Suppose 1.780g CuCl_2, and 1.773g of Na_3 PO_4, were reacted as in this experime

Question

Suppose 1.780g CuCl_2, and 1.773g of Na_3 PO_4, were reacted as in this experiment. What is the percentage yield of Cu3(PO4)2 if 0.885g of Cu_3 (PO_4) _2 was isolated? (Use 380.12g/mol for the Na_3 PO_4 and 170.48 g/mol for CuCL_2 and 434.60 g/mol for Cu_3 (PO_4)_2 Be sure to check for the limiting reactant. A mixture contains both potassium phosphate and potassium chloride. What is the percentage of potassium phosphate in this mixture if the reaction of 0.855g of this mature was reacted with excess copper(II) chloride and yielded 0.365g of copper (II) phosphate? (Assume that the potassium phosphate is not hydrated. This problem is similar to part B of the experiment.)

Explanation / Answer

(6)

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

Number of moles of CuCl2 = mass / molar mass = 1.780 / 170.48 = 0.0104 mol

Number of moles of Na3PO4 = 1.773 / 380.12 = 0.00466 mol

From the balanced eqation,

3 mol CuCl2 requires 2 mol Na3PO4

then, 0.0104 mol CuCl2 requires 2 * 0.0104 / 3 = 0.00693 mol of Na3PO4 (> 0.00466)

Hence, Na3PO4 is limiting reagent.

From the balanced equation,

2 mol Na3PO4 can form 1 mol Cu3(PO4)2

then, 0.00466 mol Na3PO4 can form 0.00466 / 2 = 0.00233 mol of Cu3(PO4)2

Theretical mass of Cu3(PO4)2 = 0.00233 * 434.60 = 2.025 g.

But Actual yield = 0.885 g.

Therefore, % yield = (actual yield / theoretical yield) * 100 = (0.885 / 2.025) * 100 = 43.7 %

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