JEE Main · 2019 · Shift-IhardRDX-074

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…

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

In order to oxidize a mixture of one mole of each of FeC2O4\mathrm{FeC_2O_4}, Fe2(C2O4)3\mathrm{Fe_2(C_2O_4)_3}, FeSO4\mathrm{FeSO_4} and Fe2(SO4)3\mathrm{Fe_2(SO_4)_3} in acidic medium, the number of moles of KMnO4\mathrm{KMnO_4} is:

Options
  1. a

    1

  2. b

    2

  3. c

    3

  4. d

    1.5

Correct Answerb

2

Detailed Solution

Electrons lost per mole of each compound:

| Compound | Fe2+Fe3+\mathrm{Fe^{2+} \rightarrow Fe^{3+}} | C2O422CO2\mathrm{C_2O_4^{2-} \rightarrow 2CO_2} | Total e⁻ | |---|---|---|---| | FeC2O4\mathrm{FeC_2O_4} (1 mol) | 1e⁻ | 2e⁻ | 3 | | Fe2(C2O4)3\mathrm{Fe_2(C_2O_4)_3} (1 mol) | 0 (Fe³⁺ already) | 3×2=63 \times 2 = 6e⁻ | 6 | | FeSO4\mathrm{FeSO_4} (1 mol) | 1e⁻ | 0 | 1 | | Fe2(SO4)3\mathrm{Fe_2(SO_4)_3} (1 mol) | 0 (Fe³⁺ already) | 0 | 0 |

Total electrons lost: 3+6+1+0=103 + 6 + 1 + 0 = 10 mol e⁻

Moles of KMnO4\mathrm{KMnO_4} (n-factor = 5 in acid): n=10/5=2 moln = 10/5 = \mathbf{2\ \mathrm{mol}}

Answer: Option (2) — 2

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In order to oxidize a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in… (JEE Main 2019) | Canvas Classes