Answer:
Heat transfer = 3564 Jolues
The same value
Step-by-step explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.