Pentaborane-9,
, is a colorless, highly reactive liquid thatwill burst into flame
when exposed to oxygen. the reactionis:

calculate the kilojoules of heat released per gram of
thecompound reacted with oxygen. the standard enthalpy of formation
ofb5 h9 is 73.2 kj/mol.

Given question is incomplete. The complete question is as follows.

Pentaborane () is a colorless highly reactive liquid that will burst into flames when exposed to oxygen.the reaction is:

Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.the standard enthalpy of formation , , and are 73.2, -1271.94, and -285.83 kJ/mol, respectively.

Explanation:

As the given reaction is as follows.

Therefore, formula to calculate the heat energy released is as follows.

Hence, putting the given values into the above formula is as follows.

     =

     = -9078.59 kJ/mol

Since, 2 moles of Pentaborane reacts with oxygen. Therefore, heat of reaction for 2 moles of Pentaborane is calculated as follows.

                  = -71.915 kJ/g

Thus, we can conclude that heat released per gram of  the compound reacted with oxygen is 71.915 kJ/g.

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