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If 402 mol of octane combusts, what volume of carbon dioxide is produced at 20.0 °C and 0.995 atm?
The combustion of octane, C8H18, proceeds according to the reaction
2C8H18(l)+25O2(g) --> 16CO2(g) + 18H2O(l)
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- Roger the MoleLv 73 years ago
(402 mol C8H18) x (16 mol CO2 / 2 mol C8H18) = 3216 mol CO2
V = nRT / P = (3216 mol) x (0.08205746 L atm/K mol) x (17.0 + 273.15) K / (0.995 atm) = 76954 L = 7.70 x 10^4 L CO2
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