Respuesta :
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 [tex]6.023\times 10^{23}[/tex] of particles.
1 mole of [tex]C_4H_{10}[/tex] weighs = 58.12 g
Thus we can say:
58.12 g of [tex]C_4H_{10}[/tex] on combustion releases = 2874 kJ
Thus 1 g of [tex]C_4H_{10}[/tex] on combustion releases =[tex]\frac{2874}{58.12}\times 1=49.45 kJ[/tex]
Thus the heat of combustion of butane, in kJ per gram is 49.45.
