Answer:
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.
