A chemistry student carries out this reaction, beginning with 2.10 grams of NaBr and 3.25 grams of Pb(NO3)2: (First, balance the reaction if needed) NaBr (aq) + Pb (NO3)2 (aq) →PbBr2 (s) + NaNO3 (aq) Which starting material is the limiting reagent? What is the theoretical yield of PbBr2, in this reaction? If the student actually isolates 3.5 g of PbBr2, what is the percent yield of the reaction?

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A chemistry student carries out this reaction, beginning with 2.10 grams of NaBr and 3.25 grams of Pb(NO3)2: (First, balance the reaction if needed)

NaBr (aq) + Pb (NO3)2 (aq) →PbBr2 (s) + NaNO3 (aq)

  • Which starting material is the limiting reagent?
  • What is the theoretical yield of PbBr2, in this reaction?
  • If the student actually isolates 3.5 g of PbBr2, what is the percent yield of the reaction?
Expert Solution
Step 1

Initially,

 mass   of   NaBr         =  2.10g

Mass of   Pb(NO3)2    =  3.25g

The balanced chemical equation is

 2NaBr (aq)  + Pb(NO3)2 (aq) PbBr2 (s)+ 2NaNO(aq)

 

from the above balanced equation it is clear that,

2 mole of NaBr require 1 mole of  Pb(NO3)

now finding their no of moles:

no of moles of NaBr=given mass of NaBrmolar mass                                                     =2.10102.894                                =0.0204 moleno of moles of Pb(NO3)2=given mass ofPb(NO3)molar mass                                                                   =3.25331.2                                  =0.0098 moleas from equation,1 mole of Pb(NO3)2 require 1 mole of NaBrtherefore, 0.0098 mole of Pb(NO3)2 require 0.0196 mole of NaBrso, NaBr is in excess amountHENCE,        Pb(NO3)2 is limiting reagent       

 

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