Limiting Reagent Experiment by s0981Z3

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									                                EXPERIMENT
                             LIMITING REAGENTS

I. INTRODUCTION

Summary
In this experiment equal amounts of copper (II) chloride and aluminum will be reacted
together. The limiting reagent will be determined and the percent yield calculated. The
amount of the unreacted excess reagent will also be determined.

Discussion
In a laboratory or industrial chemical reaction, the reactants are not added in their precise
stoichiometric ratios. Usually one reactant is consumed before the other. This reactant is
known as the limiting reagent. The other reactant(s) present are in excess, and some quantity
of them are left over after the reaction is complete. These are called excess reagents. The
amounts of the reactants are determined so that the yield of the desired product is maximized,
while the cost is kept to a minimum. The most expensive reactant is usually set as the
limiting reagent.

This experiment contains other terms that are used both in the laboratory and industry in
addition to the terms limiting reagent and excess reagent. A reactant is a compound on the
left side of the balanced chemical equation and is a starting material of the process. A
product is a compound on the right side of the balanced chemical equation, and is a
compound produced in a chemical reaction. An insoluble reaction product is called a
precipitate. When separating a precipitate from a solution in a laboratory, the solution is
called the supernatant, and it is decanted, or poured off.

The % yield of a reaction is defined as follows:

                                            actual yield
                             % yield =                      x 100
                                          theoretical yield

where the actual yield is the amount of the product experimentally collected and the
theoretical yield is the amount of the product that would be obtained if every molecule of the
limiting reagent was reacted and collected. The yield of a reaction can range from 0 to
100%.

The % purity of a compound is the percentage of the pure compound present in an impure
sample.




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                   EXPERIMENT                           LIMITING REAGENTS
Example

0.8653 g of magnesium nitrate hexahydrate and 0.7215 g of potassium hydroxide react in
aqueous solution.

   (a) Predict the products, and write the balanced molecular equation (including states) for
       this reaction.
   (b) Determine the limiting reactant and the amount of precipitate that should be formed.
   (c) Determine the amount of excess reactant present at the end of the reaction.
   (d) Determine the % yield of the reaction if 0.1841 g of the precipitate are isolated.

(a) Predict the products, and write the balanced molecular equation (including states) for
    this reaction.
To predict the products in aqueous solution, it is helpful to first break up the reactants into
their ionic components. In a double displacement reaction like this, switch the positive and
negative ions or polyatomic ions and then balance the molecule using the oxidation states. In
this example,

Mg(NO3)2 and KOH will dissociate in water to give: Mg2+ + NO3‒ and K+ + OH‒

Switching the ions from one compound to the other gives: Mg2+ + OH‒ and K+ + NO3‒

Balancing these using the oxidation states gives the products: Mg(OH)2 and KNO3

The overall balanced equation becomes: Mg(NO3)2 + 2 KOH → Mg(OH)2 + 2 KNO3

To predict the states of each component, refer to a solubility chart, as shown in Table 1.

                                          Table 1
                  Solubility Rules for Common Ionic Compounds in Water

 Soluble                                                  Insoluble
 Group I and NH4+                                         Carbonates (CO32–)
 Nitrates (NO3–)                                          Phosphates (PO43–)
 Bicarbonates (HCO3–)                                     Chromates (CrO42–)
 Chlorates (ClO3–)                                        Sulfides (S2–)
 Halides (Cl–, Br–, I–)
  Except Ag+, Hg22+, Pb2+                                   Except Group I and NH4+
 Sulfates (SO42–)
  Except Ag+, Ca2+, Sr2+, Ba2+, Hg22+, Pb2+
                                                          Hydroxides (OH–)
                                                           Except Group I and Ba2+

The balanced molecular equation with states becomes

          Mg(NO3)2 (aq) + 2 KOH (aq) → Mg(OH)2 (s) + 2 KNO3 (aq)

Therefore, magnesium hydroxide is a precipitate (a reaction product that is not soluble),
while the other compounds will all remain in solution.
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                                       EXPERIMENT                                    LIMITING REAGENTS

          (b) Determine the limiting reactant and the amount of precipitate that should be formed.
          To determine the limiting reagent, the amount of product (the magnesium hydroxide
          precipitate in this example) will be calculated using each reactant (two separate calculations).
          Whichever reactant makes the least amount of product will be the limiting reactant. Some
          quantity of the other reactant will be left over at the end of the reaction and is called the
          excess reagent.

