CH_3COOC_3H_7(tt)}=0,1.66,67\%.102=6,80034(g)\\ m_{CH_3COOC_2H_5(tt)}=0,1.66,67\%.88=5,86696(g) \end{cases}$

\(n_{C_2^{}H_5^{}OH}^{} = n_{C_3^{}H_7^{}OH}^{} = x\;{\text{mol}}\)

\(10,6 = m_{C_2^{}H_5^{}OH}^{} + m_{C_3^{}H_7^{}OH}^{} = 46x + 60x\)

\( \to x = 0,1\;{\text{mol}}\)

\({\text{n}}_{CH_3^{}COOH}^{} = \frac{9}{{60}} = 0,15\;{\text{mol}}\)

\({\text{n}}_{CH_3^{}COOH}^{} = \frac{9}{{60}} = 0,15\;{\text{mol}}\)

\(CH_3^{}COOH + ROH\xrightarrow{{H_2^{}SO_{4d}^{},{t^o}}}CH_3^{}COOR + H_2^{}O\)

\(\;{\text{   }}\frac{{0,15}}{1}{\text{     <     }}\frac{{0,2}}{1} \to \) Ancol dư, tính theo axit axetic

\(M_{Este}^{} = 53 + 60 - 18 = 95g/mol\)

\(m_{Este}^{} = 66,67\% .0,15.95 = 9,5g\)