Question

What mass of precipitate will form if 1.50 L of excess Pb(ClO3)2 is mixed with 0.250 L of 0.250 M NaI? Assume 100% yield and neglect the slight solubility of PbI2.

The balanced equation for the reaction of aqueous Pb(ClO3), with aqueous Nal is shown below. Pb(ClO3)2(aq) + 2 Nal(aq) — Pbl_(s) + 2 NaClOz (aq) What mass of precipitate will form if 1.50 L of excess Pb(ClO ), is mixed with 0.250 L of 0.250 M Nal? Assume 100% yield and neglect the slight solubility of Pbl. mass: 14.40 g

Answer 1

Ans: Given balanced chemical equation is Pb(ClO₂)₂ + 2 NaI - PI 1 + 2 Nacho, (aq) Imole amoles Imole : 2 moles PbClO₂ is the excess reagent. NaI is the limiting reagent (the first completely wed up beagent in the reaction) and it can only decide the amount of precipitate p b I formed. - Provided 100% yield of PbI a moles of Nal is stoichiometrically equal to 1 mole of Pb I Molarity (M) = Number of moles (n) volume of schntion (V) =) na = M NaIt solution (Nat) M X v Na Hal = 0.250 M X 0.250 L = n wat = 0.0625 moles. zmoles of Nal produces Imole of Pb I. 0,0625 moles of Nal ? 0.0625 moles x 1 mote = 0.03125 moles of PbI 2 motes weight of PbI = Number of moles of Pb I x Molarmass of Phi e = 0.03125 motes x 461.01g mot = 14.40 g.