Câu a: $0,1$ mol $CH_3COOH$, $0,2$ mol $C_2H_5OH$
b,
$CH_3COOH+C_2H_5OH\mathop{\Huge{\rightleftharpoons}}\limits^{\small{t^o,H_2SO_4đ}}CH_3COOC_2H_5+H_2O$
$\dfrac{0,1}{1}<\dfrac{0,2}{1}\to$ theo lí thuyết, axit hết, ancol dư.
$n_{\rm este lí thuyết}=n_{CH_3COOH}=0,1(mol)$
$H=80\%\to m_{CH_3COOC_2H_5}=0,1.88.80\%=7,04g$