a,
$CH_3COOH+C_2H_5OH\buildrel{{H_2SO_4 đ, t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$
b,
$D_{C_2H_5OH}=0,8g/ml$
$\to n_{C_2H_5OH}=\dfrac{150.45\%.0,8}{46}=1,17(mol)$
Theo PTHH:
$n_{CH_3COOC_2H_5}=n_{CH_3COOH}=n_{C_2H_5OH}= 1,17(mol)$
$\to C_{M_{CH_3COOH}}=\dfrac{1,17}{0,2}=5,85M$
c,
$H=85\%$
$\to m_{CH_3COOC_2H_5}=1,17.88.85\%=87,516g$