[Morhas]

Ok, so I am trying to find the heat of combustion of Mg. (Mg +1/2 O2 -> MgO)

So I need to use hess law to determine it. I was given these 3 equations.

1.)   MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)   ∆Hr = 104.76KJ/mol (This value found experimentally)
2.)   Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)       ∆Hr = 374.36KJ/mol (This value found experimentally)
3.)   H2(g) + ½O2(g) → H2O(l)                        ∆Hr = -285.5 kJ/mol

So 2-1+3 will give us the equation we want.

However, 374.36 - 104.76 -285.5 = -15kJ/mol.

The actual value is supposed to be -601 kJ/mol

So if 2-1+3 = - 601kJ
       2-1+(-285.5) = - 601kJ
       2-1 = -315.5

Shouldn't the value of 1>2 since we're subtracting to get a negative number? 

Any help would be appreciated beyond all comprehension. It's 2:00am here.

[Morhas]

Almost 3:00 now. I need to hand this in at 8:50.

I really need help please. 

Edit

Getting very sleepy, and very desperate.

[Borek]

1.)   MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)   ∆Hr = 104.76KJ/mol (This value found experimentally)
2.)   Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)       ∆Hr = 374.36KJ/mol (This value found experimentally)

Are you sure signs here are correct? Were these reactions exothermic, or endothermic?

[Morhas]

In the first one the temperature increased from 22.5 to 28
And in the second it went from 23.5 to 41

So they should be exothermic as heat was released or is a product of the reaction. Unless I'm horribly mistaken, which I may be....

Thanks for pointing me in the right direction (twice ). I guess the values should probably be negative, but it doesn't seem logical to me at this point in time.

EDIT

It is now understood. Infinite thanks to you good sir.