Number of moles of MnO₄^- required to oxidize 1 mole of ferrous oxalate Completely in acidic medium will be?

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Solution

Let us take the reaction between ferrous oxalate and MnO₄^- in presence of sulphuric acid as:

FeC₂O₄ + KMnO₄ +H₂SO₄ + ----> K₂SO₄ + MnSO₄ + Fe₂(SO₄)₃ + H₂O + CO₂

While, the balanced form of reaction is:

10FeC₂O₄ + 6KMnO₄ + 24H₂SO₄ ----> 3K₂SO₄+ 6MnSO₄ + 5Fe₂(SO₄)₃ + 24H₂O + 20CO₂

So, we see that, 6 moles of KMnO₄ is required to oxidize 10 moles of FeC₂O₄

So, 1 mole of FeC₂O₄ would be oxidized by = 6/10 = 0.6 moles of KMnO₄

So, the answer is a). 0.6

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