$n_{CH_3COOH} = \dfrac{12}{60} = 0,2(mol)$
$n_{C_2H_5OH} = \dfrac{1,38}{46} = 0,03(mol)$
\(CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)
Ta thấy :
$n_{CH_3COOH} : 1 > n_{C_2H_5OH}:1$ nên hiệu suất tính theo số mol $C_2H_5OH$
$n_{CH_3COOC_2H_5} = n_{C_2H_5OH\ pư} = 0,03.75\% = 0,0375(mol)$
$m_{CH_3COOC_2H_5} = 0,0375.88 =3,3(gam)$