PTHH: Fe+2HCl-----> FeCl2 +H2(1)
CaO+HCl-----> CaCl2+H2O(2)
\(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(TheoPT\left(1\right):n_{H_2}=n_{Fe}=0,15\left(mol\right)\)
\(\Rightarrow\%m_{Fe}=\frac{0,15.56}{33,6}.100=25\%\)
\(\Rightarrow\%m_{CaO}=100-25=75\%\)