How much energy will be liberated upon the combustion of 12.5 grams of ethene?
C2H4(g) + 3O2(g) ----> 2CO2(g) + 2H2O(g)
Delta H = -1387.4 kj
q = 1387.4 kJ x (12.5 g/26 g) = ? kJ liberated
To calculate the amount of energy liberated upon the combustion of ethene (C2H4), you will need to use the equation you provided and the given enthalpy change (ΔH). The molar mass of ethene is required to convert the given mass of ethene into moles.
1. Calculate the number of moles of C2H4:
- Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
- Moles of C2H4 = Mass / Molar mass = 12.5 g / 28.05 g/mol
2. Use the balanced chemical equation to determine the molar ratio between C2H4 and ΔH:
- According to the equation:
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(g)
- From the equation, you can see that the stoichiometric coefficient of C2H4 is 1, and the stoichiometric coefficient of ΔH is -1387.4 kJ.
3. Calculate the energy liberated:
- Energy liberated = Moles of C2H4 x ΔH
- Calculate: Energy liberated = Moles of C2H4 x ΔH = (12.5 g / 28.05 g/mol) x -1387.4 kJ/mol
By following these steps, you will be able to determine the amount of energy liberated upon the combustion of 12.5 grams of ethene.