Question

If a "classic" SUV features a 52 gallon (197 L) gas tank and assuming complete combustion and a molecular formulaof C8H18 for gasoline, how many L of air are consumed per tank of gasoline. Recall that air is 21% oxygen and assume a density of 0.78 g/mL for gasoline.

a. 300 Lb. 1,600 Lc. 1,000,000 Ld. 1,700,000 L

Answer 1

d. 1,700,000 L

Step-by-step explanation:

The combustion reaction of gasoline is:

C₈H₁₈ + (25/2)O₂ → 8CO₂ + 9H₂O

The mass of gasoline can be found multiplying the volume by the density:

m = 197,000 mL * 0.78 g/mL

m = 153,660 g

The molar mass of gasoline is 114.23 g/mol, thus the number of moles is:

n = mass/ molar mass

n = 153,660/114.23

n = 1,345.18 mol

By the stoichiometry of the reaction:

1 mol of C₈H₁₈ ----------------- 25/2 mol of O₂

1,345.18 mol ---------------- x

By a simple direct three rule:

x = 16,814 mol of O₂

This represents 21% of the moles of air, so, the number of moles of air is:

n = 16,814/0.21

n = 80,067 mol of air

At STP, 1 mol of gas has 22.4 L, thus the volume of air is:

V = 80,067* 22.4

V = 1,793,493.33 L

V ≅ 1,700,000 L