what mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.100 L 0.240 M NaI assume reaction goes to completion

Answer 1

Write a balanced equation for the reaction:

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

Since the Pb(ClO₃)₂ is highly concentrated, we will assume it is in excess and the NaI is limiting.

moles NaI = 0.100 L x 0.240 mol / L = 0.0240 mols NaI

moles PbI2 formed = 0.0240 mols NaI x 1 mol PbI₂ / 2 mol NaI = 0.0120 mols PbI₂

mass PbI₂ formed = 0.0120 mols PbI₂ x 461 g / mol = 5.53 g PbI₂