Both `Cr_(2)O_(7)^(2-)` (aq) and `MnO_(4)^(-)` (aq) can be used ot titrate `Fe^(2+)`(aq) if in a given titration 24-50 `cm^(3)` 0.1 M `Cr_(2)O_(7)^(2-)` were used then what volume of 0.1 M `MnO_(4)^(-)` solution would have been use for the same titration ?

Explanation

To solve the problem, we need to determine the volume of 0.1 M `MnO4^(-)` solution that would be used to titrate `Fe^(2+)` given that 24.5 cm³ of 0.1 M `Cr2O7^(2-)` was used. We will use the stoichiometry of the reactions involved in the titration.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- The first reaction involves `Cr2O7^(2-)` reacting with `Fe^(2+)` in an acidic medium:
\[ Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O \]
Here, 1 mole of `Cr2O7^(2-)` reacts with 6 moles of `Fe^(2+)`.

- The second reaction involves `MnO4^(-)` reacting with `Fe^(2+)` in an acidic medium:
\[ MnO_4^{-} + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \]
Here, 1 mole of `MnO4^(-)` reacts with 5 moles of `Fe^(2+)`.

2. **Calculate the Equivalent Moles of `Fe^(2+)`:**
- For `Cr2O7^(2-)`, the number of moles of `Cr2O7^(2-)` used can be calculated using:
\[ \text{Moles of } Cr_2O_7^{2-} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.0245 \, \text{L} = 0.00245 \, \text{mol} \]
- Since 1 mole of `Cr2O7^(2-)` reacts with 6 moles of `Fe^(2+)`, the moles of `Fe^(2+)` that reacted are:
\[ \text{Moles of } Fe^{2+} = 6 \times 0.00245 = 0.0147 \, \text{mol} \]

3. **Relate the Moles of `Fe^(2+)` to `MnO4^(-)`:**
- From the second reaction, we know that 1 mole of `MnO4^(-)` reacts with 5 moles of `Fe^(2+)`. Therefore, the moles of `MnO4^(-)` needed can be calculated as:
\[ \text{Moles of } MnO_4^{-} = \frac{0.0147 \, \text{mol}}{5} = 0.00294 \, \text{mol} \]

4. **Calculate the Volume of `MnO4^(-)` Solution:**
- Using the molarity of `MnO4^(-)` (0.1 M), we can find the volume required:
\[ \text{Volume of } MnO_4^{-} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.00294 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.0294 \, \text{L} = 29.4 \, \text{cm}^3 \]

### Final Answer:
The volume of 0.1 M `MnO4^(-)` solution required for the titration is **29.4 cm³**.