Liquid octane (CH3(CH,) CH;) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). What is the theoretical yield of water formed from the reaction of 8.00 g of octane and 42.3 g of oxygen gas? Round your answer to 3 significant figures.
Liquid octane (CH3(CH,) CH;) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). What is the theoretical yield of water formed from the reaction of 8.00 g of octane and 42.3 g of oxygen gas? Round your answer to 3 significant figures.
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
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![**Title: Theoretical Yield Calculation for the Combustion of Octane**
**Problem:**
Liquid octane (CH₃(CH₂)₆CH₃) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). What is the theoretical yield of water formed from the reaction of 8.00 g of octane and 42.3 g of oxygen gas?
Round your answer to 3 significant figures.
**Solution:**
The balanced chemical equation for the combustion of octane is:
2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
**Step-by-step calculation:**
1. Calculate the molar masses:
- Molar mass of C₈H₁₈ (octane) = 8(12.01 g/mol) + 18(1.01 g/mol) = 114.23 g/mol
- Molar mass of O₂ (oxygen) = 2(16.00 g/mol) = 32.00 g/mol
2. Convert the masses of reactants to moles:
- Moles of octane = 8.00 g C₈H₁₈ × (1 mol / 114.23 g) = 0.07002 mol
- Moles of oxygen = 42.3 g O₂ × (1 mol / 32.00 g) = 1.321875 mol
3. Determine the limiting reactant by comparing the mole ratio from the balanced equation:
- According to the balanced equation, the mole ratio of O₂ to C₈H₁₈ is 25:2 or 12.5:1.
- Moles of O₂ needed for 0.07002 mol of octane: 0.07002 mol C₈H₁₈ × 12.5 mol O₂ / 1 mol C₈H₁₈ = 0.87525 mol O₂.
- There is more O₂ available (1.321875 mol) than required, making octane the limiting reactant.
4. Calculate the theoretical yield of water:
- The balanced equation shows that 2 mol of C₈H₁₈ produces 18 mol of H₂O.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88a4d88f-5648-4e4d-a49c-10af3e87c66c%2Fce5d2747-5f68-4a59-b116-7745477c979c%2Fc82n1wf_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Theoretical Yield Calculation for the Combustion of Octane**
**Problem:**
Liquid octane (CH₃(CH₂)₆CH₃) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). What is the theoretical yield of water formed from the reaction of 8.00 g of octane and 42.3 g of oxygen gas?
Round your answer to 3 significant figures.
**Solution:**
The balanced chemical equation for the combustion of octane is:
2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
**Step-by-step calculation:**
1. Calculate the molar masses:
- Molar mass of C₈H₁₈ (octane) = 8(12.01 g/mol) + 18(1.01 g/mol) = 114.23 g/mol
- Molar mass of O₂ (oxygen) = 2(16.00 g/mol) = 32.00 g/mol
2. Convert the masses of reactants to moles:
- Moles of octane = 8.00 g C₈H₁₈ × (1 mol / 114.23 g) = 0.07002 mol
- Moles of oxygen = 42.3 g O₂ × (1 mol / 32.00 g) = 1.321875 mol
3. Determine the limiting reactant by comparing the mole ratio from the balanced equation:
- According to the balanced equation, the mole ratio of O₂ to C₈H₁₈ is 25:2 or 12.5:1.
- Moles of O₂ needed for 0.07002 mol of octane: 0.07002 mol C₈H₁₈ × 12.5 mol O₂ / 1 mol C₈H₁₈ = 0.87525 mol O₂.
- There is more O₂ available (1.321875 mol) than required, making octane the limiting reactant.
4. Calculate the theoretical yield of water:
- The balanced equation shows that 2 mol of C₈H₁₈ produces 18 mol of H₂O.
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