A mixture contains 58.50 % BaCl2 and 41.50 % Na3PO4 by weight. When dissolved in

Question

A mixture contains 58.50 % BaCl2 and 41.50 % Na3PO4 by weight. When dissolved in water, the following double displacement reaction forms the barium phosphate precipitate. If you start with 6.00 grams of the mixture, how many grams of barium phosphate precipitate should form? 3 BaCl2 (aq) + 2 Na3PO4 (aq) --> 6 NaCl (aq) + Ba3(PO4)2 (s) Use the following molar masses: BaCl2 = 208.23 g/mol Na3PO4 = 163.91 g/mol NaCl = 58.44 g/mol Ba3(PO4)2 = 601.93 g/mol Enter your answer in grams, rounded to 3 significant figures. Do not include units.

Explanation / Answer

From the data

mass of BaCl2 = 6.0x 58.5/100 = 3.51 g

and thus mass of sodium phosphate = 2.49 g

the reaction is

3BaCl2 + 2Na3PO4 --------> 6NaCl + Ba3(PO4)2

3.51/208.33 2.49/163.91

= 0.01684 =0.01519

To know the limiting reagent find the ratio BacL2 (0.01684/3=0.005616) and Na3PO4 (0.01519/2=0.007595)

Since tha ratio of BaCl2 is less BaCl2 is the limiting reagent

Thus the product is calculated from limiting reagent which is consumed completely during the reaction.

Thus 3 moles (3x208.23g/mol) of Bacl2 forms one mole (601.93g/mol) of ba3(PO4)2

3.51 g of BaCl2 can form = 3.51 x601.93/3x208.23

= 3.3821 g

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