In the reaction at left, given 75.0 milliliters of 1.50 m k2cro4 in an excess of pbcl2, how many grams of pbcro4 will precipitate out?

 The balanced equation that describes the reaction between potassium chromate and lead (II) chloride to produce potassium chloride and lead (II) chromate is expressed PbCl2 + K2CrO4 = 2KCl + PbCrO4. For every mole of K2Cr2O4, there is 1 mole of PbCrO4 produced. The number of moles of K2Cr2O4 is 0.1125 moles. The molar mass of PbCrO4  is 323.2 g/mol. THe mass of PbCrO4  then is 36.36 grams.