\(m_{CH_3COOH}=200.10\%=20\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{20}{60}=\dfrac{1}{3}\left(mol\right)\)
PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
a, \(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=\dfrac{1}{6}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{1}{6}.22,4=\dfrac{56}{15}\left(l\right)\)
b, \(n_{Fe}=\dfrac{1}{2}n_{CH_3COOH}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{1}{6}.56=\dfrac{28}{3}\left(g\right)\)
c, Ta có: m dd sau pư = 28/3 + 200 - 1/6.2 = 209 (g)
\(n_{\left(CH_3COO\right)_2Fe}=\dfrac{1}{2}n_{CH_3COOH}=\dfrac{1}{6}\left(mol\right)\)
\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{\dfrac{1}{6}.174}{209}.100\%\approx13,88\%\)
