In: Chemistry
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?
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