A5 Stoichiometry Limiting Reagent PDF

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Jespersen 7th Ed. Ch. 3

A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

Stoichiometry, Limiting Reagents & Reaction Yields Why? Stoichiometry is the process of tracking the elements, molecules, and their masses in chemical reactions to ensure that reaction equations and calculations do not violate the conservation of mass. Scientists, working with reactions perform stoichiometric calculations to quantitatively account for the amounts of material that react and that are produced in chemical reactions. You will find these calculations straightforward if you use the reaction equation to relate the quantities of the chemical compounds in moles. When reactants are combined for a chemical reaction, they are not always present in the exact amounts required to completely use up all of the reactants. Frequently, one of the reactants will be completely used up before the rest. Once this happens, no further products can form and the reaction comes to a halt. This is a useful strategy in the laboratory and in any cost-effective production process as we can add an excess of the less expensive reagents and limit the amount of the expensive reagent. As a result, it is necessary to recognize which component limits the amount of material that can be produced. Identifying the limiting reactant in a chemical reaction will strengthen your skills in dealing with moles, and reaction stoichiometry. Learning Objectives • Develop method to balance equations • Understand how to calculate the amounts of products and reactants needed in a reaction. • Determine which reactant is limiting and which is (are) the excess reactant(s) when the amounts of two or more reagents are given in a problem. Success Criteria • Successfully balance an equation • Successful identify a wide variety of stoichiometry problems. • Accurately calculate the amounts of material reacting and being produced • Correctly identify the limiting reagent in a reaction and calculate the amount of product that can be produced. Prerequisites • Atoms, Isotopes, and Ions/Periodic Table/Nomenclature • Moles, Molar Mass, and Percent Compostion • Formulas, Equations & Stoichiometry Homework • Read Models 1 – 3 • Answer Key Questions 1 – 24

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Jespersen 7th Ed. Ch. 3

A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

Model 1: Stoichiometry of Chemical Reactions Information: previously, we have focused on mole ratios between elements within a single compound. We have seen that the essential conversion factor between substances within a compound is the mole ratio obtained from the compound’s formula. In this model, we will see that the same techniques can be used to relate substances involved in a chemical reaction. The tool that relates substances involved in a reaction is the mole ratio obtained from the coefficients of the balanced chemical equation. Grams of Substance A

Grams of Substance B

Molar Mass A

Molar Mass B Formula or Balanced Chemical Equation

Moles of A Avogadro’s Number

Moles of B Avogadro’s Number

Elementary Units of A (Atoms, molecules, ions)

Elementary Units of B (Atoms, molecules, ions)

Figure 1. Stoichiometry pathways. The boxes represent the units that we start with and want to end at. Arrows between the boxes indicate the tools that provide the needed conversion factors.

For example, we can use the stoichiometric coefficients of the balanced net equation to form conversion factors that allow us to: 1. Relate the moles of one reactant or product to the moles of another reactant or product. 2. Relate the grams of one reactant to the moles of another reactant or product. 3. Relate the grams of one reactant to the grams of another reactant or product. Ex.1 How many moles of sulfuric acid, H2SO4, are needed to react with 0.366 mol of NaOH by the following balanced chemical equation? 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O Stoichiometric pathway: moles NaOH → moles H2SO4. The equation shows that there are 2 moles of NaOH for every 1 mole of H2SO4. 1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4 = 𝟎. 𝟏𝟖𝟑 𝒎𝒐𝒍 𝑯𝟐𝑺𝑶𝟒 2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 Ex.2 In the reaction 2SO2(g) + O2(g) → 2SO3(g), how many moles of O2 are needed to produce 6.76 grams of SO3? 0.366 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 ×

Stoichiometric pathway: mass SO3 → moles SO3 → moles O2. The equation shows that there are 2 moles of SO3 for every 1 mole of O2. So we use this to 2

Jespersen 7th Ed. Ch. 3

A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

form our conversion factor. We also need the molecular mass of SO 3. 6.76 𝑔 𝑆𝑂3 ×

1 𝑚𝑜𝑙 𝑆𝑂3 1 𝑚𝑜𝑙 𝑂2 = 𝟓. 𝟐𝟖 𝒎𝒐𝒍 𝑺𝑶𝟑 × 64.07 𝑔 𝑆𝑂3 2 𝑚𝑜𝑙 𝑆𝑂3

Ex.3 The thermite reaction 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l) generates large quantities of heatand is used in welding. How much aluminum metal would be needed to completely react 72.55 g of iron(III) oxide. Stoichiometric pathway: mass Fe2O3→ moles Fe2O3 → moles Al → mass Al. 72.5𝑔 𝐹𝑒2 𝑂3 ×

