Chapter 9 Stoichiometry Table of...

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Chapter 9

Table of Contents

Stoichiometry

Section 1 Introduction to Stoichiometry

Section 2 Ideal Stoichiometric Calculations

Section 3 Limiting Reactants and Percentage Yield



Chapter 9Section 1 Introduction to Stoichiometry

Lesson Starter

Mg(s) + 2HCl(aq) ? MgCl2(aq) + H2(g)

• If 2 mol of HCl react, how many moles of H2 are obtained?

• How many moles of Mg will react with 2 mol of HCl?

• If 4 mol of HCl react, how many mol of each product are produced?

• How would you convert from moles of substances to masses?




Objective

• Define stoichiometry.

• Describe the importance of the mole ratio in s toichiometric calculations .

• Write a mole ratio relating two substances in a chemical equation.




Stoichiometry Definition

• Composition stoichiometry deals with the mass relationships of elements in compounds.

• Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.



Chapter 9 Visual Concepts

Click below to watch the Visual Concept.

Stoichiometry

Visual Concept



Chapter 9

Problem Type 2: Given is anamount in moles and unknown isa mass

Amount of givensubstance (mol)

Problem Type 1: Given andunknown quantities are amountsin moles.


Reaction Stoichiometry Problems


Amount of unknown substance (mol)

Mass of unknown substance (g)




Chapter 9

Problem Type 4: Given is a massand unknown is a mass.

Mass of a given substance (g)

Problem Type 3: Given is a mass and unknown is an amount in moles.

Mass of givensubstance (g)

Reaction Stoichiometry Problems, continued




Mass of unknown substance (g)





Chapter 9

Mole Ratio

• A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction

Example: 2Al2O3(l) ? 4Al(s) + 3O2(g)

Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al 4 mol Al 3 mol O2 3 mol O2

, ,




Chapter 9

Converting Between Amounts in Moles




Chapter 9

StoichiometryCalculations






Visual Concept

Molar Mass as a Conversion Factor



Chapter 9

Lesson Starter

Acid-Base Neutralization Reaction Demonstration

• What is the equation for the reaction of HCl with NaOH?

• What is the mole ratio of HCl to NaOH?

Section 2 Ideal StoichiometricCalculations



Chapter 9

Objective

• Calculate the amount in moles of a reactant or a product from the amount in moles of a different reactant or product.

• Calculate the mass of a reactant or a product from the amount in moles of a different reactant or product.




Chapter 9

Objectives, continued

• Calculate the amount in moles of a reactant or a product from the mass of a different reactant or product.

• Calculate the mass of a reactant or a product from the mass of a different reactant or product.




Chapter 9

Conversions of Quantities in Moles






Visual Concept

Conversion of Quantities in Moles



Chapter 9

Solving Mass-Mass StoichiometryProblems




Chapter 9

Conversions of Quantities in Moles, continued

Sample Problem AIn a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation.

CO2(g) + 2LiOH(s) ? Li2CO3(s) + H2O(l)

How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day?




Chapter 9

Conversions of Quantities in Moles, continued

Sample Problem A SolutionCO2(g) + 2LiOH(s) ? Li2CO3(s) + H2O(l)

Given: amount of CO2 = 20 molUnknown: amount of LiOH (mol)Solution:

mol CO2 ?

mol LiOH mol CO2

? mol LiOH

20 mol CO2 ?

2 mol LiOH 1 mol CO2

? 40 mol LiOH

mol ratio




Chapter 9

Conversions of Amounts in Moles to Mass




Chapter 9

Solving Stoichiometry Problems with Moles or

Grams




Chapter 9

Conversions of Amounts in Moles to Mass, continuedSample Problem B

In photos ynthes is , plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water.

What mass , in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?




Chapter 9

Conversions of Amounts in Moles to Mass, continued

Sample Problem B SolutionGiven: amount of H2O = 3.00 molUnknown: mass of C6H12O6 produced (g)Solution:Balanced Equation: 6CO2(g) + 6H2O(l) ? C6H12O6(s) + 6O2(g)

mol ratio molar mass factor

mol H2O ?

mol C6H12O6

mol H2O ?

g C6H12O6

mol C6H12O6

? g C6H12O6

3.00 mol H2O ?

1 mol C6H12O6

6 mol H2O ?

