Read Study Guide for Content Mastery - Teacher Edition text version Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. Name Date Class Name Date Class CHAPTER Section 12.1 continued In your textbook, read about mole ratios. 12 STUDY GUIDE FOR CONTENT MASTERY CHAPTER 12 STUDY GUIDE FOR CONTENT MASTERY Stoichiometry Answer the questions about the following chemical reaction. Sodium 6Na(s) Fe2O3(s) 0 3Na2O(s) iron(III) oxide 0 sodium oxide iron 2Fe(s) Section 12.1 What is stoichiometry? In your textbook, read about stoichiometry and the balanced equation.

• Stoichiometry Problems Worksheet And Answers
• Chapter 11 Stoichiometry Study Guide
• Chapter 11 Stoichiometry Chemistry Answers

Chemistry chapter 12 study guide for content mastery stoichiometry answers.pdf FREE PDF DOWNLOAD NOW!!! Source #2: chemistry chapter 12 study guide for content mastery stoichiometry answers.pdf FREE PDF DOWNLOAD Study Guide for Content Mastery - Glencoe/McGraw-Hill. Chemistry chapter 12 study guide for content mastery stoichiometry answers. Start studying Stoichiometry Study Guide. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Chapter 12 Study Guide For Content Mastery Stoichiometry. Stoichiometry Study Guide Answer Key Document for Stoichiometry Study Guide Answer Key is.

For each statement below, write true or false. What is a mole ratio? The study of the quantitative relationships between the amounts of reactants used and the amounts of products formed by a chemical reaction is called stoichiometry. A mole ratio is a ratio between the numbers of moles of any two substances in a balanced chemical equation. How is a mole ratio written?

Stoichiometry is based on the law of conservation of mass. Study Guide for Content Mastery Answer Key the number of moles of another substance in the denominator. Predict the number of mole ratios for this reaction. What are the mole ratios for this reaction? In any chemical reaction, the mass of the products is less than the mass of the reactants. A mole ratio is written for two substances in a balanced chemical equation as a fraction by placing the number of moles of one substance in the numerator and true 4.

The coefficients in a chemical equation represent not only the number of individual particles but also the number of moles of particles. The mass of each reactant and product is related to its coefficient in the balanced chemical equation for the reaction by its molar mass. 12 3 mol Na2O/6 mol Na 2 mol Fe/6 mol Na Complete the table below, using information represented in the chemical equation for the combustion of methanol, an alcohol. Oxygen 0 carbon dioxide 3O2 (g) 0 2CO2(g) Number of Molecules Mass (g) Number of Moles (mol) methanol 4H2O(g) water 1 mol Fe2O3/6 mol Na 6 mol Na/1 mol Fe2O3 6 mol Na/3 mol Na2O 6 mol Na/2 mol Fe 2CH3OH(l) 3 mol Na2O/1 mol Fe2O3 1 mol Fe2O3/3 mol Na2O 1 mol Fe2O3/2 mol Fe 2 mol Fe/1 mol Fe2O3 2 mol Fe/3 mol Na2O 3 mol Na2O/2 mol Fe Substance Molar Mass (g/mol) 6. Methanol 32.05 2 3 2 4 methanol and oxygen gas carbon dioxide and water 160.10 g 160.10 g They are equal.

4 2 88.02 72.08 3 96.00 2 64.10 7. Oxygen gas 32.00 8. Carbon dioxide 44.01 19. What is the mole ratio relating sodium to iron? What is the mole ratio relating iron to sodium? Which mole ratio has the largest value?

6 mol Na/2 mol Fe 2 mol Fe/6 mol Na 6 mol Na/1 mol Fe2O3 9. Water 18.02 10.

What are the reactants? What are the products? What is the total mass of the reactants? Chemistry: Matter and Change Chemistry: Matter and Change Chapter 12 13. What is the total mass of the products? How do the total masses of the reactants and products compare? T191 Study Guide for Content Mastery 67 68 Chemistry: Matter and Change Chapter 12 Study Guide for Content Mastery T192 Date Class Name Date Class Name CHAPTER 12 STUDY GUIDE FOR CONTENT MASTERY Section 12.2 continued In your textbook, read about mole-to-mass and mass-to-mass conversions.

CHAPTER 12 STUDY GUIDE FOR CONTENT MASTERY Section 12.2 Stoichiometric Calculations In your textbook, read about mole-to-mole conversion. Read the following passage and then solve the problems. In the equation that follows each problem, write in the space provided the mole ratio that can be used to solve the problem. Complete the equation by writing the correct value on the line provided. Solving a mass-to-mass problem requires the four steps listed below.

The equations in the boxes show how the four steps are used to solve an example problem. After you have studied the example, solve the problems below, using the four steps. Example problem: How many grams of carbon dioxide are produced when 20.0 g acetylene (C2H2) is burned? Solution Step 1 Write a balanced chemical equation for the reaction. 2C2H2(g) 0 4CO2(g) 20.0 g C2H2 5O2(g) 2H2O(g) 1 mol C2H2 26.04 g C2H2 0.768 mol C2H2 0.768 mol C2H2 1.54 mol CO2 1.54 mol CO2 67.8 g CO2 44.01 g CO2 1 mol CO2 4 mol CO2 2 mol C2H2 The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas.

The following balanced chemical equation represents the reaction. 2H2O(l) 0 4NaOH(s) O2(g) 2Na2O2(s) 1.

How many moles of sodium hydroxide are produced when 1.00 mol sodium peroxide Chemistry: Matter and Change 2.00 mol NaOH Step 2 Determine the number of moles of the known substance, using mass-to-mole conversion. Reacts with water?

