Calculate the enthalpy of the reaction

4B(s) + 3O

_{2}(g)

2B

_{2}O

_{3}given the following pertinent information:

A. B

_{2}O

_{3}(s) + 3H

_{2}O(g) -> 3O

_{2}(g) + B

_{2}H

_{6}(g) :delta: H = 2035kJ

B. 2B(s) + 3H

_{2}(g) -> B

_{2}H

_{6}(g) :delta: H = 36kJ

C. H

_{2}(g) + 1/2O

_{2}(g) -> H

_{2}O(l) :delta: H = -285kJ

D. H

_{2}O(l) -> H

_{2}O(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:

6O

_{2}+2B

_{2}H

_{6} 2B

_{2}O

_{3}+ 6H

_{2}O :delta: H -4070kj

B. multiply all by 2:

4B+6H

_{2} 2B

_{2}H

_{6} :delta: H +72kj

C. multiply all by -6:

6H

_{2}0

6H

_{2}+3O

_{2} :delta: H +1710

D. leave as is

H

_{2}O(l)

H

_{2}O(g) :delta: H +44

IMO everything cancels except the original equation, 4B + 3O

_{2} 2B

_{2}O

_{3} . 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?