Question

A newspaper article about the danger of global warming from the accumulation of greenhouse gases such as carbon dioxide states that "reducing driving your car by 20 miles a week would prevent release of over 1000 pounds of CO2 per year into the atmosphere." Is this a reasonable statement? Assume that gasoline is octane (molecular formula is C8H18) and that it is burned completely to CO2 and H2O in the engine of your car. Facts (or reasonable guesses) about your car's gas mileage, the density of octane, and other factors will also be needed.

Answer 1

The newspaper article's statement is correct since the mass of CO₂ reduced = 1071.85 pounds

Step-by-step explanation:

Equation for the combustion of gasoline (octane) is given below:

2C₈H₁₈ + 25O₂ ----> 16CO₂ + 18H₂O

Density of C₈H₁₈ = 0.8 g/mL

Assuming mileage of the car is 20 miles/gallon; 1 gallon = 3.785 L

Since 20 miles is reduced in driving each week, thereby saving 1 gallon of gasoline, number of litres of gasoline reduced in a year is given by;

52 * 3.785 L = 196.82 L = 196820 mL

density = mass/volume; mass = density * volume

mass of gasoline saved = 0.8 g/mL * 196820 mL = 157456 g

number of moles = mass/molar mass

Molar mass of C₈H₁₈ = 114 g/mol

number of moles of C₈H₁₈ in 157456 g = 157456 g / 114 g/mol = 1381.2 moles

From the equation of reaction, 2 moles of C₈H₁₈ produces 16 moles of CO₂

1381.2 moles of C₈H₁₈ will produce 1381.2 * 16/2 moles of CO₂ = 11049.6 moles of CO₂

mass = number of moles * molar mass

molar mass of CO₂ = 44 g/mol

mass of CO₂ = 11049.6 moles * 44 g/mol = 486182.4 g

1 g = 0.00220462 pounds

Therefore mass of CO₂ reduced = 486182.4 g * 0.00220462 lb/g

mass of CO₂ reduced = 1071.85 pounds

Therefore, the newspaper article's statement is correct