Asked by Farah
                31. Carbon monoxide reacts with hydrogen gas to produce a mixture of methane, carbon dioxide, and water. (This mixture is known as substitute natural gas.)
4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(_)
Use the following thermochemical equations to determine the enthalpy change of the reaction.
C(graphite) + 2H2(g) → CH4(g) + 74.8 kJ
CO(g) + 12O2(g) → CO2(g) + 283.1 kJ
H2(g) + 12 O2(g) → H2O(g) + 241.8 kJ
C(graphite) + 12O2(g) → CO(g) + 110.5 kJ
H2O(l) + 44.0 kJ → H2O(g)
 
this is required to use the Hess's law, and the manipulation of the equations and mutiplying to get the coeeficients from the orginal question, however i am getting -641kJ...the answer is wrong..any suggestions of what to do..
below is my answer:
multiplied the 1st equation by 3
no change to the second equation
multiplied the 3rd equation by 2
mutiplied the 4th equation by 4 and flipped the equation
fifth equation multiplied by two and flipped
            
        4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(_)
Use the following thermochemical equations to determine the enthalpy change of the reaction.
C(graphite) + 2H2(g) → CH4(g) + 74.8 kJ
CO(g) + 12O2(g) → CO2(g) + 283.1 kJ
H2(g) + 12 O2(g) → H2O(g) + 241.8 kJ
C(graphite) + 12O2(g) → CO(g) + 110.5 kJ
H2O(l) + 44.0 kJ → H2O(g)
this is required to use the Hess's law, and the manipulation of the equations and mutiplying to get the coeeficients from the orginal question, however i am getting -641kJ...the answer is wrong..any suggestions of what to do..
below is my answer:
multiplied the 1st equation by 3
no change to the second equation
multiplied the 3rd equation by 2
mutiplied the 4th equation by 4 and flipped the equation
fifth equation multiplied by two and flipped
Answers
                    Answered by
            DrBob222
            
    It appears to me that the first equation x 3 gives 3C(graphite) and the fourth x 4 and reversed gives 4C(graphite) so those don't cancel.  I stopped there. 
    
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