Let's solve the problem step-by-step:
1. Write the balanced chemical equation:
[tex]\[ \text{MgCl}_2 + 2 \text{KOH} \rightarrow \text{Mg(OH)}_2 + 2 \text{KCl} \][/tex]
2. Identify the given quantities:
- Moles of [tex]\(\text{MgCl}_2\)[/tex]: 1 mole
- Moles of KOH: 3 moles
3. Determine the mole ratio from the balanced equation:
- For every 1 mole of [tex]\(\text{MgCl}_2\)[/tex], 2 moles of KOH are required.
4. Calculate the moles of KOH required to completely react with 1 mole of [tex]\(\text{MgCl}_2\)[/tex]:
[tex]\[ 1 \text{ mole of } \text{MgCl}_2 \times 2 \text{ moles of KOH per 1 mole of } \text{MgCl}_2 = 2 \text{ moles of KOH} \][/tex]
So, to react with 1 mole of [tex]\(\text{MgCl}_2\)[/tex], we need 2 moles of KOH.
5. Compare the required moles of KOH with the available moles:
- Required moles of KOH: 2 moles
- Available moles of KOH: 3 moles
6. Since we have more KOH (3 moles) than needed (2 moles), we have excess KOH, and the limiting reagent will not be KOH.
7. Since [tex]\(\text{MgCl}_2\)[/tex] is the reactant that will be consumed completely first, it is the limiting reagent.
Therefore, the answer is:
[tex]\[ \text{D. } \text{MgCl}_2 \][/tex]
