Question

How many moles of $KMn{O}_4$ are reduced by $1$ mole of ferrous oxalte in acidic medium?

• A. $\frac{1}{5}$
• B. $\frac{5}{3}$
• C. $\frac{1}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is D $\frac{3}{5}$

The balanced chemical equation of the reaction is-

$3Mn{O}_4^{-}+5Fe{C}_2{O}_4+24H^{+}⟶3M{n}^{2+}+5F{e}^{3+}+10C{O}_2+12H_{2}O$

Therefore $3$ mole of $KMn{O}_4$ will be required

$3$ mole of $KMn{O}_4=5$ moles of $Fe{C}_2{O}_4$

$\frac{3}{5}$ mole of $KMn{O}_4=$ 1 moles of $Fe{C}_2{o}_4$

For 1 mole $Fe{C}_2{O}_4,\frac{3}{5}KMN{O}_4$ is required

So, the correct option is $D$