Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq)

Question

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 16.54 g of PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

Explanation / Answer

Pb(NO3)2(aq) + 2NaCl(aq) ==> PbCl2(s) + 2NaNO3(aq) The molar mass of PbCl2 = Pb (207.2) + 2 x Cl (2 x 35.45) = 278.1 g/mole. 16.54 g PbCl2 x (1 mole PbCl2 / 278.1 g PbCl2) = 0.05945 moles PbCl2 The balanced equation tells us that 1 mole of Pb(NO3)2 produces 1 mole of PbCl2. 0.05945 moles PbCl2 x (1 mole Pb(NO3)2 / 1 mole PbCl2) = 0.05945 moles Pb(NO3)2 Since Pb(NO3)2 is in solution form, moles Pb(NO3)2 = Molarity Pb(NO3)2 x L Pb(NO3)2 0.05945 = M Pb(NO3)2 x 0.2000 M Pb(NO3)2 = 0.05945 / 0.2000 = 0.297 M

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