What is the mass of dry precipitate actually collected?

Question

Excess sodium Phosphate is mixed with 543.3moL of Strontium Chloride at a concentration of 0.728 mol/L. If the reactions has a 49.36% yield, what is the mass of dry precipitate actually collected?

Answer 1

Hello Raheem

 

The balanced reaction is:

 

2Na₃PO₄ (aq) + 3SrCl₂ (aq) → 6NaCl (aq) + Sr₃(PO₄)₂ (s)

 

I believe in your question, you meant to state 543.3 mL of SrCl₂ and not "moL" otherwise we would not be able to solve this problem.

 

First use the volume and molarity of the SrCl₂ solution to calculate the # of moles.

 

Molarity, M = moles of solute / Volume of solution (in Liters).  Rearrange this equation to solve for moles.

 

moles = Molarity x Volume (in L)

 

V = 543.3 mL x (1 L / 1000 mL) = 0.5433 L

 

moles = 0.728 mol/L x 0.5433 L = 0.396 mol SrCl₂

 

Now use stoichiometry to convert moles of SrCl₂ to grams of Sr₃(PO₄)₂.  This calculated mass will be the Theoretical yield.  From the chemical reaction, the mole ratio of SrCl₂ to Sr₃(PO₄)₂ is 3:1.  Molar mass of Sr₃(PO₄)₂ is 453 g/mol.

 

0.396 mol SrCl₂ x (1 mol Sr₃(PO₄)₂ / 3 mol SrCl₂) x (453 g Sr₃(PO₄)₂ / 1 mol Sr₃(PO₄)₂)

 

= 59.8 g Sr₃(PO₄)₂

 

The %-yield of Sr₃(PO₄)₂ is 49.36%.

 

%-Yield = (Actual yield / Theoretical yield) x 100%

 

The mass of dry precipitate collected refers to the actual yield.  Rearrange to solve for actual yield.

 

Actual yield = %-Yield x Theoretical yield

 

                     = 49.36% x 59.8 g

 

                     = 0.4936 x 59.8 g

 

                     = 29.5 g Sr₃(PO₄)₂ collected

Answer 2

Write a balanced chemical equation:

 

2Na3PO4 + 3SrCl2 → 6NaCl + Sr3(PO4)2

 

543.3moL SrCl2 x 1 mol Sr3(PO4)2/3 mol SrCl2 x 452.80 g Sr3(PO4)2/1 mol Sr3(PO4)2 = 82002.08 g Sr3(PO4)2

 

TY (% Y) = AY 

 

82002.08 g (.4936) = 40476.23 g

 

HOWEVER, I am pretty confident that you have mis-written this problem.  I suspect you mean "mL" and NOT "moL"

 

If so, proceed as follows: 543.3 mL x 0.728 mol/ 1000mL = .396 mol srCl2

 

Then use this, corrected molar amount to set up you stoichiometric equation  as above.