Hess's law question?

Question

The combustion of ethane can be expressed through the equation below:

C2H6(g) + 7/2O2(g) ---> 2CO2(g) + 3H2O(g)

Using Hess’ law, prove that the formation reactions of ethane, carbon dioxide and water can be used to calculated the same reaction enthalpy as using products - reactant

Answer 1

C2H6(g) + 7/2O2(g) ---> 2CO2(g) + 3H2O(g) = TARGET EQUATION

(1) Formation of ethane: 2C + 3H2 ==> C2H6

(2) Formation of CO2: C + O2 ==> CO2

(3) Formation of H2O: H2 + 1/2O2 ==> H2O

Reverse (1): C2H6 ===> 2C + 3H2

Multiply (2) by 2: 2C + 2O2 ===> 2CO2

Multiply (3) by 3H2 + 3/2O2 ===> 3H2O

Add all three to get...

C2H6 + 2C + 2O2 + 3H2 + 3/2O2 ===> 2C + 3H2 + 2CO2 + 3H2O

Cancel like terms on opposite sides of equation and add like terms on same side, and end up with ...

C2H6 + 7/2O2 ===> 2CO2 + 3H2O which is the target equation proving you can use Hess' Law.