We have the following information.
![enter image source here]()
ΔH_"rxn"^° = ΣΔH_"f"^°("p") - ΣΔH_f^°("r"),
where "p" = products and "r" = reactants.
ΣΔH_"f"^°("p") = (5 cancel("mol B"_2"O"_3) × "-1271.94 kJ"/(1cancel("mol B"_2"O"_3))) + (9 cancel("mol H"_2"O") × "-285.83 kJ"/(1cancel("mol H"_2"O"))) = "-8932.17 kJ"
ΣΔH_f^°("r") = (2 cancel("mol B"_5"H"_9) ×"73.2 kJ"/(1cancel("mol B"_5"H"_9))) + (12 cancel("mol O"_2) × "0 kJ"/(1cancel("mol O"_2))) = "146.4 kJ"
ΔH_"rxn"^° = ΣΔH_"f"^°("p") - ΣΔH_f^°("r") = "(-8932.17 - 146.4) kJ" = "-9078.57 kJ",
But this is the energy released by 2 mol of "B"_5"H"_9.
2 cancel("mol B"_5"H"_9) × ("63.13 g B"_5"H"_9)/(1 cancel("mol B"_5"H"_9)) = "126.26 g B"_5"H"_9
So ΔH_"rxn"^"o" = "-9078.57 kJ"/("126.26 g B"_5"H"_9) = "-71.90 kJ/g B"_5"H"_9"
