The combustion of octane, C8H18, proceeds according to the reaction
If 297 mol of octane combusts, what volume of carbon dioxide is produced at 20.0 °C and 0.995 atm?
297 mol Octane * 16 mol CO2 for each 2 mol Octane is equal to 2380 mol octane (since we only have three significant figures, we cannot express it with exact accuracy). Now, we apply the ideal gas law:
V=nRT/P, or volume equals number of moles times the gas constant times temperature divided by pressure. This means that V=2380*.082*(20+273)/.995=57500 liters. That's a lot of carbon dioxide!