Question

The balanced equation for the reaction of aqueous Pb(CIO), with aqueous Nal is Pb(CIO),(aq) + 2 Nal(aq) - PbI, (s) +2 Nacio, (aq) What mass of precipitate will form if 1.50 L of concentrated Pb(CIO), is mixed with 0.550 L of 0.250 M Nal? Assume the reaction goes to completion. mass of precipitate:

Answer 1

Pb(clos) can ) + 2 Nar (aq) PLT, 18) +2 Nacho, (ar) Precipitate 1.50L of Pb(C103) mixed with 0.550L of 01250m Na.I moles of Nar = volume x molarity 0:550 LX0-250 mol Nal = 0.1375 moles I mole of Pb(ClO3), react with 2 mole of Nat 0.1375 moles al NaI will react with 01375 - 0.06875 mol of РЬtt003), I mole of Pb(ClO3)2 give = 1 mole of Pb I2 0.06875 mol of 1 - 0.06875 - 1 - mass of PbI2 = 0.06875 mol x molut of PbI2 - 0.06875 mol x 461.019 mop mass of PbI2 = 31.69 g