Question

Pentaborane-9,

Answer 1

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

Pentaborane ($B_{5}H_{9}$) is a colorless highly reactive liquid that will burst into flames when exposed to oxygen.the reaction is:

  $2B_{5}H_{9}(l) + 12O_{2} \rightarrow 5B_{2}O_{2}(s) + 9H_{2}O(l)$

Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.the standard enthalpy of formation $B_{5}H_{9}(l)$ , $B_{2}O_{3}(s)$, and $H_{2}O(l)$ are 73.2, -1271.94, and -285.83 kJ/mol, respectively.

Explanation:

As the given reaction is as follows.

   $2B_{5}H_{9}(l) + 12O_{2} \rightarrow 5B_{2}O_{2}(s) + 9H_{2}O(l)$

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

       $\Delta H = \sum H_{products} - \Delta H_{reactants}$

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

         $\Delta H = \sum H_{products} - \Delta H_{reactants}$

     = $5 \times (-1271.94 kJ/mol) + 9 \times (-285.83 kJ/mol) - 2 \times (73.2 kJ/mol) - 12(0)$

     = -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.

        $\frac{\Delta H}{2 \times \text{molar mass of pentaborane}}$      

         $\frac{-9078.59 kJ/mol}{2 \times 63.12 g/mol}$

                  = -71.915 kJ/g

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