The value of y is _________.

The titration takes place in acidic medium, where the redox reaction is

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

Thus, $$1$$ mol $$KMnO_4$$ reacts with $$5$$ mol $$Fe^{2+}$$.

Step 1 : Moles of $$KMnO_4$$ used for the 25 mL aliquot
Volume of permanganate = $$12.5\text{ mL}=12.5\times10^{-3}\text{ L}$$
Molarity = $$0.03\text{ M}$$

$$\text{Moles }KMnO_4 = 0.03 \times 12.5\times10^{-3} = 3.75\times10^{-4}\text{ mol}$$

Step 2 : Moles of $$Fe^{2+}$$ in the 25 mL aliquot

$$\text{Moles }Fe^{2+} = 5 \times 3.75\times10^{-4} = 1.875\times10^{-3}\text{ mol}$$

Step 3 : Moles of $$Fe^{2+}$$ in the entire 250 mL solution
The 25 mL aliquot is $$\dfrac{25}{250}= \dfrac{1}{10}$$ of the whole solution, so

$$\text{Total moles }Fe^{2+}=1.875\times10^{-3}\times10=1.875\times10^{-2}\text{ mol}$$

Step 4 : Mass of iron present
Molar mass of Fe = $$56\text{ g mol}^{-1}$$

$$m_{Fe}=1.875\times10^{-2}\times56 = 1.05\text{ g}$$

Step 5 : Percentage of iron in the 5.6 g sample

$$y = \dfrac{1.05}{5.6}\times100 = 18.75\%$$

Hence, the weight % of iron in the sample is within $$18.70 - 18.80$$.