50.0 mL of 0.200 M KF solution is mixed with 15.0 mL of 0.250 M Pb(NO3)2 solution.

Question

Question: How many grams of precipitate would be formed?

Answer 1

First thing is to write the correctly balanced equation:

2KF(aq) + Pb(NO₃)₂(aq) ==> 2KNO3(aq) + PbF₂(s)

The precipitate is PbF₂, so we can now find the mass of PbF₂ formed under the given conditions:

Whenever given amounts of both reactants, you must find which, if any, is limiting.

Find Limiting Reactant:

moles KF present = 50.0 ml x 1 L/1000 ml x 0.200 mol/L = 0.01 mol KF present

moles Pb(NO₃)₂ present = 15.0 ml x 1 L / 1000 ml x 0.250 mol/L = 0.00375 mol Pb(NO₃)₂ present

Since it takes 2 mol KF to 1 mol Pb(NO₃)₂, the Pb(NO₃)₂ is LIMITING

The limiting reactant will dictate how much product can be formed.

grams PbF₂ formed = 0.00375 mol Pb(NO₃)₂ x 1 mol PbF₂ / 1 mol Pb(NO₃)₂ x 245 g/mol = 0.919 g PbF₂