Post lab question for Chem Lab. Synthesis of Ni(en)3 Cl2 Can someone please help

ID: 951807 • Letter: P

Question

Post lab question for Chem Lab. Synthesis of Ni(en)3 Cl2 Can someone please help. Me with these. Just a short but detailed as beer would help.

Post-Lab Questions: (Answers must be in full sentences and typed.)

1. The procedure calls for using between 0.500 to 0.650 g of the hydrate in the synthesis. Would a student using 0.500 g of hydrate obtain a smaller percent yield of TRIS than a student using 0.650 g? Explain your answer.

2. At the part when all of the limiting reactant has reacted, how many moles of the other reactant remain unreacted? Show your calculations & explain your answer clearly.

3. Based on the number of moles of en you used, how many moles of hydrate would you need for it to totally react? How would this information tell you which reactant is the limiting reactant? Show your work and explain clearly.

4. The liquid that goes through the filter paper into the filter flask is called the “filtrate.” What exactly is in your filtrate? List all the substances that you can think of that is in the filtrate

Explanation / Answer

To synthesize Ni(en)3Cl2 (tris), ethylenediamine solution was added to aqueous NiCl2•6H2O (hydrate). The subsequent reaction (provided by Hamilton et al.1) occurred as follows:

NiCl2•6H2O(s) + 3H2NCH2CH2NH2(aq) -->--> [Ni(H2NCH2CH2NH2)3]Cl2(s) + 6H2O(l)

Each nickel ion can bond to three en molecules to form the Ni(en)32+ complex ion.1 When the complex ion bonds to chloride ions in the solution, the product tris precipitates. However, the hydrate, ethylenediamine, and tris are all soluble in water, which poses a significant problem in separating tris. Tris is less soluble in an acetone/water mixture, though, so acetone was added to the reaction mixture to facilitate the precipitation of the product.

Using the measurements of the reactants and the above equation, the hydrate was found to be the limiting reactant. The theoretical yield was determined using the mass of the hydrate and the molar masses of the hydrate and of tris. After synthesizing the nickel complex, the percent yield of the product was calculated using the percent yield equation (also provided by Hamilton et al.1) below:

Percent yield = (Actual amount of product obtained /theoretical amount of product expected )×100

Consider following values for calculations;

Mass of NiCl2•6H2O:
(Obtained using the TARE feature)

0.527g

Appearance of solid NiCl2•6H2O:

Lime green, multi-sided sand form

Appearance of NiCl2•6H2O solution:

Dark green

Volume of en solution (en soln):

2.30 mL

Concentration of en solution:
(from the label on the bottle)

25.0%

Density of en solution (from step #4):

0.950 g/mL

Appearance of en solution before
addition to the hydrate:

clear

Appearance of reaction solution after
addition of en solution to hydrate:

Dark purple

Appearance of reaction mixture after
addition of acetone:

Neon purple

Color of product, NiCl2•6H2O,
after drying process:

Light purple

mass of empty vial + product:

23.1300 g

Mass of empty vial:

22.3672 g

Calculations:

= (0.7628/ 0.687)×100 = 111%

Answer 1)

Since the percent yield of an experiment does not depend on the amount used at the start, the difference between percent yields cannot be predicted. The change in the amount of hydrate that a student uses would be reflected in both the theoretical yield and the actual yield. Percent yield is a ratio that relates actual yield and theoretical yield, thus the increase or decrease in starting amount will not affect the percent yield in a perceptible manner. Even though the hydrate was the limiting reactant, the amount used at the start will not affect the percent yield because the percent yield depends on the method not the amount of reactants.

Answer 2)

When all the hydrate has reacted, there are 2.44 x 10-3 moles of en left.

Moles of en present in reaction mixture:

= [2.30 ml of en soln ×(0.950 g of en soln/1ml of en soln)]×[25 pure en/100 en soln] ×[ 1 mole en/ 60.104 g en]

= 0.00909 moles of en

Moles of en reacted

= 0.687 g tris × [1mol tris/309.902 g tris] ×[3 mol en/1 mole tris] = 0.00665 moles of en

Moles of en unreacted:

0.00909 moles of en (present ) – 0.00665 moles of en (reacted ) = 0.00244 moles of en left over

Answer 3)

In order for all 0.00909 experimental moles of ethylenediamine to react, there would need to be 0.00303 moles of hydrate. There were only 0.00222 moles of hydrate in the experiment, which is a smaller value than 0.00303 moles; thus, the hydrate was the limiting reactant.

Moles of hydrate needed to react with 0.00909 moles of en:

0.00909 moles of en × 1 mol of hydrate/ 3 mol of en = 0.000303 moles of hydrate

Answer 4)

There is leftover en solution, soluble tris, acetone, and water in the filtrate.