a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
Ta có: \(n_{H_2}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{9,2}.100\%\approx60,87\%\\\%m_{FeO}\approx39,13\%\end{matrix}\right.\)
b, Ta có: \(n_{FeO}=\dfrac{9,2-0,1.56}{72}=0,05\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{FeO}=0,15\left(mol\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)\)
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