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[answered] Lab 19: Stoichiometry Objectives Demonstrate the use of sto


I need help with pages 8, 9, and 10. ?The tables and the questions.


Lab 19: Stoichiometry

 

Objectives

 

Demonstrate the use of stoichiometry to synthesize calcium carbonate

 

Practice using a scale and proper lab techniques

 

Find the limiting reagent, the theoretical yield, and the percent yield Introduction

 

Have you ever wondered why hot dogs are sold in packages of

 

10, but hot dog buns are sold in packages of 8?

 

This is an example of a ratio. Stoichiometry is the scientific word for ratios of

 

moles. It is the study of the quantitative relationships between reactants and

 

products in a chemical reaction. These relationships are often referred to as mole

 

-to-mole ratios. Mole-to-mole ratios help scientists know how much of each reactant is needed to produce a sufficient amount of product. In this lab, you will

 

observe the importance of mole-to-mole ratios when making the main component in lime that is used in gardens! Figure 1: Buying hot dogs and buns can

 

be complicated! The manufacturers of hot dogs and hot dog buns would not make very good chemists. Hot dogs are sold in packages of 10

 

where hot dog buns are sold in packages of only eight. If one package of each is purchased, how many hot dog-on-buns

 

can you make? The answer of course is eight, meaning there are two hot dogs left over. Now examine the equation below

 

and try to fix their mistake. How many packages of hot dogs and hot dog buns must you purchase in order to not waste

 

any hot dogs? How many hot dog-on-buns does this give you? ___ 10 Hot dogs + ___ 8 Hot dog buns ? ___ Hot dog-on-buns If you buy four packages of hot dogs and five packages of buns you would get 40 hot dog-on-buns.

 

Balancing a chemical equation is done the same way: you place numbers, called coefficients, in front of each type of molecule. Just like you cannot change how many hot dogs there are in a package of hot dogs, you cannot change how many of

 

each type of atom there is in a chemical substance. You can, however, change the ratio of the chemical substances so that

 

the same number of each type of atom is on both sides of the reaction equation. In the hot dog example there are 40 hot

 

dogs and 40 hot dog buns on both sides of the arrow.

 

But what if instead you buy only two packages of hot dogs, and three packages of hot dog buns, which would you run out

 

of first? You would have a total of 20 hot dogs and 24 hot dog buns, so you would run out of the hot dogs first. This means

 

you could only make 20 of the hot dog-on-buns final product. The number of hot dogs limits the number of hot dog-onbuns you can make.

 

This is similar to what in chemistry is called the limiting reagent. By using stoichiometry it is possible to calculate the limiting reagent of a reaction. Knowing which reactant will be consumed first in a reaction allows chemists to calculate how

 

much product can be theoretically made. This can be compared to the actual amount produced by calculating the percent

 

yield. The percent yield is found by using the following equation: Percent Yield = Actual Yield

 

Theoretical Yield 100

 

195 Lab 19: Stoichiometry

 

Using the hot dog example, the percent yield would be like only making 19 hot dogs when there were enough hot dogs and

 

buns to make 20 hot dog-on-buns. In this example the percent yield would be calculated as follows: Percent Yield = 19 Actual Hotdog-on-buns

 

20 Theoretical Hotdog-on-buns 100 Percent Yield = 95%

 

Chemists will often try making a desired product many different ways. They will then compare the percent yields to determine which of the methods tried gives the greatest yield.

 

In this laboratory exercise you will explore the concepts of limiting reagents and percent yields through making calcium

 

carbonate, the main component in lime. Lime is used in many different industries such as farming, where it is used to adjust soil pH, making crops more productive.

 

Calcium carbonate, CaCO3, can be made by mixing CaCl2 and K2CO3 in water. The trick is to be sure to use the correct

 

amount of each. Look at the following reaction and balance it to determine the mole-to-mole ratio that would be necessary

 

to have the same number of each kind of atom on both sides of the equation. ___ CaCl2(aq) + ____ K2CO3(aq) ? ____ KCl(aq) + ____ CaCO3(s) When this chemical equation is balanced there is one mole of calcium chloride for every one mole of potassium carbonate

 

and together these quantities make two moles of potassium chloride and one mole of calcium carbonate. The potassium

 

chloride is soluble in water so it remains in solution, while the calcium carbonate will precipitate out as a white solid. CaCl2(aq) + K2CO3(aq) ? 2KCl(aq) + CaCO3(s) What if instead you added 2 moles of CaCl2 and 1 mole of K2CO3. How much CaCO3 could you make? The answer is still

 

only one mole of CaCO3. The K2CO3 is the limiting reagent since it limits how much of the product can be made.

