Chapter 5 Chemical Reactions and Quantities

Slides:



Advertisements
Similar presentations
Mole Relationships in Chemical Equations
Advertisements

Chemistry 103 Lecture 14. Outline I. Empirical/Molecular Formulas II. Chemical Reactions - basic symbols - balancing - classification III. Stoichiometry.
1 Chapter 5 Chemical Reactions 5.9a The Mole Relationships in Chemical Equations.
1 Chapter 5 Chemical Reactions 5.9c Mass Calculations for Reactions.
1 Chapter 5 Chemical Quantities and Reactions 5.7 Mole Relationships in Chemical Equations Copyright © 2009 by Pearson Education, Inc.
CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY Goals Chemical Equations Calculations Based on Chemical Equations The Limiting Reactant Concept.
Chapter 12 Stoichiometry part 1. Stoichiometry The study of quantitative relationships between amounts of reactants used and products formed by a chemical.
Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 9 Chemical Quantities in Reactions 9.1 Mole Relationships in Chemical Equations.
1 Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.
Glencoe Physical Science Chapter 21
Stoichiometry Chapter 10.
Section 2Chemical Reactions Describing Reactions 〉 What is a chemical equation? 〉 A chemical equation uses symbols to represent a chemical reaction and.
Chapter 11: Stoichiometry
LecturePLUS Timberlake1 Chapter 5 Chemical Reactions and Quantities Chemical Changes Balancing Chemical Equations.
Unit 12: Stoichiometry Stoicheion = element Metron = to measure.
STOICHIOMETR Y is G LAM O R O US ARE YOU READY? A. _______________________ involves the study of the relationships between ________________ and _______________.
Chapter 5 Chemical Quantities and Reactions Chapter 5 Chemical Quantities and Reactions 5.7 Mole Relationships in Chemical Equations 1.
General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 6.6 Mole Relationships in Chemical Equations Chapter 6 Chemical Reactions and.
Stoichiometry and cooking with chemicals.  Interpret a balanced equation in terms of moles, mass, and volume of gases.  Solve mole-mole problems given.
Chemical Reactions Notes. Reactants Products Substances that undergo a change New substances formed Yields or Produces.
Topic: Chemical Reactions and Equations PSSA: A/S8.C.1.1.
Balancing Chemical Equations
Chapter 5 Chemical Reactions and Quantities
Stoichiometry Coach Cox.
Day 1 - Notes Unit: Stoichiometry
Chapter 5 Chemical Reactions and Quantities
Balancing Chemical Equations
7.6 Mole Relationships in Chemical Equations
Chapter 9 Chemical Quantities in Reactions
Chapter 5 Chemical Quantities and Reactions
Stoichiometry Chemistry 11 Ms. McGrath.
Chemical Reactions: An Introduction
Chapter 9A Notes Stoichiometry
Lecture 60 Defining Stoichiometry Ozgur Unal
Chapter 5 Chemical Reactions and Quantities
Calculations with Equations
Chapter 11: Stoichiometry
Chapter 5 Chemical Reactions and Quantities
Chapter 5 Chemical Reactions and Quantities
Living By Chemistry SECOND EDITION
Calculations Based on Chemical Equations
Chapter 9 Chemical Quantities in Reactions
Chapter 5 Chemical Quantities and Reactions
Chapter 8 Chemical Quantities in Reactions
Chapter 8 Chemical Quantities in Reactions
Volume and Moles (Avogadro’s Law)
Chapter 6 Chemical Reactions and Quantities
Stoichiometry Chapter 12.
FORMING NEW SUBSTANCES
Stoichiometry Chapter 9.
Information Given by Chemical Equations
Chapter 5 Chemical Quantities and Reactions
Aim: How do chemists calculate the mass of one mole of a substance?
Chemical Reactions and Quantities
Calculations Based on Chemical Equations
Warm Up #6 Balance the following:
Chapter 6 Chemical Reactions and Quantities
Chapter 9 Chemical Quantities in Reactions
Chapter 5 Chemical Quantities and Reactions
Chapter 5 Chemical Reactions and Quantities
Day 1 - Notes Unit: Stoichiometry
Stoichiometry.
What is Polarity of a molecule?
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Calculation of Chemical Quantities
7.1 Describing Reactions In a chemical reaction, the substances that undergo change are called reactants. The new substances formed as a result of that.
Balancing Chemical Equations
Chapter 5 Chemical Quantities and Reactions
Chapter 5 Chemical Reactions and Quantities
Presentation transcript:

Chapter 5 Chemical Reactions and Quantities 5.7 Mole Relationships in Chemical Equations

Law of Conservation of Mass The Law of Conservation of Mass indicates that in an ordinary chemical reaction, Matter cannot be created nor destroyed. No change in total mass occurs in a reaction. Mass of products is equal to mass of reactants.

Conservation of Mass 2 moles Ag + 1 moles S = 1 mole Ag2S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product

Reading Equations In Moles Consider the following equation: 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) This equation can be read in “moles” by placing the word “moles” between each coefficient and formula. 4 moles Fe + 3 moles O2 2 moles Fe2O3

Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for any two substances in an equation. 4Fe(s) + 3O2(g) 2Fe2O3(s) Fe and O2 4 moles Fe and 3 moles O2 3 moles O2 4 moles Fe Fe and Fe2O3 4 moles Fe and 2 moles Fe2O3 2 moles Fe2O3 4 moles Fe O2 and Fe2O3 3 moles O2 and 2 moles Fe2O3 2 moles Fe2O3 3 moles O2

Learning Check Consider the following equation: 3 H2(g) + N2(g) 2 NH3(g) A. A mole-mole factor for H2 and N2 is 1) 3 moles N2 2) 1 mole N2 3) 1 mole N2 1 mole H2 3 moles H2 2 moles H2 B. A mole-mole factor for NH3 and H2 is 1) 1 mole H2 2) 2 moles NH3 3) 3 moles N2 2 moles NH3 3 moles H2 2 moles NH3

Solution A. A mole-mole factor for H2 and N2 is 2) 1 mole N2 3H2(g) + N2(g) 2NH3(g) A. A mole-mole factor for H2 and N2 is 2) 1 mole N2 3 moles H2 B. A mole-mole factor for NH3 and H2 is 2) 2 moles NH3

Calculations with Mole Factors How many moles of Fe2O3 can form from 6.0 mole O2? 4Fe(s) + 3O2(g) 2Fe2O3(s) Relationship: 3 mole O2 = 2 mole Fe2O3 Write a mole-mole factor to determine the moles of Fe2O3. 6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3 3 mole O2

Guide to Using Mole Factors

Learning Check How many moles of Fe are needed for the reaction of 12.0 moles O2? 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe

Solution 12.0 moles O2 x 4 moles Fe = 16.0 moles Fe 3 moles O2