a) \(n_{FeCl_2}=\dfrac{31,75}{127}=0,25\left(mol\right)\)
=> mFe = 0,25.1.56 = 14 (g)
b) \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{24}{400}=0,06\left(mol\right)\)
=> mFe = 0,06.2.56 = 6,72 (g)
c) \(n_{Fe_3O_4}=\dfrac{9,28}{232}=0,04\left(mol\right)\)
=> mFe = 0,04.3.56 = 6,72 (g)