How many mole of `FeSO_(4),H_(2)C_(2)O_(4)` and `FeC_(2)O_(4)` are oxidised separately by one mole of `KMnO_(4)` in acid medium?

Options

A. `5,5//2,5//3`
B. `2//5,2,5`
C. `5//3,3//5,2//5`
D. `3//5,2,5`

Answer

To solve the question of how many moles of `FeSO4`, `H2C2O4`, and `FeC2O4` are oxidized separately by one mole of `KMnO4` in acidic medium, we will analyze each compound's oxidation reaction with `KMnO4` and determine the stoichiometry involved.

### Step 1: Oxidation of `FeSO4`
1. **Identify the oxidation states**: In `FeSO4`, iron is in the +2 oxidation state (Fe²⁺) and will be oxidized to +3 (Fe³⁺).
2. **Write the half-reaction**:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
3. **Balance the electrons**: To balance the reaction with `KMnO4`, we need to multiply the half-reaction by 5 (since 5 moles of Fe²⁺ will be oxidized).
\[ 5 \text{Fe}^{2+} \rightarrow 5 \text{Fe}^{3+} + 5e^- \]
4. **Combine with the reduction of `KMnO4`**:
\[ \text{Mn}^{7+} + 5 \text{Fe}^{2+} \rightarrow 5 \text{Fe}^{3+} + \text{Mn}^{2+} \]
5. **Conclusion**: 1 mole of `KMnO4` oxidizes 5 moles of `FeSO4`.

### Step 2: Oxidation of `H2C2O4` (Oxalic Acid)
1. **Identify the oxidation states**: In `H2C2O4`, the carbon is in the +3 oxidation state (C²⁻) and will be oxidized to +4 (C⁴⁺).
2. **Write the half-reaction**:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2e^- \]
3. **Balance the electrons**: To balance the reaction with `KMnO4`, we need to multiply the half-reaction by 5.
\[ 5 \text{C}_2\text{O}_4^{2-} \rightarrow 10 \text{CO}_2 + 10e^- \]
4. **Combine with the reduction of `KMnO4`**:
\[ 2 \text{Mn}^{7+} + 5 \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 \]
5. **Conclusion**: 1 mole of `KMnO4` oxidizes \( \frac{5}{2} \) moles of `H2C2O4`.

### Step 3: Oxidation of `FeC2O4` (Ferrous Oxalate)
1. **Identify the oxidation states**: In `FeC2O4`, Fe is in the +2 oxidation state (Fe²⁺) and C is in the +3 oxidation state (C²⁻).
2. **Write the half-reaction**:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2e^- \]
3. **Balance the electrons**:
- For Fe: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \) (1 electron)
- For C: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2e^- \) (2 electrons)
4. **Combine the reactions**: To balance the total electrons, multiply the Fe reaction by 2:
\[ 2 \text{Fe}^{2+} + 5 \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{Fe}^{3+} + 10 \text{CO}_2 + 3 \text{Mn}^{2+} \]
5. **Conclusion**: 1 mole of `KMnO4` oxidizes \( \frac{5}{3} \) moles of `FeC2O4`.

### Final Summary
- **For `FeSO4`**: 1 mole of `KMnO4` oxidizes **5 moles** of `FeSO4`.
- **For `H2C2O4`**: 1 mole of `KMnO4` oxidizes **\( \frac{5}{2} \) moles** of `H2C2O4`.
- **For `FeC2O4`**: 1 mole of `KMnO4` oxidizes **\( \frac{5}{3} \) moles** of `FeC2O4`.