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The number of moles of $KMnO_4$ that will be needed to react completely with one mole of ferrous oxalate in acidic solution is

$(a)\;\large\frac{3}{5} \\ (b)\;\large\frac{2}{5} \\(c)\; \large\frac{4}{5} \\(d)\;1$
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$Fe^{2+}+C_2O_4^{2-} \longrightarrow Fe^{3+} +2CO_2+3e^{-}$
$MnO_4^{-}$ will oxidise $Fe^{2+}$ to $Fe^{3+}$ and $C_2O_4^{2-}$ to $CO_2$.
Thus we have,
$Fe^{2+}+C_2O_4^{2-} \longrightarrow Fe^{3+} +2CO_2+3e^{-}$(oxidation)----(i)
$MnO_4^{-}+5e^{-}+8H^{+} \longrightarrow Mn^{2+}+4H_2O$ (Reduction) -----(ii)
Multiplying (i) by 5 and (ii) by 3 and adding , we get
$5 Fe^{2+} +5 C_+2O_4^{2-}+3 MnO_4^{-} +24 H^{+} \longrightarrow 5Fe^{3+}+10CO_2+12H_2O$
From above equation
$5 FeC_2O_4=3MnO_4^{-}$
or $1 Fec_2O_4 =\large\frac{3}{5}$$MnO_4$
Hence a is the correct answer.
answered Mar 21, 2014 by meena.p

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