excess sodium phosphate is mixed with 778.6 ml of strontium chloride at a concentration of 0.309 mol/l. if the reactions has a 67.50% yield, what is the mass of dry precipitate actually collected.

Question

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Answer 1

Write a correctly balanced equation for the reaction taking place:

2Na₃PO₄(aq) + 3SrCl₂(aq) ==> 6NaCl(aq) + Sr₃(PO₄)₂(s) ... balanced equation

NOTE: the precipitate is strontium phosphate, Sr₃(PO₄)₂(s)

Next, use the moles of strontium chloride (SrCl₂) to determine theoretical yield of Sr₃(PO₄)₂ (the ppt):

mols SrCl₂ = 778.6 ml x 1 L / 1000 ml x 0.309 mol/L = 0.2405 mols

0.2405 mols SrCl₂ x 1 mol Sr₃(PO₄)₂ / 3 mols SrCl₂ = 0.0802 mols Sr₃(PO₄)₂

0.0802 mols Sr3(PO4)2 x 452.8 g / mol = 36.3 g Sr3(PO4)2 theoretical yiled

To find the actual yield, divide the theoretical yield by the percent yield:

36.3 g / 0.6750 = 53.8 g Sr₃(PO₄)₂ actually collected