Redox and Equivalent ConceptHard

Question

A quantity of 15.8 g of KMnO4 can be decolourized in acidic medium by (K = 39, Mn = 55, Fe = 56)

Options

A.18.25 g HCl
B.22.5 g H2C2O4
C.32 g SO2
D.38 g FeSO4

Solution

$KMnO_{4} = \frac{15.8}{158} \times 5 = 0.5\text{ eq}$

$HCl = \frac{18.25}{36.5} \times 1 = 0.5\text{ eq} $$${H_{2}C_{2}O_{4} = \frac{22.5}{90} \times 2 = 1.0\text{ eq} }{SO_{2} = \frac{32}{69} \times 2 = 1.0\text{ eq} }{\text{FeS}\text{O}_{4} = \frac{38}{152} \times 1 = 0.25\text{ eq}}$$

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