Đáp án:
a, \(H = \dfrac{{{m_{C{H_3}COOH}}lý thuyết}}{{{m_{C{H_3}COOH}}thực tế}} \times 100\% = 74\% \)
b, \({m_{{C_2}{H_5}OH}} = D \times V = 160g\)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COO{C_2}{H_5}}} = 0,37mol\\
\to {n_{C{H_3}COOH}} = {n_{C{H_3}COO{C_2}{H_5}}} = 0,37mol \to {m_{C{H_3}COOH}} = 22,2g\\
\to H = \dfrac{{{m_{C{H_3}COOH}}lý thuyết}}{{{m_{C{H_3}COOH}}thực tế}} \times 100\% = 74\% \\
D = 0,8g/ml \to {m_{{C_2}{H_5}OH}} = D \times V = 160g
\end{array}\)