Number of moles of MnO₄⁻ required to oxidize one mole of ferrous oxalate

Number of moles of MnO₄⁻ required to oxidize one mole of ferrous oxalate completely in acidic medium will be

Options

(a) 0.6 moles
(b) 0.4 moles
(c) 7.5 moles
(d) 0.2 moles

Correct Answer:

0.6 moles

Explanation:

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

From the above balanced equation, we observe that

6 moles of KMnO4 is required to oxidize 10 moles of  FeC2O4

Then, 1 mole of FeC2O4 would be oxidized by = 6/10=0.6

admin:

Related Questions

  1. Reaction of nitrous acid with aliphatic primary amine will give
  2. Which of the following will not be soluble in sodium hydrogen carbonate
  3. Which of the following compounds will gives positive test with Tollen’s reagent
  4. A liquid is in equilibrium with its vapour at its boiling point
  5. In acidic medium, the equivalent weight of KMnO₄ is