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5 March, 06:02

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning completely 1.00 L of octane (C8H18, density = 0.7025 g/mL), yielding carbon dioxide and water?

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  1. 5 March, 08:04
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    V = 8963 L

    Explanation:

    Our strategy here is to realize this problem involves stoichiometric calculation based on the balanced chemical equation for the combustion of octane:

    C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9 H₂O

    From the density of octane we can obtain the number of moles:

    D = m/V ⇒ m = D x V = 0.7025 g/mL x (1000 mL) = 702.5 g

    MW octane = 702.5 g / 114.23 g/mol = 6.15 mol

    Required mol oxygen to react with octane:

    6.15 mol octane x 25/2 mol O₂ / mol octane = 76.8 mol O₂

    Now mol fraction is given by mol O₂ / total number mol air ⇒

    n air = 76.8 mol O₂ / (0.2095 mol O₂ / mol air) = 366.59 mol air

    and from the ideal gas law we can compute the volume of air:

    PV = nRT ⇒ V = nRT/P

    V = 366.59 mol air x 0.08205 Latm/Kmol x (25 + 273) K / 1 atm

    = 8,963 Lts

    Note we treat here air as a compund which is allowed in combustion problems.
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