In order to oxidise a mixture of one mole of each of FeC_{2}O_{4},Fe_{2}(C_{2}O_{4})_{3},FeSO_{4} and Fe_{2}(SO_{4})_{3} in acidic medium, the number of moles of KMnO_{4} required is :

  • Option 1)

    1.5

  • Option 2)

    3

  • Option 3)

    1

  • Option 4)

    2

Answers (1)

\\FeC_{2}O_{4}+KMnO_{4}\rightarrow Fe^{+3}+CO_{2}+Mn^{2+}\\V.F=3\; \; \;\; \; \; V.F=5

\Rightarrow +3 is the n=factor of FeC_{2}O_{4} and not Fe

\Rightarrow In this compound, carbon is in +3 oxidation state and Fe is in +2 oxidation state.

\RightarrowIn acidic medium , Fe^{2+} is converted to Fe^{+3} and C_{2}O_{4}^{2-} is  converted to CO_{2}. So oxidation number of carbon changes from +3\; to\; +4. As, there are two carbon atoms, the net change in O. Number is +2 and adding to this, the increase in oxidation state of iron, the final increase in O. Number is +3.

Hence n-factor of FeC_{2}O_{4} is +3.

1\times 3= mole of KMnO_{4}\times5\Rightarrow mole=\frac{3}{5}

\\Fe_{2}(C_{2}O_{4})_{3}+KMnO_{4}\rightarrow Fe^{3+}+CO_{2}+Mn^{2+}\\ \; \; V.F=6\; \; \; \; \; \; \; \; \; \; \; V.F=5

1\times 6=mole \times5\; of KMnO_{4}\; \; \; \; \Rightarrow mole\: of\; KMnO_{4}=\frac{6}{5}

\\\rightarrow FeSO_{4}+KMnO_{4}\rightarrow Fe^{+3}+SO_{4}^{2-}+Mn^{2+}\\V.F=1\; \! \; \; \; \; \; V.F=5

1\times 1 = mole of KMnO_{4}\times 5

mole of KMnO_{4}=\frac{1}{5}

\rightarrow Fe_{2}(SO_{4})_{3} doesn't oxidise

Total moles of KMnO_{4}=\frac{3}{5}+\frac{6}{5}+\frac{1}{5}=2


Option 1)

1.5

Option 2)

3

Option 3)

1

Option 4)

2

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