The complete combustion of octane, a component of gasoline, is represented by the equation: 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(l)
How many liters of CO₂(g) , measured at 63.1°C and 688 mmHg, are produced for every gallon of octane burned? (1 gal = 3.785 L; density of C₈H1₈(l) = 0.703 g/mL)

Question

Quizplus · Accepted Answer

First, calculate the mass of octane in 1 gallon: 3.785 L × 0.703 g/mL × 1000 mL/L = 2661.755 g. The molar mass of C8H18 is 114.22 g/mol, so moles of octane = 2661.755 g / 114.22 g/mol = 23.31 mol. From the balanced equation, 2 mol of C8H18 produces 16 mol of CO2, so 23.31 mol of C8H18 produces 186.48 mol of CO2. Using the ideal gas law, PV = nRT, convert to liters at 63.1°C (336.25 K) and 688 mmHg (0.905 atm): V = nRT/P = (186.48 mol × 0.0821 L·atm/mol·K × 336.25 K) / 0.905 atm = 5680.68 L.

The Complete Combustion of Octane, a Component of Gasoline, Is