The reaction is:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
Based on stoichiometry:
1 mole of Ba(OH)2 reacts with 2 moles of HCl to form 2 moles of H2O
Based on the given data:
# moles of Ba(OH)2 present = 0.050 L * 0.320 moles/L = 0.016 moles
# moles of HCl present = 0.050 L * 0.640 moles/L = 0.032 moles
Thus, in accordance with the 1:2 (Ba(OH)2:HCl) stoichiometry, all reactants would combine to form 0.032 moles of H2O
Now,
total volume of solution = 50 + 50 = 100 ml
Since the density = density of water = 1 g/ml
Total mass, m = 100 ml* 1 g.ml-1 = 100 g
The enthalpy change or heat of the reaction is the energy released (Q),
Q = -mc(T2-T1) = 100 * 4.18*(28.12-23.76) = -1822.48 J
Therefore, enthalpy change per mole of H2O =
= -1822.48 J/0.0320 moles = -56952.5 J/mol
Ans: ΔH/mole of H2O = -56.95 kJ/mol
