Đáp án:
b, \({m_{C{H_3}COO{C_2}{H_5}}} = 88g\)
c, \(H = \dfrac{{5,5}}{{88}} \times 100\% = 6,25\% \)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = 1mol\\
{n_{{C_2}{H_5}OH}} = 2mol\\
\to {n_{{C_2}{H_5}OH}} > {n_{C{H_3}COOH}}
\end{array}\)
\({C_2}{H_5}OH\) dư
\(\begin{array}{l}
\to {n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 1mol \to {m_{C{H_3}COO{C_2}{H_5}}} = 88g\\
\to H = \dfrac{{5,5}}{{88}} \times 100\% = 6,25\%
\end{array}\)