JEE Main 2019ChemistryRedox ReactionsHardMCQ

JEE Main 2019 — Redox Reactions Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

In order to oxidize a mixture of one mole of each of FeC2O4,Fe2(C2O4)3,FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B2

Step-by-step explanation

Eq.KMnO4⇒Eq.FeSO4+Eq.Fe2SO43+EqFeC2O4+Eq.Fe2C2O43
KMnO4⇒MnO4-
5e-+8H++MnO4-⇒Mn+2+4H2O V.F=5

So Eq.KMnO4⇒Eq.FeSO4+Eq.FeC2O4+Eq.Fe2C2O43
n×5=1×1+1×0+1×3+1×6
n×5=1+3+6
n=105=2 mole

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About this question

This is a previous-year question from JEE Main 2019, covering the Redox Reactions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.