Having some issues with this. "What is the standard enthalpy of combustion of C2H6 in kj/mol? H2(g) + 1/2O2(g) --> H2O(l) H0= –286 kJ C2H4(g) + H2(g) --> C2H6(g) H0= –137 kJ C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(l) H0= –1412 kJ" Now, the answer is presumedly -31561 kj. But I'm getting no where a number that close. Since C2H6 is combusted, I want the end-result to be: (2)C2H6 + (7)O2 --> (6)H2O + (4)CO2 (right?) So I would turn around the 2nd one to: C2H6(g) --> C2H4(g) + H2(g) and then: 2 * [C2H6(g) --> C2H4(g) + H2(g) ] 2 * [H2(g) + (1/2)O2 --> H2O(l)] 2 * [C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(l)] Which gets me 3122 kJ, or 1561kj/mol. Did I do something wrong? Miss a step? Where could the other 30,000 kj come from?