26.98 𝑔 𝐴𝑙 2 𝑚𝑜𝑙 𝐴𝑙 1 𝑚𝑜𝑙 𝐹𝑒2 𝑂3 × × = 𝟑𝟕. 𝟕 𝒎𝒐𝒍 𝑨𝒍 103.8 𝑔 𝐹𝑒2 𝑂3 1 𝑚𝑜𝑙 𝐹𝑒2 𝑂3 1 𝑚𝑜𝑙 𝐴𝑙

Key Questions: 1. Based on the Model above, what relates the moles of substance A to the moles of substance B?

For 2 – 5, consider the reaction between sulfuric acid and sodium hydroxide from Model 2 Ex. 1 how many moles of sodium sulfate can be made from 0.240 mol of sodium hydroxide? 2.

Provide the balanced equation here.

3.

Based on Figure 1, what is the stoichiometric pathway for this question?

4.

From the balanced equation above, what is the mole ratio between sodium hydroxide and sodium sulfate?

5.

Solve for the number of moles of sodium sulfate using this mole ratio.

Exercises 1. Zinc metal is used to extract gold from pre-treated gold ore: ___Au(CN)2–(aq) + ___Zn(s)

→

___Au(s) + ___Zn(CN)42– (aq)

a.

Balance the equation above using the lines provided.

b.

How many moles of Zn are needed to react with 0.25 mol of Au(CN)2–?

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Jespersen 7th Ed. Ch. 3

2.

A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

c.

How many grams of Zn are needed to react with 0.25 mol of Au(CN)2–?

d.

If you start with 1.00 mole of Zn, how many grams of Au can form? (Assume that Au(CN)2– is present in excess.)

e.

How many grams of Au can form if you start with 10.8 g of Zn metal and an excess of Au(CN)2– ?

f.

How many gold atoms are in 0.25 moles of Au(CN)2– ?

g.

How many total atoms are in 36.9 g of Au(CN)2– ?

The body combines glucose, C6H12O6, and oxygen, O2, in a process called metabolism, to give carbon dioxide and water. How many grams of oxygen must the body take in to completely process 1.00 g of glucose? a. Write the balanced equation for this combustion reaction.

b.

How many moles of glucose are in 1.00 grams?

c.

How many moles of molecular oxygen are needed to completely react all of the glucose?

d.

How many grams of molecular oxygen does the answer in part c. correspond to?

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A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

Oxygen gas can be produced in the laboratory by decomposition of potassium chlorate (KClO): ___KClO3 (s) → ___KCl (s)+ ___O2(g) How many kg of O2 can be produced from 1.0 kg of KClO3?

4.

Barium sulfate, BaSO4, is made by the following reaction. Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq) How many grams of BaSO4 can be produced from a reaction where 75.00 g of Ba(NO3)2 and was allowed to react completely with an excess of Na2SO4.

Model 2A: Limiting Reagents in Everyday Life A simple Scottish Shortbread recipe to make a batch of cookies requires: 2 cups of softened butter 1 cup packed brown sugar 4 cups of all purpose flour. You need to make a large number of cookies for a party. You have 6 cups of butter, 5 cups of packed brown sugar and 15 cups of flour in your kitchen. You will run out of one of these ingredients first—this is the limiting ingredient or reagent. It also means that some of the other ingredients are present in excess, or they will be left over. Key Questions: 6. What would be the cooking “equation” or stoichiometry?

7.

How many batches of cookies can you make from the butter you have?

8.

How many batches of cookies can you make from the sugar you have?

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A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

How many batches of cookies can you make from the flour you have?

10. You cannot make more than the smallest number from (12) through (14) above, that quantity is called the limiting reagent. Why can you only make that many batches of cookies?

11. How much of each non-limiting reagent are left? Explain how you arrived at this value.

Model 2B: Limiting Reagents in Chemistry Similarly in chemistry we also have limiting reagents. Whenever you are given two amounts of reagents, it is a limiting reagent problem. Suppose that 12.0 moles of molecular oxygen, and 10.0 moles of molecular hydrogen are combined in a vessel and allowed to react as follows: 2 H2(g) + O2(g) → 2 H2O (ℓ). Key Questions 12. How many moles of water can you make from the oxygen?

