How many grams of precipitate forms when 250 mL of a 2.5 M solution of potassium carbonate reacts with 500 mL of a 1.7 M solution of barium chloride? Assume 100% yield, round your answer to one decimal place, and make sure to include the name of the solid compound.
How many grams of precipitate forms when 250 mL of a 2.5 M solution of potassium carbonate reacts with 500 mL of a 1.7 M solution of barium chloride? Assume 100% yield, round your answer to one decimal place, and make sure to include the name of the solid compound.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question:**
How many grams of precipitate forms when 250 mL of a 2.5 M solution of potassium carbonate reacts with 500 mL of a 1.7 M solution of barium chloride? Assume 100% yield, round your answer to one decimal place, and make sure to include the name of the solid compound.
**Answer:**
To solve this problem, we first write the balanced chemical equation for the reaction:
\[ K_2CO_3 (aq) + BaCl_2 (aq) \rightarrow 2KCl (aq) + BaCO_3 (s) \]
1. **Determine the moles of each reactant:**
- Potassium carbonate (\( K_2CO_3 \)):
\[
\text{Moles of } K_2CO_3 = M \times L = 2.5 \, \text{mol/L} \times 0.250 \, \text{L} = 0.625 \, \text{mol}
\]
- Barium chloride (\( BaCl_2 \)):
\[
\text{Moles of } BaCl_2 = M \times L = 1.7 \, \text{mol/L} \times 0.500 \, \text{L} = 0.850 \, \text{mol}
\]
2. **Identify the limiting reactant:**
The balanced equation indicates a 1:1 mole ratio between \( K_2CO_3 \) and \( BaCl_2 \). Therefore, \( K_2CO_3 \) is the limiting reactant.
3. **Calculate the moles of precipitate formed (barium carbonate, \( BaCO_3 \)):**
Since \( K_2CO_3 \) is the limiting reagent, it will determine the amount of \( BaCO_3 \).
- Moles of \( BaCO_3 \) formed = 0.625 mol
4. **Calculate the mass of \( BaCO_3 \):**
- Molar mass of \( BaCO_3 \) = \( 137 (Ba) + 12 (C) + 3 \times 16 (O) = 197 \, \text{g/mol} \)
- Mass = moles \(\times\) mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb077f2ab-172b-438d-8d7b-87f013063300%2F15d6bcbd-eb0c-4d3f-8ea7-ce5c0f6b3d94%2Fy209tqb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
How many grams of precipitate forms when 250 mL of a 2.5 M solution of potassium carbonate reacts with 500 mL of a 1.7 M solution of barium chloride? Assume 100% yield, round your answer to one decimal place, and make sure to include the name of the solid compound.
**Answer:**
To solve this problem, we first write the balanced chemical equation for the reaction:
\[ K_2CO_3 (aq) + BaCl_2 (aq) \rightarrow 2KCl (aq) + BaCO_3 (s) \]
1. **Determine the moles of each reactant:**
- Potassium carbonate (\( K_2CO_3 \)):
\[
\text{Moles of } K_2CO_3 = M \times L = 2.5 \, \text{mol/L} \times 0.250 \, \text{L} = 0.625 \, \text{mol}
\]
- Barium chloride (\( BaCl_2 \)):
\[
\text{Moles of } BaCl_2 = M \times L = 1.7 \, \text{mol/L} \times 0.500 \, \text{L} = 0.850 \, \text{mol}
\]
2. **Identify the limiting reactant:**
The balanced equation indicates a 1:1 mole ratio between \( K_2CO_3 \) and \( BaCl_2 \). Therefore, \( K_2CO_3 \) is the limiting reactant.
3. **Calculate the moles of precipitate formed (barium carbonate, \( BaCO_3 \)):**
Since \( K_2CO_3 \) is the limiting reagent, it will determine the amount of \( BaCO_3 \).
- Moles of \( BaCO_3 \) formed = 0.625 mol
4. **Calculate the mass of \( BaCO_3 \):**
- Molar mass of \( BaCO_3 \) = \( 137 (Ba) + 12 (C) + 3 \times 16 (O) = 197 \, \text{g/mol} \)
- Mass = moles \(\times\) mol
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