\(CH_3COOH+C_2H_5OHCH_3⇌COOHC_2H_5+H_2O\) ( \(dk:H_2SO_4\left(đặc\right),t^o\))
\(n_{CH_3COOH}=\dfrac{m}{M}=\dfrac{23}{60}\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{m}{M}=\dfrac{120}{46}=\dfrac{60}{23}\left(mol\right)\)
\(\dfrac{23}{60}\left(mol\right)< \dfrac{60}{23}\left(mol\right)\Rightarrow\) \(C_2H_5OH\) dư
Cho hiệu suất là \(100\%\Rightarrow n_{COOHC_2H_5}=n_{C_2H_5OH}=\dfrac{60}{23}\left(mol\right)\)
Hiệu suất là \(90\%\Rightarrow n_{COOHC_2H_5}=\dfrac{\dfrac{60}{23}\times90}{100}=\dfrac{54}{23}\left(mol\right)\)
Vậy \(m_{COOHC_2H_5}=n.M=\dfrac{54}{23}.88=\dfrac{4752}{23}\approx207\left(g\right)\)