Equivalent weights of `KMnO_(4)` in acidic medium, alkaline medium and neutral (dilute alkaline) medium respectively are `M/5, M/1 , M/3` . Reduced products can be :

Explanation

To solve the problem regarding the equivalent weights of `KMnO4` in different media and the reduced products, we can follow these steps:

### Step 1: Understand the Equivalent Weights
The equivalent weight of a substance is defined as the mass of the substance that will combine with or displace 1 mole of hydrogen or 1 mole of electrons in a redox reaction.

- In acidic medium, the equivalent weight of `KMnO4` is given as `M/5`.
- In alkaline medium, the equivalent weight is `M/1`.
- In neutral (dilute alkaline) medium, the equivalent weight is `M/3`.

### Step 2: Identify the Oxidation and Reduction Reactions
In acidic medium, `KMnO4` acts as an oxidizing agent. The manganese in `KMnO4` is reduced from +7 to +2 oxidation state.

In the case of oxalic acid (C2O4^2-), carbon is oxidized from +3 to +4 oxidation state.

### Step 3: Calculate the Valency Factors
1. **For Carbon in Oxalic Acid:**
- The oxidation state of carbon changes from +3 to +4.
- Each carbon atom loses 1 electron. Since there are 2 carbon atoms in oxalic acid, the total loss is 2 electrons per molecule of oxalic acid.
- For 1 mole of `Fe2(C2O4)3`, the total loss of electrons is 6 (since there are 3 moles of C2O4^2-).
- Therefore, the valency factor for carbon is 6.

2. **For Manganese in KMnO4:**
- The oxidation state of manganese changes from +7 to +2.
- Manganese gains 5 electrons.
- Therefore, the valency factor for manganese is 5.

### Step 4: Set Up the Equivalence Equation
In a redox reaction, the number of equivalents of the oxidized species must equal the number of equivalents of the reduced species:

\[ \text{Equivalents of oxidized species} = \text{Equivalents of reduced species} \]

For our reaction:
- Moles of `Fe2(C2O4)3` = 1
- Moles of `KMnO4` = X

Using the valency factors:
\[ 1 \times 6 = X \times 5 \]

### Step 5: Solve for X
\[ 6 = 5X \implies X = \frac{6}{5} = 1.2 \]

### Conclusion
The value of X, which represents the moles of `KMnO4` required to oxidize 1 mole of `Fe2(C2O4)3`, is 1.2.