a,
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
$\Rightarrow n_{CH_3COOH}=n_{NaOH}=0,05.2=0,1 mol$
$m_{CH_3COOH}=0,1.60=6g$
$\Rightarrow \%m_{CH_3COOH}=\frac{6.100}{12,9}= 46,5\%$
$\%m_{C_2H_5OH}=100-46,5=53,5\%$
b,
$n_{C_2H_5OH}=\frac{12,9-6}{46}=0,15 mol$
$CH_3COOH+C_2H_5OH\rightleftharpoons CH_3COOC_2H_5+H_2O$
Theo lí thuyết tạo 0,1 mol este.
$\Rightarrow H=\frac{7,04.100}{88.0,1}=80\%$