In order to oxidise a mixture of one mole of each of `FeC_(2)O_(4).Fe_(2)(C_(2)O_(4))_(3),FeSO_(4)` and `Fe_(2)(SO_(4))_(3)` in acidic medium, the number of moles of `KMnO_(4)` required is:

Explanation

To determine the number of moles of KMnO4 required to oxidize a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4, and Fe2(SO4)3 in acidic medium, we will follow these steps:

### Step 1: Identify the Oxidation States and Changes
1. **FeC2O4 (Iron(II) oxalate)**:
- Iron changes from +2 to +3 (1 electron).
- Oxalate (C2O4^2-) is oxidized to CO2 (2 electrons).
- Total change = 1 (from Fe) + 2 (from oxalate) = 3 electrons.

2. **Fe2(C2O4)3 (Iron(III) oxalate)**:
- Iron is already in +3 state, so it does not change.
- Oxalate (C2O4^2-) is oxidized to CO2 (2 electrons).
- Total change = 2 (from 3 moles of oxalate) = 6 electrons.

3. **FeSO4 (Iron(II) sulfate)**:
- Iron changes from +2 to +3 (1 electron).
- Total change = 1 electron.

4. **Fe2(SO4)3 (Iron(III) sulfate)**:
- Iron is already in +3 state, so it does not change.
- Total change = 0 electrons.

### Step 2: Calculate the Total Number of Electrons Transferred
- For FeC2O4: 3 electrons
- For Fe2(C2O4)3: 6 electrons
- For FeSO4: 1 electron
- For Fe2(SO4)3: 0 electrons

Total electrons = 3 + 6 + 1 + 0 = 10 electrons.

### Step 3: Determine the n-factor of KMnO4
In acidic medium, the reaction of KMnO4 can be represented as:
\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \]
Thus, the n-factor of KMnO4 is 5 (it accepts 5 electrons).

### Step 4: Calculate the Number of Moles of KMnO4 Required
Using the relationship:
\[ \text{Moles of KMnO}_4 \times \text{n-factor of KMnO}_4 = \text{Total electrons transferred} \]
Let \( n \) be the number of moles of KMnO4 required:
\[ n \times 5 = 10 \]
\[ n = \frac{10}{5} = 2 \]

### Conclusion
The number of moles of KMnO4 required to oxidize the mixture is **2 moles**.

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