Determine Weight Percentage of Iron in Sample

To find the weight percentage of iron ($y$) in the sample, we use redox titration equivalence between $\text{Fe}^{2+}$ and $\text{KMnO}_4$.

The reaction between $\text{Fe}^{2+}$ and $\text{MnO}_4^-$ in an acidic medium follows the stoichiometric mole ratio: $n_{\text{Fe}^{2+}} = 5 \times n_{\text{MnO}_4^-}$

Calculating the moles of $\text{KMnO}_4$ consumed in the $25.0\text{ mL}$ aliquot: $n_{\text{MnO}_4^-} = 0.03\text{ M} \times 12.5 \times 10^{-3}\text{ L} = 3.75 \times 10^{-4}\text{ mol}$

Thus, the total moles of $\text{Fe}^{2+}$ present in the full $250\text{ mL}$ solution are: $n_{\text{Fe}^{2+}, \text{total}} = 10 \times (5 \times 3.75 \times 10^{-4}\text{ mol}) = 1.875 \times 10^{-2}\text{ mol}$

The mass of iron present in the sample is calculated using its molar mass ($56\text{ g mol}^{-1}$): $\text{Mass of Fe} = 1.875 \times 10^{-2}\text{ mol} \times 56\text{ g mol}^{-1} = 1.05\text{ g}$

Finally, the weight percentage of iron ($y$) in the $5.6\text{ g}$ sample is: $y = \left( \frac{1.05\text{ g}}{5.6\text{ g}} \right) \times 100 = 18.75$