Xét phần 1:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(m_{Fe}=0,2.56=11,2g\)
Xét phần 2:
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
\(m_{Fe\left(pứ\right)}=33,6-11,2=22,4g\)
\(n_{Fe}=\dfrac{22,4}{56}=0,4mol\)
\(\rightarrow m_{Fe_2O_3}=\left(\dfrac{0,4.1}{2}\right).160=32g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{11,2+32}.100=25,92\%\\\%m_{Fe_2O_3}=100\%-25,92\%=74,08\%\end{matrix}\right.\)
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