How many moles of `KMnO_(4)` are reduced by 1 mole of ferrous oxalate in acidic medium :-

Options

A. `(1)/(5)`
B. `(5)/(3)`
C. `(1)/(3)`
D. `(3)/(5)`

Explanation

To determine how many moles of \( KMnO_4 \) are reduced by 1 mole of ferrous oxalate (\( FeC_2O_4 \)) in acidic medium, we need to analyze the redox reaction taking place.

### Step-by-Step Solution:

1. **Identify the Oxidation State Changes**:
- In \( KMnO_4 \), manganese (\( Mn \)) is in the +7 oxidation state.
- In the reaction, \( MnO_4^- \) is reduced to \( Mn^{2+} \), which has a +2 oxidation state.
- The change in oxidation state for manganese is:
\[ \Delta \text{Oxidation State} = +7 \text{ to } +2 = 5 \]

2. **Determine the Oxidation of Ferrous Oxalate**:
- Ferrous oxalate (\( FeC_2O_4 \)) contains iron in the +2 oxidation state.
- During the reaction, iron is oxidized to +3 oxidation state (\( Fe^{3+} \)).
- The change in oxidation state for iron is:
\[ \Delta \text{Oxidation State} = +2 \text{ to } +3 = 1 \]

3. **Balance the Reaction**:
- The number of moles of \( KMnO_4 \) reduced must equal the number of moles of \( FeC_2O_4 \) oxidized, taking into account their respective changes in oxidation states.
- Let \( n \) be the number of moles of \( KMnO_4 \) reduced by 1 mole of \( FeC_2O_4 \).
- The relationship can be established as:
\[ n \times \Delta \text{Oxidation State of } KMnO_4 = 1 \times \Delta \text{Oxidation State of } FeC_2O_4 \]
- Substituting the values:
\[ n \times 5 = 1 \times 1 \]

4. **Solve for \( n \)**:
\[ n = \frac{1}{5} \]

5. **Conclusion**:
- Therefore, the number of moles of \( KMnO_4 \) reduced by 1 mole of ferrous oxalate in acidic medium is:
\[ n = \frac{1}{5} \text{ moles of } KMnO_4 \]

### Final Answer:
The number of moles of \( KMnO_4 \) reduced by 1 mole of ferrous oxalate is \( \frac{1}{5} \) moles.