To solve this problem, let's go through it step-by-step:
1. Conversion of Volume from mL to Liters:
- Given that we have 250 mL of KOH solution, we need to convert this volume to liters.
- Since 1 liter (L) is equal to 1000 milliliters (mL), we have:
[tex]\[
\text{Volume of KOH solution} = \frac{250 \text{ mL}}{1000} = 0.25 \text{ L}
\][/tex]
2. Calculation of Moles of KOH:
- The molarity (concentration) of the KOH solution is given as 2.0 M, which means there are 2.0 moles of KOH in 1 liter of solution.
- To find the moles of KOH in 0.25 L of solution, we use the molarity equation:
[tex]\[
\text{Moles of KOH} = \text{Molarity} \times \text{Volume (in liters)} = 2.0 \text{ M} \times 0.25 \text{ L} = 0.50 \text{ mol}
\][/tex]
3. Stoichiometry of the Reaction:
- The balanced chemical equation for the reaction is:
[tex]\[
Mg(NO_3)_2 (aq) + 2 KOH (aq) \rightarrow 2 KNO_3 (aq) + Mg(OH)_2 (s)
\][/tex]
- According to the stoichiometry of the balanced equation, 2 moles of KOH react to produce 1 mole of [tex]\(Mg(OH)_2\)[/tex].
4. Calculation of Moles of [tex]\(Mg(OH)_2\)[/tex]:
- Since we have determined that 0.50 moles of KOH are reacting, we can now determine the moles of [tex]\(Mg(OH)_2\)[/tex] produced based on the stoichiometric ratio:
[tex]\[
\text{Moles of \(Mg(OH)_2\)} = \frac{\text{Moles of KOH}}{2} = \frac{0.50 \text{ mol}}{2} = 0.25 \text{ mol}
\][/tex]
Therefore, the moles of [tex]\(Mg(OH)_2\)[/tex] formed during the reaction is 0.25 mol.
