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The complete combustion of octane, a component of gasoline, is represented by the equation: 2 C8H18(l) + 25 O2(g) →16 CO2(g) + 18 H2O(l) How many liters of CO2(g), measured at 63.1°C and 688 mmHg, are produced for every gallon of octane burned? (1 gal = 3.785 L; density of C8H18(l) = 0.703 g/mL)

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Edufirst

Answer:

  • 5670 liter

Explanation:

1) Chemical equation (given):

  • 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(l)

2) Mole ratio:

  • 2 mol C₈H₁₈(l) : 16 mol CO₂(g)

3) C₈H₁₈ (l) moles:

Molar mass: 114.2285 g/mol (taken from a table or internet)

  • Volume C₈H₁₈ = 1 galon = 3.785 liter (given)

  • density = mass / volume ⇒ mass = density × volume

  • mass = 0.703 g/ ml × 3785 ml  = 2,661 g

  • moles = mass in grams / molar mass = 2,661 g / 114.2285 g/mol = 23.3 mol

4) Proportion:

  • 2 mol C₈H₁₈(l) / 16 mol CO₂(g) = 23.3 mol C₈H₁₈(l) / x

  • x = 186 mol CO₂ (g)

5) Ideal gas equation:

  • pV = nRT

Substitute with:

  • n = 186 mol
  • R = 0.08206 atm-liter / mol-K
  • T = 63.1 + 273.15 K = 336.25 K
  • p = 688 mmHg × 1 atm/760 mmHg = 0.905 atm

Solve for V:

  • V = 186 mol × 0.08206 atm-liter / K-mol × 336.25K / 0.905 atm

  • V = 5671 liter = 5670 liter (using 3 significant figures) ← answer

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