When an excess of CaO is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 4.01C is observed.

Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
Delta Hrxn, for the reaction of

CaO(s) + 2H+(aq) Ca2+(aq) + H2O(l)

Here's what I did but it isn't right and I don't know what to do anymore..

-q=mcDeltaT
-q = (200)(4.184)(4.01) = 3355.568
(-3355.568)(56.0778)/1000 = -188.17 kJ/mol

3355.568 J is heat generated. You used 0.2L x 0.500M = 0.1 mol HCl which is 0.05 mol CaO. I would divide 3355.568 J/0.05 mol CaO and round for significant figures.