Question

What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.600 L of 0.280 M NaI? Assume the reaction goes to completion.

Answer 1

Pb(ClO3)2 + 2NaI -------> PbI2 + 2NaClO3

Pb(ClO3)2 is exces reagent

no of moles of NaI =molarity * volume in L

                          = 0.28*0.6 = 0.168 moles

from balanced equation

2 moles of NaI react with Pb(ClO3)2 to form 1 mole of PbI2

0.168 moles of NaI react with Pb(ClO3)2 to form = 1*0.168/2 = 0.084 moles of PbI2

mass of PbI2 = no of moles * molar mass

                 = 0.084*461 = 38.724gm of PbI2