\(n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{100}{46}=\dfrac{50}{23}\left(mol\right)\)
PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)
Xét tỉ lệ: \(\dfrac{1}{1}< \dfrac{\dfrac{50}{23}}{1}\), ta được C2H5OH dư.
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=1\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=1.88=88\left(g\right)\)
\(\Rightarrow H=\dfrac{55}{88}.100\%=62,5\%\)