What is the mass in grams of CO2 that can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C4H10(9) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
What is the mass in grams of CO2 that can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C4H10(9) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
Chemistry
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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
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![**Transcription for Educational Website:**
Title: Calculating the Mass of CO₂ from Butane Combustion
**Question:**
What is the mass in grams of CO₂ that can be produced from the combustion of 2.76 moles of butane according to this equation:
\[ 2 \, \text{C}_4\text{H}_{10}(g) + 13 \, \text{O}_2(g) \rightarrow 8 \, \text{CO}_2(g) + 10 \, \text{H}_2\text{O}(g) \]
**Explanation:**
This question involves calculating the mass of carbon dioxide (CO₂) produced when a specific amount of butane (C₄H₁₀) undergoes combustion in the presence of oxygen (O₂). The chemical equation provided is balanced and indicates the stoichiometry of the reaction:
- **Reactants:** 2 moles of butane react with 13 moles of oxygen.
- **Products:** 8 moles of carbon dioxide and 10 moles of water (H₂O) are formed.
To find the mass of CO₂ produced, use stoichiometry based on the balanced equation:
1. **Determine Moles of CO₂:** From the equation, 2 moles of C₄H₁₀ produce 8 moles of CO₂. Thus, 2.76 moles of C₄H₁₀ will produce:
\[
\text{Moles of CO}_2 = \frac{8}{2} \times 2.76 = 11.04 \text{ moles of CO}_2
\]
2. **Calculate Mass of CO₂:** Using the molar mass of CO₂ (approximately 44.01 g/mol), the mass is:
\[
\text{Mass of CO}_2 = 11.04 \text{ moles} \times 44.01 \text{ g/mol} = 485.85 \text{ grams}
\]
This calculation assumes complete combustion and no side reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeef920a-b7fc-42c4-a405-16f92aebf3b1%2Fe631e3ac-72ca-4907-9cef-4e7edca4d60f%2F0wu3d0q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website:**
Title: Calculating the Mass of CO₂ from Butane Combustion
**Question:**
What is the mass in grams of CO₂ that can be produced from the combustion of 2.76 moles of butane according to this equation:
\[ 2 \, \text{C}_4\text{H}_{10}(g) + 13 \, \text{O}_2(g) \rightarrow 8 \, \text{CO}_2(g) + 10 \, \text{H}_2\text{O}(g) \]
**Explanation:**
This question involves calculating the mass of carbon dioxide (CO₂) produced when a specific amount of butane (C₄H₁₀) undergoes combustion in the presence of oxygen (O₂). The chemical equation provided is balanced and indicates the stoichiometry of the reaction:
- **Reactants:** 2 moles of butane react with 13 moles of oxygen.
- **Products:** 8 moles of carbon dioxide and 10 moles of water (H₂O) are formed.
To find the mass of CO₂ produced, use stoichiometry based on the balanced equation:
1. **Determine Moles of CO₂:** From the equation, 2 moles of C₄H₁₀ produce 8 moles of CO₂. Thus, 2.76 moles of C₄H₁₀ will produce:
\[
\text{Moles of CO}_2 = \frac{8}{2} \times 2.76 = 11.04 \text{ moles of CO}_2
\]
2. **Calculate Mass of CO₂:** Using the molar mass of CO₂ (approximately 44.01 g/mol), the mass is:
\[
\text{Mass of CO}_2 = 11.04 \text{ moles} \times 44.01 \text{ g/mol} = 485.85 \text{ grams}
\]
This calculation assumes complete combustion and no side reactions.
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