Calculate the percentage yield for the following and SHOW FULL WORKING . 1. Reac

Question

Calculate the percentage yield for the following and SHOW FULL WORKING.

1. Reaction of Part A: [Co(en)2Cl2]Cl

Note that (en) is ethylenediamine or 1,2-diaminoethane

Actual Yield obtained = 2.69 g

The procedure for A is shown below:

2. Reaction of Part B: [Ni(en)3]Cl2.2H2O

Note that (en) is ethylenediamine or 1,2-diaminoethane

Actual Yield obtained = 2.68 g

The procedure for Part B is shown below:

3. Reaction of Part C: Ni(acac)2

Note that (acac) is acetylacetonato

Actual Yield obtained = 1.33 g

The procedure for Part C is shown below:

4. Reaction of Part D: K3[Fe(ox)3].3H2O

Note that (ox) is oxalato

Actual Yield obtained = 1.41 g

The procedure for Part D is shown below:

A. trans-Dichloridobis(ethylenediamine)cobalt(III) chloride, [Co(en)2C2|Cl 1.Dissolve 2.0 g of cobalt(II) chloride hexahydrate in 15 mL of water, and add this solution to a mixture of 1.85 mL ethylenediamine (measured in the fume hood using a small measuring cyclinder) and 8 mL water. Do not use more water than this. 2.Transfer the solution to a side-arm test tube fitted with a stopper and glass tube and clamp this at the joint. Draw air through the liquid for 60 minutes. At this stage you can start on your next preparation Transfer the solution to a beaker and add dropwise 5.0 mL of conc. hydrochloric acid. 4. Heat the beaker on hot plate stirrer to evaporate the solution down to about 5 mL (or until a green colour is observed). 5 Cool the solution in ice and filter to collect the product. Wash with cold ethanol and air dry. Record the yield.

Explanation / Answer

1) Moles of Cobalt salt = 0.015

Moles of Ethylene diamine = 0.022

So, Cobalt salt is limiting reagent.

Actual yield is = 2.69 g

Theoretical yield = M.Wt of complex * 0.015 = 285.48*0.015 = 4.2822 g

%Yield = (Actual yield/Thereotical yield)*100

%Yield = (2.69/4.2822)*100 = 62.8%

2) Nickel salt moles = 0.01543

Actual yield = 2.68 g

Theoretical yield = 345.908*0.01543 = 5.3373

%Yield = (2.69/5.3373)*100 = 50.4%

3) Moles of Nickel acetate salt = 0.0075

Actual Yield = 1.33 g

Theoretical Yield = 258.927*0.0075 = 1.9419 g

%Yield = (1.33/1.9419)*100 = 68.48%

4) Moles of Iron chloride = 0.0074

Actual Yield = 1.41 g

Theoretical Yield = 494.263*0.0074 = 3.6575 g

%Yield = (1.41/3.6575)*100 = 38.55%

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