Limiting Reactants. Copyright 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Similar documents
3/6/2018. Limiting Reactant. Reacting Amounts. Reacting Amounts. Limiting Reactants. Example of Everyday Limiting Reactant.

Chapter 5 Chemical Reactions Law of Conservation of Mass and Quantities 5.7 Mole Relationships in Chemical Equations

EXTRA CREDIT REMINDER

Math-tastic! Lesson 9.5 Limiting Reagent & Percent Yield 2/21/2015. Identify the limiting reagent in a reaction. Limiting Reactants OBJECTIVES:

Unit 6: Stoichiometry. How do manufacturers know how to make enough of their desired product?

9.1 Information Given by Chemical Equations 9.2 Mole Mole Relationships 9.3 Mass Calculations 9.4 The Concept of Limiting Reactants 9.

Stoichiometry Ch. 11. I. Stoichiometric Calculations

Stoichiometry. The study of quantities of substances in chemical reactions

CHAPTER 11 Stoichiometry Defining Stoichiometry

9.2 Chemical Calcualtions. Chapter 9 Stoichiometry. 9.1 The Arithmetic of Equations. 9.2 Chemical Calculations. 9.3 Limiting Reagent and Percent Yield

Stoichiometry. Before You Read. Chapter 10. Chapter 11. Review Vocabulary. Define the following terms. mole. molar mass.

How many molecules are in 0.25 moles of CH 4?

Stoichiometry. Please take out your notebooks

Name Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages )

Practice Problems Stoich!

Chapter 4: Chemical and Solution Stoichiometry

Chapter 9. Calculations from Chemical Equations. to patients Introduction to General, Organic, and Biochemistry 10e throughout the

Sample Problem Set. Limiting Reactants

Unit 10: Stoichiometry. Stoichiometry= the process of using a to determine the relative amounts of reactants and products involved in a reaction.

Problem Solving. Limiting Reactants

Chapter 7: Stoichiometry in Chemical Reactions

STOICHIOMETRY. Engr. Yvonne Ligaya F. Musico 1

Study Guide: Stoichiometry

phet: Molarity Go to: simulation/molarity Click on Run in HTML5

3.7 Chem. Eqs & 3.8 Balancing Chem. Eqs

Chem. I Notes Ch. 11 STOICHIOMETRY NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics.

Notes: Stoichiometry (text Ch. 9)

Ch. 10 Notes STOICHIOMETRY NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics.

Chemical Equations. shows the results of a chemical process

Notes 2: Stoichiometry

Limiting Reactants. In other words once the reactant that is present in the smallest amount is completely consumed the reaction will stop.

UNIT 1 Chemical Reactions Part II Workbook. Name:

Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq)

Calculations From Chemical Equations

Chemical Quantities. Conversion factors. Definition of mole. 1 mole = x items. (Example)

Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Lecture Presentation

UNIT 9. Stoichiometry

Unit 7: Stoichiometry Homework Packet (85 points)

Chapter 3. Mass Relations in Chemistry; Stoichiometry

Stoichiometry Dry Lab

Unit 9 Stoichiometry Notes

2.9 The Mole and Chemical Equations:

STOICHIOMETRY. Measurements in Chemical Reactions

Stoichiometry study of the relationships in a

This reaction might be scaled up to make a larger quantity of product. For example, to make three times as much:

Chemical Quantities: Stoichiometry and the Mole

Chapter 9. Stoichiometry. Mr. Mole. NB page 189

Chapter 9 STOICHIOMETRY

7.1 Describing Reactions. Burning is a chemical change. When a substance undergoes a chemical change, a chemical reaction is said to take place.

Stoichiometry. Homework EC. cincochem.pbworks.com. Academic Chemistry DATE ASSIGNMENT

A. Correct. You successfully completed the stoichiometry problem. B. Incorrect. There are 2 moles of AgCl produced for each mole of CaCl 2 reacted.

Unit IV: Stoichiometry

Stoichiometry is the relationship between the amount of reactants used and/or the amount of products produced in a chemical reaction.

Stoichiometry is the relationship between the amount of reactants used and/or the amount of products produced in a chemical reaction.

Spring Semester Final Exam Study Guide

THE MOLE - PART 2. Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.

Honors Chemistry Unit 6 Moles and Stoichiometry Notes. Intro to the mole 1. What is the chemical mole? 2. What is Avogadro s number?

