To determine how much \( \text{Mg(OH)}_2 \) is produced when 3 moles of \( \text{MgCl}_2 \) are added to 4 moles of \( \text{KOH} \), we need to consider the stoichiometry of the reaction and identify the limiting reagent.
The balanced chemical equation for the reaction is:
[tex]\[ \text{MgCl}_2 + 2 \text{KOH} \rightarrow \text{Mg(OH)}_2 + 2 \text{KCl} \][/tex]
### Step-by-Step Solution:
1. Identify the moles of reactants:
- Moles of \( \text{MgCl}_2 \): 3
- Moles of \( \text{KOH} \): 4
2. Determine the stoichiometric ratio required for the reaction:
- According to the balanced equation, 1 mole of \( \text{MgCl}_2 \) reacts with 2 moles of \( \text{KOH} \).
- Therefore, to react completely with 3 moles of \( \text{MgCl}_2 \), the required moles of \( \text{KOH} \) would be:
[tex]\[
\text{Required moles of } \text{KOH} = 3 \text{ moles of } \text{MgCl}_2 \times 2 \frac{\text{moles of KOH}}{\text{mole of MgCl}_2} = 6 \text{ moles of KOH}
\][/tex]
3. Compare the available moles of \( \text{KOH} \) with the required moles:
- Available moles of \( \text{KOH} \): 4
- Required moles of \( \text{KOH} \) for 3 moles of \( \text{MgCl}_2 \): 6
4. Identify the limiting reagent:
- Since the available moles of \( \text{KOH} \) (4 moles) are less than the required moles (6 moles), \( \text{KOH} \) is the limiting reagent.
5. Determine the amount of \( \text{Mg(OH)}_2 \) produced:
- According to the stoichiometry of the reaction, 2 moles of \( \text{KOH} \) produce 1 mole of \( \text{Mg(OH)}_2 \).
- Since \( \text{KOH} \) is the limiting reagent, the amount of \( \text{Mg(OH)}_2 \) produced is determined by the available moles of \( \text{KOH} \):
[tex]\[
\text{Moles of } \text{Mg(OH)}_2 \text{ produced} = \frac{\text{4 moles of KOH}}{2} = 2 \text{ moles of } \text{Mg(OH)}_2
\][/tex]
### Conclusion:
The amount of \( \text{Mg(OH)}_2 \) produced is determined by the amount of \( \text{KOH} \), which is the limiting reagent in this reaction. Therefore, the correct answer is:
C. The amount of [tex]\( \text{KOH} \)[/tex]
