Question

determine the limiting reactant

express your answer ss a chemical formual

A 27.0 mL sample of a 1.88 M potassium chloride solution is mixed with 15.0 mL of a 0.890 M lead(II) nitrate solution and this precipitation reaction occurs: 2KCl(aq) + Pb(NO3)2(aq) + PbCl2 (s) + 2KNO3(aq) The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

Balanced precipitation reaction's 2 kcl (aq) + Pb(NO3), (aq) PbClg (3) + 2 KNO₃ (99) Moles of KCl used = molarity & volume in litres = 1.88 M2 0.027L=0.0508 Moles of 0.01335 mol. Pb(NO do used = 0,890M X 0.0156 = for 0.01335 mol. Pb(NO3), we require = (amorkce 1x 0.01335 mothe (Nools ( I mol-Pb(NO3)2 = 0.0267 mol-kce for complete reaction. And we have already 0.0508 mol-kll available, thus Pb(NO3)2 will consumed all while kell is still remained and Hence [Pb(NO3)2 is the limiting reagent. 0.01335 mol Pb(NO3)2 will produce = 0.01335 mol of Pb Clg (S) tence theoretical mass of PbCl₂ will produce = moles x molar mass = 0.01335 mot x 278.1g. mol = 3.7139. Actual mass of PbCly formed : 2.45g 7 yield = actual mass x100 theoretical mass ng x100 165.98% yield 3.7139 111111 Kindly rate the answer!!!