CHAPTER 9: STOICHIOMETRY

Stoichiometry CHAPTER 12

Proportional Relationships

**continued on next page**

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

Funsheet 3.0 [WRITING & BALANCING EQUATIONS] Gu/R. 2017

Slide 1 / 90. Stoichiometry HW. Grade:«grade» Subject: Date:«date»

Stoichiometry CHAPTER 12

Outcomes: Interpret a balanced chemical equation in terms of moles, mass and volume of gases. Solve stoichiometric problems involving: moles, mass,

Calculations with Chemical Formulas and Equations

UNIT 9 - STOICHIOMETRY

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

Chapter 9. Table of Contents. Stoichiometry. Section 1 Introduction to Stoichiometry. Section 2 Ideal Stoichiometric Calculations

Stoichiometry World of Chemistry: Chapter 9

Reading Balanced Chemical Equations (see MHR Text p )

INTRODUCTORY CHEMISTRY Concepts and Critical Thinking

CHEMICAL REACTIONS. Introduction. Chemical Equations

STOICHIOMETRY. STOICHIOMETRY Chemists use balanced chemical equations to calculate how much reactant is needed or how much product is formed.

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

AP Chemistry Multiple Choice Questions - Chapter 4

Basic Concepts of Chemistry Notes for Students [Chapter 7, page 1] D J Weinkauff - Nerinx Hall High School

Chapter 9: Stoichiometry The Arithmetic ti Of Equations

Chapter 5 Chemical Reactions

3. Carbon disulfide is an important organic solvent. It can be produced from sulfur dioxide, a byproduct of buring coal.

Chapter 4 Chemical Quantities and Aqueous Reactions

Chapter 12 Stoichiometry. Mr. Mole

Lab #5 - Limiting Reagent

TOPIC 10. CHEMICAL CALCULATIONS IV - solution stoichiometry.

20 trolls and 15 hobbits go to a dance. What is the greatest # of dancing pairs on the dance floor at any one time?

Moles Revisited Name Date Molar Mass How do you calculate the formula mass of a compound? Examples Potassium fluoride Strontium nitrate Aluminum nitri

Solutions to the Extra Problems for Chapter 7

Quantitative Relationships in Chemical Reactions Chapter 7

key content vocabulary next to definitions (sometimes #2 and #3 will be the same and in that case I expect to see a box AND DEF)

AP CHEMISTRY THINGS TO KNOW

The Mole. Relative Atomic Mass Ar

Note: coefficients of 1 can be omitted, and are only shown here for clarity. S 2 O 6 charge 0 Check: Al 4 Mn 3 O 6. charge 0. Pb 2 O 4.

What does this equation tell you? 1. 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen gas to produce 2 molecules of ammonia gas.

Stoichiometric Calculations

INTRO AND BACKGROUND: Reactions, Moles, Stoichiometry, and Solutions. Chemical Reaction Atoms are REARRANGED to form a different substance

Chapter 3. Stoichiometry

Chemistry I Chapter 9 Stoichiometry Objective Sheet. Equation 1. Objectives: 1. Define stoichiometry

Unit VI Stoichiometry. Applying Mole Town to Reactions

Stoichiometric Calculations

Transcription:

Limiting Reactants Copyright 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Limiting Reactants Limiting Reactant used up in a reaction Determines/limits the amount of product Stops the reaction Ecess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? Copyright 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings What is the limiting item?

Reacting Amounts Four table settings can be made. Initially Use Left over plates 5 4 1 forks 6 4 2 spoons 4 4 0 knives 7 4 3 The limiting item is the spoon.

Eample of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. Copyright 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

Eample of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. Copyright 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

Limiting Reactants When 4.00 mol H 2 is mied with 2.00 mol Cl 2,how many moles of HCl can form? H 2 (g) + Cl 2 (g) 2HCl (g) 4.00 mol 2.00 mol??? Mol Calculate the moles of product from each reactant, H 2 and Cl 2. The limiting reactant is the one that produces the smaller amount of product.

