Question

Calculate the number of pounds of CO2 released into the atmosphere when a 14.0 gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18, and that the density of gasoline is 0.692 g⋅mL−1. This assumption ignores additives. Also, assume complete combustion.

Answer 1

14.0 gallon tank contains 14.0·3.785=52.99 L of gasoline. Making use of the density, we calculate the mass of the gasoline:

$m = d·V = 0.692(\text{g/mL)}·52.99·10^3(\text{mL}) = 36.67·10^3$ g.

As the gasoline is primarily octane, the mass of octane is 36.67·10³ g. The number of the moles of octane is its mass divided by its molar mass $M = 114.23$ g/mol:

$n = \frac{m}{M} = \frac{36.67·10^3 (g)}{114.23 (g/mol)} = 321$ mol.

The equation of the reaction between octane and oxygen is:

2C₈H₁₈ + 25O₂ $\rightarrow$ 16CO₂ + 18H₂O.

As one can see, when 2 mol of octane reacts, 16 mol of CO₂ is produced:

$\frac{n(C_8H_{18})}{2} = \frac{n(CO_2)}{16}$.

Therefore, the number of the moles of CO₂ produced is:

$n(CO_2) = 321·16/2 = 2568$ mol.

The molar mass of CO₂ is 44.01 g/mol. Thus:

$m(CO_2) = n·M =2568·44.01 = 1.13·10^{5}$ g.

In pounds, the mass of CO₂ will be:

$\frac{1.13·10^5}{453.59} = 249$ lb.

Answer: 249 lb of CO₂ is released into the atmosphere when a 14.0 gallon tank of gasoline is burned in an automobile engine.