Đáp án:

\(\% {m_{C{H_3}COOH}}  = 66,2\%  ;\% {m_{{C_2}{H_5}OH}} = 33,8\% \)

\( {m_{C{H_3}COO{C_2}{H_5}}} = 7,04{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)

Ta có:

\({n_{NaOH}} = 0,15.1 = 0,15{\text{ mol}}\)

\( \to {n_{C{H_3}COOH}} = {n_{NaOH}} = 0,15{\text{ mol}}\)

\( \to {m_{C{H_3}COOH}} = 0,15.60 = 9{\text{ gam}}\)

\( \to {m_{{C_2}{H_5}OH}} = 13,6 - 9 = 4,6{\text{ gam}}\)

\( \to {n_{{C_2}{H_5}OH}} = \frac{{4,6}}{{46}} = 0,1{\text{ mol}}\)

\( \to \% {m_{C{H_3}COOH}} = \frac{9}{{13,6}} = 66,2\%  \to \% {m_{{C_2}{H_5}OH}} = 33,8\% \)

Este hóa

\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)

\( \to {n_{C{H_3}COOH}} > {n_{{C_2}{H_5}OH}}\) nên axit dư

\( \to {n_{C{H_3}COO{C_2}{H_5}}} = {n_{{C_2}{H_5}OH}}.80\%  = 0,1.80\%  = 0,08{\text{ mol}}\)

\( \to {m_{C{H_3}COO{C_2}{H_5}}} = 0,08.88 = 7,04{\text{ gam}}\)