Sprint ?? 10:18 37%) a session.masteringchemistry.com C Exercise 8.65 MasteringC

Question

Sprint ?? 10:18 37%) a session.masteringchemistry.com C Exercise 8.65 MasteringChemistry 28 of 57 Constants | Periodic Table Lead ions can be precipitated from solution with NaCl according to the following r Pb2 (a) 2 NaCl(aq)PbCl2(s) +2 Na (aq) When 135.8 g of NaCl are added to a solution containing 95.7 g of Pb2+,a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 251.4 g Part A eaction. Determine the limiting reactant for the reaction. Express your answer as an ion. ABC | 123-10; AZd q w e rt yui o

Explanation / Answer

Pb^2+ (aq) + 2NaCl(aq) ----------> PbCl2(s) + 2Na^+ (aq)
   no of moles of NaCl = W/G.M.Wt
                        = 135.8/58.5   = 2.32moles
no of moles of Pb^2+ = W/G.A.Wt
                        = 195.7/207 = 0.945 moles
   2 moles of NaCl react with 1 mole of Pb^2+
   2.32 moles of NaCl react with = 1*2.32/2   = 1.16 moles of Pb^2+ is required
Pb^2+ is limiting reactant
1 mole of Pb^2+ react with NaCl to gives 1 mole of PbCl2
0.945 moles of Pb^2+ react with NaCl to gives = 1*0.945/1    = 0.945 moles of PbCl2
mass of PbCl2 = no of moles * gram molar mass
              = 0.945*278   = 262.71g

Theoretical yiled of PbCl2 = 262.71g
percent yiled = actual yiled*100/theoretical yiled
              = 251.4*100/262.71   = 95.7% >>>>answer

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