$a\big)CH_3COOH+NaOH\to CH_3COONa+H_2O$
$b\big)$
Theo PT: $n_{CH_3COOH}=n_{CH_3COONa}=\dfrac{8,2}{82}=0,1(mol)$
$\to m_{CH_3COOH}=0,1.60=6(g)$
$\to m_{C_2H_5OH}=10,6-6=4,6(g)$
$\to \begin{cases}\%m_{CH_3COOH}\approx 56,6\%\\ \%m_{C_2H_5OH}=43,4\% \end{cases}$
