$n_{CH_3COOH}=\dfrac{90}{60}=1,5\ (mol)$
$n_{C_2H_5OH}=\dfrac{150}{46}=3,26\ (mol)$
$CH_3COOH+C_2H_5OH\mathop{\huge{\rightleftharpoons}}\limits^{\small{H_2SO_4đ,t^o}} CH_3COOC_2H_5+H_2O$
Vì $n_{CH_3COOH}<n_{C_2H_5OH}$
$\Rightarrow CH_3COOH$ hết
Ta có: $n_{CH_3COOC_2H_5}=n_{CH_3COOH}=1,5\ (mol)$
$\Rightarrow m_{CH_3COOC_2H_5\ lt}=1,5.88=132\ (g)$
$\Rightarrow H=\dfrac{82,5}{132}.100\%=62,5\%$