Đáp án:
21,8g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
b)\\
n{H_2} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
hh:{C_2}{H_5}OH(a\,mol),C{H_3}COOH(b\,mol)\\
46a + 60b = 15,2\\
0,5a + 0,5b = 0,15\\
\Rightarrow a = 0,2;b = 0,1\\
\% m{C_2}{H_5}OH = \dfrac{{0,2 \times 46}}{{15,2}} \times 100\% = 60,53\% \\
\% mC{H_3}COOH = 100 - 60,53 = 39,47\% \\
c)\\
n{C_2}{H_5}ONa = n{C_2}{H_5}OH = 0,2\,mol\\
nC{H_3}COONa = nC{H_3}COOH = 0,1\,mol\\
m = 0,2 \times 68 + 0,1 \times 82 = 21,8g
\end{array}\)