1.00 mol Na2O2 4 mol NaOH/2 mol Na2O2 2. How many moles of oxygen gas are produced when 0.500 mol Na2O2 reacts with water? 0.500 mol Na2O2 mol O2 1 mol O2/2 mol Na2O2 0.250 Step 3 Determine the number of moles of the unknown substance, using mole-to-mole conversion. Step 4 Determine the mass of the unknown substance, using moleto-mass conversion. How many moles of sodium peroxide are needed to produce 1.00 mol sodium hydroxide? 1.00 mol NaOH mol Na2O2 2 mol Na2O2/4 mol NaOH 0.500 4.

How many moles of water are required to produce 2.15 mol oxygen gas in this reaction? In some mole-to-mass conversions, the number of moles of the known substance 2.15 mol O2 mol H2O 2 mol H2O/1 mol O2 4.30 is given. In those conversions, which step of the above solution is not necessary?

In a blast furnace, iron and carbon monoxide are produced from the reaction of iron(III) Step 2 5. How many moles of water are needed for 0.100 mol of sodium peroxide to react com- pletely in this reaction? Oxide (Fe2O3) and carbon. How many grams of iron are formed when 150 g iron(III) oxide reacts with an excess of carbon? 0.100 mol Na2O2 mol H2O 2 mol H2O/2 mol Na2O2 0.100 6. How many moles of oxygen are produced if the reaction produces 0.600 mol sodium hydroxide?

Fe2O3(s) 3C(s) 0 2Fe(s) 3CO(g) 150 g Fe2O3 1 mol Fe2O3/159.7 g Fe2O3 0.939 mol Fe2O3 0.939 mol Fe2O3 2 mol Fe/1 mol Fe2O3 1.88 mol Fe 1.88 mol Fe 55.85 g Fe/1 mol Fe 105 g Fe 9. Solid sulfur tetrafluoride (SF4) and water react to form sulfur dioxide and an aqueous 0.600 mol NaOH mol O2 1 mol O2/4 mol NaOH 0.150 solution of hydrogen fluoride. How many grams of water are necessary for 20.0 g sulfur tetrafluoride to react completely? SF4(s) 2H2O(l) 0 SO2(g) 4HF(aq) 20.0 g SF4 1 mol SF4/108.07 g SF4 0.185 mol SF4 0.185 mol SF4 2 mol H2O/1 mol SF4 0.370 mol H2O 0.370 mol H2O 18.02 g H2O/1 mol H2O 6.67 g H2O Study Guide for Content Mastery Answer Key Chemistry: Matter and Change Chapter 12 Study Guide for Content Mastery 69 70 Chemistry: Matter and Change Chapter 12 Study Guide for Content Mastery Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. Name Date Class Name Date Class CHAPTER 12 STUDY GUIDE FOR CONTENT MASTERY CHAPTER 12 STUDY GUIDE FOR CONTENT MASTERY Section 12.3 In your textbook, read about the yields of products.

Stoichiometry Problems Worksheet And Answers

Limiting Reactants Section 12.4 Study the diagram and the example problem. Percent Yield mass of product from experimental measurement In your textbook, read about why reactions stop and how to determine the limiting reactant. Study the diagram showing a chemical reaction and the chemical equation that represents the reaction.

Then complete the table. Show your calculations for questions 25­27 in the space below the table. Percent yield actual yield theoretical yield 100% 0 O2 2NO 0 2NO2 mass of product predicted from stoichiometric calculation using Study Guide for Content Mastery Answer Key Amount of NO2 none NO 2 molecules O2 none Limiting Reactant Amount and Name of Excess Reactant The molar masses of O2, NO, and NO2 are 32.00 g/mol, 30.01 g/mol, and 46.01 g/mol, respectively.

Amount of O2 Amount of NO 1 molecule 2 molecules 2 molecules 4 molecules 4 molecules 4 molecules a. Mass of reactant b. 4-step mass-to-mass conversion 1. Write the balanced chemical equation. Calculate the number of moles of reactant, using molar mass.

Calculate the number of moles of product, using the appropriate mole ratio. Calculate the mass of product, using the reciprocal of molar mass.

2 molecules 8 molecules 1. 4 molecules 4. 2.00 mol NO 18. 0.400 mol O2 21.

50.12 g NO 27. 1.24 g NO 17. 1.50 mol O2 8. 2.00 mol O2 5. 4 molecules NO 1.00 mol 2.00 mol Example Problem: The following chemical equation represents the production of gallium oxide, a substance used in the manufacturing of some semiconductor devices.

4Ga(s) 3O2(g) 0 2Ga2O3(s) 4.00 mol 4.00 mol 5.00 mol 7.00 mol 10. 7.00 mol 1.00 mol 4.00 mol 13. 2.00 mol 0.500 mol 0.200 mol 16.

0.200 mol In one experiment, the reaction yielded 7.42 g of the oxide from a 7.00-g sample of gallium. Determine the percent yield of this reaction.

Chapter 11 Stoichiometry Study Guide

The molar masses of Ga and Ga2O3 are 69.72 g/mol and 187.44 g/mol, respectively. 32.00 g 60.02 g 19. 92.02 g 16.00 g 80.00 g 22. 46.01 g Use the information in the diagram and example problem to evaluate each value or expression below. If the value or expression is correct, write correct. If it is incorrect, write the correct value or expression.

Actual yield: unknown 10.00 g 20.00 g 25. 28.76 g 7.42 g Ga2O3 2. Mass of reactant: 7.00 g Ga correct 3.

Chapter 11 Stoichiometry Chemistry Answers

Number of moles of reactant: 7.00 g Ga 4. Number of moles of product: 0.100 mol Ga 5. Theoretical yield: 0.0500 mol Ga2O3 6.