Question

Assuming gasoline is 89.0% isooctane, with a density of 0.692 g/mL, what is the theoretical yield (in grams) of CO2 produced by the combustion of 1.62 x 1010 gallons of gasoline (the estimated annual consumption of gasoline in the U.S.)

Answer 1

1.164 × 10¹⁴ g

Step-by-step explanation:

1.62 × 10¹⁰ gallons = 1.62 × 10¹⁰ × 3.785 litres = 6.1317 × 10¹⁰ L

mass = density × volume = 6.1317 × 10¹⁰ × 1000 ml × 0.692 g/ml = 4.24 × 10¹³ g

4.24 × 10¹³ g × 0.89 = 3.776 × 10¹³ g

2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

molar mass of isooctane = 114. 22 × 2 moles = 228.44 g

molar mass of CO₂ = 44.01 × 16 moles = 704.16 g

228.44 yields 704.16 g

3.776 × 10¹³ g will yield (3.776 × 10¹³ g × 704.16 g) / 228.44 = 1.164 × 10¹⁴ g