[nikita]

Calculate the enthalpy of the reaction
4B(s) + 3O₂(g)  2B₂O₃

given the following pertinent information:
A. B₂O₃(s) + 3H₂O(g) -> 3O₂(g) + B₂H₆(g)  :delta: H = 2035kJ
B. 2B(s) + 3H₂(g) -> B₂H₆(g)  :delta: H = 36kJ
C. H₂(g) + 1/2O₂(g) -> H₂O(l)  :delta: H = -285kJ
D. H₂O(l) -> H₂O(g)  :delta: H = 44kJ

Express your answer numerically in kilojoules per mole.

i just want to make sure I am doing this correctly.  I am going to write what the equations become and the new  :delta:H.

A.  multiply all by -2:
     6O₂+2B₂H₆ 2B₂O₃+ 6H₂O   :delta: H  -4070kj
B.  multiply all by 2:
      4B+6H₂ 2B₂H₆  :delta: H +72kj
C.  multiply all by -6:
      6H₂0 6H₂+3O₂   :delta: H +1710
D.  leave as is
       H₂O(l)  H₂O(g)  :delta: H +44

IMO everything cancels except the original equation, 4B + 3O₂  2B₂O₃ . I am not getting the correct answer.  Have I been looking at this so long that I am missing something obvious?  I am starting to not be able to do it anymore, but Ive done it several times and always come out with -2244kj.  Any insight as to what I am doing wrong?

[enahs]

You have to reverse D and multiply times 6 as well, other wise your waters do not cancel.

[nikita]

I cannot believe that I can look at a problem that many times and not see the liquid and the gas.  thank you, i needed some fresh eyes.