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.

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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.
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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