180.18 g C6H12O6

1 mol C6H12O6

=

90.1 g C6H12O6




Chapter 9

Conversions of Mass to Amounts in Moles






Visual Concept

Mass and Number of Moles of an Unknown



Chapter 9

Conversions of Mass to Amounts in Moles, continuedSample Problem D

The firs t s tep in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.

NH3(g) + O2(g) ? NO(g) + H2O(g) (unbalanced)

The reaction is run using 824 g NH3 and excess oxygen.a. How many moles of NO are formed?b. How many moles of H2O are formed?




Chapter 9

Conversions of Mass to Amounts in Moles, continued

Sample Problem D SolutionGiven: mass of NH3 = 824 gUnknown: a. amount of NO produced (mol)

b. amount of H2O produced (mol)Solution:Balanced Equation: 4NH3(g) + 5O2(g) ? 4NO(g) + 6H2O(g)

a.

b.

g NH3 ?

mol NH3

g NH3

? mol NOmol NH3

? mol NO

g NH3 ?

mol NH3

g NH3

? mol H2Omol NH3

? mol H2O

molar mass factor mol ratio




Chapter 9

Conversions of Mass to Amounts in Moles, continuedSample Problem D Solution, continued

a.

b.

824 g NH3 ?

1 mol NH3

17.04 g NH3

? 4 mol NO 4 mol NH3

? 48.4 mol NO

824 g NH3 ?

1 mol NH3

17.04 g NH3

? 6 mol H2O4 mol NH3

? 72.5 mol H2O

molar mass factor mol ratio




Chapter 9

Mass-Mass to Calculations






Visual Concept

Mass-Mass Calculations



Chapter 9

Solving Mass-Mass Problems




Chapter 9

Mass-Mass to Calculations, continued

Sample Problem ETin(II) fluoride, SnF2, is used in some toothpastes . It is made by the reaction of tin with hydrogen fluoride according to the following equation.

Sn(s) + 2HF(g) ? SnF2(s) + H2(g)

How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?




Chapter 9

g HF ?

1 mol HF20.01 g HF

? 1 mol SnF2

2 mol HF ?

156.71 g SnF2 1 mol SnF2

Mass-Mass to Calculations, continued

Sample Problem E SolutionGiven: amount of HF = 30.00 gUnknown: mass of SnF2 produced (g)Solution:

g HF ?

mol HFg HF

? mol SnF2

mol HF ?

g SnF2 mol SnF2

? g SnF2

molar mass factor mol ratio molar mass factor

= 117.5 g SnF2




Chapter 9Section 2 Ideal StoichiometricCalculations

Solving Various Types of Stoichiometry Problems



Chapter 9

Solving Various Types of Stoichiometry Problems




Chapter 9

Solving Volume-Volume Problems




Chapter 9

Solving Particle Problems




Chapter 9

Objectives

• Describe a method for determining which of two reactants is a limiting reactant.

• Calculate the amount in moles or mass in grams of a product, given the amounts in moles or masses in grams of two reactants, one of which is in excess.

• Distinguish between theoretical yield, actual yield, and percentage yield.

• Calculate percentage yield, given the actual yield and quantity of a reactant.




Chapter 9

Limiting Reactants

• The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction.

• The excess reactant is the substance that is not used up completely in a reaction.






Visual Concept

Limiting Reactants and Excess Reactants



Chapter 9

Limited Reactants, continued

Sample Problem FSilicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to thefollowing equation.

SiO2(s) + 4HF(g) ? SiF4(g) + 2H2O(l)

If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant?




Chapter 9


Sample Problem F SolutionSiO2(s) + 4HF(g) ? SiF4(g) + 2H2O(l)

Given: amount of HF = 6.0 molamount of SiO2 = 4.5 mol

Unknown: limiting reactantSolution:

mol HF ?

mol SiF4

mol HF ? mol SiF4 produced

mol SiO2 ?

mol SiF4

mol SiO2

? mol SiF4 produced

mole ratio




Chapter 9


Sample Problem F Solution, continuedSiO2(s) + 4HF(g) ? SiF4(g) + 2H2O(l)

6.0 mol HF ?

1 mol SiF4

4 mol HF ? 1.5 mol SiF4 produced

4.5 mol SiO2 ?

1 mol SiF4

1 mol SiO2

? 4.5 mol SiF4 produced


HF is the limiting reactant.



Chapter 9

Percentage Yield

• The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant.

• The actual yield of a product is the measured amount of that product obtained from a reaction.

