3Pb(NO3)2(aq) Al2(CrO4)3(aq)=PbCrO4(s) 2Al(NO3)3(aq)

Question

how much solid precipitate will be produced from the reaction of 4.512 kg of lead (II) nitrate with 5.000 kg aluminum chromate? what is the limiting reactant?

Answer 1

3Pb(NO3)2 + Al2(CrO4)3 >>3 PbCrO4 + 2Al(NO3)3

is the balanced equation

Moles Lead nitrate = 4512 g / 331.2 g/mol = 13.6

Moles Aluminium chromate = 5000 g / = 402 = 12.4

the ratio between reactants is 3 : 1 so lead nitrate is the limiting reactant ( 12.4 x 3 = 37.2 moles Pb(NO3)2 are needed to react with 12.4 moles of aluminium chromate)

The ratio between lead nitrate amd lead chromate is 3 : 3 ( or 1 : 1)

moles lead chromate = 13.6

Mass = 13.6 mol x 323 g/mol =4395 g