Câu 4:
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
b, Ta có: \(n_{H_2}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{21,6}.100\%\approx25,93\%\\\%m_{Fe_2O_3}\approx74,07\%\end{matrix}\right.\)
c, Ta có: \(n_{Fe_2O_3}=\dfrac{21,6-0,1.56}{160}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=0,8\left(mol\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{0,8.36,5}{200}.100\%=14,6\%\)
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