Question

Number of moles of acidic $KMn{O}_4$ needed for complete oxidation of $1$ mole equimolar mixture of $FeS{O}_4$ and $Fe{C}_2{O}_4$ is:

• A. $\frac{2}{5}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. none of the above

Answer 1

The correct option is A $\frac{2}{5}$

Moles of acidic $KMn{O}_4$ needed for complete oxidation of $1$ mole equimolar mixture of $FeS{O}_4$ and $Fe{C}_2{O}_4$ is $\frac{2}{5}$.
$1$ mole equimolar mixture of $FeS{O}_4$ and $Fe{C}_2{O}_4$ will contain $0.5$ moles of $FeS{O}_4$ and $0.5$ moles of $Fe{C}_2{O}_4$.
Thus, it contains $1$ mol of $F{e}^{2+}$ ions which will be oxidized to $F{e}^{3+}$ ions and $0.5$ moles of $C_2{O}_4^{2-}$ ions which will be oxidized to $C{O}_2$.
The reactions are as follows:
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
$Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$
Thus, $2$ moles $KMn{O}_4=$ $5$ moles $C_2{O}_4^{2-}$
$1$ mole $KMn{O}_4=$ $5$ moles $F{e}^{2+}$
Hence, moles of acidic $KMn{O}_4$ needed for complete oxidation of $1$ mole equimolar mixture of $FeS{O}_4$ and $Fe{C}_2{O}_4$ is $\frac{0.5}{5}+\frac{0.5}{5}+\frac{1}{5}=\frac{2}{5}$.