Final answer:
Burning a 17.0-gallon tank of gasoline releases approximately 302.72 pounds of CO2 into the atmosphere, assuming complete combustion of gasoline, with an approximation based on the density of gasoline and the stoichiometry of the combustion reaction of octane.
Step-by-step explanation:
To calculate the number of pounds of CO2 released when a 17.0-gallon tank of gasoline is burned, we'll first convert the volume to mass using the given density, then use the stoichiometry of the combustion reaction to find the mass of CO2 produced.
First, convert gallons to liters (1 gallon = 3.78541 L), and then to grams using the density (0.692 g/mL), and finally to moles of octane (C8H18). The balanced combustion reaction for octane:
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
From this, we calculate that 16 moles of CO2 are produced per 2 moles of C8H18. Convert the moles of CO2 to grams, and then to pounds (1 pound = 453.592 g).
Here's the calculation broken down:
- 17.0 gallons x 3.78541 L/gallon x 1000 mL/L x 0.692 g/mL = 44537.57 g of gasoline
- 44537.57 g / 114.23 g/mol (molar mass of C8H18) = 389.87 moles of C8H18
- 389.87 moles C8H18 x (16 moles CO2 / 2 moles C8H18) = 3118.96 moles of CO2
- 3118.96 moles CO2 x 44.01 g/mol (molar mass of CO2) = 137241.97 g of CO2
- 137241.97 g / 453.592 g/pound = 302.72 pounds of CO2
So, burning a 17.0-gallon tank of gasoline will release approximately 302.72 pounds of CO2 into the atmosphere, assuming complete combustion.