Question

Write the balanced equation for the reaction of aqueous Pb (ClO3)2 with aqueous Nal. Include phases

What mass of precipate will form if 1.50L of highly concentrated Pb(ClO3)2 is mixed with 0.400L of 0.110 M Nal?

Answer 1

The balanced chemical equation is : Pb(ClO₃)₂(aq) + 2 NaI(aq) PbI₂(s) + 2 NaClO₃ (aq)

Given volume of NaI is , V = 0.400 L

Molarity of NaI is ,M = 0.110 M

We know that MOlarity , M = number of moles / volume of solution in L

So number of moles of NaI is , n = M x V

                                                    = 0.110 M x 0.400 L

                                                    = 0.044 moles

Accordin g to the balanced equation ,

1 mole Pb(ClO₃)₂ of reacts with 2 moles of NaI produces 1 mole of solid PbI₂

Since we have to taken high concentration of Pb(ClO₃)₂ so it is the excess reactant & the limiting reactant is NaI.

The mass of PbI₂ formed depends upon NaI.

From the equation, 2 moles of NaI produces 1 mole of PbI₂

                              0.044 moles of NaI produces N mole of PbI₂

                    N = ( 0.044 x 1 ) / 2

                       = 0.022 moles of PbI₂

Molar mass of PbI₂ is = ( 1 x At.mass of Pb ) + ( 2 x At.mass of I )

                                   = (1x207.2) + ( 2 x 127) g/mol

                                   = 461.2 g/mol

So the mass of PbI₂ , m = number of moles x molar mass

                                       = 0.022 mol x 461.2 g/mol

                                       = 10.15 g

So the mass of the precipitate formed is 10.15 g