Answer:
ΔH of reaction is 62,3 kJ/mol
Explanation:
For the reaction:
HCl + NaOH → H₂O + NaCl
The heat transfer in the reaction (q) =
q = Specific heat × mass × ΔT
The mass is 40,6 mL + 71,3 mL × (1g/mL) = 111,9 g
And ΔT is 28,57 - 25,00 = 3,57°C
q = 4,184 J/gK × 111,9 g × 3,57°C = <em>1671 J</em>
The moles of NaOH are:
0,0406 L×0,659 M = 0,0268 moles
And the moles of HCl are:
0,0713 L ×0,659 M = 0,0470 moles
The moles of reaction are the moles of NaOH because are the limiting reactant. 0,0268 moles
ΔH of reaction is q/moles of reaction. Thus:
ΔH of reaction = 1671J/0,0268 moles = 62350 J/mol ≡ 62,3 kJ/mol
I hope it helps!