                                    1 mole Mg(NO 3 ) 2  6 H 2 O     1 mole Mg(NO 3 ) 2        1 mole Mg(OH) 2    58.3 g Mg(OH) 2
0.8653 g Mg(NO 3 ) 2  6 H 2 O x                                 x                           x                  x                  0.1968 g Mg(OH) 2
                                   256.3 g Mg(NO 3 ) 2  6 H 2 O 1 mole Mg(NO 3 ) 2  6 H 2 O 1 mole Mg(NO 3 ) 2 1 mole Mg(OH) 2



                                             1 mole KOH   1 mole Mg(OH) 2 58.3 g Mg(OH) 2
                          0.7215 g KOH x                x                x                  0.3749 g Mg(OH) 2
                                             56.1 g KOH     2 moles KOH    1 mole Mg(OH) 2


          Comparing these two results, the reactant that makes the least amount of product is
          Mg(NO3)2 • 6 H2O. Therefore, it is the limiting reagent, and the maximum amount of
          Mg(OH)2 that can be produced is 0.1968 g. Never assume that the reactant with the smaller
          initial mass is the limiting reagent!

          Often, the reactants are in solution. To determine the limiting reagent in these examples,
          calculate moles of each reactant based on their volume and molarity, and continue through to
          the desired product.

          When determining the limiting reactant using each reactant amount, if the product produced
          from both reactions is less than the amounts of both starting materials, usually the reactant
          that gives the highest percentage yield is determined to be the limiting reactant.


          (c) Determine the amount of excess reactant present at the end of the reaction.
          Determining the amount of excess reactant present at the end of the reaction is a two step
          process. First, the amount of excess reactant consumed is calculated. When doing this, it is
          helpful to start with the limiting reagent amount and then calculate the amount of excess
          reagent needed to react with this. Then, the amount consumed is subtracted from the original
          reactant amount to obtain the amount left unreacted.

                Amount of excess reactant consumed in the reaction:

                                                   1 mole Mg(NO 3 ) 2  6 H 2 O        2 moles KOH            56.1 g KOH
               0.8653 g Mg(NO 3 ) 2  6 H 2 O x                                 x                           x             0.3788 g KOH
                                                  256.3 g Mg(NO 3 ) 2  6 H 2 O 1 mole Mg(NO 3 ) 2  6 H 2 O 1 mole KOH


                Amount of excess reactant present after the reaction:

                0.7215 g KOH at start – 0.3788 g KOH reacted = 0.3427 g KOH in excess


          (d) Determine the % yield of the reaction if 0.1841 g of the precipitate are isolated.
              Using the formula
                                                 actual yield
                            % yield =                            x 100      % = 0.1841 g x 100 = 93.55%
                                               theoretical yield                       0.1968 g
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                  EXPERIMENT                            LIMITING REAGENTS

II. EXPERIMENTAL

A. Reagents                                                         All aqueous waste can
    Copper (II) chloride dihydrate            CuCl2 • 2 H2O         be disposed down the
    Aluminum foil                             Al                    laboratory sink. Cu can
    Hydrochloric acid, 6 M                    HCl                   be disposed in trash.

B. Procedure
1. Weigh and accurately record approximately 0.5 g of CuCl2 • 2 H2O. Place in a 100
     mL beaker.

2.   Weigh and accurately record approximately 0.5 g of aluminum foil. (~4” x 4”)

3.   Weigh an evaporating dish.

4.   Add approximately 30 mL of DI water to the CuCl2 • 2 H2O. Swirl to dissolve the
     solid. It may be necessary to use a stirring rod to break up the solid. Use the DI water
     bottle to rinse off the stirring rod to the beaker. Note the color of the solution.

5.   Add the aluminum foil, roughly torn and crumpled in loose balls to the beaker.

6.   After 3 minutes, add approximately 20 mL of 6 M HCl.

7.   Wait until all the bubbling has subsided. Gently break up any floating copper with
     a stirring rod to get all of the copper to settle. Do not pulverize it. Record the color of
     the solution.

8.   When all the copper has settled, decant about half of the supernatant to a 400 mL
     beaker using this technique: put the stirring rod at a 90° angle in the lip of the 100 mL
     beaker, with the stirring rod extending into the 400 mL beaker. Pour the liquid out.

9.   Add about 30 mL of water from the DI water bottle to the contents in the 100 mL
     beaker by squirting water on the side of the beaker (not directly into the solution). Let
     the copper settle again and decant most of the supernatant.

10. Repeat step 9 five more times, until the water with the copper has a neutral pH.

11. Decant most of the water and transfer the copper to an evaporating dish. Use a
    small amount of water from the DI water bottle to completely transfer the copper.