 

The following calculation illustrates how the limiting reagent and percent yield are determined for the above reaction. Determining the limiting reagent can be complex, so breaking the calculation into several steps can be useful. The following

 

example demonstrates how to determine the limiting regent for a typical reaction. 196 Lab 19: Stoichiometry

 

Example:

 

A procedure calls for 8.25 g of CaCl2 to be dissolved in water and reacted with 5.00 g of K2CO3. How many grams of CaCO3 can

 

theoretically be made? If 3.25 g of CaCO3 were actually made, what is the percent yield?

 

a. First write the balanced reaction equation: CaCl2 + K2CO3

 

b. Calculate the formula weight for each reactant:

 

Ca = 40.08 g/mol

 

+ 2 Cl = 70.90 g/mol

 

= 110.98 g/mol CaCl2 c. d. 2 K = 78.20 g/mol

 

3 O = 48.00 g/mol

 

+

 

C = 12.01 g/mol

 

= 138.20 g/mol K2CO3 Calculate the number of moles used of each reactant:

 

CaCl2 : (8.25 g) (1 mol/110.98 g) = 0.0743 mol K2CO3: (5.00 g) (1 mol/138.21 g) = 0.0362 mol Determine how many moles of the product could be made by each of the reactants. CaCl2 : K2CO3: e. 2KCl + CaCO3 ? (0.0743 mol CaCl2 ) (0.0362 mol K2 CO3 ) 1 mol CaCO3

 

1 mol CaCl2 = 0.0743 mol CaCO3 1 mol CaCO3

 

1 mol K2 CO3 = 0.0362 mol CaCO3 Determine the limiting reagent. It is the reactant that will make the least number of moles of CaCO3.

 

K2CO3 is the limiting reagent!

 

This means that only 0.0362 moles of CaCO3 can be made when 8.25 g of CaCl2 and 5.00 g of K2CO3

 

are mixed in water. In this example, the reactant that had the lowest mass in the procedure turned

 

out to be limiting. However, this is not always true! The only way to know for sure which reactant is

 

limiting is to calculate it. 197 Lab 19: Stoichiometry

 

f. Determine how many grams of CaCO3 can theoretically be produced:

 

First calculate the formula weight for the product, CaCO3: Ca = 40.08 g/mol

 

3 O = 48.00 g/mol

 

+

 

C = 12.01 g/mol

 

= 100.09 g/mol CaCO3 Next calculate the theoretical yield by calculating how many grams there are in 0.0362 moles of

 

CaCO3: Theoretical Yield = (0.0362 mol CaCO3 ) g. 100.09 g CaCO3

 

1 mol CaCO3 = 3.62 g CaCO3 Calculate the percent yield: Percent Yield = 3.25 g

 

3.62 g 100 = 89.8% This means that in this example only 89.8% of what theoretically could have been made was actually made. 198 Lab 19: Stoichiometry

 

Pre-lab Questions

 

1. What is a limiting reagent? 2. A student used 7.15 g of CaCl2 and 9.25 g of K2CO3 to make CaCO3. The actual yield was 6.15 g of CaCO3. Calculate the limiting reagent and the percent yield.

 

1 mole CaCl2 requires 1 mole K2CO3 to fully react

 

A 1 : 1 reaction ratio

 

moles CaCl2 = mass / molar mass

 

= 7.15 g / 110.98 g/mol

 

= 0.064426 moles

 

moles K2CO3 = 9.25 g / 138.21 g/mol

 

= 0.066927 mol

 

The balanced equation shows that 1 mole CaCl2 reacts to form 1 mole CaCO3. Another 1 : 1 ratio

 

So 0.064426 moles CaCl2 can form up to 0.064426 moles CaCO3

 

mass CaCO3 possible = moles x molar mass

 

= 6.45 g

 

ANSWER:

 

% yield = actual mass / theoretical yield x 100/1

 

= 6.15 g / 6.45 g x 100/1

 

= 95.3 %

 

the reaction is 1:1, so the limiting reagent is going to be whatever you have less of, in this case the

 

CaCl2. 199 Lab 19: Stoichiometry

 

Experiment: Synthesis of Garden Lime

 

Materials

 

Safety Equipment: Eye goggles, gloves

 

Calcium Chloride (CaCl2)

 

Potassium Carbonate (K2CO3)

 

Ethanol

 

100 mL Graduated Cylinder

 

250 mL Erlenmeyer flask

 

250 mL beaker

 

50 mL beaker

 

Watch glass Scale

 

Funnel

 

Filter paper

 

Spatula

 

Stirring rod

 

Wash bottle

 

Distilled water*

 

Oven*

 

*You must provide Procedure

 

1. An example set of data has been provided for 1.0 g CaCl2. 2. For Trial 1, weigh into a 250 mL beaker the amount of calcium chloride (CaCl2) shown in Table 1. Record the exact

