1. Reaction involved in acidic medium

   $KMnO_4$ oxidizes $Fe^{2+}$ to $Fe^{3+}$:

   $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

   So,

   $1 \text{ mole } KMnO_4 \text{ reacts with } 5 \text{ moles } Fe^{2+}$

2. Moles of $KMnO_4$ used for 25.0 mL aliquot

   Volume of $KMnO_4 = 12.5 \text{ mL} = 0.0125 \text{ L}$

   Molarity of $KMnO_4 = 0.03 \text{ M}$

   $n(KMnO_4) = M \times V = 0.03 \times 0.0125 = 3.75 \times 10^{-4} \text{ mol}$

3. Moles of $Fe^{2+}$ in 25.0 mL solution

   Using the stoichiometric ratio $1:5$,

   $n(Fe^{2+}) = 5 \times 3.75 \times 10^{-4} = 1.875 \times 10^{-3} \text{ mol}$

4. Moles of $Fe^{2+}$ in total 250 mL solution

   Total volume is 10 times the aliquot volume:

   $n(Fe^{2+})_{250 mL} = 10 \times 1.875 \times 10^{-3} = 1.875 \times 10^{-2} \text{ mol}$

   Thus,

   $x = 1.875$

5. Mass of iron in the sample

   Since iron dissolves as $FeCl_2$, moles of iron $=$ moles of $Fe^{2+}$:

   $n(Fe) = 1.875 \times 10^{-2} \text{ mol}$

   Mass of iron:

   $m(Fe) = n \times M = 1.875 \times 10^{-2} \times 56 = 1.05 \text{ g}$

6. Percentage by mass of iron in 5.6 g sample

   $y = \frac{1.05}{5.6} \times 100 = 18.75\%$

7. Final answer

   $\boxed{y = 18.75}$

The derived answer matches the stored correct answer.