1) \(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
\(n_{H_2}=\dfrac{2.24}{22,4}=0,1\left(mol\right)\)
PTHH : Fe2O3 + 3CO -----to----> 2Fe + 3CO2 ( 1 )
0,1 0,2
PTHH : Fe + 2HCl -> FeCl2 + H2 ( 2 )
0,1 0,1
\(m_{Fe_2O_3}=0,1.160=16\left(g\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
2 ) \(m_{hh}=16+5,6=21,6\left(g\right)\)
\(\%m_{Fe}=\dfrac{5.6}{21,6}.100\%=25,92\%\)
\(\%m_{Fe_2O_3}=100\%-25,92\%=74,08\%\)