Question

During an experiment, a student adds 1.23 g of CaO to 200.0 mL of 0.500 M HCl. The student observes a temperature increase of 5.10 °C. Assuming the solution\'s final volume is 200.0 mL, the density if 1.00 g/mL, and the heat capacity is 4.184 J/(g·°C), calculate the heat of the reaction, ΔHrxn.i know i need to use the equations q=mcdeltaT 
and delta H= q/n but i keep getting the wrong answer. also which one is the limiting reactant in the equation CaO(s) +2H+(aq)-->Ca^2+(aq)+H2(g)

MY WORK

Q=10*4.184*5.1=213.384
delta H=213.384/1.23=173.483

Answer 1

Where do you get the 10?
mass H2O = 200 g
sp.h. = 4.184
delta T = 5.10
CaO is the limiting reagent.

Answer 2

Where do you get the H2(g)? There is no H2 gas liberated.
CaO + 2HCl ==> CaCl2 + 2H2O
CaO + 2H^+ ==> H2O + Ca^2+