Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution.

Question

A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ⁢ ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 16.79 g PbCl2(s) 16.79 g PbCl 2 ⁢ ( s ) is obtained from 200.0 mL 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution.

Answer 1

The reaction taking place is ...

2NaCl(aq) + Pb(NO₃)₂(aq) ==> 2NaNO₃(aq) + PbCl₂(s)

molar mass PbCl₂ = 278.1 g / mol

moles PbCl₂ obtained = 16.79 g PbCl₂ x 1 mol / 278.1 g = 0.0604 mols PbCl₂

moles Pb(NO₃)₂ originally present = 0.0604 mols PbCl₂ x 1 mol PbCl₂/mol Pb(NO₃)₂ = 0.0604 mols Pb(NO₃)₂

Molarity of Pb(NO₃)₂ = moles/L = 0.0604 mols / 0.200 L = 0.302 M

Answer 2

Molarity is g/ml

M= g/ml

If you get the grams that correlates to this reaction based in a balanced reaction you can find the Molarity for the PB(NO3)2.

The reaction would look something like this:

NaCl(aq) + Pb(NO3)2(aq) ==> PbCl2(s) + NaNO3 (aq)

Now that you have a balanced equation you can use stoichiometry to figure out what the grams would be.

The ratio for everything is:

1:1:1:1

There are no numbers in front of all of the species in the reaction.

So if:

1molPb(NO3)2 =1 mol if PbCl2

16.79g. = 16.79g

16.79g/200mL. = 0.08395M