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Q.
Number of moles of $KMnO_{4}$ required to oxidise one mole of $Fe(C_{2}O_{4})$ in acidic medium is
Redox Reactions
Solution:
$MnO^{-}_{4}+8H^{+}+5e^{-} \to Mn^{2+}+4H_{2}O$
$Fe(C_{2}O_{4}) \to Fe^{2+}+C_{2}O^{2-}_{4}$
$Fe^{2+} \to Fe^{3+}+e^{-}, C_{2}O^{2-}_{4} \to 2CO_{2}+2e^{-}$
We can see that one mole of $KMnO_{4}$ accepts 5 electrons whereas one mole of $Fe(C_{2}O_{4})$ loses 3 electrons
$\therefore $ No. of moles of $KMnO_{4}$ required to oxidise one mole of $Fe(C_{2}O_{4}) =\frac{3}{5}=0.6$ mole