Limiting Reactants aluminum + chlorine gas aluminum chloride Al(s) + Cl 2 (g) AlCl 3 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 100 g 100 g? g A. 200 g B. 125 g C. 667 g D. 494 g

Limiting Reactants aluminum + chlorine gas aluminum chloride 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 100. g 100. g g How much product would be made if we begin with 100 g of aluminum? g AlCl 3 = 100 g Al 1 mol Al 2 mol AlCl 3 27 g Al 2 mol Al 133.5 g AlCl 3 1 mol AlCl 3 = 494 g AlCl 3 Al AlCl 3 How much product would be made if we begin with 100 g of chlorine gas? 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 g AlCl 3 = 100 g Cl 2 71 g Cl 2 = 125 g AlCl 3 3 mol Cl 2 1 mol AlCl 3 Cl 2 AlCl 3

Limiting Reactants Using Mass If 4.80 mol Ca mied with 2.00 mol N 2, which is the limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Moles of Ca 3 H 2 from Ca 4.80 mol Ca 1 mol Ca 3 N 2 = 1.60 mol Ca 3 N 2 3 mol Ca (Ca used up) Moles of Ca 3 H 2 from N 2 2.00 mol N 2 1 mol Ca 3 N 2 = 2.00 mol Ca 3 N 2 1 mol N 2 (not possible) All Ca is used up when 1.60 mol Ca 3 N 2 forms. Thus, Ca is the limiting reactant.

Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l)

Limiting Reactants Using Mass Calculate the grams of H 2 for each reactant. H 2 : 8.00 g H 2 1 mol H 2 2 mol H 2 O 18.02 g H 2 O 2.016 g H 2 2 mol H 2 1 mol H 2 O = 71.5 g H 2 O (not possible) O 2 : 24.0 g O 2 1 mol O 2 2 mol H 2 O 18.02 g H 2 O 32.00 g O 2 1 mol O 2 1 mol H 2 O = 27.0 g H 2 O (smaller) O 2 is the limiting reactant.

Eample: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. There are two solution methods you could use. In both methods, the first step is to convert the mass to moles.

Method 1 Use the moles of one reactant to calculate the necessary moles of the other reactant to fully react. Compare the calculated value with the actual value to see if this reagent is ecess or limiting.

Eample: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. Answers using method 1: Convert mass to moles: 0.038 mol O 2, 0.5 mol H 2 Calculate H 2 moles necessary to react with O 2 : 0.076 mol H 2. Compare 0.076 mol H 2 to actual mol of H 2 (0.5mol H 2 ), Since 0.5 mol H 2 is more than 0.076 mol H 2, H 2 is the ecess reagent and O 2 is the limiting reagent.

Method 2 Use the moles of each of the reactant to calculate one of the products. The reagent that gave the smaller calculated value of product is the limiting reagent. The actual value of the amount of product is the smaller of the calculated values.

Eample: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. Answers using Method 2: Convert mass to moles: 0.038 mol O 2, 0.5 mol H 2 Calculate H 2 O moles produced by using each of the reactants: Using O 2 : 0.076 mol H 2 O. Using H 2 : 0.5 mol H 2 O. The actual amount H 2 O produced is the smaller one of the two values(0.076mol H 2 O). O 2 is the limiting reagent, since O 2 was used in the calculation of the 0.076mol H 2 O.

Practice questions 1. 2Al + 6HCl 2AlCl 3 + 3H 2 If 25 g of aluminum was added to 90 g of HCl, what mass of H 2 will be produced (try this two ways with a chart & using the shortcut)? 2. N 2 + 3H 2 2NH 3 : If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 3. What mass of aluminum oide is formed when 10.0 g of Al is burned in 20.0 g of O 2? 4. When C 3 H 8 burns in oygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 5. How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

1 #mol Al = 25 g Al 1 mol Al = 0.926 mol 27.0g Al # mol HCl = 90 g HCl 1 mol HCl = 2.466 mol 36.5g HCl What we have What we need # g H 2 = 90 g HCl Al HCl 0.926 2.466 0.926/0.926 2.466/0.926 = 1 mol = 2.7 mol 2 6 2/2 = 1 mol 6/2 = 3 mol HCl is limiting. 1 mol HCl 3 mol H 2 2.0 g H 2 = 2.47 g H 2 36.5 g HCl 6 mol HCl 1 mol H 2

Question 1: shortcut 2Al + 6HCl 2AlCl 3 + 3H 2 If 25 g aluminum was added to 90 g HCl, what mass of H 2 will be produced? # g H 3 mol H 2 2 = 25 g Al 1 mol Al 2.0 g H 2 = 2.78 g H 2 27.0 g Al 2 mol Al 1 mol H 2 # g H 3 mol H 2 2 =90 g HCl 1 mol HCl 2.0 g H 2 = 2.47 g H 2 36.5 g HCl 6 mol HCl 1 mol H 2

