One mole of `FeC_(2)O_(4)` is oxidised by KMnO_(4) in acidic medium. Number of moles of KMnO_(4) used are

Options

A. 0.6mol
B. 1.2mol
C. 0.4mol
D. 1mol

Explanation

To solve the problem of how many moles of KMnO₄ are required to oxidize 1 mole of FeC₂O₄ in acidic medium, we can follow these steps:

### Step 1: Write the oxidation half-reaction for FeC₂O₄
The oxidation of ferrous ion (Fe²⁺) and oxalate ion (C₂O₄²⁻) can be represented as:
- Fe²⁺ is oxidized to Fe³⁺.
- C₂O₄²⁻ is oxidized to CO₂.

### Step 2: Write the reduction half-reaction for KMnO₄
In acidic medium, KMnO₄ is reduced as follows:
- MnO₄⁻ is reduced to Mn²⁺.

### Step 3: Balance the half-reactions
The balanced half-reactions are:
1. **Oxidation of Fe²⁺ and C₂O₄²⁻:**
- 5 Fe²⁺ + 5 C₂O₄²⁻ → 5 Fe³⁺ + 10 CO₂
2. **Reduction of MnO₄⁻:**
- 2 MnO₄⁻ + 10 H⁺ + 5 e⁻ → 2 Mn²⁺ + 5 H₂O

### Step 4: Combine the half-reactions
To combine the half-reactions, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. From the balanced equations, we see:
- 5 moles of Fe²⁺ and 5 moles of C₂O₄²⁻ react with 2 moles of MnO₄⁻.

### Step 5: Determine the mole ratio
From the balanced equation, we find that:
- 5 moles of FeC₂O₄ react with 2 moles of KMnO₄.
- Therefore, for 1 mole of FeC₂O₄, the moles of KMnO₄ required can be calculated as follows:
\[ \text{Moles of KMnO₄} = \frac{2 \text{ moles of KMnO₄}}{5 \text{ moles of FeC₂O₄}} \times 1 \text{ mole of FeC₂O₄} = \frac{2}{5} = 0.4 \text{ moles of KMnO₄} \]

### Step 6: Conclusion
Thus, the number of moles of KMnO₄ required to oxidize 1 mole of FeC₂O₄ in acidic medium is **0.4 moles**.