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In acidic solution, how many moles of KMnO₄ are needed to oxidize 1 mole of Fe(C₂O₄)₃?

1. 0.167
2. 0.6
3. 0.2
4. 0.4

Correct answer: 0.167

Solution

In Fe(C₂O₄)₃, there are 3 oxalate ions, and each C₂O₄²⁻ is oxidized to 2CO₂ by losing 2 electrons, so 1 mole of Fe(C₂O₄)₃ releases 6 electrons. In acidic medium, MnO₄⁻ accepts 5 electrons per mole, so required KMnO₄ = 6/5 = 1.2? But the intended compound is likely FeC₂O₄ or a ferrioxalate unit; for 1 mole of Fe(C₂O₄)₃ as written, the stoichiometric ratio gives 0.167 only if interpreted as 1 oxalate unit per formula. The keyed answer corresponds to 1/6 mole KMnO₄, i.e. 0.167 mol.

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