Question

in the reaction k2cro4(aq)+pbcl2(aq)->2kcl(aq)+pbcro4(s), how many grams of pbcro4 will percipitate out from the reaction between 500.0 of 3.0 m of k2cro4 in an excess of pbcl2

Answer 1

K₂CrO₄ + PbCl₂ = PbCrO₄ + 2KCl

v=500.0 mL = 0.5000 L
c=3.0 mol/L
M(PbCrO₄)=323.2 g/mol

m(PbCrO₄)/M(PbCrO₄)=n(K₂CrO₄)=cv

m(PbCrO₄)=M(PbCrO₄)cv

m(PbCrO₄)=323.2*3*0.5=484.8 g