Đáp án:
\(\begin{array}{l}
b)\\
{m_{C{H_3}COO{H_d}}} = 25,5g\\
c)\\
H = 80\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_5}OH + C{H_3}COOH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
b)\\
{n_{C{H_3}COOH}} = \dfrac{{36}}{{60}} = 0,6mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{{8,05}}{{46}} = 0,175mol\\
\dfrac{{0,175}}{1} < \dfrac{{0,6}}{1} \Rightarrow C{H_3}COOH\text{ dư}\\
{n_{C{H_3}COO{H_d}}} = {n_{C{H_3}COOH}} - {n_{{C_2}{H_5}OH}} = 0,6 - 0,175 = 0,425mol\\
{m_{C{H_3}COO{H_d}}} = 0,425 \times 60 = 25,5g\\
c)\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{{C_2}{H_5}OH}} = 0,175mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,175 \times 88 = 15,4g\\
H = \dfrac{{12,32 \times 100\% }}{{15,4}} = 80\%
\end{array}\)