June 9th, 2006, 5:40 pm
a generic saturated hydrocarbon is C(n) H(2n+2) which produces n CO2 and n+1 H2O molecules on burning. assuming:(a) the hydrocarbons present are either saturated or don't have too many double/triple bonds(b) they are complex which means n is high(c) burning is almost complete then we can see that there is about 1 carbon atom for every 2 hydrogen atoms. so C is 12/14 by weight of gasoline.Now 1 gallon = 3.8 litres => 3.8*800 g = 3040 g, so 20 gallon = about 60kg of gasoline. weight of C should be about 60*12/14 kg = 51 kg approx. but when this gets converted to CO2, the weight of O2 is added and so this becomes 51 * (12+2*16)/12 = 51*44/12 = 187kg.on the other extreme, suppose all of gasoline was benzene C6H6. then the molecular ratio will be C:H = 1:1 and weight ratio is 12/13. this shouldn't make much difference to the calculation. in otherwords, condition (a) is not very important.so, i believe the answer should be about 187 kg which is 187/0.45 lb = 416 lb approx.it does seem quite a bit of CO2 to unleash with just one full tank... (edited)
Last edited by
needaclue on June 8th, 2006, 10:00 pm, edited 1 time in total.