chemistry precipitate

Question

A student mixes 45.0 mL of 3.50 M Pb(NO3)2(aq) with 20.0 mL of 0.00241 M NaI(aq) . 

How many moles of PbI2(s) precipitate from the resulting solution? The 𝐾sp of PbI2(s) is 9.8×10−9.

Answer 1

Pb(NO₃)₂(aq) + 2NaI(aq) ==> PbI₂(s) + 2NaNO₃(aq) ... balanced equation

moles Pb(NO₃)₂ = 0.045 L x 3.50 mol/L = 0.1575 mols

moles NaI =0.02 L x 0.00241 mol/L = 2.41x10^-5 mols

LIMITING reactant = NaI

Theoretical yield of PbI₂ = 2.41x10^-5 mol NaI x 1 mol PbI₂ / 2 mol NaI = 1.21x10^-5 mol PbI₂ formed

Final [PbI₂] = 1.21x10^-5 mol/0.065 L = 1.86x10^-4 M > Ksp = 9.8x10^-9 so ppt will form