Ask Question
10 July, 21:28

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.)

+4
Answers (1)
  1. 10 July, 22:38
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers