Question

when a solution of lead(ii) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation: Pb(NO3)2 (aq) + K2CrO4 (aq) -> 2 KNO3 (aq) + PbCrO4. Volume of 0.105 M lead(ii) nitrate react with 100.0 ml of 0.120 M potassium chromate. What mass of PbCrO4 will be formed?

Answer 1

Given reaction:

Pb(NO3)2 (aq) + K2CrO4 (aq) -> 2 KNO3 (aq) + PbCrO4.

No. of moles of Pb(NO3)2 = Molarity x volume in L
                          = 0.105 mol/L x 0.100 L
                          = 0.0105 mol
No. of moles of K2CrO4 = Molarity x volume in L
                          = 0.120 mol/L x 0.100 L
                          = 0.0120 mol
From the stoichiometry, 1 mol of Pb (NO3)2 reacts with 1 mol of K2CrO4 and produced 1 mol of PbCrO4. Therefore, in this reaction consumes equal moles of potassium chromate and of lead(II) nitrate.

Since we have fewer moles of Pb(NO3)2 than no. of moles of K2CrO4; Pb(NO3)2 reactant is the limiting reagent.

Therefore, the no. of moles of PbCrO4 = 0.0105 mol
mass of PbCrO4 = no. of moles x mol.wt
               = 0.0105 mol x 323.2 g/mol
               = 3.3936 g
Hence, the mass of PbCrO4 = 3.4 g