In order to oxidise a mixture of 1 mole each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 required is
- 3
- 2
- 5
- 7
Answer
(2) 2
Solution
In acidic medium: MnO4- + 8H+ + 5e- Mn2+ + 4H2O (each KMnO4 takes 5 e-)
Oxidisable parts: Fe2+ Fe3+ + e- and C2O42- 2CO2 + 2e-
FeC2O4: 1 ( Fe2+) + 2 (one oxalate) = 3 mol e-
Fe2(C2O4)3: Fe already +3; 3 oxalates × 2 = 6 mol e-
FeSO4: 1 mol e-
Fe2(SO4)3: nothing to oxidise = 0
⇒ Total electrons lost = 3 + 6 + 1 + 0 = 10 mol
∴ Moles of KMnO4 = 105 = 2
Answer: (2)