What mass of precipitate will form if 1.50 L of concentrated Pb ( ClO 3)2  is mixed with 0.750 L of 0.150 M NAI? Assume the reaction goes to completion.

Answer 1

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

Assuming Pb(ClO₃)₂ is in excess and NaI is limiting...

moles NaI present = 0.750 L x 0.150 mol / L = 0.1125 moles NaI

moles PbI₂ formed = 0.1125 mols NaI x 1 mol PbI₂ / 2 mols NaI = 0.05625 moles PbI₂ formed

molar mass PbI₂ = 461 g / mole

grams PbI₂ formed = 0.05625 moles x 46 1 g / mole = 25.9 g PbI₂