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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.
`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`
`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.
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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