Answer : 49.45 kJ per gram.
Explanation:
Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.
Given :
Amount of heat released on combustion of butane = 2874 kJ/mol
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number
of particles.
1 mole of
weighs = 58.12 g
Thus we can say:
58.12 g of
on combustion releases = 2874 kJ
Thus 1 g of
on combustion releases =

Thus the heat of combustion of butane, in kJ per gram is 49.45.