In the given reaction, `FeC_(2) O_(4) + KMnO_(4) overset( H^(+))(rarr) Fe^(3+) + CO_(2) + Mn^(2+)`, if the equivalents of `Fe^(3+)` are 'a', then equivalent of `FeC_(2) O_(4)` will be

Options

A. a
B. 2a
C. 3a
D. none of these

Answer

To solve the problem, we need to determine the equivalents of `FeC2O4` in the given reaction. The reaction is:

\[ \text{FeC}_2\text{O}_4 + \text{KMnO}_4 \overset{H^+}{\rightarrow} \text{Fe}^{3+} + \text{CO}_2 + \text{Mn}^{2+} \]

### Step-by-Step Solution:

1. **Identify the Change in Oxidation States**:
- In the reaction, `FeC2O4` is oxidized to `Fe^{3+}`. The oxidation state of `Fe` changes from `+2` to `+3`.
- The carbon in `FeC2O4` is in the `+3` oxidation state and is converted to `CO2`, where carbon is in the `+4` oxidation state.

2. **Calculate the n-factor for `FeC2O4`**:
- The n-factor is defined as the total number of moles of electrons lost or gained per mole of the substance.
- For `FeC2O4`, we have:
- `Fe` changes from `+2` to `+3`: This is a change of `1` for `1` atom of `Fe`.
- Each `C` in `FeC2O4` is `+3` and changes to `+4` in `CO2`. There are `2` carbon atoms, so the total change for carbon is `2 * 1 = 2`.
- Thus, the total change in oxidation states (n-factor) for `FeC2O4` is:
\[ n_{\text{FeC}_2\text{O}_4} = 1 + 2 = 3 \]

3. **Calculate the n-factor for `KMnO4`**:
- `Mn` changes from `+7` to `+2`, which is a change of `5` for `1` atom of `Mn`.
- Thus, the n-factor for `KMnO4` is `5`.

4. **Calculate the n-factor for `Fe^{3+}`**:
- `Fe^{3+}` does not change in this reaction, so its n-factor is `1` (as it does not gain or lose electrons).

5. **Using the Law of Equivalents**:
- According to the law of equivalents:
\[ \text{Equivalents of } \text{FeC}_2\text{O}_4 = \text{Equivalents of } \text{Fe}^{3+} \]
- Since the equivalents of `Fe^{3+}` are given as `A`, we can express the equivalents of `FeC2O4` as:
\[ \text{Equivalents of } \text{FeC}_2\text{O}_4 = \frac{\text{Moles of } \text{FeC}_2\text{O}_4 \times n_{\text{FeC}_2\text{O}_4}}{1} \]
- Since we found that the n-factor for `FeC2O4` is `3`, we can set up the equation:
\[ \text{Equivalents of } \text{FeC}_2\text{O}_4 = \text{Moles of } \text{FeC}_2\text{O}_4 \times 3 \]

6. **Conclusion**:
- Since the equivalents of `Fe^{3+}` are `A`, and the equivalents of `FeC2O4` must equal `A` as well, we conclude that:
\[ \text{Equivalents of } \text{FeC}_2\text{O}_4 = A \]

### Final Answer:
The equivalents of `FeC2O4` will be **A**.