What would be the number of moles of MnO_4^ - required to oxidize one mole of ferrous oxalate completely in acidic medium?A) 7.5 molesB) 0.2 molesC) 0.6 molesD) 0.4 moles

To determine the number of moles of \(\text{MnO}_4^-\) required to oxidize one mole of ferrous oxalate \((\text{FeC}_2\text{O}_4)\) completely in an acidic medium, we need to balance the redox reaction between \(\text{MnO}_4^-\) and ferrous oxalate.

1. **Write the Oxidation and Reduction Half-Reactions:**

- **Oxidation Half-Reaction:** Ferrous oxalate \((\text{FeC}_2\text{O}_4)\) is oxidized to ferric ion \((\text{Fe}^{3+})\) and carbon dioxide \((\text{CO}_2)\).

\[\text{FeC}_2\text{O}_4 (s) \rightarrow \text{Fe}^{3+} (aq) + 2 \text{CO}_2 (g) + 2 \text{e}^-\]

- **Reduction Half-Reaction:** The permanganate ion \((\text{MnO}_4^-)\) is reduced to manganese(II) ion \((\text{Mn}^{2+})\) in acidic medium.

\[\text{MnO}_4^- (aq) + 8 \text{H}^+ (aq) + 5 \text{e}^- \rightarrow \text{Mn}^{2+} (aq) + 4 \text{H}_2\text{O} (l)\]

2. **Balance the Electrons:**

To combine the two half-reactions, we need to balance the number of electrons transferred. In the oxidation half-reaction, each mole of ferrous oxalate loses 2 moles of electrons. In the reduction half-reaction, each mole of permanganate ion gains 5 moles of electrons.

To equalize the electron transfer, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

- **Oxidation Reaction (multiplied by 5):**

\[5 \text{FeC}_2\text{O}_4 (s) \rightarrow 5 \text{Fe}^{3+} (aq) + 10 \text{CO}_2 (g) + 10 \text{e}^-\]

- **Reduction Reaction (multiplied by 2):**

\[2 \text{MnO}_4^- (aq) + 16 \text{H}^+ (aq) + 10 \text{e}^- \rightarrow 2 \text{Mn}^{2+} (aq) + 8 \text{H}_2\text{O} (l)\]

3. **Combine the Balanced Half-Reactions:**

Add the oxidation and reduction half-reactions together:

\[5 \text{FeC}_2\text{O}_4 (s) + 2 \text{MnO}_4^- (aq) + 16 \text{H}^+ (aq) \rightarrow 5 \text{Fe}^{3+} (aq) + 10 \text{CO}_2 (g) + 2 \text{Mn}^{2+} (aq) + 8 \text{H}_2\text{O} (l)\]

From the balanced equation, we see that 2 moles of \(\text{MnO}_4^-\) are required to oxidize 5 moles of \(\text{FeC}_2\text{O}_4\).

4. **Calculate Moles of \(\text{MnO}_4^-\) Required for 1 Mole of Ferrous Oxalate:**

Since 2 moles of \(\text{MnO}_4^-\) oxidize 5 moles of \(\text{FeC}_2\text{O}_4\), 1 mole of \(\text{FeC}_2\text{O}_4\) would require:

\[
\frac{2 \text{ moles MnO}_4^-}{5 \text{ moles FeC}_2\text{O}_4} = 0.4 \text{ moles MnO}_4^-
\]

So, the number of moles of \(\text{MnO}_4^-\) required to oxidize one mole of ferrous oxalate completely is \(0.4\) moles.

**Answer: D) 0.4 moles**