Is There Enough of Each Chemical Reactant to Make a Desired Amount of Product?

Limiting and Excess Reactants (H) Is there enough of each chemical reactant to make a desired amount of product? Why? You have spent a lot of time studying the various types of reactions that can occur in chemistry. You have also become experts in balancing chemical equations. One thing you will do in this activity is practice stoichiometry skills while solving a problem with a familiar campfire snack. Stoichiometry is the chemical term to describe calculations that allow us to find the amounts of chemicals involved in a given reaction. We will also look at the idea of limiting reactants. Two atoms or molecules must come together in just the right way in order for them to react. As a result, it is virtually impossible to obtain 100% yield in a chemical reaction by combining the reactants in exact proportions. In order to increase the odds that at least one reactant will react completely, we often add more than is needed of another reactant. In this activity, you will look at several situations where the process or reaction is stopped because one of the required components has been used up.

Model 1 – Assembling a S’more In this activity, you will use a recipe for S’mores as an analogy for a chemical equation in which reactants and products are in set proportions to each other. You will be given varying amounts of each reactant. One of these reactants will limit the number of S’mores you can produce. The other reactants will be in excess. After working with this culinary “reaction,” you will be able to identify the limiting and excess reactants in chemical reactions.

PROCEDURE: 1. Obtain a plastic bag of s’mores ingredients. Record your group number. ______Record the amount of each ingredient.

graham cracker chocolate marshmallows______quarters______squares______

2. Write a balanced equation for the synthesis reaction using the chemical symbols given in the recipe. Recipe for 1 S’more (G2CsM) 2 graham cracker quarters (chemical symbol G) 1 chocolate square (chemical symbol Cs) 1 marshmallow (chemical symbol M)

Balanced equation:

3. Use the following recipe to perform a synthesis reaction using your ingredients. (i.e: cause the reaction to go to completion by forming as many S’mores as possible).

4. How many products (S’MORES) were you able to form? ______

(This is known as your actual yield)

*YOU MAY START ROASTING AND EATING BUT CONTINUE WITH ANSWERING THE LAB QUESTIONS Chem Is Try: Critical Thinking Questions

5. What was your limiting reactant (which ingredient limited how many s’mores your group could make)?

6. Define, in your own words, what a limiting reactant is.

7. Which starting materials were in excess?

8. Define, in your own words, what an excess reactant is.

10. Suppose you had a bag of 11 graham crackers, 12 chocolate squares, and 7 marshmallows.

a. How many s’mores can you make? This is known as your theoretical yield, or amount of product that should be produced given the quantity of reactants. Explain your reasoning.

b. What (if anything) will be left over, and how much of that item will there be? Explain your reasoning.

c. What item limits the number of s’mores you can make? Explain your reasoning.

Model 2- Manufacturing a S’more

Molar Masses of the Substances:

Substance Molar Mass (mass of 1 piece or 1 mole) 3.69 g/mole Molar mass of 1 piece of Chocolate (Cs) Molar mass of a Marshmallow (M) 7.34 g/mole Molar mass of a Graham cracker (G) 7.75 g/mole

Molar mass of a S’more (G2MCs) You figure this out!

Chem Is Try: Critical Thinking

Questions

11. Calculate the Molar mass of the S’more (G2MCs) below and put in the table above:

12. If you have 4 moles (pieces) of M, how many grams of M do you have? Show your work.

13. If you have 30 grams of Cs, approximately how many moles (pieces) do you have? Show your work.

14. If you have 33.21 grams of Cs, a. How many moles of M and G will be necessary to make the maximum number of S’mores possible? Show your work.

b. How many grams of S’mores can be made? Show your work.

Model 3 – Assembling & Manufacturing a Race Car Chem Is Try: Critical Thinking Questions

15. How many of each race car part are needed to construct one complete race car?

Body (B) Cylinder (Cy) Engine (E) Tire (Tr)

16. How many of each part would be needed to construct three complete race cars? Show your work.

Body (B) Cylinder (Cy) Engine (E) Tire (Tr)

17. Assuming that you have 15 cylinders and an unlimited supply of the remaining parts:

a. How many complete race cars can you make? Show your work.

b. How many of each remaining part would be needed to make this number of cars? Show your work.

18. Count the number of each race car part present in Container A of Model 3.

Body (B) Cylinder (Cy) Engine (E) Tire (Tr) 19. Complete Model 3 by determining the number of cars that can be made from the parts in Container A. What excess parts do you have? How many of each part? Draw the cars next to Container A in Model 3.

