To a solution containing 3 g of PbCl2 was added a solution containing 0.02 mol of K2CrO4. What is the mass of the PbCrO4 precipitate?

We write down the reaction equation.
PbCl2 + K2CrO4 = 2KCL + PbCrO4.
Next, we find the amount of lead chloride substance. We use the following formula.
n = m / M.
Let us determine the molar mass of lead chloride.
M (PbCl2) = 278 g / mol.
n = 3 g / 278 g / mol = 0.01 mol of lead chloride.
Further, since potassium chromate was taken in excess, it means that we consider the amount of PbCrO4 substance in terms of lead chloride. We write down the solution.
0.01 mol PbCl2 – x mol PbCrO4.
1 mol PbCl2 – 1 mol PbCrO4.
Find the unknown value of x.
X = 0.01 × 1 ÷ 1 = 0.01 mol of lead chromate.
Find the mass of lead chromate.
m = n × M.
M (PbCrO4) = 323.20 g / mol.
m = 323.20 g / mol × 0.01 mol = 3.232 g.
Answer: m = 3.232 g.