Question

How do I find the moles of solid used for MgO and Mg?

Data and Results Data Sheet Reaction 2 Reaction 1 Data 2H Volume of 1.00 MHC /00m Final teraperature,T? initial Demperature Change in temp. AT 20.5 Mass of solid, Heat 35.5 Moles of solid mol Mg mol AH, kJ/mole Equation Mgat hozo Mgok) Experimental enthalpy or formation, AHE The actual enthalpy of formation, AH e, found in Appendix of your textbook, is the enthalpy of formation for Mgo, since: Actual AHF Percentage Euror Experimental AH Actual Actual AH' x 100%

I had to calculate delta H and I wanted to know if I did it correctly

Answer 1

So as to get the moles of the solid, you must use the formula for moles:

n = mass/ Molecular Weight.

If you want the moles of Mg: n= 0.494g/ 24.305g/mol = 0.0203moles of Mg

Now for MgO: n= 0.494g/(24.305g/mol + 15.9994g/mol) = 0.0122 moles of MgO.

Then for the enthalpy:

So as to get the moles of the solid, you must use the formula for moles: n = mass/Molecular Weight. If you want the moles of Mg: n = 0.494g/ 24.305g/mol = 0.0203moles of Mg Now for MgO: n = 0.494 g/(24.305 g/mol + 15.9994g/mol) = 0.0122 moles of MgO. Then for the enthalpy: (1) Mgo + 2 Hcl rightarrow MgCl_+ 2 + H_2 o invert Delta H_1 (2) Mg + 2 Hcl rightarrow Mgcl_2 + H_2 Delta H_2 (3) H_2 + 1/2 O_2 rightarrow H_2 o (4) Mg + 1/2 O_2 rightarrow Mgo

As you inverted reaction 1, your enthalpy must be inverted as well, by multiplying times -1. To get your enthalpy for reaction 4: -\delta H1 + \delta H2 + \delta H3 = \delta H4.
The first two enthalpies are those that you already calculated, whilst the third is the one for water formation, found in tables. (-285.8kJ/mol) .