Question

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is shown below. What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.200 L of 0.190 M NaI? Assume the reaction goes to completion.

Answer 1

The reaction is:
Pb(ClO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaClO3 (aq)

Since Pb(ClO3)2 is fully concentrated, it will not be limiting reagent

Number of moles of NaI added = M*V
       = 0.19 M * 0.2 L
       = 0.038 mol

2 mol of NaI forms 1 mol of PbI2

So,
moles of PbI2 formed = 0.5* number of moles of NaI
   = 0.5*0.038
   = 0.019 mol

Molar mass of PbI2= 461 g/mol

So,
mass of PbI2 formed = molar mass * number of moles
                                           = 461 g/mol * 0.019 mol
                                            = 8.76 g
Answer: 8.76 g