Answer:
Heat realesed is - 2192.7 kJ
Explanation:
First lets have the chemical equation balanced, and then solve the question based on the fact that the enthalpy change for a reaction is the sum of the enthalpies of formation of products minus reactants.
B₅H₉ (l) + O₂ (g) ⇒ B₂O₃ (s) + H₂O(l)
B atoms we have 5 reactants and 2 products, so the common mltiple is 10 and we have
2 B₅H₉ (l) + O₂ (g) ⇒ 5 B₂O₃ (s) + H₂O(l)
Now balance H by multiplying by 9 the H₂O
2 B₅H₉ (l) + O₂ (g) ⇒ 5 B₂O₃ (s) + 9 H₂O(l)
Finally, balance the O since we have 24 in products by multiplying by 12 the
O₂ ,
2 B₅H₉ (l) + 12 O₂ (g) ⇒ 5 B₂O₃ (s) + 9 H₂O
ΔHrxn = 5 x ΔHºf B₂O₃ + 9 x ΔHºf H₂O - ( 2 x ΔHºf B₅H₉ + 12 x ΔHºf O₂ )
We have all the ΔHºf s except oxygen but remember the enthalpy of formation of a pure element in its standard estate is cero.
ΔHrxn = 5 mol x ( -1272 kJ/mol )+ 9 mol x ( -285.4 kJ/mol ) - ( 2mol x 732 kJ/mol )
= -8928.6 kJ - 1464 kJ = -10,392 kJ
Now this enthalpy change was based in 2 mol reacted according to the balanced equation, so for 0.211 mol of B₅H₉ we will have:
-10,392 kJ/ 2 mol B₅H₉ x 0.211 mol B₅H₉ = - 2192.7 kJ