• The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

percentage yield ?

actual yieldtheorectical yield

? 100






Visual Concept

Comparing Actual and Theoretical Yield



Chapter 9

Percentage Yield, continued

Sample Problem HChlorobenzene, C6H5Cl, is used in the production of many important chemicals , such as aspirin, dyes , and dis infectants . One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine, as represented by the following equation.

C6H6 (l) + Cl2(g) ? C6H5Cl(l) + HCl(g)

When 36.8 g C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g.What is the percentage yield of C6H5Cl?




Chapter 9


Sample Problem H SolutionC6H6 (l) + Cl2(g) ? C6H5Cl(l) + HCl(g)

Given: mass of C6H6 = 36.8 gmass of Cl2 = excessactual yield of C6H5Cl = 38.8 g

Unknown: percentage yield of C6H5ClSolution:Theoretical yield

g C6H6 ?

mol C6H6

g C6H6

? mol C6H5Clmol C6H6

? g C6H5Cl

mol C6H5Cl ? g C6H5Cl

molar mass factor mol ratio molar mass




Chapter 9

percentage yield ?

38.8 g53.0 g

? 100 ? 73.2%


Sample Problem H Solution, continuedC6H6(l) + Cl2(g) ? C6H5Cl(l) + HCl(g)

Theoretical yield

36.8 g C6H6 ? 1 mol C6H6

78.12 g C6H6

? 1 mol C6H5Cl1 mol C6H6

? 112.56 g C6H5Cl

1 mol C6H5Cl

? 53.0 g C6H5Cl

percentage yield C6H5Cl ?

actual yieldtheorectical yield

? 100


Percentage yield



End of Chapter 9 Show



Chapter 9

Multiple Choice

1. In s toichiometry, chemis ts are mainly concerned with

A. the types of bonds found in compounds.B. mass relationships in chemical reactions .C. energy changes occurring in chemical reactions .D. the speed with which chemical reactions occur.

Standardized Test Preparation



Chapter 9

Multiple Choice

1. In s toichiometry, chemis ts are mainly concerned with

A. the types of bonds found in compounds.B. mass relationships in chemical reactions .C. energy changes occurring in chemical reactions .D. the speed with which chemical reactions occur.




Chapter 9

2. Assume ideal s toichiometry in the reactionCH4 + 2O2 ? CO2 + 2H2O. If you know the mass of CH4, you can calculate

A. only the mass of CO2 produced.B. only the mass of O2 reacting.C. only the mass of CO2 + H2O produced.D. the mass of O2 reacting and CO2 + H2O produced.


Multiple Choice



Chapter 9

Multiple Choice

2. Assume ideal s toichiometry in the reactionCH4 + 2O2 ? CO2 + 2H2O. If you know the mass of CH4, you can calculate

A. only the mass of CO2 produced.B. only the mass of O2 reacting.C. only the mass of CO2 + H2O produced.D. the mass of O2 reacting and CO2 + H2O produced.




