Question

In the reaction K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 500.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?

Answer 1

v=500.0 mL = 0.5 L
c=3.0 mol/L
M(PbCrO₄)=323.2 g/mol

m(PbCrO₄)=M(PbCrO₄)n(PbCrO₄)

m(PbCrO₄)=M(PbCrO₄)n(K₂CrO₄)=M(PbCrO₄)vc

m(PbCrO₄)=323.2*0.5*3.0=484.8 g