Respuesta :
Given question is incomplete. The complete question is as follows.
Pentaborane ([tex]B_{5}H_{9}[/tex]) is a colorless highly reactive liquid that will burst into flames when exposed to oxygen.the reaction is:
[tex]2B_{5}H_{9}(l) + 12O_{2} \rightarrow 5B_{2}O_{2}(s) + 9H_{2}O(l)[/tex]
Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.the standard enthalpy of formation [tex]B_{5}H_{9}(l)[/tex] , [tex]B_{2}O_{3}(s)[/tex], and [tex]H_{2}O(l)[/tex] are 73.2, -1271.94, and -285.83 kJ/mol, respectively.
Explanation:
As the given reaction is as follows.
[tex]2B_{5}H_{9}(l) + 12O_{2} \rightarrow 5B_{2}O_{2}(s) + 9H_{2}O(l)[/tex]
Therefore, formula to calculate the heat energy released is as follows.
[tex]\Delta H = \sum H_{products} - \Delta H_{reactants}[/tex]
Hence, putting the given values into the above formula is as follows.
[tex]\Delta H = \sum H_{products} - \Delta H_{reactants}[/tex]
= [tex]5 \times (-1271.94 kJ/mol) + 9 \times (-285.83 kJ/mol) - 2 \times (73.2 kJ/mol) - 12(0)[/tex]
= -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.
[tex]\frac{\Delta H}{2 \times \text{molar mass of pentaborane}}[/tex]
[tex]\frac{-9078.59 kJ/mol}{2 \times 63.12 g/mol}[/tex]
= -71.915 kJ/g
Thus, we can conclude that heat released per gram of the compound reacted with oxygen is 71.915 kJ/g.
