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?

No. mols CO2 = 114

mol H2O = 110
so dHrxn = [(114*-393.5) + (110*-285.84)] - [ etc.

-75520 exothermic?

-35120 exothermic?

Yes, you are on the right track in terms of using the enthalpy of formation (DHf) values to calculate the DHrxn (change in enthalpy of the reaction). However, it seems like there are a couple of errors in your calculations.

First, let's double-check the balanced equation:
2C57H110O6 (s) + 163O2(g) → 114CO2(g) + 110H2O(l)

Now, let's calculate the DHrxn using the DHf values:
DHrxn = (114 × DHf[CO2]) + (110 × DHf[H2O]) - (2 × DHf[C57H110O6]) - (163 × DHf[O2])

Plug in the DHf values:
DHrxn = (114 × (-393.5 kJ/mol)) + (110 × (-285.840 kJ/mol)) - (2 × (-390.70 kJ/mol)) - (163 × 0 kJ/mol)

Simplifying the equation:
DHrxn = (-44811 kJ/mol) - (-781.4 kJ/mol) - (-781.4 kJ/mol)
DHrxn = -44811 kJ/mol + 1562.8 kJ/mol + 1562.8 kJ/mol
DHrxn = -44811 kJ/mol + 3125.6 kJ/mol
DHrxn = -41685.4 kJ/mol

Therefore, the correct DHrxn for the reaction C57H110O6 (s) + O2(g) → CO2(g) + H2O(l) using the given DHf values is -41685.4 kJ/mol.

It seems like you made an error in the calculation when determining the moles involved in the reaction. Please check your calculations to find where the discrepancy occurred.