13. How many moles of water can you make from the hydrogen?

14. How many moles of water can you make given the available amounts of hydrogen and oxygen?

15. At the end of the reaction, will anything be present in the vessel besides water? If so, what is left over and how much of it is there?

Model 2C: Limiting Reagent Calculations in Chemistry When setting up a reaction, chemists usually add more of one reactant to help insure that the reaction goes to completion or that all of the other reactant gets used up. Now we must do two calculations to determine how much product we can make in theory. Ex. Calculate the mass of silver carbonate, Ag2CO3(s) (MM = 275.8 g/mol) produced by mixing 123 g of sodium carbonate, Na2CO3(aq) and 76.5 g of silver nitrate, AgNO3(aq). We are using a slightly alternative approach than the textbook uses: 6

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A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

Analysis: 1. Write the balanced equation. 2. Identify the limiting reagent. A. Choose one reagent, A, and calculate the amount of product formed. Mass of reactant A → mol reactant A → mol product → mass product produced B. Choose the other reagent, B, and calculate the amount of product formed. Mass of reactant B → mol reactant B → mol product → mass product produced The limiting reagent is the one that produces the least amount of product. 3. Calculate the amount of product formed (the smaller amount of two, A and B). Solution: First, write the balanced equation for the reaction. Na2CO3(aq) + 2AgNO3(aq) → Ag2CO3(s) + 2NaNO3(aq) Next calculate the amount of the desired product that each of the reactants could make 123 𝑔 𝑁𝑎2 𝐶𝑂3 ×

1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3 1 𝑚𝑜𝑙 𝐴𝑔2 𝐶𝑂3 275.8 𝑔 𝐴𝑔2 𝐶𝑂3 × = 320. 𝑔 𝐴𝑔2 𝐶𝑂3 × 105.988 𝑔 𝑁𝑎2 𝐶𝑂3 1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3 1 𝑚𝑜𝑙 𝐴𝑔2 𝐶𝑂3

We find that, 123 g of Na2CO3 produces 320 g of Ag2CO3. 76.5 𝑔 𝐴𝑔𝑁𝑂3 ×

1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3 1 𝑚𝑜𝑙 𝐴𝑔2 𝐶𝑂3 275.8 𝑔 𝐴𝑔2 𝐶𝑂3 × × = 𝟔𝟐. 𝟏 𝒈 𝑨𝒈𝟐𝑪𝑶𝟑 169.872 𝑔 𝐴𝑔𝑁𝑂3 2 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3 1 𝑚𝑜𝑙 𝐴𝑔2 𝐶𝑂3

We produce less Ag2CO3 product with the AgNO3, therefore it is the limiting reagent! 62.1 g Ag2CO3 of are produced! We have a limiting reagent, so we must also have a reagent in excess. How much Na3CO3 is left over after the reaction? The amount left over is simply the amount you started with minus the amount used to make 62.1 g of Ag3CO3. To calculate the amount of Na2CO3 that was used to react all of the AgNO3 we can start with either the grams of AgNO3 we started with or the grams of Ag2CO3 we formed. 62.1 𝑔 𝐴𝑔2 𝐶𝑂3 ×

1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3 105.988 𝑔 𝑁𝑎2 𝐶𝑂3 1 𝑚𝑜𝑙 𝐴𝑔2 𝐶𝑂3 × × = 23.9 𝑔 𝑁𝑎2 𝐶𝑂3 275.8 𝑔 𝐴𝑔2𝐶𝑂 3 1 𝑚𝑜𝑙 𝐴𝑔2 𝐶𝑂3 1 𝑚𝑜𝑙 𝑁𝑎 2 𝐶𝑂3

123 g Na2CO3 (to start) – 23.9 g Na2CO3 (used) = 99 g Na2CO3 (left over or “in excess”) Key Questions: 16. Why is writing the balanced reaction equation an important part of the methodology?

17. Why is AgNO3 and not Na2CO3 the limiting reactant? 18. What insights about solving stoichiometry problems did your group gain from the Model and the Key Questions?

