\(CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right);n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\Rightarrow C_2H_5OH.dư\\ n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\\ n_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{11}{88}=0,125\left(mol\right)\\ \Rightarrow H=\dfrac{0,125}{0,2}.100\%=62,5\%\)