1.12 g of solid potassium hydroxide (KOH) is added to 50 mL of a...

Molar mass of $\text{KOH} = 39 + 16 + 1 = 56 \text{ g mol}^{-1}$

Moles of $\text{KOH} = \dfrac{1.12}{56} = 0.02 \text{ mol} = 20 \text{ mmol}$

Moles of $\text{H}_2\text{SO}_4 = 50 \times 0.25 = 12.5 \text{ mmol}$

The balanced chemical equation is:

$2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$

From the stoichiometry, $2 \text{ moles}$ of $\text{KOH}$ react with $1 \text{ mole}$ of $\text{H}_2\text{SO}_4$.

Moles of $\text{H}_2\text{SO}_4$ required to react with $20 \text{ mmol}$ of $\text{KOH} = \dfrac{20}{2} = 10 \text{ mmol}$

Since $12.5 \text{ mmol}$ of $\text{H}_2\text{SO}_4$ is present, $\text{KOH}$ is the limiting reagent and $\text{H}_2\text{SO}_4$ is in excess.

Unreacted $\text{H}_2\text{SO}_4 = 12.5 - 10 = 2.5 \text{ mmol}$

Molar mass of $\text{H}_2\text{SO}_4 = 2(1) + 32 + 4(16) = 98 \text{ g mol}^{-1}$

Mass of unreacted $\text{H}_2\text{SO}_4 = 2.5 \times 10^{-3} \times 98 = 0.245 \text{ g} = 245 \text{ mg}$

Answer: 245 mg