B5h9(l) is a colorless liquid that will explode when exposed to oxygen. how much heat is released when 0.211 mol of b5h9 reacts with excess oxygen where the products are b2o3(s) and h2o(l). the standard enthalpy of formation of b5h9(l) is 73.2 kj/mol, the standard enthalpy of formation of b2o3(s) is -1272 kj/mol and that of h2o(l) is -285.4 kj/mol. express your answer in kj.

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

your answer is c) plasma

~*"auba14"*~

1 is the answer to this problem