A nickel coordination compound, Ni(NH_3)6CI_2, can be prepared by a procedure si

Question

A nickel coordination compound, Ni(NH_3)6CI_2, can be prepared by a procedure similar to the one you will use to prepare Cu(NH_3)4SO_4H_2O. The reaction can be represented stoichiometrically as NiCl_2.6H_2O + 6 NH_4OH rightarrow Ni(NH_3)6C1_2 + 12 H_2O. 4.850 g of NiCl_2 6H_2O is treated with 22.0 mL of 15 M NH_4OH solution. 4.230 g of product is obtained. What is the limiting reagent?() What is the ratio molesNH3 in solution? (i. e, actually present in the reaction beaker) moles Ni^2+ Calculate theoretical yield and percent yield. (Give set-ups and numerical answers) Briefly describe how you are going to filter the crystals you will have obtained using vacuum filtration.

Explanation / Answer

Balanced equation:
NiCl2*6H2O + 3 NH4OH ====> Ni(NH3)3Cl2 + 9 H2O

4.852 gm of NiCl2 = 4.852 / 237.69 = 0.0204 Moles

22 ml of 15M NH4OH = 22 x 15 /1000 = 0.33 Moles

One mole NiCl2.6H2O needs 6 Moles of NH4OH. In the same way 0.0204 Moles of NiCl2.6H2O needs 0.1224 Moles of NH4OH. But we have 0.33 Moles of NH4OH which excess reagent and NiCl2.6H2O is limiting reagent.

Question b

Moles ratio of NH3 in soution = 15 M

Moles ratio of NiCl2 = 4.85 x 1000/22 x 237.69 = 0.9274 M

Moles ratio of NH3 and NiCL2.6H2O = 15 M : 0.9274 M

Theoretical yield in Moles = 0.0204 M

Theoretical yield in grams = 0.0204 x 231.78 = 4.731 gm

Percentage yield = obtained yield x 100 / Theoretical yield

Percentage yield = 4.23 x 100 / 4.731 = 89.41%

Question 3

Prepare to filter your sample by placing a filter paper in the Buchner funnel and wetting it with clean solvent. You should see the paper being sucked down against the holes in thefunnel and the solvent should quickly pass through into the filter flask. To filter your sample, slowly pour into the center of the filter paper. Use more clean solvent to rinse your beaker, so that all the solid is collected.

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