[Plato@space]

Calculate ΔH°com [Mg(s)] from the following thermochemical equations.

1. Mg(s) + 2HCl(aq) --> MgCl2 (s) +H2 (g)      ΔH°= -501 kJ
2. MgO(s) + 2HCl(aq) --> MgCl2 (s) +H2O (l)    ΔH°= -151 kJ
3.2H2(g) +O2(g) --> 2H2O(l)                         ΔH°= -572 kJ
I think the ans is:
Mg + O2  MgO
Mg(s) + 2HCl(aq) -> MgCl2 (s) +H2 (g)        ΔH°= -501 kJ
MgCl2 (s) +H2O (l) -> MgO(s) + 2HCl(aq)     ΔH°= 151 kJ
H2(g) +1/2O2(g) ->   H2O(l)                      ΔH°= -572 kJ /2 = -286

Since Product – Reactant,
(-286) + (-501) – 151 = -938kJ

BUT the Answer should be -636 kJ 
that means the equation should be (-286) + (-501) - (-151) = 636kJ
why? And the Hint is Hess's Law....How is this question related to Hess's Law?

[Yggdrasil]

The way you have set up the equations is correct.  All that you need to do, once you have the series of three equations is just to add the standard enthalpies of reaction:

(-501 kJ/mol ) + (151 kJ/mol) + (-286kJ/mol) = -636kJ/mol

The question is related to Hess's law in that both are based on the principle that you can add the standard enthalpies of reaction of a series of reactions to find the standard enthalpy of reaction of the overall reaction.

[Plato@space]

The way you have set up the equations is correct.  All that you need to do, once you have the series of three equations is just to add the standard enthalpies of reaction:

(-501 kJ/mol ) + (151 kJ/mol) + (-286kJ/mol) = -636kJ/mol

The question is related to Hess's law in that both are based on the principle that you can add the standard enthalpies of reaction of a series of reactions to find the standard enthalpy of reaction of the overall reaction.

Thank you for your reply. 
According to Hess’ law, for any reaction that can be written in a series of steps, the standard enthalpy change for the reaction is the same as the sum of the standard enthalpy changes of all the steps. How can I determine the steps are in series or not?
For example, if the question given
CH3OH(l)  ΔH°(f)= -239 kJ
CO2(g)    ΔH°= -394 kJ
H2O(l)    ΔH°= -286 kJ
Why Hess’s Law is not suitable in calculate the heat of combustion of CH3OH?
 I’m very confused about the meaning of “any reaction that can be written in a series of steps.”

[Yggdrasil]

You can apply Hess's law to determine the heat of combusion of methanol given the data you have.  To solve this problem, you first need to write out the chemical equation for the combustion of methanol and then write out the chemical equations that apply to the heats of formation of methanol, carbon dioxide and water.

[Plato@space]

You can apply Hess's law to determine the heat of combusion of methanol given the data you have.  To solve this problem, you first need to write out the chemical equation for the combustion of methanol and then write out the chemical equations that apply to the heats of formation of methanol, carbon dioxide and water.

I see, thank you very much! 

CO2 + 2H2O  -->  CH3OH + 3/2O2      ΔH° = 239
C  +  O2 ------->  CO2                    ΔH° = -394
2H2  +  O2 -----> 2H2O                   ΔH° = -286 x 2

239 + (-394) + (-286 x 2) = -727 kJ/mol

Do you mean Hess's Law can apply in all calculation of Enthalpy change ?
then when we should use the formula:  ΔH° = ΣΔH°(products) - ΣΔH°(reactants)