To determine the number of moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] produced during the reaction, we can follow these steps:
1. Write the balanced chemical equation:
[tex]\[
\text{Mg(NO}_3\text{)}_2(aq) + 2 \text{KOH}(aq) \rightarrow 2 \text{KNO}_3(aq) + \text{Mg(OH)}_2(s)
\][/tex]
2. Interpret the stoichiometry of the reaction:
The balanced equation indicates that 2 moles of [tex]\( \text{KOH} \)[/tex] react with 1 mole of [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex] to produce 1 mole of [tex]\( \text{Mg(OH)}_2 \)[/tex].
3. Given information:
- Volume of [tex]\( \text{KOH} \)[/tex] solution = 250 mL = 0.250 L
- Molarity of [tex]\( \text{KOH} \)[/tex] solution = 2.0 M
- Moles of [tex]\( \text{KOH} \)[/tex] reacted = 0.50 mol
4. Determine the moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] produced:
Based on the stoichiometric relationship from the balanced chemical equation, [tex]\( 2 \)[/tex] moles of [tex]\( \text{KOH} \)[/tex] yield [tex]\( 1 \)[/tex] mole of [tex]\( \text{Mg(OH)}_2 \)[/tex]. Therefore, the number of moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] produced is half the number of moles of [tex]\( \text{KOH} \)[/tex] that reacted.
Moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] produced:
[tex]\[
\text{Moles of Mg(OH)}_2 = \frac{\text{Moles of KOH reacted}}{2} = \frac{0.50 \text{ mol}}{2} = 0.25 \text{ mol}
\][/tex]
The moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] formed during the reaction is:
[tex]\[
\boxed{0.25 \text{ mol}}
\][/tex]
