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Pentaborane B5H9 (s) burns vigorously in O2 to give B2O3 and H2O.

Pentaborane B5H9 (s) burns vigorously in O2 to give B2O3 and H2O. Calculate H*rxn for the combustion of 1 mol of B5H9.
H*f [B2O3(s)] = -1273.5 kJ/mol
H*f [B5H9(s)] = 73.2 kJ/mol
H*f [H2O(l)] = -285.8 kJ/mol

I believe the answer is -4543 kJ/mol , but I don't know how to get it.
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Answer #1

5x-127 3.5+98-285,8)-(2x732+,2x0 6367.S2572.2-146-4 +O = .9086, 1 kJ a2

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Answer #2

B5H9 + 6 O2 => 5/2 B2O3 + 9/2 H2O

ΔH(combustion) = ΔHf(products) - ΔH(reactants)

= 5/2 x ΔHf(B2O3) + 9/2 x ΔHf(H2O) - ΔHf(B5H9) - 6 x ΔHf(O2)

= 5/2 x (-1273.5) + 9/2 x (-285.8) - (73.2) - 6 x (0.0)

= -4543.05 kJ ≈ -4543.1 kJ

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