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22 August, 05:52

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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  1. 22 August, 09:48
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    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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