Question

What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.600 L of 0.100 M NaI? Assume the reaction goes to completion.

Answer 1

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

no of moles of NaI   = molarity * volume in L

                                = 0.1*0.6 = 0.06 moles

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

0.06 moles of NaI react with Pb(ClO3)2 to gives = 1*0.06/2 = 0.03 moles of PbI2

mass of PbI2 = no of moles * gram molar mass

                         = 0.03*461   = 13.83g of PbI2 >>>>answer