The problem requires us to find the initial quantity of $MnCl_2$ in a solution. The method involves a two-step process: first, all the $MnCl_2$ is converted to $KMnO_4$, and second, the resulting $KMnO_4$ is titrated with a known amount of oxalic acid. We can solve this problem by working backwards from the titration step.

Step 1: Write the balanced chemical equation for the titration reaction.

The titration involves the reaction of potassium permanganate ($KMnO_4$) with oxalic acid ($H_2C_2O_4$) in an acidic medium (provided by $HCl$). This is a redox reaction.

• Oxidation half-reaction: Oxalic acid is oxidized to carbon dioxide. $H_2C_2O_4 \to 2CO_2 + 2H^+ + 2e^-$ The oxidation state of Carbon changes from +3 in $H_2C_2O_4$ to +4 in $CO_2$. The n-factor for oxalic acid is 2.

• Reduction half-reaction: Permanganate ion ($MnO_4^−$) is reduced to manganese(II) ion ($Mn^{2+}$) in an acidic medium. $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ The oxidation state of Manganese changes from +7 in $MnO_4^−$ to +2 in $Mn^{2+}$. The n-factor for $KMnO_4$ is 5.

To get the overall balanced equation, we multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the electrons transferred:

$5H_2C_2O_4 \to 10CO_2 + 10H^+ + 10e^-$ $2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$

Adding these two half-reactions gives the net ionic equation: $2MnO_4^- + 5H_2C_2O_4 + 6H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$

From this balanced equation, we see that 2 moles of $KMnO_4$ react with 5 moles of $H_2C_2O_4$.

Step 2: Calculate the moles of oxalic acid used.

First, we need the molar mass of oxalic acid ($H_2C_2O_4$).

• Molar Mass of $H_2C_2O_4 = 2 \times (1.0) + 2 \times (12.0) + 4 \times (16.0) = 2 + 24 + 64 = 90 \, g/mol$.

The mass of oxalic acid used is given as $225 \, mg$.

• Mass in grams = $225 \, mg = 0.225 \, g$.

Now, we can calculate the moles of oxalic acid:

• Moles of $H_2C_2O_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.225 \, g}{90 \, g/mol} = 0.0025 \, mol$.

Step 3: Calculate the moles of $KMnO_4$ that reacted.

Using the stoichiometry from the balanced equation ($2MnO_4^- : 5H_2C_2O_4$):

• Moles of $KMnO_4 = \text{Moles of } H_2C_2O_4 \times \frac{2}{5}$
• Moles of $KMnO_4 = 0.0025 \, mol \times \frac{2}{5} = 0.001 \, mol$.

Step 4: Relate the moles of $KMnO_4$ to the moles of $MnCl_2$.

The first reaction states that $MnCl_2$ was completely converted to $KMnO_4$. This means that every mole of $Mn$ in the initial $MnCl_2$ is converted into one mole of $Mn$ in $KMnO_4$. By the principle of conservation of atoms (specifically, Manganese atoms), the number of moles of $MnCl_2$ must be equal to the number of moles of $KMnO_4$ produced.

• Moles of $MnCl_2$ = Moles of $KMnO_4 = 0.001 \, mol$.

Step 5: Calculate the mass of $MnCl_2$ in the initial solution.

First, we calculate the molar mass of $MnCl_2$.

• Atomic weight of $Mn = 55 \, g/mol$.
• Atomic weight of $Cl = 35.5 \, g/mol$.
• Molar Mass of $MnCl_2 = 55 + 2 \times (35.5) = 55 + 71 = 126 \, g/mol$.

Now, we can find the mass of $MnCl_2$:

• Mass of $MnCl_2 = \text{Moles} \times \text{Molar Mass}$
• Mass of $MnCl_2 = 0.001 \, mol \times 126 \, g/mol = 0.126 \, g$.

The question asks for the quantity in milligrams (mg).

• Mass of $MnCl_2 = 0.126 \, g \times 1000 \, mg/g = 126 \, mg$.

Thus, the quantity of $MnCl_2$ present in the initial solution is 126 mg.