a) PTHH: Fe + 2HCl ===> FeCl2 + H2
Fe2O3 + 6HCl ===> 2FeCl3 + 3H2O
Ta có: nH2 = \(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=> nFe = 0,1(mol)
=> mFe = \(0,1\cdot56=5,6\left(gam\right)\)
=> %mFe = \(\dfrac{5,6}{8}\cdot100\%=70\%\)
=> %mFe2O3 = \(100\%-70\%=30\%\)