Question

Consider stoichiometric combustion of gasoline and air. Assume a full tank of gasoline holds 50 L (approximately 13.2 gallons), determine mass of air in kg required to combustion it. You can use isooctane C8H18 to approximate gasoline

Answer 1

Answer:

520 kg

Explanation:

Let's consider the combustion of isooctane.

C₈H₁₈(l) + 12.5 O₂(g) → 8 CO₂(g) +  9 H₂O(l)

We can establish the following relations:

• 1 mL of C₈H₁₈ has a mass of 0.690 g (ρ = 0.690 g/mL).
• The molar mass of C₈H₁₈ is 114.22 g/mol.
• The molar ratio of C₈H₁₈ to O₂ is 1:12.5.
• The mole fraction of O₂ in air is 0.21.
• The molar mass of air is 28.96 g/mol.

50 L of isooctane require the following mass of air.

$50 \times 10^{3}mLC_{8}H_{18}.\frac{0.690gC_{8}H_{18}}{1mLC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.22gC_{8}H_{18}} .\frac{12.5molO_{2}}{1molC_{8}H_{18}} .\frac{1molAir}{0.21molO_{2}} .\frac{28.96 \times 10^{-3}kgAir}{1molAir} =520kgAir$