Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooctane is 0.792 g/mLg/mL, what mass of CO2CO2 (in kilograms) is produced each year by the annual U. S. gasoline consumption of 4.6*1010L4.6*1010L?

1.12*10¹¹ kg of CO₂ are produced with 4.6*10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density. Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6*10¹⁰ L. 1000 mL / 1L = 4.6*10¹³ mL

0.792 g/mL. 4.6*10¹³ mL = 3.64 * 10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 * 10¹³ g. 1mol / 114 g = 3.19*10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19*10¹¹ mol would produce (3.19*10¹¹ mol. 8) = 2.55*10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles. molar mass

2.55*10¹² mol. 44 g/mol = 1.12*10¹⁴ g of CO₂

If we convert to kg 1.12*10¹⁴ g / 1000 = 1.12*10¹¹ kg