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
![**Predicting the Theoretical Yield of a Solid Product**
**Problem Statement:**
When 15.11 mL of 0.102 M K₂SO₄ solution reacts with 35.00 mL of 0.114 M Pb(C₂H₃O₂)₂, predict the theoretical yield of the solid product.
**Detailed Solution:**
To predict the theoretical yield, we first need to identify the balanced chemical equation for the reaction, then calculate the moles of each reactant, determine the limiting reagent, and finally calculate the theoretical yield of the solid product.
1. **Balanced Chemical Equation:**
\[
K₂SO₄ (aq) + Pb(C₂H₃O₂)₂ (aq) \rightarrow PbSO₄ (s) + 2 K(C₂H₃O₂) (aq)
\]
2. **Moles of Reactants:**
- For K₂SO₄:
\[
\text{Moles of K₂SO₄} = \text{Volume (L)} \times \text{Molarity} = 0.01511 \, \text{L} \times 0.102 \, \text{M} = 0.00154 \, \text{mol}
\]
- For Pb(C₂H₃O₂)₂:
\[
\text{Moles of Pb(C₂H₃O₂)₂} = \text{Volume (L)} \times \text{Molarity} = 0.03500 \, \text{L} \times 0.114 \, \text{M} = 0.00399 \, \text{mol}
\]
3. **Identify the Limiting Reagent:**
Using the stoichiometry from the balanced equation, 1 mole of K₂SO₄ reacts with 1 mole of Pb(C₂H₃O₂)₂ to produce 1 mole of PbSO₄.
- From the moles calculated:
\[
0.00154 \, \text{mol K₂SO₄}
\]
\[
0.00399 \, \text{mol Pb(C₂H₃O₂)₂}
\]
Since 0.00154 mol of K₂](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b0aed3-745e-4879-8577-ea35399669d0%2F6563e63c-6706-4154-973f-4de105fcb69f%2F58dgd4p_processed.png&w=3840&q=75)
Transcribed Image Text:**Predicting the Theoretical Yield of a Solid Product**
**Problem Statement:**
When 15.11 mL of 0.102 M K₂SO₄ solution reacts with 35.00 mL of 0.114 M Pb(C₂H₃O₂)₂, predict the theoretical yield of the solid product.
**Detailed Solution:**
To predict the theoretical yield, we first need to identify the balanced chemical equation for the reaction, then calculate the moles of each reactant, determine the limiting reagent, and finally calculate the theoretical yield of the solid product.
1. **Balanced Chemical Equation:**
\[
K₂SO₄ (aq) + Pb(C₂H₃O₂)₂ (aq) \rightarrow PbSO₄ (s) + 2 K(C₂H₃O₂) (aq)
\]
2. **Moles of Reactants:**
- For K₂SO₄:
\[
\text{Moles of K₂SO₄} = \text{Volume (L)} \times \text{Molarity} = 0.01511 \, \text{L} \times 0.102 \, \text{M} = 0.00154 \, \text{mol}
\]
- For Pb(C₂H₃O₂)₂:
\[
\text{Moles of Pb(C₂H₃O₂)₂} = \text{Volume (L)} \times \text{Molarity} = 0.03500 \, \text{L} \times 0.114 \, \text{M} = 0.00399 \, \text{mol}
\]
3. **Identify the Limiting Reagent:**
Using the stoichiometry from the balanced equation, 1 mole of K₂SO₄ reacts with 1 mole of Pb(C₂H₃O₂)₂ to produce 1 mole of PbSO₄.
- From the moles calculated:
\[
0.00154 \, \text{mol K₂SO₄}
\]
\[
0.00399 \, \text{mol Pb(C₂H₃O₂)₂}
\]
Since 0.00154 mol of K₂
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