Assuming gasoline is 100% isooctane, with a density of 0.692 g/mL, what mass of carbon dioxide is produced by the combustion of 1.2 x 1010 gal of gasoline?

1 Answer
Dec 14, 2015

Answer:

I use a US gallon, #3.758*L#.

Explanation:

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

For each equiv of octane consumed, clearly 8 equiv carbon dioxide are produced.

So how many moles of octane?

#(1.2xx10^(10)cancel(gallons)xx3.758*cancel(L)/(cancel(gallon))xx10^3*cancel(mL)*cancel(L^-1)xx0.692*cancel(g)*cancel(mL^(-1)))/(114.23*cancel(g)*mol^-1) = (??mol)#.

So this is how many moles of octane? And clearly, from the stoichiometrically balanced equation, 8 mol carbon dioxide result from the combustion of 1 mole of octane. You are still not done in that you to multiply this number in moles by the molecular mass of carbon dioxide, #44.0*g*mol^-1#.

I make no guarantees of my formatting or of my arithmetic. This is the problem with using ridiculous units such as pounds or ounces or furlongs or gallons. I don't even know whether you are using a US gallon or an Imperial gallon, #4.8*L#; I have assumed the former.

PS I have to go to the bank this afternoon to get some readies (cash) for Xmas. Do you think I will get served if instead of asking for £100-00, I ask for 100 guineas?