If 313 mol of octane combusts, what volume of carbon dioxide is produced at 13.0 °C and 0.995 atm?

Question

Roger the Mole · Accepted Answer

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

(313 mol C8H18) x (16 mol CO2 / 2 mol C8H18) = 2504 mol CO2

V = nRT / P = (2504 mol) x (0.08205746 L atm/K mol) x (13.0 + 273.15) K / (0.995 atm) = 59091 L = 59.1 kL

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