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Calculate the mass of CO2 produced by thecombustion of a tank of gasoline. Assume...

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rif1191 | eNotes Newbie

Posted March 25, 2013 at 9:21 PM via web

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Calculate the mass of CO2 produced by the
combustion of a tank of gasoline. Assume the
mass of a tank of gasoline is 46.0 kg and that
gasoline is pure octane (C8H18).

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted March 26, 2013 at 12:56 AM (Answer #1)

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`C_8H_18+27/2O_2 rarr 8CO_2+9H_2O`


When you have a combustion of hydrocarbon as results we will get `CO_2` and water.

Mole mass

`C_8H_18 = 114g/(mol)`

`CO_2 = 44g/(mol)`


Amount of `C_H_18` reacted `= 46kg`

Amount of `C_H_18` reacted `= 46000/114 mol`


Mole ratio

`C_8H_18:CO_2 = 1:8`


Amount of` CO_2` formed `= 46000/114xx8 moles`

Amount of `CO_2` formed `= 46000/114xx8xx44 g`

Amount of `CO_2` formed `= 142035g`


So 142.035kg of `CO_2` will be formed.



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llltkl | College Teacher | Valedictorian

Posted March 26, 2013 at 1:50 AM (Answer #2)

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Assuming complete combustion (in sufficient supply of oxygen or air), the balanced chemical equation for the combustion of pure octane (C8H18) is,

2C8H18 + 25O2 = 16CO2 + 18H2O

The stoichiometry of this reaction shows that, 2 moles of the gasoline upon combustion produces 16 moles of CO2

Or, 2× (8×12+18×1) i.e. 228 gm gasoline upon combustion produces 16*44 gms = 704 gms of CO2.

Then, 46 kg, i.e. 46000 gms gasoline should produce 704*46000/228 = 142035.1 gms = 313.134 lbs of CO2 on complete burning.

Hence, 313.13 lbs of CO2 will be formed on burning 46 kg of that gasoline.


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