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.)

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Chemistry

Kyra Huffman

21 July, 12:32

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.)

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Joseph Tyler · Answer 1

1.164 * 10¹⁴ g

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