# 2B5H9(l)+ 12 O2(g)= 5B2O3(s) + 9H2O(l) Calculate the kilojoules of heat released per gram of the compound reacted with oxygen. The standard enthalpy of formations of B5H9(l), B2O3(s), and H2O(l) are 73.2, -1271.94, and -285.83 kJ/mol, respectively ?

The enthalpy of reaction is -69.58 kJ per gram of ##”B”_5″H”_9##.

We have the following information.

##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″##

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