Calculate the DHrxn of

C57H110O6 (s) + O2(g) --> CO2(g) + H2O(l)
with DHf.
C57H110O6(s) -390.70
O2(g) 0
CO2(g) -393.5
H2O(l) -285.840

I balanced the equation:
2C57H110O6 (s) + 163O2(g) --> 114CO2(g) + 110H2O(l)
and my work:
DHrxn = [224mol(-393.5+(-285.840))] - [165mol(-390.70+0)] = -87706.66 KJ/mol

I posted this early without the balanced equation written, but my work is the same. What am I doing wrong with this? You add the moles, right?

3 answers

No. mols CO2 = 114
mol H2O = 110
so dHrxn = [(114*-393.5) + (110*-285.84)] - [ etc.

-75520 exothermic?

-35120 exothermic?