20. A student says, “I can see that we have three car bodies in Container A, so we should be able to build three complete race cars.” Explain why this student is incorrect in this case.

21. The number of complete cars you predicted that could be built is called the theoretical yield. Theoretical yield is defined as the maximum amount of product produced during ideal conditions in a chemical reaction. It is what we mathematically think will be produced. What is the theoretical yield of cars in this scenario?

22. Fill in the table below with the maximum number of complete race cars that can be built from each container of parts (A–E), and indicate which part limits the number of cars that can be built. Divide the work evenly among group members. Space is provided below the table for each group member to show their work. Have each group member describe to the group how they determined the maximum number of complete cars for their container. Container A from Model 3 is already completed as an example. 1 B + 3 Cy + 4 Tr + 1 E = 1 car

23. You are planning activities for a younger sibling’s birthday party. You decide that each of the 10 partygoers will build a toy race car to take home. You have almost enough toy race car parts to build 100 cars, except you only have 10 car bodies and 46 tires. a. Which component is the limiting reactant? Explain how you came to this conclusion. Show any math you did.

b. Do you relax or do you run to the toy store to buy more car pieces? Why?

24. Look back at questions 20 and 21 with your group. Is the component with the smallest number of pieces always the one that limits production? Explain your group’s reasoning.

Model 4 – Assembling Water Molecules Chem Is Try: Critical Thinking Questions

25. Refer to the chemical reaction in Model 4.

a. How many moles of water molecules are produced if one mole of oxygen molecules completely reacts? (Show your work.)

b. How many moles of hydrogen molecules are needed to react with one mole of oxygen molecules? (Show your work.)

26. Complete Model 4 by drawing the maximum moles of water molecules that could be produced from the reactants shown, and draw any remaining moles of reactants in the container after the reaction as well.

a. Which reactant (oxygen or hydrogen) limited the production of water in Container Q?______

b. Which reactant (oxygen or hydrogen) was present in excess and remained after the production of water was complete?______27. Fill in the table below with the maximum moles of water that can be produced in each container (Q-U). Indicate which reactant limits the quantity of water produced—this is the limiting reactant. Also show how much of the other reactant—the reactant in excess—will be left over. Divide the work evenly among group members. Space is provided below the table for each group member to show their work. Have each group member describe to the group how they determined the maximum number of moles of water produced and the moles of reactant in excess. Container Q from Model 4 is already completed as an example.

2H2 + O2 2H2O Container Moles of Moles of Max. Moles of Limiting Reactant in Hydrogen Oxygen Water Reactant Excess Produced

Q 7 3 6 O2 1 mole H2 R 8 3 S 10 5 T 5 5 U 8 6

28. Look back at Questions 27. Is the reactant with the smaller number of moles always the limiting reactant? Explain your group’s reasoning. Model 5 – Limiting Reactant Mathematical

You know you have a limiting reactant problem if you are given starting quantities of two reactants. Below are two examples of mathematical calculations that could be performed to find the limiting reactant for Container U in Question 26, which has 8 mol H2 and 6 mol O2

Balanced Equation: 2H2 + O2 2H2O

Example #1 Example #2

Chem Is Try: Critical Thinking Questions

Use example #1 or example #2 to solve to find the limiting reactant. Show all work.

29. Iron and oxygen combine to produce iron (III) oxide. Write a balanced equation below, and then complete the columns of the data table. balanced equation:

Show work Have (Given amounts)

30 moles iron

25 moles oxygen

30. Oxygen and hydrogen gas combine to produce water. Write a balanced equation below then complete the data table. balanced equation:

Show work Have (Given amounts) 12 moles oxygen

15 moles hydrogen

Example Starting with Mass

Consider the following reaction:

2 Al + 6HBr  2AlBr3 + H2

When 5.45 g of Al and 5.45 g of HBr are reacted, what is the theoretical yield of AlBr3? (Remember… theoretical yield the predicted mass of the product if the reaction occurs perfectly).

Step 1: Do a mass to mass Stoichiometry calculation from each reactant to the product.

5.45 g Al 1 mol Al 2 mol AlBr3 266.7 g AlBr3 = 53.8 g AlBr3 27 g Al 2 mol Al 1 mol AlBr3

5.45 g HBr 1 mol HBr 2 mol AlBr3 266.7 g AlBr3 = 5.99 g AlBr3 80.9 g HBr 6 mol HBr 1 mol AlBr3

Step 2: Determining the Limiting Reactant

The limiting reactant limits the amount of product produced…it has run out! You can figure out the limiting reactant by choosing the reactant that produced less product.

