$nH_{2}=\frac{2,24}{22,4}=0,1$
$C_{2}H_{5}OH +Na \to C_{2}H_{5}ONa+1/2H_{2}$
$CH_{3}COOH+Na \to CH_{3}COONa+1/2H_{2}$
Gọi số mol $C_{2}H_{5}OH$: a mol
$CH_{3}COOH$: b mol
$46a+60b=10,6$
$1/2a+1/2b=0,1$
⇒$a=b=0,1$
$\%mC_{2}H_{5}OH=\frac{46.0,1}{10,6}.100=43,4\%$
$\%mCH_{3}COOH=100-43,4=56,6\%$
$mCH_{3}COONa=0,1.82=8,2g$