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Week 5 CHE1212R

19. Is it true that whichever reactant I have less moles of initially is always the limiting reagent? Why, or why not?

Exercises: 5. The propane burns according to the following equation C3H8 + 5O2 → 3CO2 + 4H2O a. Verify that the chemical equation above is balanced. What does “balanced” mean?

b. If 10 moles of propane (C3H8) is reacted with 35 moles of oxygen (O2), which is the limiting reagent?

6.

c.

How many moles of carbon dioxide (CO2) can be produced?

d.

How many moles of water can be produced?

e.

Which reagent is in excess?

f.

How many moles of the excess reagent are left at the end of the reaction?

In the synthesis of aspirin we react salicylic acid with acetic anhydride. The balanced chemical equation is: 2HOOCC6H4OH + C4H6O → 2HOOCC6H4O2C2H3 + H2O salicylic acid

acetic anhydride

aspirin

water

If we mix together 28.2 grams of salicylic acid with 15.6 grams of acetic anhydride in this reaction, how many grams of aspirin can be made and which reactant is limiting?

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A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

Gold(III) hydroxide can be made by the following reaction.

2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O → 2Au(OH)3(aq) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) a.

If 45.67 g of KAuCl4 is reacted with 24.32 g of Na2CO3 (both dissolved in a large excess of water), what is the maximum number of grams of Au(OH)3 that can form and which reagent is limiting ?

b.

How many grams of which reagent are left over?

Model 3: Actual, Theoretical and Percentage Yield During chemical synthesis, the actual amount of product obtained is often much less than the calculated maximum amount. Losses can occur for several reasons. 1. Mechanical losses such as chemicals sticking to the glassware. 2. Evaporation of a gaseous or volatile (has a boiling point close to the reaction temperature) product. 3. Presence of another or competing reaction that produces a byproduct which competes with the main reaction thus lowering the amount of product. Main reaction: S(s) + O2(g) → SO2 (g) Competing Reaction: 2 SO2 (g) + O2(g) → 2 SO3 (g) Actual Yield—the amount of desired product that is isolated. Theoretical Yield—the amount of product that would be obtained if no losses occurred. (Based on stoichiometry and limiting reagent. Percentage Yield—the actual yield as a percentage of the theoretical yield. 𝐏𝐞𝐫𝐜𝐞𝐧𝐭𝐚𝐠𝐞 𝐲𝐢𝐞𝐥𝐝 =

𝐚𝐜𝐭𝐮𝐚𝐥 𝐲𝐢𝐞𝐥𝐝 × 𝟏𝟎𝟎% 𝐭𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐲𝐢𝐞𝐥𝐝

Ex. Suppose we carry out the reaction in Model 2C, and obtain 45.9 g of Ag2CO3 solid. Based on our calculations in Model 2C, what are the actual, theoretical and percentage yields for this reaction? Actual yield Ag2CO3 = 45.9 g Theoretical yield Ag2CO3 = 62.1 g 45.9 𝑔 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑌𝑖𝑒𝑙𝑑 = × 100 = 𝟕𝟑. 𝟗% 62.1 𝑔 9

Jespersen 7th Ed. Ch. 3

A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

Key Questions: 20. What is the difference between the actual yield and the theoretical yield?

21. Why is the percentage yield useful?

22. Explain why having the chemicals stick to the glassware results in an actual yield less than the theoretical yield?

Exercises 8. Based on the grams of Au(OH)3 that could be formed from the limiting reagent in Exercise 7, calculate the percentage yield of Au(OH)3 if 19.75 g of Au(OH)3 was obtained.

9.

64.45 g of barium sulfate, BaSO4, was obtained from the synthesis in Exercise 4. Based on the theoretical yield of BaSO4 calculated in Exercise 4, what is the percentage yield of BaSO4.

10. If we obtain 30.7 grams of aspirin from the synthesis in Exercise 6, calculate the percentage yield for this aspirin synthesis.

Extra Practice 1. 150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. 2AsF3 + 3CCl4 → 2AsCl3 + 3CCl2F2 a. How many grams of CCl2F2 (Freon) can be made and which reactant is limiting?

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A5—Stoichiometry & Reaction Yields

Week 5 CHE1212R

b. If there are 125.4 g of CCl2F2 obtained when the reaction is actually carried out, what is the percentage yield?

2. The (unbalanced) reaction between the limestone and hydrochloric acid produces carbon dioxide as shown in the reaction. ___CaCO3(s) + ___HCl (aq) → ___CO2(g) + ___CaCl2(aq) + ___H2O a. If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?

b. How many grams of which reactant are left over?

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