Oh. I didn't know that sorry.

Well I believe that the equation of the combustion of propane is :
C3h8 + 5O2 = 3Co2 + 4H20
I think I have to find the heat of the propane combustion first and it would be (according to my calculs, so still unsure) :
2,312kj/C° x 10,0 C° = -23,12KJ
then :
0,500/44,11 = 0,0113 mol
so :
? KJ/ 1mol = -23,12KJ/0,0113mol = 23,12 x 1mol/0,0113mol = -2046,02KJ
But then I'm totally lost, how do I integrate the 3kg factor in this equation? to find the energy produced.
