Chemistry class

Question

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is

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

What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.850 L of 0.170 M NaI? Assume the reaction goes to completion.

Answer 1

Pb(ClO₃)₂(aq) + 2NaI(aq)==> PbI₂(s) + 2NaClO₃(aq) ... balanced equation

Assuming that Pb(ClO3)2 is in excess (concentrated solution), NaI will be the limiting reactant.

moles NaI present = 0.850 L x 0.170 moles/L = 0.1445 moles NaI

moles PbI₂ formed = 0.1445 mol NaI x 1 mol PbI₂ / 2 moles NaI = 0.07225 moles PbI₂ formed

mass PbI₂ formed = 0.07225 moles PbI₂ x 461 g/mol = 33.3 g PbI₂ formed