Answered

In the reaction [tex]$MgCl_2 + 2 KOH \rightarrow Mg(OH)_2 + 2 KCl[tex]$[/tex], if 3 moles of [tex]$[/tex]MgCl_2[tex]$[/tex] are added to 4 moles of [tex]$[/tex]KOH[tex]$[/tex], what determines how much [tex]$[/tex]Mg(OH)_2$[/tex] is made?

A. The amount of [tex]$MgCl_2$[/tex]
B. The amount of [tex]$KCl$[/tex]
C. The amount of [tex]$KOH$[/tex]
D. The amount of [tex]$MgCl_2 + KOH$[/tex]



Answer :

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]