The colour of `KMnO_(4)` solution is decolourised by `Fe^(+2)` solution, one mole of `Fe^(2+)` reacts with x moles of `KMnO_(4)` Find x.

Explanation

To solve the problem, we need to determine how many moles of potassium permanganate (KMnO₄) react with one mole of ferrous ion (Fe²⁺) in an acidic medium.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between KMnO₄ and Fe²⁺ in an acidic medium can be represented as:
\[ \text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} \]

2. **Determine Oxidation States**:
- In KMnO₄, manganese (Mn) is in the +7 oxidation state.
- In the product, Mn²⁺, manganese is in the +2 oxidation state.
- For Fe²⁺, it is oxidized to Fe³⁺, changing from +2 to +3.

3. **Calculate Change in Oxidation States**:
- For Mn: The change in oxidation state is \( +7 \) to \( +2 \), which is a change of \( 5 \) (7 - 2 = 5).
- For Fe: The change in oxidation state is \( +2 \) to \( +3 \), which is a change of \( 1 \) (3 - 2 = 1).

4. **Determine the n-factor**:
- The n-factor for Fe²⁺ is \( 1 \) (since it changes by 1).
- The n-factor for MnO₄⁻ is \( 5 \) (since it changes by 5).

5. **Set Up the Equivalent Equation**:
According to the concept of equivalents:
\[ \text{Equivalents of Fe}^{2+} = \text{Equivalents of KMnO}_4 \]
This can be expressed as:
\[ 1 \text{ mole of Fe}^{2+} \times 1 \text{ (n-factor of Fe)} = x \text{ moles of KMnO}_4 \times 5 \text{ (n-factor of KMnO}_4) \]

6. **Solve for x**:
\[ 1 = 5x \]
\[ x = \frac{1}{5} = 0.2 \]

### Final Answer:
Thus, one mole of Fe²⁺ reacts with **0.2 moles of KMnO₄**.