[guayguay]

Petrol is 100% octane C8H18(l)
E10 is 10% ethanol C2H5OH(l) and 90% C8H18(l)

Density of octane is 703g/L, ethanol is 789g/L
Octane has ΔH -250.1KJ/mol
Ethanol has ΔH -235.2KJ/mol

Which is the better fuel?

I can't figure this out. Currenty reading over notes.

[Borek]

Quite an ambiguous question, as it is not clear what is meant by a "better fuel". There are many ways to compare two fuels, for some applications one can be better, for some applications the other.

[billnotgatez]

I agree with @Borek
but let us just guess at what they might want
If you had 1 liter of each how much KJ do you have in each

[guayguay]

This is what I did

2C8H18 + 25O2  16CO2 + 18H2O
ΔH = ΣnΔH°f(product) - ΣnΔH°f(reactant)
= [(16 x -393.5KJ/mol) + (18 x -285.8KJ/mol)] - (2 x -250.1KJ/mol)
= -10940.2KJ

C2H5OH + 3O2  2CO2 + 3H2O
ΔH = ΣnΔH°f(product) - ΣnΔH°f(reactant)
= [(2 x -393.5KJ/mol) + (3 x-285.8KJ/mol)] - (1 x -235.2KJ/mol)
= -1409.2KJ

From here I'm not really sure what to do. The units are also confusing me. How did KJ/mol become KJ?

[Corribus]

Most likely they want you to compare energy density, that is, energy per unit volume.  Billnotgatez already gave you a good hint.  But take 1 liter of each of your fuels and compare how much energy you would have if you combusted all of it with 100% efficiency.  Remember, one of your fuels is 90% octane, 10% ethanol, not 100% ethanol as your attempt showed.

You'll have to assume that the mixture is ideal (that is, the two substances stay essentially separate even when they're mixed).

[guayguay]

Ok I did some more calculations

For octane:
(-10940.2KJ/2mol) x (1molC8H18/114g) x (703g/L) = -33731.67KJ/L

For ethanol:
(-1409.2KJ/mol) x (1molC2H5OH/46g) x (789g/L) = -24170.84KJ/L

Octane is 100% x -33731.67KJ/L = -33731.67KJ/L
Ethanol is (10% x -24170.84KJ/L) + (90% x -33731.67KJ/L) = -32775.59KJ/L

If we compare ΔH between petrol (-33731.67KJ/L) with E10 (-32775.59KJ/L) then we can say that petrol is the better fuel because it produces more energy per L.

Am I right?