Calculate the DHrxn of

Question

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 = [(114x -393.5)+(110x -285.840)] - [(2x -390.7)+(163x0)] = -35120KJ/mol
It is exothermic

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

Answer 1

The equation isn't balanced. You have (163*2)+(2*6) = 338 O on the left and (114*2) + 110*2) = 448 on the right. You can make it balance with 55 H2O and not 110 H2O.

Answer 2

If I use 55 for H2O, then the H isn't balanced

Answer 3

Yes it is.
2C57H110O6 (s) + 163O2(g) --> 114CO2(g) + 55H2O(l)

You have 110 H on the left and 110 on the right.

Answer 4

55x2 = 110 on the right
but have 220 on the left

Answer 5

I am really embarrassed. I completely ignored that coefficient of 2 for the first material. Ok so the equation is ok. Let me recalculate the dH.

Answer 6

For this one I actually think it's -75520 KJ/mol

Answer 7

Also, could you please check my question with 578 g of C57H110O6 on page two?

Answer 8

I don't get 35,120. I get 75,520 for the balanced equation. If you take the question literally you divide by 2 but that gives my number still higher. I normally don't divide by 2 UNLESS the problem states that they want kJ/mol but I don't know the context of your questions so you may be right to divide by 2. However, we still differ by about 2600.

Answer 9

I don't understand the dividing by 2. Could you explain more?