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?

We first calculate the number of moles of K2CrO4 by multiplying the volume with the molarity.
                                     n = (25/1000) x (3 M) = 0.075 mol K2CrO4
Then, from the balanced chemical equation, each mole of K2CrO4 will produce 1 mole PbCrO4 with molar mass of 323.2 g/mol.
                          (0.075 mol K2CrO4)(1 mol PbCrO4/1 mol K2CrO4) x (323.2 g/mol)
                             = 24.24 grams PbCrO4