Equation for the reaction
         Ba(OH)2 +
2 HCl → BaCl2 + 2 H2O
The barium hydroxide and hydrochloric acid are present in
stoichiometric amounts, so either can be considered the limiting
reactant:
Given- 60.0 mL of 0.320 M Ba(OH)2
             60.0
mL of 0.640 M HCl
Mol of H2O =Â Â
        Â
(60ml/1000 L) x (0.640 mol/L HCl) x (2 mol H2O / 2 mol HCl) =
0.0384 mol H2O
Now we can calculate heat gained by solution
          Â
q = mxs xt
          Â
   = (60.0g +60.0g ) x (4.186 J/g·°C x (26.06 -
21.70)°C
          Â
   = 2186.976 J
        ΔH for per mole
of H2O = (2186.976 J) / (0.0384 mol H2O) =56952.5 J
                                         Â
= 56.9525 kJ
