\(3CO\left(0,1485\right)+Fe_2O_3\left(0,0495\right)\rightarrow2Fe\left(0,099\right)+3CO_2\left(,1485\right)\)
\(Fe\left(0,099\right)+H_2SO_4\left(0,099\right)\rightarrow FeSO_4\left(0,099\right)+H_2\left(0,099\right)\)
\(n_{H_2}=\dfrac{2,22}{22,4}=0,099\)
\(\Rightarrow m_{Fe_2O_3}=0,0495.160=7,92\left(g\right)\)
\(\Rightarrow\%Fe_2O_3=\dfrac{7,92}{10}.100\%=79,2\%\)