PTHH \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4]{t^o}CH_3COOC_2H_5+H_2O\)
(mol) 0,05→ 0,05
\(n_{C2H5OH}=\frac{2,3}{46}=0,05\left(mol\right)\\ n_{CH3COOH}=\frac{3,6}{60}=0.06\left(mol\right)\)
⇒So sánh tỉ số \(\frac{n_{C2H5OH}}{1}\) và \(\frac{n_{CH3COOH}}{1}\)
⇒\(\frac{0,05}{1}< \frac{0,06}{1}\)⇒Sau phản ứng CH3COOH dư
Theo PTHH ta có:
\(n_{CH3COOH2H5}=0,05\left(mol\right)\\ =>m_{CH3COOC2H5}=0,05.88=4,4\left(g\right)\)