a)
1,5 lít = 1500 ml
$V_{rượu} = 1500.\dfrac{20}{100} = 300(ml)$
b)
$m_{rượu} = D.V = 0,8.300 = 240(gam)$
$n_{C_2H_5OH} = \dfrac{240}{46} = \dfrac{120}{23}(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$n_{CH_3COOH} = n_{C_2H_5OH} = \dfrac{120}{23}(mol)$
$m_{CH_3COOH} = \dfrac{120}{23}.60 = 313,04(gam)$