Question

Number of moles of $KMn{O}_4$ required to oxidise one mole of $Fe(C_2{O}_4)$ in acidic medium is

• A. $0.6$
• B. $0.4$
• C. $0.2$
• D. $1.67$

Answer 1

Answer: (A)

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
$Fe(C_2{O}_4)\to F{e}^{2+}+C_2{O}_4^{2-}$
$F{e}^{2+}\to F{e}^{3+}+e^{-},C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$
We can see that one mole of $KMn{O}_4$ accepts 5 electrons whereas one mole of $Fe(C_2{O}_4)$ loses 3 electrons
$\therefore$ No. of moles of $KMn{O}_4$ required to oxidise one mole of $Fe(C_2{O}_4)=\frac{3}{5}=0.6$ mole