Answer: HBr is the limiting reactant because it produces less product that Al.

Step 3: Determining the Theoretical Yield

The mass produced by the limiting reactant is the theoretical yield.

Answer: 5.99 g AlBr3 is the theoretical yield because that is the maximum quantity able to be produced by the limiting reactant.

Exercises 31. Consider the synthesis of water as shown in Model 5. A container is filled with 10.0 g of H2 and 5.0

g of O2.

a. Write the balanced equation b. What is the theoretical yield of water? Show your work. Once this is determined, what is the limiting reactant?

c. Which reactant is the limiting reactant? Which reactant is the excess reactant?

32. Given the balanced chemical equation:

2NO(g) + O2(g)  2NO2(g) a. Calculate the mass of nitrogen dioxide that can be made from 30.0 grams of NO and 30.0 grams of O2 (Hint: do two mass-to mass stoichiometry)

b. Which produced the least amount of nitrogen dioxide? Which reactant is the limiting reactant? Which is the excess reactant?

Information: Calculating Percent Yield

Equation: Percent Yield= ___ Actual Yield x100 Theoretical Yield

Percent Yield is the actual yield (what you measured in the lab) divided by the theoretical yield (the mass you should have obtained in the lab based on Stoichiometry) multiplied by 100 to get the percent. This is used a way to measure how good your experiment was. The closer to 100% yield you are, the better your experiment was.

Answer the following question using the data from question 32.

33. What is the percent yield of NO2 if your actual yield was 42 g?

Model 6—Determing the Amount of Excess Reactant Remaining

EXAMPLE: Consider the following reaction:

2 Al + 6HBr  2AlBr3 + H2

When 5.45 g of Al and 5.45 g of HBr are reacted, how much excess reactant remains?

Step 1: Determine the limiting reactant

5.45 g Al 1 mol Al 2 mol AlBr3 266.7 g AlBr3 = 53.8 g AlBr3 27 g Al 2 mol Al 1 mol AlBr3

5.45 g HBr 1 mol HBr 2 mol AlBr3 266.7 g AlBr3 = 5.99 g AlBr3 80.9 g HBr 6 mol HBr 1 mol AlBr3

Answer: HBr is the limiting reactant because it produces less product that Al. This means that Al is the excess reactant.

Step 2: Determining the Excess Reactant Used

You have to determine the mass of the excess reactant USED first by doing MassMass Stoich from the amount of limiting reactant you start with to the mass of excess reactant. This is the mass of excess reactant that is USED DURING THE REACTION.

5.45 g HBr 1 mol HBr 2 mol Al 27 g Al = 0.61 g Al used 80.9 g HBr 6 mol HBr 1 mol

Step 3: Determining the Excess Reactant Remaining Subtract the amount used from the amount you started with.

5.45 g Al – 0.61 g Al = 4.84 g Al remaining You Try!

34. Given the following reaction:

Pb(NO3) 2 + 2NaIà2NaNO3 +PbI2 If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide…

a. What is the theoretical yield of sodium nitrate that can be formed?

b. How many grams of the excess reactant will be left over?

35. When 2.60 g of hydrogen react with 10.20 g of oxygen, water is formed.

Balanced Eq:

(a) What mass of water will be produced?

(b) How many grams of the excess reactant will be left over?

(c) What is the percent yield if your actual yield was 9.46 g H2O? a. Teacher Notes: For a class of 26 and groups of 3 (with 2 groups of 4) Bag M G Cs Limiting Reactant 1 4 6 4 G 2 4 6 4 G 3* 5 8 5 G 4 4 8 3 Cs 5 4 8 3 Cs 6 4 8 3 Cs 7* 4 10 5 M 8 3 8 4 M *group of four

Cs(chocolate squares) You need 128 pieces for 4 classes. I used 32 pieces for each class. I bought 2 bags (18.50 oz) of the Hershey miniatures (not nuggets) but I can only find them in assorted flavors so warn you student about nut allergies. You may need a few extra pieces. I liked how they are individually wrapped.

M (marshmallows) You will need 124 marshmallows for 4 classes. I used ¾ of a 10 oz bags (big ones) for each class (so 3 bags in total with about ½ bag left over).

G(graham crackers) You will need 248 half(quarter) pieces for 4 classes. I used 62 half(quarter) pieces for each class. Honey maid sells them in the half pieces, “Honey Maid Fresh Stacks” so they are in stacks of 8 quarters (half pieces) wrapped. So I used about 1 and 1/3 box for each class (so 5 boxes in total).