Hess's Law Problem. PLEASE HELP

Question

Use Hess’s Law to calculate the enthalpy of combustion for ethane gas, C2H6 given the following data:

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g) ΔH = -1323 kJ/mole

C2H4 (g) + H2 (g) → C2H6 (g) ΔH = -137 kJ/mole

for H2O (g) ΔHOf = -241.8 kJ/mole

H2O (l) → H2O (g) ΔH = +44

Answer 1

2C2H6 + 7O2 ==> 4CO2 +6H2O  This is the equation for combustion of ethane

Reverse the second equation and multiply by 2:

2C2H6 ==> 2C2H4 + 2H2  ∆H = 2x-137 = -274 kJ

Multiply first equation by 2:

2C2H4 + 6O2 ==> 4CO2 + 4H2O  ∆H = 2x-1323 = -2646 kJ

Add these 2 together to get:

2C2H6 + 2C2H4 + 6O2 ==> 2C2H4 + 2H2 + 4CO2 + 4H2O  ∆H = -2929 kJ

C2H4 cancel leaving 2C2H6 + 6O2 ==> 4CO2 + 4H2O + 2H2

Note that we need an additional O on the left and we need to cancel 2H2 on the right.

Using H2 + 1/2O2 ==> H2O  ∆H = -241.8 kJ.  Multiply this by 2 and add it to the final equation above:

2C2H6 + 6O2 ==> 4CO2 + 4H2O + 2H2  ∆H = -2929 kJ

2H2 + O2 ==> 2H2O  ∆H = 2x-241.8 = -483.6

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2C2H6 + 7O2 ==> 4CO2 + 6H2O  ∆H = -483.6 + -2929 = -3412.6 kJ