# 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?

Dec 14, 2015

I use a US gallon, $3.758 \cdot L$.

#### Explanation:

${C}_{8} {H}_{18} \left(l\right) + {O}_{2} \left(l\right) + \left(\frac{25}{2}\right) {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(g\right)$

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, $44.0 \cdot g \cdot m o {l}^{-} 1$.

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, $4.8 \cdot L$; 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?