Question

Write the balanced equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide and water. Use the smallest possible integers to balance the equation. Also, separate the + sign with 1 space.

Remember that a combustion reaction requires oxygen.

Use the smallest whole number coefficients in your balanced equation.

Write the reactants in the order isooctane then oxygen. Write the products in the order carbon dioxide then water.

Assuming gasoline is 90.0% isooctane, with a density of 0.692 g/mL, what is the theoretical yield (in grams) of CO₂ produced by the combustion of 1.66 x 10¹⁰ gallons of gasoline (the estimated annual consumption of gasoline in the U.S.)?

Remember, there are 3.785 liters in 1 gallon and assume that isooctane is the only carbon containing component of gasoline.

Answer 1

2C8H18 (l) + 25O2 (g) => 16CO2 (g) + 18H2O (g)

volume of gasoline = 1.66*10^10 gallon
= 1.66*10^10 * 3.785 L
= 6.2831*10^10 L

density of gasoline = 0.692 g/mL = 0.692 Kg/L

mass of gasoline = density * volume
= 0.692 kg/L * 6.2831*10^10 L
=4.3479*10^10 Kg

mass of octance = 90% of 4.3479*10^10 kg
= 3.913*10^10 kg
= 3.913*10^13 g

number of moles of isooctane = mass/molar mass
= (3.913*10^13)/(8*12 + 18)
=3.4326*10^11 mol
2 mole of isooctane produces 16 moles of CO2
so,
moles of CO2 produced = 8*3.4326*10^11
=2.746*10^12 mol

mass of CO2 = moles of CO2 * Molar mass of CO2
= 2.746*10^12 mol*(44 g/mol)
=1.208*10^14 g
Answer: 1.208*10^14 g