\(n_{CH_3COOH}=\dfrac{120}{60}=2\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{46}{46}=1\left(mol\right)\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(ĐK:H_2SO_{4\left(đ\right)},t^0\right)\)
\(Bđ:\) \(2.........................1\)
\(Pư:1.......................1.....................1\)
\(KT:1.....................0...................1\)
\(m_{CH_3COOC_2H_5}=1\cdot88=88\left(g\right)\)
\(H\%=\dfrac{52.8}{88}\cdot100\%=60\%\)
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