a. Show how much CO2 is produced from each if only 1 gram of gasoline fuel is burned: Molecular weight of Carbon (C) = 12 g/mol; Oxygen (0) = 16 g/mol; Hydrogen (H) = 1 g/mol C8H18 12.5 02 8 CO2 + 9 H20 +
a. Show how much CO2 is produced from each if only 1 gram of gasoline fuel is burned: Molecular weight of Carbon (C) = 12 g/mol; Oxygen (0) = 16 g/mol; Hydrogen (H) = 1 g/mol C8H18 12.5 02 8 CO2 + 9 H20 +
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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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Calculation of CO₂ Production from Burning 1 Gram of Gasoline
#### Problem Statement:
Determine how much CO₂ is produced from burning 1 gram of gasoline fuel based on the provided chemical reaction and molecular weights.
#### Given Information:
1. **Molecular Weights**:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol
2. **Chemical Reaction**:
\[
\text{C}_8\text{H}_{18} + 12.5 \text{O}_2 \rightarrow 8 \text{CO}_2 + 9 \text{H}_2\text{O}
\]
#### Steps to Solution:
1. **Molecular Weight of Gasoline (C₈H₁₈)**:
\[
\text{C}_8\text{H}_{18}:
\]
\[
8 \times 12 \, (\text{C atoms}) + 18 \times 1 \, (\text{H atoms}) = 96 \, \text{g/mol} + 18 \, \text{g/mol} = 114 \, \text{g/mol}
\]
2. **Molecular Weight of CO₂**:
\[
\text{CO}_2:
\]
\[
1 \times 12 \, (\text{C atom}) + 2 \times 16 \, (\text{O atoms}) = 12 \, \text{g/mol} + 32 \, \text{g/mol} = 44 \, \text{g/mol}
\]
3. **Mole Ratio**:
In the given balanced chemical equation:
\[
1 \, \text{mol} \, \text{C}_8\text{H}_{18} \rightarrow 8 \, \text{mol} \, \text{CO}_2
\]
4. **Grams to Moles Conversion for C₈H₁₈**:
\[
\text{Number of moles of} \, \text{C}_8\text{H}_{18} = \frac{1 \, \text{g}}{114 \, \text{g/mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7af8d86f-3d07-4c75-9fb6-689055daa42a%2F86b76223-655e-4fb8-a6f4-7d7a0f2999f8%2F7wfwh0g_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculation of CO₂ Production from Burning 1 Gram of Gasoline
#### Problem Statement:
Determine how much CO₂ is produced from burning 1 gram of gasoline fuel based on the provided chemical reaction and molecular weights.
#### Given Information:
1. **Molecular Weights**:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol
2. **Chemical Reaction**:
\[
\text{C}_8\text{H}_{18} + 12.5 \text{O}_2 \rightarrow 8 \text{CO}_2 + 9 \text{H}_2\text{O}
\]
#### Steps to Solution:
1. **Molecular Weight of Gasoline (C₈H₁₈)**:
\[
\text{C}_8\text{H}_{18}:
\]
\[
8 \times 12 \, (\text{C atoms}) + 18 \times 1 \, (\text{H atoms}) = 96 \, \text{g/mol} + 18 \, \text{g/mol} = 114 \, \text{g/mol}
\]
2. **Molecular Weight of CO₂**:
\[
\text{CO}_2:
\]
\[
1 \times 12 \, (\text{C atom}) + 2 \times 16 \, (\text{O atoms}) = 12 \, \text{g/mol} + 32 \, \text{g/mol} = 44 \, \text{g/mol}
\]
3. **Mole Ratio**:
In the given balanced chemical equation:
\[
1 \, \text{mol} \, \text{C}_8\text{H}_{18} \rightarrow 8 \, \text{mol} \, \text{CO}_2
\]
4. **Grams to Moles Conversion for C₈H₁₈**:
\[
\text{Number of moles of} \, \text{C}_8\text{H}_{18} = \frac{1 \, \text{g}}{114 \, \text{g/mol
Transcribed Image Text:**Question: Show the equivalent of 1 kW-hr in Joules**
On an educational website, this question would be presented to help students understand the conversion between kilowatt-hours and Joules, two units of energy measurement. The solution involves using the relationship that 1 kilowatt-hour (kW-hr) equals 3,600,000 Joules (J).
**Explanation:**
1 kilowatt-hour = 1 kW × 1 hour
= 1000 watts × 3600 seconds
= 1000 Joules/second × 3600 seconds
= 3,600,000 Joules
Hence, the equivalent of 1 kW-hr in Joules is 3,600,000 J.
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