Đáp án:
\( \% {m_{C{H_3}COOH}} = 60\%; \% {m_{{C_2}{H_5}OH}} = 40\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + 2Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{n}}_{C{H_3}COOH}} = 2{n_{{H_2}}} = 0,2{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,2.60 = 12{\text{ gam}}\)
\( \to \% {m_{C{H_3}COOH}} = \frac{{12}}{{20}} = 60\% \to \% {m_{{C_2}{H_5}OH}} = 40\% \)