How many grams of yellow PbI2 precipitate (MM= 461.0 g/mol) should form by mixing together 35 mL of 0.27 M Pb(NO3)2 and 50 mL of 0.36 M KI?

Answer 1

First write a balanced chemical equation for the reaction:

Pb(NO3)2 + 2KI ---> 2PbI + 2 KNO3

Then determine the amount of moles of each reactant. This is done by multiplying the volume x the molarity.

.035 l (.27M) = .00945 moles of Pb(NO3)2

.050 l (.36M) = .018 moles KI the limiting reactant is KI therefore KI and PBI are in a 1 to 1 ratio meaning you would produce .018 moles of PbI

Convert to grams by multiplying by molar mass:

.018 moles ( 461g/mol) = 8.3 grams PbI ( 2 sig figs due to the volumes)