$\rm CH_3COOH + NaOH \to CH_3COONa + H_2O$

$\rm n_{NaOH} = \dfrac{160.5\%}{40} = 0,2$ mol

a)$\rm n_{CH_3COOH} = n_{NaOH} = 0,2$ mol

$\rm \to m = m_{CH_3COOH} = 0,2 . 60 = 12g$

b)$\rm C_2H_5OH + O_2 \xrightarrow[]{\text{men giấm}} CH_3COOH + H_2O$

$\rm n_{C_2H_5OH} = n_{CH_3COOH} = 0,2$ mol

$\rm \to m_{C_2H_5OH} = 0,2 . 46 = 9,2g$