CH4(g) + 2O2(g) => CO2(g) + 2H2O(l) + Q
ΔfH°=ΣnpHp-ΣnrHr
ΛfH°=[(ΔfH°CO2(g)+2ΔfH°H2O(l))-(ΔfH°CH4(g)+2ΔfH°O2(g))] = (-393,5)+2(-241,8) - ((-74,8)+2*0) = -877,1+74,8=802,3 Kj/mol
1 mol CH4.....................802,3 Kj
5 moli CH4......................x Kj
x=802,3*5=