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.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.