Đáp án:
\(\begin{array}{l}
\% {m_{C{H_3}COOH}} = 39,47\% \\
\% {m_{{C_2}{H_5}OH}} = 60,53\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{C{H_3}COOH}} = 2{n_{C{O_2}}} = 0,2\,mol\\
\% {m_{C{H_3}COOH}} = \dfrac{{0,2 \times 60}}{{30,4}} \times 100\% = 39,47\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 39,47 = 60,53\%
\end{array}\)
