3. Potassium chlorate is the prime ingredient in most Chinese fireworks. If I have 25 grams of potassium chlorate, how many moles of oxygen can I produce?
3. Potassium chlorate is the prime ingredient in most Chinese fireworks. If I have 25 grams of potassium chlorate, how many moles of oxygen can I produce?
Living By Chemistry: First Edition Textbook
1st Edition
ISBN:9781559539418
Author:Angelica Stacy
Publisher:Angelica Stacy
ChapterU4: Toxins: Stoichiometry, Solution Chemistry, And Acids And Bases
Section: Chapter Questions
Problem 11RE
Related questions
Question
100%
![### Chemical Stoichiometry Problem
**Question 3:**
Potassium chlorate (KClO₃) is the prime ingredient in most Chinese fireworks. If you have 25 grams of potassium chlorate, how many moles of oxygen gas (O₂) can be produced?
#### Solution Explanation:
To solve this problem, follow these steps:
1. **Find the Molar Mass of Potassium Chlorate (KClO₃):**
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol each, and there are 3 oxygen atoms
- Molar mass of KClO₃ = 39.10 + 35.45 + (3 × 16.00) = 122.55 g/mol
2. **Convert Grams to Moles:**
Use the molar mass to convert the mass of potassium chlorate to moles:
\[
\text{Moles of KClO₃} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} = \frac{25 \, \text{grams}}{122.55 \, \text{g/mol}} \approx 0.204 \, \text{moles}
\]
3. **Chemical Equation:**
The decomposition reaction for potassium chlorate is:
\[
2 KClO₃ (s) \rightarrow 2 KCl (s) + 3 O₂ (g)
\]
According to the balanced chemical equation, 2 moles of potassium chlorate produce 3 moles of oxygen gas.
4. **Determine the Moles of Oxygen:**
From the balanced equation:
\[
\text{Moles of O₂} = 0.204 \, \text{moles KClO₃} \times \frac{3 \, \text{moles O₂}}{2 \, \text{moles KClO₃}} = 0.306 \, \text{moles O₂}
\]
Therefore, 25 grams of potassium chlorate can produce approximately 0.306 moles of oxygen gas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd909c482-4f4e-4400-a93e-251a85b7679b%2F70f6b587-1e5f-4084-86d4-580515c12653%2F9y754oa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Stoichiometry Problem
**Question 3:**
Potassium chlorate (KClO₃) is the prime ingredient in most Chinese fireworks. If you have 25 grams of potassium chlorate, how many moles of oxygen gas (O₂) can be produced?
#### Solution Explanation:
To solve this problem, follow these steps:
1. **Find the Molar Mass of Potassium Chlorate (KClO₃):**
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol each, and there are 3 oxygen atoms
- Molar mass of KClO₃ = 39.10 + 35.45 + (3 × 16.00) = 122.55 g/mol
2. **Convert Grams to Moles:**
Use the molar mass to convert the mass of potassium chlorate to moles:
\[
\text{Moles of KClO₃} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} = \frac{25 \, \text{grams}}{122.55 \, \text{g/mol}} \approx 0.204 \, \text{moles}
\]
3. **Chemical Equation:**
The decomposition reaction for potassium chlorate is:
\[
2 KClO₃ (s) \rightarrow 2 KCl (s) + 3 O₂ (g)
\]
According to the balanced chemical equation, 2 moles of potassium chlorate produce 3 moles of oxygen gas.
4. **Determine the Moles of Oxygen:**
From the balanced equation:
\[
\text{Moles of O₂} = 0.204 \, \text{moles KClO₃} \times \frac{3 \, \text{moles O₂}}{2 \, \text{moles KClO₃}} = 0.306 \, \text{moles O₂}
\]
Therefore, 25 grams of potassium chlorate can produce approximately 0.306 moles of oxygen gas.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 4 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:
9781559539418
Author:
Angelica Stacy
Publisher:
MAC HIGHER
Introductory Chemistry: A Foundation
Chemistry
ISBN:
9781337399425
Author:
Steven S. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:
9781559539418
Author:
Angelica Stacy
Publisher:
MAC HIGHER
Introductory Chemistry: A Foundation
Chemistry
ISBN:
9781337399425
Author:
Steven S. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:
9781305079243
Author:
Steven S. Zumdahl, Susan A. Zumdahl
Publisher:
Cengage Learning
Introductory Chemistry: An Active Learning Approa…
Chemistry
ISBN:
9781305079250
Author:
Mark S. Cracolice, Ed Peters
Publisher:
Cengage Learning