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?

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.