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

520 kg

Step-by-step 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 * 10^(3)mLC_(8)H_(18).(0.690gC_(8)H_(18))/(1mLC_(8)H_(18)) .(1molC_(8)H_(18))/(114.22gC_(8)H_(18)) .(12.5molO_(2))/(1molC_(8)H_(18)) .(1molAir)/(0.21molO_(2)) .(28.96 * 10^(-3)kgAir)/(1molAir) =520kgAir`