Question

In the reaction K2CrO4 (aq) + PbCl2 (aq) 2KCl (aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 25.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?

Answer 1

Lead(II) chromate PbCrO4 Ksp= 3×10-13
the Ksp it is very small, because the problem doesn't said the final volume of the mix, it is not possible calculate how much PbCrO4 will be soluble, We can assume that is very little and calculate that all the K2CrO4 will react.

calculate how many moles of K2CrO4 are in 25.0 milliliters of 3.0 M K2CrO4

M= n/V(L) => n = M x V(L) = 3.0 x 0.025 L = 0.075 moles K2CrO4

these moles is going to be equal to the moles of PbCrO4 because the stoiquiometry is 1:1.

then calculate the mass of 0.075 moles of PbCrO4

n= m / MM => m = n x MM = 0.075 x 323.19 g/mol = 24.24 g