a,
Gọi a, b là mol rượu và axit.
$\Rightarrow 46a+60b=10,6$ (1)
$n_{H_2}=\frac{2,24}{22,4}=0,1 mol$
$2CH_3COOH+2Na\to 2CH_3COONa+H_2$
$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$
$\Rightarrow 0,5a+0,5b=0,1$ (2)
(1)(2) $\Rightarrow a=b=0,1$
$\%m_{CH_3COOH}=\frac{0,1.60.100}{10,6}=56,6\%$
$\%m_{C_2H_5OH}= 43,4\%$
b,
$n_{CH_3COOH}= n_{CH_3COONa}=0,1 mol$
$n_{C_2H_5OH}=n_{C_2H_5ONa}=0,1 mol$
$\Rightarrow m_{muối}=0,1.82+0,1.68=15g$