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?
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 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.
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