If equal volumes of `0.1 M KMnO_(4)` and `0.1 M K_(2)Cr_(2)O_(7)` solutions are allowed to oxidise `Fe^(2+)` to `Fe^(3+)` in acidic medium, then `Fe^(2+)` oxidised will be:

Options

A. More by `K_(2)Cr_(2)O_(7)`
B. More by `KMnO_(4)`
C. Equal in both cases
D. The data is insufficient to predict the answer

Explanation

To solve the problem, we need to determine how much `Fe^(2+)` is oxidized when equal volumes of `0.1 M KMnO4` and `0.1 M K2Cr2O7` are used in an acidic medium. We will analyze the oxidation-reduction reactions involved and calculate the equivalents of `Fe^(2+)` oxidized by each oxidizing agent.

### Step-by-Step Solution:

1. **Identify the Reducing Agents and Their Reactions**:
- The two oxidizing agents are `KMnO4` and `K2Cr2O7`.
- In acidic medium:
- `KMnO4` reduces from `MnO4^-` (Mn in +7 oxidation state) to `Mn^(2+)` (Mn in +2 oxidation state), which involves the gain of 5 electrons.
- `K2Cr2O7` reduces from `Cr2O7^(2-)` (Cr in +6 oxidation state) to `Cr^(3+)` (Cr in +3 oxidation state), which involves the gain of 6 electrons.

2. **Calculate the n-factor for Each Oxidizing Agent**:
- For `KMnO4`, the n-factor is 5 (since it gains 5 electrons).
- For `K2Cr2O7`, the n-factor is 6 (since it gains 6 electrons).

3. **Calculate the Normality of Each Solution**:
- Normality (N) is calculated as:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
- For `KMnO4`:
\[ N_{KMnO4} = 0.1 \, M \times 5 = 0.5 \, N \]
- For `K2Cr2O7`:
\[ N_{K2Cr2O7} = 0.1 \, M \times 6 = 0.6 \, N \]

4. **Determine the Total Equivalents of Each Oxidizing Agent**:
- Since equal volumes of both solutions are used, we can compare their normalities directly.
- Let’s assume we take 1 L of each solution:
- Equivalents of `KMnO4` = Normality × Volume = \(0.5 \, N \times 1 \, L = 0.5 \, eq\)
- Equivalents of `K2Cr2O7` = Normality × Volume = \(0.6 \, N \times 1 \, L = 0.6 \, eq\)

5. **Determine the Amount of `Fe^(2+)` Oxidized**:
- Since `K2Cr2O7` has a higher normality, it will oxidize more `Fe^(2+)` than `KMnO4`.
- The total equivalents of `Fe^(2+)` oxidized will be equal to the equivalents provided by the limiting reagent, which in this case is `K2Cr2O7`.

6. **Conclusion**:
- The amount of `Fe^(2+)` oxidized will be determined by the `K2Cr2O7` solution, which is 0.6 equivalents.

### Final Answer:
The amount of `Fe^(2+)` oxidized will be more by `K2Cr2O7`.