Step-by-step Solution:

1. Write the balanced redox reaction equation.

The titration involves the reaction between potassium permanganate ($KMnO_4$) and ferrous ions ($Fe^{2+}$) in an acidic medium (provided by $HCl$). Iron in the sample dissolves in $HCl$ to form $FeCl_2$, which contains $Fe^{2+}$ ions.

The permanganate ion ($MnO_4^-$) is a strong oxidizing agent, and the ferrous ion ($Fe^{2+}$) is a reducing agent.

• Oxidation half-reaction: The ferrous ion is oxidized to the ferric ion. $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
• Reduction half-reaction: In an acidic medium, the permanganate ion is reduced to the manganous ion. $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

To obtain the overall balanced equation, we need to balance the electrons. We multiply the oxidation half-reaction by 5 and add it to the reduction half-reaction: $5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^{-}$ $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ Adding these two gives the net ionic equation: $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

From the stoichiometry, we see that 1 mole of $MnO_4^-$ reacts with 5 moles of $Fe^{2+}$.

2. Calculate the moles of $KMnO_4$ used in the titration.

The titration of 25.0 mL of the iron solution required 12.5 mL of 0.03 M $KMnO_4$ solution.

• Molarity of $KMnO_4$ solution ($M_{KMnO_4}$) = 0.03 mol/L
• Volume of $KMnO_4$ solution used ($V_{KMnO_4}$) = 12.5 mL = 0.0125 L

Moles of $KMnO_4$ used = $M_{KMnO_4} \times V_{KMnO_4}$ $\text{Moles of } KMnO_4 = 0.03 \text{ mol/L} \times 0.0125 \text{ L} = 0.000375 \text{ mol}$

3. Calculate the moles of $Fe^{2+}$ in the 25.0 mL aliquot.

Using the stoichiometric ratio from the balanced equation (1 mole $MnO_4^-$ : 5 moles $Fe^{2+}$):

Moles of $Fe^{2+}$ in 25.0 mL = $5 \times$ Moles of $KMnO_4$ $\text{Moles of } Fe^{2+} \text{ (in 25.0 mL)} = 5 \times 0.000375 \text{ mol} = 0.001875 \text{ mol}$

4. Calculate the total moles of $Fe^{2+}$ in the original 250 mL solution.

The 0.001875 moles of $Fe^{2+}$ were found in a 25.0 mL aliquot taken from a 250 mL stock solution. To find the total moles in the original solution, we scale up the amount.

Total moles of $Fe^{2+}$ in 250 mL = (Moles in 25.0 mL) $\times \frac{\text{Total Volume}}{\text{Aliquot Volume}}$ $\text{Total moles of } Fe^{2+} = 0.001875 \text{ mol} \times \frac{250 \text{ mL}}{25.0 \text{ mL}} = 0.001875 \times 10 = 0.01875 \text{ mol}$

5. Determine the value of x.

The problem states that the number of moles of $Fe^{2+}$ present in the 250 mL solution is equal to $x \times 10^{-2}$.

$x \times 10^{-2} = 0.01875$

To find x, we solve the equation: $x = \frac{0.01875}{10^{-2}} = 0.01875 \times 100$ $x = 1.875$

The value of x is 1.875.