Question

Assume that an automobile burns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) \((d=0.692 \mathrm{~g} / \mathrm{mL})\) (a) Write a balanced equation for the combustion of octane to carbon dioxide and water. (b) A car has a fuel efficiency of 22 mi/gal of octane. What volume of carbon dioxide at \(25^{\circ} \mathrm{C}\) and one atmosphere pressure is generated by the combustion when that car goes on a 75 -mile trip.

Short answer

Answer: Approximately 15,498.72 L of carbon dioxide is generated during the 75-mile trip.

Step-by-step solution

Write a balanced equation for the combustion of octane (C8H18)

To write a balanced chemical equation for the combustion of octane, we first need to identify the reactants and products. Combustion occurs when a hydrocarbon (in this case, octane) reacts with oxygen to produce carbon dioxide and water. The general equation looks like this: C8H18 + O2 → CO2 + H2O Now we will balance the equation. First, balance carbon atoms: C8H18 + O2 → 8CO2 + H2O Now balance hydrogen atoms: C8H18 + O2 → 8CO2 + 9H2O Finally, balance oxygen atoms: C8H18 + 12.5O2 → 8CO2 + 9H2O The balanced equation for the combustion of octane is: C8H18 + 12.5O2 → 8CO2 + 9H2O

Calculate the moles of octane consumed during the 75-mile trip

Now we need to find how many moles of octane are consumed during the 75-mile trip, considering that the car has a fuel efficiency of 22 mi/gal. First, find the number of gallons of octane consumed during the trip: Gallons of octane = (75 mi) / (22 mi/gal) = 3.4091 gal Now convert the gallons of octane to grams using the density of octane (0.692 g/mL): 1 gal = 3785 mL Mass of octane = (3.4091 gal) * (0.692 g/mL) * (3785 mL/gal) = 8980.98 g Now convert grams of octane to moles using the molar mass of octane (C8H18 = (12.01*8) + (1.01*18) = 114.23 g/mol): Moles of octane = (8980.98 g) / (114.23 g/mol) = 78.615 moles

Calculate the moles of carbon dioxide generated from the combustion

Using the stoichiometric coefficients from the balanced equation (1 mole of octane produces 8 moles of carbon dioxide), calculate the moles of carbon dioxide generated during the trip: Moles of CO2 = (78.615 moles of octane) * (8 moles of CO2 / 1 mole of octane) = 628.92 moles of CO2

Calculate the volume of carbon dioxide generated

Now we can calculate the volume of carbon dioxide generated at 25°C and one atmosphere pressure using the ideal gas law (PV=nRT). In this case, the ideal gas constant R = 0.0821 L atm/mol K, and the temperature T = 298 K (25°C + 273): V = (nRT) / P V = (628.92 moles * 0.0821 L atm/mol K * 298 K) / 1 atm = 15498.72 L The volume of carbon dioxide generated during the 75-mile trip is approximately 15498.72 L.