\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{21,2}.100\%\approx56,6\%\\\%m_{C_2H_5OH}\approx43,4\%\end{matrix}\right.\)