Find the equivalent weights of each oxidant in the following reaction.FeSO4+KClO3⟶Fe2SO43+KCl

Writing the given equation in the ionic form and keeping those ions only whose oxidation number changes:Fe2++Cl+5O3−⟶Fe3++Cl−(i) Fe2+⟶Fe3++1e−∴ Eq. wt. of FeSO4 (reductant) = Mol. wt. of FeSO4 no. of e-s lost =56+32+(4×16)1=1521=152 Ans. (ii) Cl+5 (of ClO−) +6e−⟶Cl− ∴ Eq. wt. of KClO3 (oxidant)  = Mol. wt. of KClO3 no. of e s gained =39+35.5+(3×16)6=122.56=20.42 Ans.