Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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]
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]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.