Question

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

What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO₃)₂ is mixed with 0.900 L 0.100 M NaI? Assume the reaction goes to completion.

mass of precipitate in grams:

Answer 1

moles of NaI reacting = M(NaI)*V(NaI)

= 0.100 M * 0.900 L

= 0.090 mol

From reaction,

mol of PbI2 formed = (1/2)*number of moles of NaI reacted

= (1/2)*0.090 mol

= 0.045 mol

Molar mass of PbI2,

MM = 1*MM(Pb) + 2*MM(I)

= 1*207.2 + 2*126.9

= 461 g/mol

use:

mass of PbI2,

m = number of mol * molar mass

= 4.5*10^-2 mol * 4.61*10^2 g/mol

= 20.75 g

Answer: 20.8 g