Question

Assuming that gasoline is C8H18 and has a density of 0.703 g/ml, how many liters of CO2 will be produced from the consumption of 1.8 gallons of gasoline at standard conditions ?

Answer 1

`C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)`

Explanation:

Is the equation above stoichiometrically balanced? (And what does stoichiometric mean?). This equation tells us that for each mole of octane (each 116 g!), 8 moles of carbon dioxide (`8 xx 44 *g`) will result. Now the question tells us that 1.8 gallons of gas have been consumed.

And now I have a problem. Where I live, a gallon = 8 imperial pints, 4.8 L (in fact gasoline is sold in litres, it was > £1-10 per litre at the local gas station; ouch!!). If you live in the States or South America, a gallon might be something else. (This is 1 reason why most scientists insist on the use of sensible, consistent units; so that they can talk to each other across continents and compare their results). Anyway, the equation tells you that for each 114 g (each mole) of hydrocarbon that is combusted, 8 moles of carbon dioxide will result (about 200 L of gas at standard temperature and pressure).