Question

how to calculate delta h for the reaction
2B(s)+3H2(g)arrow B2H6(g)given the following data:
2B(s)+3/2O2(g)arrowB2O3(s) deltaH=-1273kj
B2H6(g)+3O2(g)arrowB2O3(s)+3H2O(g) deltaH=-2035kj
H2(g)+1/2O2(g)arrowH2O(l) deltaH=-286kj
H2O(l)arrowH2O(g) deltaH=+44kj
Use equation 1, reverse equation 2, multiply equation 3 by 3 and multiply equation 4 by 3. Add them. Don't forget to reverse the sign of equations you reverse.
Why multiply equation 4 by 3?
Wouldn't there be a 3/2 left over?

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