Question

Question 13 of 13 > A precipitation reaction occurs when 745 mL of 0.648 M Pb(NO), reacts with 459 mL of 0.851 M KI, as shown by the following equation. Pb(NO3)2(aq) + 2 Kl(aq) — Pl_(s) + 2 KNO, (aq) Identify the limiting reactant. OKI O KNO, Pb(NO) Pol, Calculate the theoretical yield of Pbl, from the reaction. mass of Pbl :

Answer 1

Pb(NO3)2 Lag) + 2 KI Laas PbI215) + 2K NO lag) Volume of Pb(NO3) = 745 ml = 0.745 L con of Pb(NO3)2 = 0.648 M = 0.648 mol/L no. of mode of Pb(NO3)2 = 0.45Lx 0.648 mol/ 0.48276 mol Nol of kl = 459 ml = 0.459 L lonen of KI = 0.851 m = 0.851 mooi no. of mole of kI = 0.459 L× 0.851 mol/L = 0.390609 mol Acc to balance cans I mole Pb(NO3), react with 2 mol of KI, then 0.48276 mole Pb(NO), react with 2x0.48276 = 0.96552 mol of kz but we have only 0.39060g med kl. hence, limiting reactant is KI. no. of mole of pb I producted from 0.390609 mol of KI = x 0.390609 = 0.1953045 mol Thoritical yield of Ph Iz = nx malas mais = 0.1953045 modx 461 g mol = 90.08 - Mence, mass of PhT₂ = 90.0 g in leo 78.39 x100 90.og Experimental yield of PI = 78.39. 1 yield of Pb I = Exp.yield Theoriticed yield = 877. Henie, 7 yield of PbT₂ = 871...