PTHH : \(C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)
\(n_{C_2H_5OH}=\frac{m}{M}=\frac{6,9}{46}=0,15\left(mol\right)\)
\(m_{ddCH_3COOH}=D.V=100.1,05=105\left(g\right)\)
Lại có : \(C\%_{CH_3COOH}=\frac{m_{CH_3COOH}}{m_{dd}}.100\%=36\%\)
=> \(m_{CH_3COOH}=37,8\left(g\right)\)
=> \(n_{CH_3COOH}=\frac{m}{M}=\frac{37,8}{60}=0,63\left(mol\right)\)
Ta có : \(\frac{n_{C_2H_5OH}}{n_{CH_3COOH}}=\frac{0,15}{0,63}~0,23< 1\)
=> C2H5OH phản ứng hết, CH3COOH còn dư .
- Theo PTHH : \(n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,15\left(mol\right)\)
=> \(m_{CH_3COOC_2H_5\left(pt\right)}=n.M=88.0,15=13,2\left(g\right)\)
=> \(m_{CH_3COOC_2H_5}=H.m=11,88\left(g\right)\)