A solution containing 15.2 g of BaBr2 is reacted with a solution containing excess Na3PO4 to form 9.50 g of precipitate. What is the percent yield of the reaction?

Explanation:

  3BaBr₂       +       2Na₃PO₄     =       6BaBr + Ba₃(PO₄)₂ (s)

3 x 297 = 891 gm                                          1 x 602 = 602 gm

891 gram of BaBr₂ yields     602 gram of precipitate

15.2 gram of BaBr₂ yields     602 x 15.2 / 891 gram of precipitate

= 10.23 gram

actual yield = 9.5 gram

percent yield = 9.5 x 100 / 10.23

= 92.86 % .

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