Determination of the enthelpy of combustion of Magnesium using Hess's law

[Sekminara]

Homework Statement

ok so I'm trying to find the change in enthalpy for this reaction:
Mg(s) + (1/2)O₂--> MgO(s)

I've done the lab to find the temperatures of certain reactions, these were:
9.70 Celsius for the reaction Mg(s) + 2HCl(aq)--> MgCl₂(aq) + H₂(l)

7.60 Celsius for the reaction MgO(s) + 2HCl --> MgCl₂(aq) + H₂O(l) reaction

(and we know the enthalpy of change for the reaction
H₂(g) + (1/2)O₂(g) --> H₂O(l) is 285.8 kJ)

Homework Equations

change in enthalpy = mass (100ml/1000ml) * specific heat (4.18) * change in temp (above)

The Attempt at a Solution

using the (above) equation i found that the enthalpy of change for the first reaction was -405 kJ per mol and the second was -318 kJ per mol.

so i built a Hess's law chart and found that [Mg(s) + 2HCl(aq)--> MgCl₂(aq) + H₂(l)] + [H₂(g) + (1/2)O₂(g) --> H₂] = [MgO(s) + 2HCl --> MgCl₂(aq) + H₂O(l)] + [Mg(s) + (1/2)O₂--> MgO(s)]

so basically the equation look like this: -405 + (-285.8) = -318 + x

therefore x should be -372.8 kJ per mol

but all the answers I've found on the net so far say that the enthalpy of combustion is approximately 602 kJ per mol

help?

 

[Borek]

-372.8 kJ/mol agrees with your experimental data, so if it doesn't agree with tables - you have to look for experimental error.

 

[Sekminara]

dang...ouch 38% error

well...i guess that's what you get when you use Styrofoam insulators...