I. C(s) + O2(g) → CO2(g)

ΔHf = −394 kJ

II. H2(g) + O2(g) → H2O(l)
ΔHf = −242 kJ

III. 2C(s) + 3H2(g) → C2H6(g)
ΔH =−84 kJ

The combustion of C2H6 is shown by the following equation:

C2H6(g) + O2(g) → 2CO2(g) + 3H2O(l)

Which option correctly gives the enthalpy of combustion of 0.2 moles of C2H6(g)?

I came up with -286 kJ, is this correct?

In the future please show how you cam up with this answer. Yes, -286 is correct but I spent considerable time putting all of this together to work the problem and to check your answer. I could hve checked your equatiions much faster and helped others.

To determine the enthalpy of combustion of 0.2 moles of C2H6(g), we can use the given enthalpy values and apply the stoichiometry of the reaction.

Given:
ΔHf of CO2(g) = -394 kJ
ΔHf of H2O(l) = -242 kJ
ΔH of the reaction 2C(s) + 3H2(g) → C2H6(g) = -84 kJ

We can break down the combustion of C2H6(g) into individual steps using the given reactions and their stoichiometric coefficients:

1. Multiply Reaction I by 2 to obtain 2CO2(g):
2C(s) + 2O2(g) → 2CO2(g)
ΔH = 2(-394 kJ) = -788 kJ

2. Multiply Reaction II by 3 to obtain 3H2O(l):
3H2(g) + 3O2(g) → 3H2O(l)
ΔH = 3(-242 kJ) = -726 kJ

3. Add Reaction I and II to get the formation of CO2(g) and H2O(l):
2C(s) + 2O2(g) + 3H2(g) + 3O2(g) → 2CO2(g) + 3H2O(l)
ΔH = -788 kJ + (-726 kJ) = -1,514 kJ

4. Subtract Reaction III to cancel out the formation of C2H6(g):
2C(s) + 2O2(g) + 3H2(g) + 3O2(g) - 2C(s) - 3H2(g) → 2CO2(g) + 3H2O(l)
ΔH = -1,514 kJ - (-84 kJ) = -1,430 kJ

5. Divide the enthalpy change by the stoichiometric coefficient of C2H6(g) to calculate the enthalpy of combustion of 1 mole:
ΔHcombustion = -1,430 kJ / 2 = -715 kJ

Finally, multiply the enthalpy of combustion by the number of moles to get the enthalpy of combustion of 0.2 moles:
ΔHcombustion (0.2 moles) = -715 kJ * 0.2 = -143 kJ

Therefore, the correct enthalpy of combustion for 0.2 moles of C2H6(g) is -143 kJ, not -286 kJ.