Answer:
1280.71 kJ/mol
Explanation:
First we need to determine the limit reagent in order to how many initial compound reacted and later and much heat that reacted quantity produced:
[tex]0.0180 g Fe*\frac{1 mole Fe}{55.845 g Fe} *\frac{1}{4 moles Reacting Fe} =0.00008058[/tex]
[tex]0.0100 g O2*\frac{1 mole O2}{31.98 g O2} *\frac{1}{3 moles Reacting O2} =0.00010423[/tex]
The limit reagent is Iron.
Now to determine how much heat was generated by the reaction of those 0.0180 g Fe we take into account that the calorimeter is adiabatic and the heat produced by the reaction is the heat absorbed by the calorimeter and the solution:
[tex]Q(Calorimeter and solution)=-Q(Rxn)[/tex]
C*ΔT=-ΔH(rxn)*n
Where C is the Heat Capacity of the entire calorimeter and the solution and n is the number of moles reacting:
Then we solve for n:
[tex]H=\frac{C * T}{n} =\frac{12.0 J/C*8.6 C}{0.00008058 mol Fe} = 1280.71 kJ/mol[/tex]
