# If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate...If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate, a precipitate forms....

If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate...

If 125 mL of 0.223 M barium chloride is mixed with 50.0 mL of 0.544 M Sodium Phosphate, a precipitate forms. Assume that the reaction has a 63.2% yield.

What is the mass and formula of the precipitate that forms?

What is the concentration of each of the ions remaining in solution?

ndnordic | High School Teacher | (Level 2) Associate Educator

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1. Write a balanced chemical equation:

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

2. Calculate moles of starting materials to see if one reactant is limiting:

125 mL * 0.223 M = 27.88 mmoles of BaCl2

50 mL* 0.544 M  = 27.2 mmoles of Na3PO4

From the balanced equation, you see that you that BaCl2 is limiting reactant.

Theoretically, at 100 % yield, for every 27.88 mmoles of BaCl2, you will use 18.59 mmoles of Na3PO4.  Since you produce one mmole of barium phosphate for every 3 mmoles of BaCl2, at 100% yield you would get 9.29 mmoles of barium phosphate.

Given that you have only a 63.2% yield, that means that you will use:

27.88 * 0.632 = 17.62 mmoles of BaCl2, with 10.26 mmoles remaining.

18.59 * 0.632 = 11.75 mmoles of Na3PO4 used, 15.45 mmoles remaining.

9.29 * 0.632 = 5.87 mmoles of Barium phosphate.

Convert mmoles to mass:

formula mass of Barium Phosphate is:

Ba:  3 * 137.33 = 411.99

P: 2  30.97 = 61.94

O: 8 * 16 =128

total = 601.94 g/mole or 601.94 mg/mmole

601.94 mg/mmole * 5.87 mmoles = 3.53 g of barium phosphate

What is left?

You have10.26 mmoles of Ba+2, 20.52 mmoles of Cl-1, 46.45 mmmoles of Na+1, 15.45 mmoles of PO4(-3) still in solution.

The concentrations (M) are found by dividing moles by liters of total solution.

For Ba+2:   0.01026/.175 = 0.059 M

For Cl -1: 0.02052/0.175 = 0.117 M

For Na+1: 0.04645/0.175 = 0.265 M

For PO4(-3): 0.01545/0.175 = 0.088 M