Chem Exp 11

LEARNING OBJECTIVES:

1. Students will be able to predict the amount of product produced from a double displacement reaction and determine the percent yield of a chemical reaction.

2. Students will be able to identify the limiting and excess reactants.

BACKGROUND.

Stoichiometry is the mole to mole relationships between the reactants and the products of a chemical reaction. Stoichiometric calculations are about the amounts of substances that react (reactants) and form (products) in chemical reactions.

Chemical reactions are represented by equations which are balanced using the smallest wholenumber coefficients. A balanced chemical equation helps the chemist to calculate the amounts of product formed in a chemical reaction. This is the theoretical yield of the reaction. In other words, it is the maximum amount of product that can be obtained in a particular reaction from a known quantity of reactants. This may not be always the case and the amount of product obtained in a chemical reaction is usually less than the theoretical amount. The amount of product obtained in a chemical reaction performed in the laboratory is called the actual yield. The percent yield of a reaction is then given by the following expression:

Actual Yield

Percent yield = × 100

Theoritical Yield

In this experiment, you will react Na2CO3 and CaCl2. You will prepare calcium carbonate, CaCO3, from the reaction of aqueous solution of sodium carbonate, Na2CO3, with the aqueous solution of calcium chloride, CaCl2. From the amounts of the reactants, you will determine which reactant is the limiting reactant, and from this amount, will calculate the theoretical yield of calcium carbonate produced. From the actual amount of calcium carbonate obtained, you can then calculate your percent yield of reaction.

The balanced equation for this chemical reaction is: CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2NaCl(aq)

The net ionic equation of the reaction is:

Ca2+(aq) + CO32-(aq)  CaCO3(s)

(aq) and (s) mean aqueous and precipitate or solid, respectively.

1. Wear your eye goggles at all times.

2. All chemicals used in these experiments are generally recognized as safe. But, you MUST wash your hands after you finish the experiment.

3. If your instructor recommends you use vacuum filtration, proceed with caution. The side arm of the Erlenmeyer can break.

MATERIALS

Centigram balance, ring stand, O-ring, funnel, filter paper, clay triangle, spatula, stirring rod, 50-mL graduated cylinder, two 150-mL beakers, watch glass, pencil, solid sodium carbonate, 0.50 M sodium carbonate solution.

PRE-LABORATORY WORKED EXAMPLE.

A student adds 30.0 mL of 0.10 M potassium chromate into a beaker containing 100.0 mL 0.010 M silver nitrate solution. The precipitate is filtered, washed and dried. The mass of the solid product is 0.15 gram. Cr = 52; Ag = 108; O = 16; N = 14; K = 39.

Note: The molarity of a solution is the number of moles of solute in 1.0 Liter of solution.

Note: The number of moles of solute is, mole = Molarity × Volume

a. Write the balanced equation representing the chemical reaction.

2AgNO3(aq) + K2CrO4(aq)  Ag2CrO4(s) + 2KNO3(aq)

b. Write the net ionic equation of the chemical reaction.

2Ag+(aq) + CrO42-(aq)  Ag2CrO4(s)

c. Determine the number of moles of each reactant.

nAgNO3 = M × V = 0.010 mol.L-1 × 0.100 L = 0.0010 mol AgNO3 nK2CrO4

= M × V = 0.10 mol.L-1 × 0.030 L = 0.0030 mol K2CrO4

d. Identify the limiting reactant. Note: Lesser amount is the limiting reactant.

1mol Ag CrO 332 g Ag CrO

1mol K CrO2 4 1mol Ag2CrO4

The limiting reactant is AgNO3.

e. Determine the theoretical yield of the reaction.

The theoretical yield of the reaction is 0.166 g.

f. Calculate the percent yield of the reaction.

Actual Yield 0.15

Percent yield = × 100 = × 100 = 88%

Theoritical Yield 0.17

A student weighs 1.00 gram of zinc powder and places it into a beaker containing 25.00 mL of a 0.50

M CuCl2 solution. After the reaction is completed, the solid (metallic copper) is recovered from the

bottom of the beaker, filtered, washed with hydrochloric acid solution, and dried completely. The mass of the precipitate is 0.72 gram. Zn = 65, Cu = 63.5C; Cl = 35.5.

1. Write the equation representing the chemical reaction.

2. Determine the number of moles of zinc reactant present.

3. Determine the number of moles of CuCl2 reactant present

4. Calculate the theoretical yield of the reaction.

5. Calculate the percent yield of this reaction.

1. Using a pencil, write your name on the edges of the filter paper, and place it on a watch glass.Weigh the filter paper and the watch glass together.

2. Weigh an empty 150-mL beaker and record its mass.

3. Place about 1.00 gram of the powdered sodium carbonate, Na2CO3, into the beaker, and weigh it again. Record the mass.

4. Dissolve the solid using about 25 to 30 mL of deionized water.

5. Using the 50-mL graduated cylinder, measure 50.0 mL of 0.50 M CaCl2 solution.

6. Add the CaCl2 solution into the beaker containing the Na2CO3 solution. Observe the precipitation reaction. Mix the mixture with a stirring rod.

7. Assemble the stand, O-ring, funnel, and filter paper according to the figure described below. Filter the mixture.

8. Wash the residue twice with 5.0 mL of deionized water. Rinse with isopropyl alcohol and dry in theoven for half an hour.

9. Carefully remove the filter paper and precipitate and place it on the watch glass.

10. Give the watch glass to your instructor.

11. Weigh the filter paper, dried residue, and watch glass. Record the mass.

12. Calculate the percent yield of the reaction.

#Use the following stoichiometric calculations to determine the theoretical yield.

POSTLAB QUESTIONS NAME:

1. A sample of 30.00 grams of potassium phosphate was added to a solution containing 100.0 mL of 0.25 M calcium nitrate. Calcium phosphate was precipitated according to the following reaction.

2K3PO4(s) + 3Ca(NO3)2(aq)  Ca3(PO4)2(s) + 6KNO3(aq)

2.25 grams of Calcium phosphate were recovered. Calculate the theoretical yield and the percent yield.

2. A student adds 50.0 mL of 0.25 M K2SO4 solution to 25.0 mL 0.50 M BaCl2 solution. What is the mass of the precipitate that can be recovered if the percent yield of the reaction is 75%? Ba = 137; Cl = 35.5; S = 32 O = 16; K = 39. Note: You must write the balanced equation first.

BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2KCl(aq)

3. A 4.00-gram sample of sodium chloride was added to a beaker containing 40.0 mL 0.033 M

lead(II) nitrate solution. Lead(II) chloride was precipitated according to the following reaction.

2NaCl(s) + Pb(NO3)2 (aq)  PbCl2(s) + 2NaNO3(aq)

Calculate the mass of lead(II) chloride that can be recovered. Pb = 207; Na = 23; Cl = 35.5; N = 14; O

= 16.

4. A student adds 50.0 mL of 0.25 M KCl solution to 25.0 mL 0.50 M AgNO3 solution. What is the mass of the precipitate that can be recovered if the percent yield of the reaction is 90%? Ag = 108; Cl

= 35.5; N = 14; O = 16; K = 39. Note: You must write the balanced equation first.