Ta có: \(n_{Na_2CO_3}=\dfrac{14,84}{106}=0,14\left(mol\right)\)
PT: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2+H_2O\)
Theo PT: \(n_{CH_3COOH}=2n_{Na_2CO_3}=0,28\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,28.60}{20}.100\%=84\%\\\%m_{C_2H_5OH}=16\%\end{matrix}\right.\)