0.8100g MgO(s)(40.3g/mol) was reacted with 52.3 ml 1.00M HCl(density= 1.00g/ml) in the coffee cup calorimeter, and the observed temperature change was +10.4 degree Celsius. write the equation for this reaction, determine the limiting reactant and find difference H for the reaction in kj/mol.

Answer 1

I'm not sure what you mean by "diffference in H" but I'll assume you mean to find delta H rxn in kJ/mol.
MgO + 2HCl ==> H2O + MgCl2
mols MgO = 0.8100/40.3 = about 0.02 but you may want more accuracy than that.(meaning you do it too)
mols HCl = M x L = about 0.052 (again, you do it more accurately).
Convert mols MgO to mols MgCl2. That's 1:1 ratio; therefore, mols MgCl2 = 0.02.
mols HCl converted to mols MgCl2 = 0.052/2 = about 0.026. Therefore, MgO is the limiting reagent.
q = mass H2O x specific heat H2O x dT.
q = 52.3g x 4.184 x 10.4 = about 2276 J/0.02 mol = ? J/mol and convert to kJ/mol