Lead ions can be precipitated from solution with NaCl according to the reaction:
Pb2+ (aq) + 2 NaCl(aq) → PbCl2(s) + 2 Na+(aq)
When 135.8 g of NaCl are added to a solution containing 195.7 g of Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252.4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction.
In this question , the given mass of NaCl = 135.8 g
Mass of lead = 195.7g
The mass of PbCl2 precipitate = 252.4g
Pb2++ 2 NaCl → PbCl2(s) + 2 Na+
The molecular mass of NaCl = 58.4430 g/mol and atomic weight of
Pb = 207.2100g/mol
Hence number of mol of NaCl
=(135.8 g NaCl) / (58.4430 g NaCl/mol) = 2.3236 mol NaCl
Number of mol of Pb
= (195.7 g Pb) / (207.2100 g Pb/mol) = 0.944452 mol Pb
0.944452 mole of Pb would react completely with 0.944452 (2/1)...