Answer:
52,000 kJ
Step-by-step explanation:
The combustion reaction is the following:
2 C₂H₆(g) + 7 O₂(g) → 4CO₂(g) + 6H₂O(l) ΔHm= –3120 kJ
From the chemical reaction and the value of ΔH we can see that 2 moles of ethane gas (C₂H₆(g)) release 3120 kJ of heat energy. So, we have the conversion factor: 3120 kJ/2 moles C₂H₆.
To calculate how many kJ are released from the combustion of 1 kg of ethane, we have to first convert the mass to moles by using the molecular weight (MW) of ethane (C₂H₆):
MW(C₂H₆) = (12 g/mol x 2 C) + (1 g/mol x 6 H) = 30 g/mol
mass = 1 kg x 1000 g/1 kg = 1000 g
moles of C₂H₆= mass/MW = 1000 g/(30 g/mol) = 33.3 mol C₂H₆
Finally, we multiply the conversion factor by the number of moles of C₂H₆ in 1 kg:
heat energy released = 33.3 mol C₂H₆ x 3120 kJ/2 moles C₂H₆ = 52,000 kJ
Therefore, 51,000 kJ are released when 1 kg of ethane is burned.