Write an equation for the formation of [Ni(NH3)6]Cl2 and calculate its percentag

Question

Write an equation for the formation of [Ni(NH3)6]Cl2 and calculate its percentage yield in your experiment. Given that Mass of [Ni(NH3)6]Cl2 used is 4.0141g Mass of the product [Ni(NH3)6]Cl2 is 1.87g. with the procedure here

Procedure for the synthesis of hexaamminenickel(II) chloride

1. Dissolve 4.0 g hydrated nickel(II) chloride in 6 mL distilled water in a 50-mL flask.

2. In the fume cupboard, add 12 mL concentrated ammonia to the above solution and warm the

mixture for 10 minutes on a hot plate.

3. Cool the solution in an ice-bath, and while stirring it with a glass rod, add 6 mL ethanol. Note the

formation of a solid.

4. When all of the solid has formed, filter it under suction and wash it with a few mL of cold

concentrated ammonia solution, followed with ethanol, and finally with acetone.

5. Record the weight of the solid.

Explanation / Answer

The chemical equation is-

[Ni(H2O)6]Cl2 (aq) + 6 NH3 (aq) [Ni(NH3)6]Cl2 (aq) + 6 H2O (l)

[Ni(NH3)6]Cl2

Mol. Wt.: 231.78

[Ni(H2O)6]Cl2

Mol. Wt.: 237.69

Mass of starting material = 4.0141 g and that of product = 1.87 g.

Now, 231.78 g product expected from 237.69 g of starting material.

So, 4.0141 g will give = 4.0141*231.78/237.69 g = 3.9143 g.

So, percentage yield = 100*1.87/3.9143 = 47.77%

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