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)

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