Final answer:
To calculate the grams of CO2 produced from the combustion of 17.0 gallons of octane, we convert gallons to liters, then to grams, and use the stoichiometry of the balanced chemical equation to find the mass of CO2.
Step-by-step explanation:
To determine how many grams of CO2 are produced when 17.0 gallons of C8H18 (octane) are combusted, we need to first convert gallons to liters, then to grams using the density of octane, and, finally, use the balanced chemical equation for the combustion of octane to find the mass of CO2 produced.
The balanced chemical equation for the combustion of octane is:
2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
This equation tells us that 2 moles of octane produce 16 moles of carbon dioxide. To find the mass of CO2 produced, follow these steps:
- Convert gallons of C8H18 to liters (1 gallon = 3.78541 liters).
- Convert liters of C8H18 to grams (density of C8H18 = 0.703 g/mL, and 1 L = 1000 mL).
- Convert grams of C8H18 to moles (molecular weight of C8H18 = 114.23 g/mol).
- Use the molar ratio from the balanced equation to find moles of CO2 produced.
- Convert moles of CO2 to grams (molecular weight of CO2 = 44.01 g/mol).
This chemical equation is used as a stoichiometric conversion factor between moles of reactants and moles of products in a chemical reaction.