The number of moles of `KMnO_(4)` that will be needed to react completely with one mole of ferrous oxalate in acidic solution is:

Options

A. `3//5`
B. `2//5`
C. `4//5`
D. `1`

Explanation

To determine the number of moles of \( KMnO_4 \) needed to react completely with one mole of ferrous oxalate in an acidic solution, we need to analyze the redox reaction involved.

### Step-by-Step Solution:

1. **Identify the Oxidation States:**
- In \( KMnO_4 \), manganese (Mn) is in the +7 oxidation state.
- In ferrous oxalate (\( FeC_2O_4 \)), iron (Fe) is in the +2 oxidation state, and the oxalate ion (\( C_2O_4^{2-} \)) has carbon in the +3 oxidation state.

2. **Determine the Reduction of \( KMnO_4 \):**
- \( KMnO_4 \) is reduced to \( Mn^{2+} \), changing from +7 to +2. This means it gains 5 electrons:
\[ MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \]

3. **Determine the Oxidation of Ferrous Oxalate:**
- The oxalate ion (\( C_2O_4^{2-} \)) is oxidized to carbon dioxide (\( CO_2 \)). Each carbon goes from +3 to +4, resulting in a total of 2 electrons lost for each \( C_2O_4^{2-} \):
\[ C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-} \]
- The ferrous ion (\( Fe^{2+} \)) is oxidized to ferric ion (\( Fe^{3+} \)), losing 1 electron:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^{-} \]

4. **Combine the Half-Reactions:**
- For each mole of \( C_2O_4^{2-} \), 2 electrons are produced, and for each mole of \( Fe^{2+} \), 1 electron is produced. Therefore, for 1 mole of \( C_2O_4^{2-} \) and 5 moles of \( Fe^{2+} \) (to balance the electrons), we have:
\[ 5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^{-} \]
- The total electrons lost in this reaction is 5 from \( Fe^{2+} \) and 2 from \( C_2O_4^{2-} \), giving a total of 7 electrons lost.

5. **Balancing the Electrons:**
- To balance the electrons, we need to multiply the oxalate reaction by 5 (to match the 5 electrons gained by \( KMnO_4 \)):
\[ 5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^{-} \]
- Now we have:
- 5 \( KMnO_4 \) (gaining 25 electrons)
- 5 \( C_2O_4^{2-} \) (losing 10 electrons)
- 5 \( Fe^{2+} \) (losing 5 electrons)

6. **Final Reaction:**
- The balanced overall reaction is:
\[ 3MnO_4^{-} + 5Fe^{2+} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 8H_2O \]

7. **Calculate Moles of \( KMnO_4 \):**
- From the balanced equation, we see that 3 moles of \( KMnO_4 \) react with 5 moles of ferrous oxalate. Therefore, for 1 mole of ferrous oxalate:
\[ \text{Moles of } KMnO_4 = \frac{3}{5} \text{ moles} \]

### Conclusion:
The number of moles of \( KMnO_4 \) needed to react completely with one mole of ferrous oxalate in acidic solution is \( \frac{3}{5} \) moles.