Câu 12 :
\(n_{CH3COOH}=\dfrac{3}{60}=0,05\left(mol\right)\)
\(n_{CH3COOC2H5}=\dfrac{2,2}{88}=0,025\left(mol\right)\)
a) Pt : \(CH_3COOH+C_2H_5OH\xrightarrow[]{H_2SO_{4đặc}}CH_3COOC_2H_5+H_2O\)
b) \(H\%=\dfrac{0,025}{0,05}.100\%=50\%\)
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