$n_{H_2}=2,24/22,4=0,1mol$
$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$
$2CH_3COOH+2Na\to 2CH_3COONa+H_2$
$a/Gọi n_{C_2H_5OH}=a;n_{CH_3COOH}=b$
$\text{Ta có :}$
$m_{hh}=46a+60b=10,6$
$n_{H_2}=0,5a+0,5b=0,1$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
46a+60b=10,6 & \\
0,5a+0,5b=0,1 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,1 & \\
b=0,1 &
\end{matrix}\right.$
$⇒\%m_{C_2H_5OH}=\dfrac{0,1.46.100\%}{10,6}≈43,4\%$
$\%m_{CH_3COOH}=100\%-43,4\%=56,6\%$
$b/n_{C_2H_5ONa}=n_{C_2H_5OH}=0,1$
$⇒m_{C_2H_5ONa}=0,1.68=6,8g$
$n_{CH_3COONa}=n_{CH_3COOH}=0,1mol$
$⇒m_{CH_3COONa}=0,1.82=8,2g$
$⇒m_{\text{muối}}=8,2+6,8=15g$