Đáp án:
\({m_{C{H_3}COO{C_2}{H_5}}} = \)13,2g
Giải thích các bước giải:
\(\begin{array}{l}
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = 0,2mol\\
{n_{{C_2}{H_5}OH}} = 0,26mol\\
{n_{C{H_3}COOH}} < {n_{{C_2}{H_5}OH}}
\end{array}\)
\({C_2}{H_5}OH\) dư
\(\begin{array}{l}
\to {n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2mol\\
\to {m_{C{H_3}COO{C_2}{H_5}}} = 17,6g\\
\to {m_{C{H_3}COO{C_2}{H_5}}}thực= \dfrac{{17,6 \times 75}}{{100}} = 13,2g
\end{array}\)