Chapter 9

Multiple Choice

3. Which mole ratio for the equation 6Li + N2 ? 2Li3N is incorrect?

A. C.

B. D.

6 mol Li2 mol N2

2 mol Li3N

1 mol N2

1 mol N2

6 mol Li2 mol Li3N

6 mol Li




Chapter 9

Multiple Choice

3. Which mole ratio for the equation 6Li + N2 ? 2Li3N is incorrect?

A. C.

B. D.

6 mol Li2 mol N2

2 mol Li3N

1 mol N2

1 mol N2

6 mol Li2 mol Li3N

6 mol Li




Chapter 9

Multiple Choice

4. For the reaction below, how many moles of N2 are required to produce 18 mol NH3?

N2 + 3H2 ? 2NH3

A. 4.5 B. 9.0 C. 18D. 36




Chapter 9

Multiple Choice

4. For the reaction below, how many moles of N2 are required to produce 18 mol NH3?

N2 + 3H2 ? 2NH3

A. 4.5 B. 9.0C. 18D. 36




Chapter 9

Multiple Choice

5. What mass of NaCl can be produced by the reaction of 0.75 mol Cl2?

2Na + Cl2 ? 2NaCl

A. 0.75 gB. 1.5 gC. 44 gD. 88 g




Chapter 9

Multiple Choice

5. What mass of NaCl can be produced by the reaction of 0.75 mol Cl2?

2Na + Cl2 ? 2NaCl

A. 0.75 gB. 1.5 gC. 44 gD. 88 g




Chapter 9

Multiple Choice

6. What mass of CO2 can be produced from 25.0 g CaCO3 given the decomposition reaction

CaCO3 ? CaO + CO2

A. 11.0 g B. 22.0 gC. 25.0 gD. 56.0 g




Chapter 9

Multiple Choice

6. What mass of CO2 can be produced from 25.0 g CaCO3 given the decomposition reaction

CaCO3 ? CaO + CO2

A. 11.0 gB. 22.0 gC. 25.0 gD. 56.0 g




Chapter 9

Multiple Choice

7. If a chemical reaction involving substances A and B s tops when B is completely used up, then B is referred to as the

A. excess reactant. B. primary reactant. C. limiting reactant.D. primary product.




Chapter 9

Multiple Choice

7. If a chemical reaction involving substances A and B s tops when B is completely used up, then B is referred to as the

A. excess reactant. B. primary reactant. C. limiting reactant.D. primary product.




Chapter 9

Multiple Choice

8. If a chemis t calculates the maximum amount of product that could be obtained in a chemical reaction, he or she is calculating the

A. percentage yield.B. mole ratio.C. theoretical yield.D. actual yield.




Chapter 9

Multiple Choice

8. If a chemis t calculates the maximum amount of product that could be obtained in a chemical reaction, he or she is calculating the

A. percentage yield.B. mole ratio.C. theoretical yield.D. actual yield.




Chapter 9

Multiple Choice

9. What is the maximum number of moles of AlCl3 that can be produced from 5.0 mol Al and 6.0 mol Cl2?

2Al + 3Cl2 ? 2AlCl3

A. 2.0 mol AlCl3B. 4.0 mol AlCl3C. 5.0 mol AlCl3D. 6.0 mol AlCl3




Chapter 9

Multiple Choice

9. What is the maximum number of moles of AlCl3 that can be produced from 5.0 mol Al and 6.0 mol Cl2?

2Al + 3Cl2 ? 2AlCl3

A. 2.0 mol AlCl3B. 4.0 mol AlCl3C. 5.0 mol AlCl3D. 6.0 mol AlCl3




Chapter 9

Short Answer

10. Why is a balanced equation necessary to solve a mass -mass s toichiometry problem?




Chapter 9

Short Answer

10. Why is a balanced equation necessary to solve a mass -mass s toichiometry problem?

Answer: The coefficients of the balanced equation are needed for the mole-mole ratio that is necessary to solve s toichiometric problems involving two different subs tances .




Chapter 9

Short Answer


11. What data are necessary to calculate the percentage yield of a reaction?



Chapter 9

Short Answer

11. What data are necessary to calculate the percentage yield of a reaction?

Answer: the theoretical yield and the actual yield of the product




Chapter 9

Extended Response

12. A student makes a compound in the laboratory and reports an actual yield of 120%. Is this result possible? Assuming that all masses were measured correctly, give an explanation.




Chapter 9

Extended Response

12. A student makes a compound in the laboratory and reports an actual yield of 120%. Is this result possible? Assuming that all masses were measured correctly, give an explanation.

Answer: The product was impure, so the mass contained the product and other substances. For example, if NaClis made from HCl and NaOH but is not dried, the mass of NaCl can include water, which would result in a yield greater than 100%.




Chapter 9

Extended Response13. Benzene, C6H6, is reacted with bromine, Br2, to

produce bromobenzene, C6H5Br, and hydrogen bromide, HBr, as shown below. When 40.0 g of benzene are reacted with 95.0 g of bromine, 65.0 g of bromobenzene is produced.

C6H6 + Br2 ? C6H5Br + HBra. Which compound is the limiting reactant?b. What is the theoretical yield of bromobenzene?c. What is the reactant in excess , and how much remains

after the reaction is completed?d. What is the percentage yield?




Chapter 9

Extended Response13. Benzene, C6H6, is reacted with bromine, Br2, to

produce bromobenzene, C6H5Br, and hydrogen bromide, HBr, as shown below. When 40.0 g of benzene are reacted with 95.0 g of bromine, 65.0 g of bromobenzene is produced.

C6H6 + Br2 ? C6H5Br + HBra. Which compound is the limiting reactant? benzeneb. What is the theoretical yield of bromobenzene? 80.4 gc. What is the reactant in excess , and how much remains

after the reaction is completed? bromine, 13.2 gd. What is the percentage yield? 80.8%


Chapter 9 Stoichiometry Table of...

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