Đáp án:
11,88g
Giải thích các bước giải:
\(\begin{array}{l}
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
n{H_2} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
hh:{C_2}{H_5}OH(a\,mol),C{H_3}COOH(b\,mol)\\
46a + 60b = 20,5\\
0,5a + 0,5b = 0,2\\
\Rightarrow a = 0,25;b = 0,15\\
{C_2}{H_5}OH + C{H_3}COOH \xrightarrow{xt,t^0} C{H_3}COO{C_2}{H_5} + {H_2}O\\
\dfrac{{0,25}}{1} > \dfrac{{0,15}}{1} \Rightarrow \text{$C_2H_5OH$ dư}\\
nC{H_3}COO{C_2}{H_5} = nC{H_3}COOH = 0,15\,mol\\
H = 90\% \Rightarrow mC{H_3}COO{C_2}{H_5} = 0,15 \times 88 \times 90\% = 11,88g
\end{array}\)