$n_{H_2}=1,344/22,4=0,06mol$
$a/2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2↑$
$ 2CH_3COOH+2Na\to 2CH_3COONa+H_2↑$
b/Gọi $n_{C_2H_5OH}=a;n_{CH_3COOH}=b$
$\text{Ta có :}$
$m_{hh}=42a+60b=6,64g$
$n_{H_2O}=0,5a+0,5b=0,6mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
46a+60b=6,64 & \\
0,5a+0,5b=0,6 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,04 & \\
b=0,08 &
\end{matrix}\right.$
$⇒\%m_{C_2H_5OH}=\dfrac{46.0,04.100\%}{6,64}=27,71\%$
$\%m_{CH_3COOH}=100\%-27,71\%=72,29\%$