The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is:

2C(s)+2H2O(g)--->CH4(g)+CO2(g)
Determine the standard enthalpy change for this reactionf rom the following standard enthalpies of reactions:

C(s)+H2O(g)--->CO(g)+H2(g) ΔH=131.3 kJ
CO(g)+H2O(g)--->CO2(g)+H2(g) ΔH=-41.2 kJ
CH4(g)+H2O(g)--->3H2(g)+CO(g) ΔH=206.1 kJ

ΔH = 15.3kJ

Explanation:

It is possible to obtain standard enthalpy change of a reaction from the sum of half-reactions (Hess's law). Thus:

(1) C(s) + H₂O(g) ⇄ CO(g) + H₂(g) ΔH = 131.3kJ

(2) CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g) ΔH = -41.2kJ

(3) CH₄(g) + H₂O(g) ⇄ 3H₂(g) + CO(g) ΔH = 206.1kJ

The sum of 2×(1) + (2) gives:

2C(s) + 3H₂O(g) ⇄ CO₂(g) + 3H₂(g) + CO(g) ΔH = 131.3kJ×2 + -41.2kJ  = 221.4kJ

Now, the sum of this reaction with -(3) gives:

2C(s) + 2H₂O(g) ⇄ CH₄(g) + CO₂(g) ΔH = 221.4kJ - 206.1kJ = 15.3kJ

I hope it helps!

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"