Solution Stoichiometry

 

98) m = 1.50 g Pb(NO3)2          [Na2SO4] = 0.100 M          V = 125 mL

       

      (a) Pb(NO3)2(aq) + Na2SO4(aq) 2NaNO3(aq) + PbSO4(s)

            Pb2+(aq) + SO42-(aq) PbSO4(s)

 

      (b) [SO42-] = n/V

            nSO42- = 0.100 mol Na2SO4/L x 125 mL x 1 L/103 mL x 1 mol SO42-/1 mol Na2SO4 = 1.25 x 10-2 mol SO42-

            nPb2+ = 1.50 g Pb(NO3)2 x 1 mol Pb(NO3)2/331.22 g Pb(NO3)2 x 1 mol Pb 2+/1 mol Pb(NO3)2 = 4.53 x 10-3 mol Pb2+

      

            nPbSO4 = 1.25 x 10-2 mol SO42- x 1 mol PbSO4/1 mol SO42- = 1.25 x 10-2 mol PbSO4

            nPbSO4 = 4.53 x 10-3 mol Pb2+ x 1 mol PbSO4/1 mol Pb2+ = 4.53 x 10-3 mol PbSO4

            The limiting reactant produces the smallest amount of product, therefore the Pb(NO3)2 is the limiting reactant.

 

      (c) Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + SO42-(aq) 2Na+(aq) + 2NO3-(aq) + PbSO4(s)

 

           [Pb2+] = n/V = 0

 

           [NO3-] = 4.53 x 10-3 mol Pb2+ x 1 mol Pb(NO3)2/1 mol Pb2+ x 2 mol NO3-/1 mol Pb(NO3)2/(125 mL x 1 L/103 mL)

           [NO3-] = 7.25 x 10-2 M

 

           [Na+] = 0.100 mol Na2SO4/L x 2 mol Na+/ 1 mol Na2SO4 = 0.200 M

 

           nSO42-(start) = 0.100 mol Na2SO4/L x 125 mL x 1 L/103 mL = 0.0125 mol SO42-

           nSO42-(reacted) = 4.53 x 10-3 mol Pb2+ x 1 mol SO42- /1 mol Pb2+ = 4.53 x 10-3 mol SO42-

           [SO42-] = nSO42-(start) - nSO42-(reacted)/V

           [SO42-] = (0.0125 mol SO42- - 4.53 x 10-3 mol SO42-)/(125 mL x 1 L/103mL) = 0.064 M