Assuming the amount of the metal salt you used as a starting material in each ca

Question

Assuming the amount of the metal salt you used as a starting material in each case was the limiting reagent, calculate the % yield of Ni(NH3)6Cl2?

procedure: weigh 2g of solid nickel(II) chloride hexahydrate in a 100ml beaker, added 2 ml of water and heated. After dissolution, 10ml of conc. NH3 was added. The solution was cooled later on with addition of 7ml of cold denatured ethanol added to complete precipitate the product. At the end use suction filter the product and rinse twice with 5 ml ethanol, once with 5ml of acetone. Weigh the final product.

initia : 2.0150g final:1.8146g

Explanation / Answer

Let us consider the reaction as below

Ni(H2O)6Cl2 + 6NH3 = Ni(NH3)6Cl2 +3Cl2

limiting reagent is the Ni(H2O)6Cl2, hence for 2.0150g of initial weight 2.0150g of theoretical yield should come.

actual yield in g

----------------------- x   100 % =   Percent Yield

theoretical yield in g

Percent Yield   = 1.8146g/2.0150g*100

                           = 90.05 %

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