12. When 50.0 mL of 1.00 M Pb(NO3)2 (d = 1.30 g/mL) and 50.0 mL of 1.50 M KI (d

Question

12. When 50.0 mL of 1.00 M Pb(NO3)2 (d = 1.30 g/mL) and 50.0 mL of 1.50 M KI (d = 1.18 g/mL) are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture Increases from 19.0 degree C to 23.3 degree C. (a) Indicate If the precipitation reaction Is endothermic or exothermic. (b) Write both the complete and net ionic chemical equations, Including phase symbols, for the reaction that occurs when the two solutions are mixed. complete: net Ionic: (c) Calculate the moles of the limiting reactant. (d) Calculate AH rxn in kilojoules for the precipitation reaction, assuming that the heat capacity of the calorimeter is 15 J/degree C in this case and the specific heat capacity of the solution after mixing Is the same as that of water.

Explanation / Answer

(a) reaction is endothermic

moles of Pb(NO3)2 = 1 x 50 / 1000 = 0.05

moles of KI = 50 x 1.5 / 1000 = 0.075

(b) complete reaction :

Pb(NO3)2   + 2 KI -----------------------------> PbI2 + 2 KNO3

net ionic :

Pb+2 (aq) + 2 I- (aq) ---------------------------> PbI2 (s)

(c) limiting reagent is KI. and KI moles = 0.075

explanation : Pb(NO3)2   + 2 KI -----------------------------> PbI2 + 2 KNO3

1 mole Pb(NO3)2   = 2 moles KI

0.05 moles Pb(NO3)2 needs 0.1 moles KI

but we have only 0.075 moles of KI

so limiting reagent is KI

(d) moles of precipitate formed = 0.075

          PbI2 moles = 0.075

           weight /461 = 0.075

         weight= 34.56 g

Q = m Cp dT

    detla Hrxn = m Cp dT

                      = 34.56 x 15 x (23.3-19)

                      =2230 J

                     = 2.23 kJ

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