Ta có: \(n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)
PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (to, H2SO4 đặc)
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,1.88=8,8\left(g\right)\)
Mà: mCH3COOC2H5 (TT) = 4,4 (g)
\(\Rightarrow H\%=\dfrac{4,4}{8,8}.100\%=50\%\)