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?

Question

Mills

Chemistry

28 August, 03:54

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?

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Aimee Holder · Answer 1

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.