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? 75 grams 0.024 grams 24 grams 0.075 grams 3.0 grams

Answer 1

 K₂CrO₄  +  PbCl₂  → 2KCl   +   PbCrO₄

   If      1000 cm³ of  K₂CrO₄ contain 3 moles
then let   25 cm³ of  K₂CrO₄ contain   x

⇒ x = (25 cm³ × 3 mol) ÷ 1000 cm³
       = 0.075 mol

Since mole ratio of  K₂CrO₄  :  PbCrO₄ is   1  :  1 then
moles of PbCrO₄ that will precipitate = 0.075 mol

Mass of PbCrO₄ that precipitate = 0.075 mol × (molar mass)
                                                   = 0.075 mol × [(207 × 1) + (52 × 1) + (16 × 4)] g/mol
                                                   = 24.23 g

∴ Mass of PbCrO₄ that precipitate is 24 g