C8H18(g) + 25/2(O2)(g) -> 8CO2(g) + 9H2O(g)
STANDARD ENTHALPY VALUES (H):
02(g) = 0 kj/mol
CO2 (g) = -393.5 kj/mol
H20(g) = -241.8 kj/mol
H total = -5094 kJ
H of C8H18(g) is your unknown, so call that x.
use: H of reaction = H of products - H of reactants
so, -5094 kJ = [8(CO2) + 9(H2O)] - [x +12.5(O2)]
plug in your standard enthalpy values.
-5094kJ = [8(-393.5) + 9(-241.8)] - [X + 12.5(0)]
-5094 kJ = [-3148 + (-2176.2)] - [x + 0]
-5094 kJ = -5324.2 - x
add -5324.2 to -5094
to get +230.2 = -x
move the negative to the other side
and you get -230 kj/mol
that is the correct answer.