Calculate the volume of 0.5 M KMnO₄ required to oxidize 0.2 mol of KNO₂ to KNO₃ in an acidic medium.
- 40 ml
- 80 ml
- 160 ml
- 200 ml
📝 Detailed explanation
2KMnO₄ + 5KNO₂ + 3H₂SO₄ → 2MnSO₄ + 5KNO₃ + K₂SO₄ + 3H₂OMolar ratio KMnO₄ : KNO₂ is 2 : 5.
Moles KMnO₄ = 0.2 × (2 / 5) = 0.08 mol.
Volume = 0.08 / 0.5 = 0.16 L = 160 ml.
Wait, calculation: 0.2 / 2.5 = 0.08. 0.08/0.5 = 0.16. Correct is 160 ml.