Question

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

What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.300 L of 0.180 M NaI? Assume the reaction goes to completion.

Answer 1

The balanced equation is

Pb(ClO₃)₂ (aq) + 2NaI(aq) PbI₂ (s) + 2 NaClO₃ (aq)

Number of moles of NaI , n = Molarity x volume in L

                                           = 0.180 M x 0.300 L

                                           = 0.054 mol

From the balanced equation ,

2 moles of NaI produces 1 mole of PbI₂ as precipitate

0.054 moles of NaI produces M mole of PbI₂ as precipitate

M = ( 0.054 x 1 ) / 2

    = 0.027 moles

Molar mass of PbI₂ = At.mass of Pb + ( 2xAt.mass of I )

                                = 207 + ( 2x127)

                                = 461 g/mol

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

                                                 = 0.027 mol x 461 (g/mol)

                                                 = 12.4 g

Therefore the mass of precipitate formed is 12.4 g