Calculate the DHrxn of

Question

C6H12O6 (s) + O2 (g) --> CO2 (g) + H2O (l)
with DHf.
C6H12O6(s) -1260.04
O2(g) 0
CO2(g) -393.5
H2O(l) -285.840

I balanced the equation:
C6H12O6 (s) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)
and my work:
DHrxn = [12mol(-393.5+(-285.840))] - [7mol(-1260.04+0)] = 668.2 KJ/mol
It is endothermic

Is this right?

Answer 1

Go back to the C57 compound and see how I did that. You are using TOTAL mols and that isn't the right way to do it.

Answer 2

-2816 Exothermic?

Answer 3

or -2551.916?