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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)

S solutionqc

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

$\Rightarrow +3$ 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.

$\Rightarrow$ In 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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