Question 2: shortcut N 2 + 3H 2 2NH 3 If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? # g NH 3 = 20 g N 1 mol N 2 mol NH 3 2 17.0 g NH 2 3 = 24.3 g H 2 28.0 g N 2 1 mol N 2 1 mol NH 3 # g NH 3 = 5.0 g H 1 mol H 2 mol NH 3 2 17.0 g NH 2 = 3 28.3 g H 2 2.0 g H 2 3 mol H 2 1 mol NH 3 N 2 is the limiting reagent

Question 3: shortcut 4Al + 3O 2 2 Al 2 O 3 What mass of aluminum oide is formed when 10.0 g of Al is burned in 20.0 g of O 2? # g Al 2 O 3 = 10.0 g Al 1 mol Al 2 mol Al 2 O 3 102.0 g Al O 2 3 = 18.9 g Al 2 O 3 27.0 g Al 4 mol Al 1 mol H 2 # g Al 2 O 3 = 20.0 g O 1 mol O 2 mol Al 2 O 3 2 2 102.0 g Al 2 O 3 = 42.5 g Al 2 O 3 32.0 g O 2 3 mol O 2 1 mol H 2

Question 4: shortcut C 3 H 8 + 5O 2 3CO 2 + 4H 2 O When C 3 H 8 burns in oygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? # g CO 2 = 15.0 g C 3 mol CO 2 3 H 1 mol C 44.0 g CO 3 H 8 8 2 = 45.0 g CO 2 44.0 g C 3 H 8 1 mol C 3 H 8 1 mol CO 2 # g CO 2 = 60.0 g O 1 mol O 3 mol CO 2 2 2 44.0 g CO 2 = 49.5 g CO 2 32.0 g O 2 5 mol O 2 1 mol CO 2 5. Limiting reagent questions give values for two or more reagents (not just one)

Question 2 # mol N 1 mol N 2 2 = 20 g N 2 = 0.714 mol N 2 28 g N 2 # mol H 1 mol H 2 2 = 5.0 g H 2 = 2.5 mol H 2 2 g H 2 What we have What we need N 2 H 2 0.714 mol 2.5 mol 0.714/0.714 = 1 mol 2.5/0.714 = 3.5 mol 1 mol 3 mol We have more H 2 than what we need, thus H 2 is in ecess and N 2 is the limiting factor.

3 4Al + 3O 2 2 Al 2 O 3 # mol Al = 10 g Al 1 mol Al 0.37 mol Al = 27 g Al # mol O 1 mol O 2 2 = 20 g O 0.625 mol O 2 = 2 32 g O 2 There is Al O 2 0.37 mol 0.625 mol more What we than have 0.37/.37 0.625/0.37 = 1 mol = 1.68 mol enough What we 4 mol 3 mol O 2 ; Al is need 4/4 = 1 mol 3/4 = 0.75 mol limiting # g Al 2 mol Al 2 O 3 2 O 3 = 0.37 mol Al 102 g Al 2 O 3 4 mol Al 1 mol Al 2 O 3 = 18.87 g Al O

4 C 3 H 8 + 5O 2 3CO 2 + 4H 2 O # mol C 1 mol C 3 H 8 3 H 8 =15 g C 3 H 0.34 mol 8 = 44 g C 3 H 8 C 3 H 8 # mol O 1 mol O 2 2 = 60 g O 1.875 mol O 2 = 2 32 g O 2 C 3 H 8 O 2 We have What we 0.34 mol 1.875 mol more than 0.34/.34 1.875/0.34 enough O have 2, = 1 mol = 5.5 mol C 3 H 8 is limiting Need 1 mol 5 mol # g CO 2 = 0.34 mol C3 H 8 3 mol CO 2 1 mol C 3 H 8 = 44 g CO 2 1 mol CO 2 45 g CO 2

Limiting Reagents: shortcut MgCl 2 + 2AgNO 3 Mg(NO 3 ) 2 + 2AgCl If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced? # g AgCl= 25 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 2 mol AgCl 1 mol MgCl 2 = 143.3 g AgCl 1 mol AgCl 75.25 g AgCl # g AgCl= 68 g AgNO 3 For more lessons, visit www.chalkbored.com 1 mol AgNO 3 169.88 g AgNO 3 2 mol AgCl 143.3 g AgCl 2 mol AgNO 3 1 mol AgCl = 57.36 g AgCl

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback