\(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{m}{60}\left(mol\right)\\n_{C_2H_5OH}=\dfrac{m}{46}\left(mol\right)\end{matrix}\right.\)
\(n_{CH_3COOC_2H_5}=\dfrac{m}{88}\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
Xét tỉ lệ: \(\dfrac{\dfrac{m}{60}}{1}< \dfrac{\dfrac{m}{46}}{1}\) => Hiệu suất tính theo CH3COOH
\(n_{CH_3COOH\left(pư\right)}=n_{CH_3COOC_2H_5}=\dfrac{m}{88}\left(mol\right)\)
=> \(H=\dfrac{\dfrac{m}{88}}{\dfrac{m}{60}}.100\%=68,182\%\)