Đáp án:

 5,28g

Giải thích các bước giải:

\(\begin{array}{l}
P1:\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
P2:\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
P3:\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
P1:\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow nX = 0,15 \times 2 = 0,3\,mol\\
P2:\\
nC{O_2} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol \Rightarrow nC{H_3}COOH = 0,05 \times 2 = 0,1\,mol\\
 \Rightarrow n{C_2}{H_5}OH = 0,3 - 0,1 = 0,2\,mol\\
P3:\\
\dfrac{{0,1}}{1} < \dfrac{{0,2}}{1} \Rightarrow nC{H_3}COO{C_2}{H_5} = nC{H_3}COOH = 0,1\,mol\\
H = 60\%  \Rightarrow m = 0,1 \times 60\%  \times 88 = 5,28g
\end{array}\)