1- Suppose 0.976g CuCl2 and 0.775g of Na3Po4 were reacted. What is the percentag

Question

1- Suppose 0.976g CuCl2 and 0.775g of Na3Po4 were reacted. What is the percentage yield of Cu3(Po4)2 if 0.430g of Cu3(PO4)2. was isolated? (Use 380.12g/mol for Na3PO4 and 170.48g/mol for CuCl2 and 434.60 g/mol  for CU3(PO4)2 ).Be sure to check for the limting reactant.

2- A mixture contains both potassium phosphate and potassium chloride. What is the percentage of potassium phosphate in this mixture if reaction of 0.401g of this mixture with excess copper (II) chloride yields 0.213g of copper (II) phosphate? (Assume that the potassium phosphate is not hydrated.

Explanation / Answer

a) 3 CuCl2 + 2 Na3PO4 -----------> Cu3(PO4)2 + 6 NaCl

Here, 3*170.48 g CuCl2 = 340.96 g, reacts with 2*380.12 g Na3PO4 = 760.24 g

So, 0.976 g CuCl2 will require = 2.176 g Na3PO4 > available.

Hence Product formed according to Na3PO4

760.24 g Na3PO4 gives 434.6 g Cu3(PO4)2

So, 0.775 g will give = 0.443 g Cu3(PO4)2.

% yield = 0.43*100/0.443 = 97.05 %

b) Conserving Phosphate:

All the phosphate must have come from potassium phosphate.

Moles of Cu3(PO4)2 = 0.213 / 434.6 = 0.49 milli-mole

2 K3PO4 + 3 CuCl2 -----> Cu3(PO4)2 + 6 KCl

moles of K3PO4 present = 2 * moles of Cu3(PO4)2 formed = 0.98 milli-mole

Mass of K3PO4 = 0.98*10^-3*212.27 = 0.208 g

% K3PO4 = 0.208 *100 / 0.401 = 51.87 %

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