 

mass you weigh out in the Trial 1 column of the Data section. Example About 1.0 g CaCl2 Trial 1 About 2.0 g CaCl2 3. Measure 50.0 mL of distilled water into a 100 mL graduated

 

cylinder. Pour the water into the 250 mL beaker with the

 

calcium chloride. Trial 2 About 3.0 g CaCl2 4. Stir the solution with a stirring rod until all of the calcium chloride is dissolved. 5. Weigh out 2.5 g of potassium carbonate (K2CO3) in a 50 mL beaker. Record the exact mass in the Data section. 6. Measure 25.0 mL of distilled water into a 100 mL graduated cylinder. Add the water into the 50 mL beaker

 

containing the potassium carbonate. 7. Stir the potassium carbonate in the distilled water with a stirring rod until it is all dissolved. 8. Pour the K2CO3 solution into the 250 mL beaker that has the CaCl2 solution. Rinse the beaker that contained

 

the K2CO3 with a few mL of water and add this to the CaCl2 solution. Stir the mixture. 9. As soon as the reaction begins, record your observations in the Data section. Continue stirring until you see

 

no more precipitate forming. Table 1: Approximate CaCl2 amounts 10. Set up the funnel in the Erlenmeyer flask as shown in Figure 2. HINT: Do NOT begin filtering yet!

 

11. Zero the scale and weigh a piece of filter paper and a watch glass. Record the masses of both items in the

 

Data section. 200 Lab 19: Stoichiometry

 

a b c d e f Figure 2: Filtering funnel preparation diagram 12. Prepare a filtering funnel as shown in Figure 2: fold a piece of filter paper in half twice to make quarters, and

 

open the paper to make a small cone (three quarters are open on one side and one quarter is on the opposite

 

side). Place the paper cone into the funnel and hold it in place with your fingers. Pour a small amount of distilled

 

water through the paper to secure it inside the funnel.

 

13. Filter the mixture by pouring it into the filter paper in the funnel. Use the stirring rod and distilled water in a wash bottle to

 

transfer the entire solid into the filter paper. HINT: For best

 

results, be sure to transfer all of the precipitate into the filter

 

paper. Use a rubber policeman if it is available to help with the

 

transfer.

 

14. Rinse the remaining solid in the filter paper twice with distilled

 

water from a wash bottle to rinse off excess sodium chloride

 

(NaCl). After all the liquid has filtered through, rinse the product with approximately 5 mL of ethanol to aid in its drying. Allow the ethanol to completely finish filtering through the paper.

 

15. Remove the filter paper carefully so as to not lose any product.

 

Gently unfold the filter paper and lay it flat on the pre-weighed

 

watch glass to dry.

 

16. Allow the product to air dry completely. This may take 24 hours

 

or more. Once dry, weigh the dry product on the filter paper

 

and watch glass. Record the total mass in the Data section.

 

Calculate the mass of the product.

 

17. Repeat the above procedure for Trial 2 using the amount of

 

CaCl2 mass indicated in Table 1. Filter Paper

 

Funnel Flask Figure 3: Filter Setup 18. To clean up, wash any dirty glassware, pour liquids down the

 

drain, and throw the product on the filter paper in the trash. 201 Lab 19: Stoichiometry

 

Data

 

1. Record your data for each of the trials in Table 2. 2. Record your reaction observations (Step 8) below: Table 2: Reaction product data Mass (g) Example Mass of CaCl2 1.0 g Mass of K2CO3 2.5 g Mass of filter paper 0.8 g Mass of watch glass 38.5 g Combined mass of product, filter paper, and

 

watch glass 40.2 g Mass of dry product 0.9 g 202 Trial 1 Trial 2 Lab 19: Stoichiometry

 

Calculations

 

1. Determine the limiting reagent for each trial. Show your calculations. (Hint: See the example in the Introduction.)

 

Example: Remember: the limiting reagent is going to be whatever you have less of, in this case the CaCl2. Trial 1: Note: These should be about the same and either CaCl2 or K2CO3 can be the limiting reagent

 

depending on their initial masses. Trial 2: 203 Lab 19: Stoichiometry

 

Table 4: Comparison of theoretical and actual yields for CaCO3 Trial # Limiting Reagent Theoretical Yield

 

of CaCO3 Actual Yield of

 

CaCO3 % Yield Trial 1 Trial 2 Trial 3 204 2. Calculate the theoretical yield of CaCO3 that could be produced by each trial and then fill in Table 2. 3. Find the percent yield each trial obtained for the CaCO3. Lab 19: Stoichiometry

 

Post-lab Questions

 

1. Compare the results of the different trials. How does the amount of grams of CaCO3 compare? 2. Were the results of the trials as you expected? Why or why not? 3. Predict what would happen if 6.0 grams of CaCl2 were used for the reaction and the amount of K2CO3 remained the same. 205

 


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