Bài 25:
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{CH_3COOH}=2n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{16,6}.100\%\approx72,29\%\\\%m_{C_2H_5OH}\approx27,71\%\end{matrix}\right.\)
Bài 26:
Ta có: \(46n_{C_2H_5OH}+60n_{CH_3COOH}=21,2\left(g\right)\left(1\right)\)
PT: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
\(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+\dfrac{1}{2}n_{CH_3COOH}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,2.46}{21,2}.100\%\approx43,4\%\\\%m_{CH_3COOH}\approx56,6\%\end{matrix}\right.\)
