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Calculate the DHrxn of

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 = [(6x -393.5)+(6x -285.840)] - [(1x -1260.04)+(6x0)] = -2552 KJ/mol
It is exothermic

I ma having problems with this question. I keep getting different answers. Is this one right?

6 answers

I think it is set up right but I don't get that answer. It is exothermic and the equation is balanced.
how would I enter it into my calculator? I've entered it multiple ways and gotten different answers for each
If you show each step perhaps I can figures out what you're doing wrong.
For example:
6*-393.5 = ?
6*-285.84 = ?
Total = ?

etc.
-2816 KJ/mol?
That's what I obtained.
Thank you. Sorry for all the questions.
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