Đáp án:
\(\begin{array}{l}
m = 1,66g\\
\% {m_{{C_2}{H_5}OH}} = 27,71\% \\
\% {m_{C{H_3}COOH}} = 72,29\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{KOH}} = 0,2 \times 0,1 = 0,02\,mol\\
{n_{C{H_3}COOH}} = {n_{KOH}} = 0,02\,mol\\
{n_{{H_2}}} = \dfrac{{0,336}}{{22,4}} = 0,015\,mol\\
{n_X} = 2{n_{{H_2}}} = 0,03\,mol\\
{n_{{C_2}{H_5}OH}} = 0,03 - 0,02 = 0,01\,mol\\
m = 0,01 \times 46 + 0,02 \times 60 = 1,66g\\
\% {m_{{C_2}{H_5}OH}} = \dfrac{{0,01 \times 46}}{{1,66}} \times 100\% = 27,71\% \\
\% {m_{C{H_3}COOH}} = 100 - 27,71 = 72,29\%
\end{array}\)
