\(n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt, to)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được C2H5OH dư.
Theo PT: \(n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5}=0,1.88=8,8\left(g\right)\)
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