Question

Lead ions can be precipitated from solution with KCl according to the following reaction: Pb2+(aq)+2KCl(aq) →PbCl2(s)+2K+(aq) When 28.5 g KCl is added to a solution containing 25.5 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.3 g. Determine the limiting reactant. Determine the theoretical yield of PbCl2. Determine the percent yield for the reaction.

Answer 1

Pbht caq, + 2kel lag. Pbiles) + 2 kt lag, weight of ko = 28.5g. no. of mole of ko = 28.59 = 0.3826 mol 74.5 g/mol weight of Phet in roon = 25.5g no. of mole of Phet = 25.5g - 207g Imus 0.1232 ml I mole Pb²+ reart with 2 mol ke, then, 0. 1232 mol Phat react with 2x0.1232 = 0.2464 molte But we have exters of kle Hence the limiting reactant is peet 0.1232 mol Phet prodline 0. 1232 mol Pole Henie no. of mole of Poll produced = 0.1232 mue mas of Polly produced = 0.1232 mod x 278 goud = 34.85g Hence, Theoritical yield of Poll = 34.25.. Experimental yield of Phl= 29.39. % yield = Exp. yield Theoritical field 2 29.39 X100 = .552 34.25g Hence percent yield for x = 85.55%