Answer to Question #77030 in General Chemistry for Nishat Nahar

Question

What mass of precipitate will form if 1.50 L of excess Pb(ClO3)2 is mixed with 0.500 L of 0.180 M NaI? Assume 100% yield and neglect the slight solubility of PbI2.

Answer 1

Answer: Pb(ClO3)2 + 2 NaI → PbI2 + 2 NaClO3

Clearly Pb(ClO3)2 is in excess, leaving NaI to be the limiting reactant.

(0.500 L) x (0.180 mol/L NaI) x (1 mol PbI2 / 2 mol NaI) x (461.0189 g PbI2/mol) = 20.745 g PbI2