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In the reaction [tex]MgCl_2 + 2KOH \rightarrow Mg(OH)_2 + 2KCl[/tex], if 1 mole of [tex]MgCl_2[/tex] is added to 3 moles of [tex]KOH[/tex], what is the limiting reagent?

A. [tex]KCl[/tex]
B. [tex]Mg(OH)_2[/tex]
C. [tex]KOH[/tex]
D. [tex]MgCl_2[/tex]

Sagot :

GinnyAnsweridn
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]
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