12. Place the evaporating dish on a wire gauze and heat the copper to dryness using a
    Bunsen burner. Heat for 2 minutes after the water has been evaporated. If the copper
    starts turning dark, remove the heat source and continue with the next step.

13. Shut off the Bunsen burner and remove the evaporating dish from the wire gauze
    with the hot hands to the laboratory bench.

14. When the evaporating dish is at room temperature, weigh the dish and copper.
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                   EXPERIMENT                          LIMITING REAGENTS

                        LABORATORY REPORT SHEET
Name___________________________________Date________________
Partner _____________________________

Mass of copper (II) chloride dihydrate
Mass of aluminum
Mass of empty evaporating dish
Mass of evaporating dish plus copper
Mass of copper

1.   Write the balanced molecular reaction for this reaction in aqueous solution. Include states.


2.   Write the balanced ionic and net ionic reactions for this reaction. Include states.

Ionic:


Net Ionic:


3.   Calculate the theoretical amount of copper produced based on the mass of the copper
     (II) chloride dihydrate.




4.   Calculate the theoretical amount of copper produced based on the mass of the
     aluminum.




5.   What is the limiting reagent? _____________________________

6.   How much excess reagent is left over?




7.   What is the % yield of copper?


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                  EXPERIMENT                           LIMITING REAGENTS
Conclusion:
In a short paragraph written in a technical manner, summarize in one or two sentences the
general lab procedure. Next, in sentence form, detail relevant data from the laboratory report
sheet, in logical order, including units where appropriate. Also include physical observations
that happened during the reaction.

Questions and Error Analysis:
1. What physical parameter confirms the reaction is complete?



2.   What evidence gives a clue as to which reactant is the limiting reagent?



3.   Why is the HCl added to the solution?



4.   Why are the exact amounts of water and HCl relatively unimportant?



5.   Give two sources of error as to why the percent yield in this experiment may be greater
     than 100%.



6.   Give two sources of error as to why the percent yield in this experiment may not be
     exactly 100%.



7.   Assuming the same amount of copper (II) chloride dihydrate as used in this experiment
     was used in another experiment, what is the maximum mass of aluminum that would
     make aluminum the limiting reagent?




8.   What physical evidence would show that aluminum is the limiting reagent?


9.   In industry, what would be a factor in making one reactant the limiting reagent and
     another reactant the excess reagent?


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                  EXPERIMENT                          LIMITING REAGENTS


                             HOMEWORK EXERCISES

1. 17.98 g of calcium chloride and 18.05 g of potassium hydroxide are added to 1.000 L of
   water and react to form calcium hydroxide and potassium chloride.
   (a) Write the balanced molecular equation for this reaction. Include states.
   (b) Write the ionic and net ionic equations for this reaction. Include states.
   (c) Determine the limiting reagent and how many grams of calcium hydroxide should be
       formed.
   (d) How much excess reagent is left over?
   (e) If 10.56 g of calcium hydroxide are obtained in an experiment, what is the % yield of
       the experiment?

2. 25.64 g of calcium sulfate pentahydrate and 12.69 g of sodium phosphate are added to
   1.000 L of water.
   (a) Predict the products and write the balanced molecular equation for this reaction.
       Include states.
   (b) Write the ionic and net ionic equations for this reaction. Include states.
   (c) Determine the limiting reagent and how many grams of the precipitate are formed.
   (d) How much excess reagent is left over?
   (e) If 9.95 g of the product were recovered, what is the % yield of the reaction?

3. 200.0 mL of 0.350 M lead nitrate are mixed with 300.0 mL of 0.250 M sodium iodide.
   (a) Predict the products and write the balanced molecular equation for this reaction.
       Include states.
   (b) Write the ionic and net ionic equations for this reaction. Include states.
   (c) Determine the limiting reagent and how many grams of the precipitate are formed.
   (d) How much (in g) of the excess reagent is left over?
   (e) What is the concentration of the excess reagent in solution?

4. In the industrial synthesis of aspirin, 200 kg of salicylic acid, C7H6O3, and 100 kg of
   acetic anhydride, (CH3CO)2O, form acetylsalicylic acid, C9H8O4, (aspirin) and acetic
   acid, C2H4O2.
   (a) Write the balanced chemical reaction for this reaction.
   (b) In a manufacturing process, 200 kg of aspirin are produced from 200 kg of impure
       salicylic acid and 100 kg of acetic anhydride. The salicylic acid is the limiting
       reagent. Calculate the purity of the salicylic acid.
   (c) How much excess reagent is left over, assuming there are no side reactions between
       the impurities in the limiting reagent and the excess reagent?




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