## 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