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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![### Stoichiometry and Chemical Reactions
**Problem Statement:**
**2.** You start with 2.46 g of KClO<sub>3</sub> and heat it. Use stoichiometry to calculate the mass (g) of KCl produced by the reaction.
---
**Step-by-Step Solution:**
1. **Write the balanced chemical equation for the decomposition of KClO<sub>3</sub>:**
\[2KClO_3 \rightarrow 2KCl + 3O_2\]
2. **Calculate the molar mass of KClO<sub>3</sub> and KCl:**
- KClO<sub>3</sub>: \( 39.10 (\text{K}) + 35.45 (\text{Cl}) + 3 \times 16.00 (\text{O}) = 122.55 \, \text{g/mol} \)
- KCl: \( 39.10 (\text{K}) + 35.45 (\text{Cl}) = 74.55 \, \text{g/mol} \)
3. **Convert grams of KClO<sub>3</sub> to moles:**
Moles of KClO<sub>3</sub>:
\[
\frac{2.46 \, \text{g KClO}_3}{122.55 \, \text{g/mol}} = 0.0201 \, \text{mol KClO}_3
\]
4. **Use the stoichiometric ratios from the balanced equation to find moles of KCl produced:**
According to the balanced equation, \(2 \, \text{mol KClO}_3\) produces \(2 \, \text{mol KCl}\):
\[
0.0201 \, \text{mol KClO}_3 \times \frac{2 \, \text{mol KCl}}{2 \, \text{mol KClO}_3} = 0.0201 \, \text{mol KCl}
\]
5. **Convert moles of KCl to grams:**
\[
0.0201 \, \text{mol KCl} \times 74.55 \, \text{g](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c5734d7-71be-4626-8cb5-227ca6c7fb69%2F1f977eb3-813b-49f1-a905-0d3888d395e2%2Fqlizp2_processed.png&w=3840&q=75)
Transcribed Image Text:### Stoichiometry and Chemical Reactions
**Problem Statement:**
**2.** You start with 2.46 g of KClO<sub>3</sub> and heat it. Use stoichiometry to calculate the mass (g) of KCl produced by the reaction.
---
**Step-by-Step Solution:**
1. **Write the balanced chemical equation for the decomposition of KClO<sub>3</sub>:**
\[2KClO_3 \rightarrow 2KCl + 3O_2\]
2. **Calculate the molar mass of KClO<sub>3</sub> and KCl:**
- KClO<sub>3</sub>: \( 39.10 (\text{K}) + 35.45 (\text{Cl}) + 3 \times 16.00 (\text{O}) = 122.55 \, \text{g/mol} \)
- KCl: \( 39.10 (\text{K}) + 35.45 (\text{Cl}) = 74.55 \, \text{g/mol} \)
3. **Convert grams of KClO<sub>3</sub> to moles:**
Moles of KClO<sub>3</sub>:
\[
\frac{2.46 \, \text{g KClO}_3}{122.55 \, \text{g/mol}} = 0.0201 \, \text{mol KClO}_3
\]
4. **Use the stoichiometric ratios from the balanced equation to find moles of KCl produced:**
According to the balanced equation, \(2 \, \text{mol KClO}_3\) produces \(2 \, \text{mol KCl}\):
\[
0.0201 \, \text{mol KClO}_3 \times \frac{2 \, \text{mol KCl}}{2 \, \text{mol KClO}_3} = 0.0201 \, \text{mol KCl}
\]
5. **Convert moles of KCl to grams:**
\[
0.0201 \, \text{mol KCl} \times 74.55 \, \text{g
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