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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### Chemical Reaction Problem
**Problem:** What mass of MnO₂ will completely react with 56.19 g of Al according to the reaction below?
\[ \text{3 MnO}_2 + \text{4 Al} \rightarrow \text{3 Mn} + \text{2 Al}_2\text{O}_3 \]
**Solution:** Use the following reaction stoichiometry for the calculation:
\[
\text{3 MnO}_2 + \text{4 Al} \rightarrow \text{3 Mn} + \text{2 Al}_2\text{O}_3
\]
#### Steps to Solve the Problem:
1. **Calculate the molar mass of Aluminum (Al):**
\[ M(\text{Al}) = 26.98 \, \text{g/mol} \]
2. **Determine the number of moles of Al given the mass:**
\[
\text{n(Al)} = \frac{\text{mass}}{\text{molar mass}}
\]
\[
\text{n(Al)} = \frac{56.19 \, \text{g}}{26.98 \, \text{g/mol}}
\]
\[
\text{n(Al)} \approx 2.08 \, \text{mol}
\]
3. **Use the stoichiometric coefficients from the balanced equation:**
For every 4 moles of Al, 3 moles of MnO₂ are required.
4. **Find the number of moles of MnO₂ needed:**
\[
\text{n(MnO}_2\text{)} = \left( \frac{3}{4} \right) \times 2.08 \, \text{mol}
\]
\[
\text{n(MnO}_2\text{)} \approx 1.56 \, \text{mol}
\]
5. **Calculate the molar mass of MnO₂:**
\[
M(\text{MnO}_2\text{)} = 54.94 \, (\text{Mn}) + 2 \times 16.00 \, (\text{O})
\]
\[
M(\text{Mn](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b0aed3-745e-4879-8577-ea35399669d0%2F2e822320-44ec-4a95-837c-c08601ddc7de%2F419n2aj_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Chemical Reaction Problem
**Problem:** What mass of MnO₂ will completely react with 56.19 g of Al according to the reaction below?
\[ \text{3 MnO}_2 + \text{4 Al} \rightarrow \text{3 Mn} + \text{2 Al}_2\text{O}_3 \]
**Solution:** Use the following reaction stoichiometry for the calculation:
\[
\text{3 MnO}_2 + \text{4 Al} \rightarrow \text{3 Mn} + \text{2 Al}_2\text{O}_3
\]
#### Steps to Solve the Problem:
1. **Calculate the molar mass of Aluminum (Al):**
\[ M(\text{Al}) = 26.98 \, \text{g/mol} \]
2. **Determine the number of moles of Al given the mass:**
\[
\text{n(Al)} = \frac{\text{mass}}{\text{molar mass}}
\]
\[
\text{n(Al)} = \frac{56.19 \, \text{g}}{26.98 \, \text{g/mol}}
\]
\[
\text{n(Al)} \approx 2.08 \, \text{mol}
\]
3. **Use the stoichiometric coefficients from the balanced equation:**
For every 4 moles of Al, 3 moles of MnO₂ are required.
4. **Find the number of moles of MnO₂ needed:**
\[
\text{n(MnO}_2\text{)} = \left( \frac{3}{4} \right) \times 2.08 \, \text{mol}
\]
\[
\text{n(MnO}_2\text{)} \approx 1.56 \, \text{mol}
\]
5. **Calculate the molar mass of MnO₂:**
\[
M(\text{MnO}_2\text{)} = 54.94 \, (\text{Mn}) + 2 \times 16.00 \, (\text{O})
\]
\[
M(\text{Mn
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