Part i. (10 points) when 50.0 m L of 0.400 M Ca(OH)2(aq) is added to 50.0 mL of
ID: 1072817 • Letter: P
Question
Part i. (10 points) when 50.0 m L of 0.400 M Ca(OH)2(aq) is added to 50.0 mL of 0.400 M H3PO4(aq), a precipitate of Ca3(POJ2(s) forms. The balanced equation is below. The initial temperature of both solutions is 25.0°C. After the reaction the final solution 6.30C. If the total mass of solution is 100.0 g and the specific heat of the 63°C lft what m H for the reaction in ganothe specific heat of the temperature is 2 final solution is 4.18 J/g-°C, what is H for the reaction in kJ/mol? 2 H3PO4(aq) + 3 Ca(OH)2(aq)-> Ca3(PO4)2(s) + 6 H20(1)Explanation / Answer
Q = m x c x Delta T
m = 100 gm
c= 4.18 j /gm 0C
Delta T = 26.3-25 1.3 0C
Q = 100 x 4.18 x 1.3 = 543.4 Joules
From the given concentration Ca(OH)2 is the limiting reagent
The product formed in moles = 50 x0.4 / 3 x 1000 = 0.006667 Moles
Hence enthalpy of reaction = 543.4 / 0.006667 = 81509 Joules or 81.51 Kilo Joules