Chapter 12 Stoichiometry 12.2 Chemical Calculations

Slides:



Advertisements
Similar presentations
Chemical Quantities Chapter 9
Advertisements

Chapter 12 Stoichiometry 12.2 Chemical Calculations
What quantities are conserved in chemical reactions? grams and atoms.
Chapter 12 Stoichiometry part 1. Stoichiometry The study of quantitative relationships between amounts of reactants used and products formed by a chemical.
Can’t directly measure moles Measure units related to moles: –Mass (molar mass) –Number of particles (6.02 x ) –Liters of gas (22.4 Liters at STP)
When gases react, the coefficients in the balanced chemical equation represent both molar amounts and relative volumes. Section 3: Gas Stoichiometry K.
Stoichiometry. Stoichiometry- mass and quantity relationships among reactants and products in a chemical reaction Chemists use balanced chemical equations.
Section 9.1 Using Chemical Equations Steven S. Zumdahl Susan A. Zumdahl Donald J. DeCoste Gretchen M. Adams University of Illinois at Urbana-Champaign.
Chapter 12: Stoichiometry
12.3 Limiting Reagent and Percent Yield > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry 12.1.
Stoichiometry. Stoichiometry Stoichiometry – the process of using a balanced chemical equation to calculate the relative amounts of reactants and products.
12.2 Chemical Calculations > 12.2 Chemical Calculations > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry.
Stoichiometry: Calculations with Chemical Formulas and Equations
Chapter 9 Chemical Quantities.
Chem 1A Chapter 3 Lecture Outlines
Chapter 12 Review “Stoichiometry”
Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield
Solving a Stoichiometry Problem
Chemistry 200 Fundamentals E Limiting Reagents.
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Chapter 9 Chemical Quantities in Reactions
Stoichiometry Mole-to-Mole:
Stoichiometry: Chapter 9.
Gas Stoichiometry Molar volume can be used in stoichiometry calculations to convert quantities of chemicals in a chemical reaction. This ONLY works if.
Ideal Gas Law and Stoichiometry
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Calculations with Equations
Chapter 12 Review “Stoichiometry”
Calculations with Equations
Chapter 12 Stoichiometry
Chapter 12 Review.
Chapter 5 Chemical Reactions and Quantities
Chapter 12 Review “Stoichiometry”
Stoichiometry Review.
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
12.1/12.2 Equations and Chemical Calculations
Chapter 9 “Stoichiometry”
Calculations Based on Chemical Equations
Section 11.3 – Stoichiometry of Gases
2.50g C2H4 x 1 mole C2H4 x 2 mol CO2 x 22.4 L at STP = 4.00 L at STP
Chapter 12 Stoichiometry 12.2 Chemical Calculations
The Arithmetic of Equations
Chapter 8 Chemical Quantities in Reactions
Gas Volumes and the Ideal Gas Law
Calculations Based on Chemical Equations
Chapter 10 Chemical Equation Calculations by Christopher Hamaker
Quantities In Chemical Reactions
Chapter 6 Chemical Reactions and Quantities
Ch. 11: Molecular Composition of Gases
Conversions of Quantities in Moles
Chemical Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chapter 9.2 Ideal Stoichiometric Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Volume-Mass Relationships of Gases
Stoichiometry Chapter 9.
Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield
Stoichiometry and the combustion of fuels 2
Calculations Based on Chemical Equations
Section 3 Gas Volumes and the Ideal Gas Law
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Stoichiometry.
Stoichiometry Chapter 12.
Stoichiometry.
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Stoichiometry volume -to- volume:
Aim # 11: What do the coefficients of a chemical equation represent?
Mass to Mass Conversions
Presentation transcript:

Chapter 12 Stoichiometry 12.2 Chemical Calculations 12.1 The Arithmetic of Equations 12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is the mole ratio of Fe2O3 to Fe? Do Now 4Fe + 3O2  2Fe2O3 What is the mole ratio of Fe2O3 to Fe? What is the mole ratio of Fe2O3 to O2? How many moles of Fe2O3 will form from 8 moles of Fe in excess oxygen? If there are 127 moles of Fe, how many moles of Fe2O3 will form? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Writing and Using Mole Ratios Steps for Solving a Mass-Mass Problem Grams Known Moles Known Moles Unknown Grams Unknown Using Molar Mass Using Mole Ratios Using Molar Mass Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What mass of phosphorus will be needed to produce 3.25 mol of P4O10? Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) + 5O2(g)  P4O10(s) What mass of phosphorus will be needed to produce 3.25 mol of P4O10? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What mass of phosphorus will be needed to produce 3.25 mol of P4O10? Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) + 5O2(g)  P4O10(s) What mass of phosphorus will be needed to produce 3.25 mol of P4O10? 3.25 mol P4O10   = 403 g P 31.0 g P 1 mol P 1 mol P4O10 4 mol P Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How many grams of carbon dioxide will be made when 100 grams of CH4 combusts with sufficient oxygen? CH4 + 2O2  CO2 + 2H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How many grams of carbon dioxide will be made when 100 grams of CH4 combusts with sufficient oxygen? CH4 + 2O2  CO2 + 2H2O 44 g CO2 1 mol CO2 100 g CH4  16.0 g CH4 1 mol CH4  = 275 g CO2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Other Stoichiometric Calculations The figure below summarizes the steps for a typical stoichiometric problem. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Molecules of a Product Sample Problem 12.5 Calculating the Molecules of a Product How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2H2O(l) 2H2(g) + O2(g) electricity Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

= 4.88  1023 molecules O2 Sample Problem 12.5 1 mol O2  2 mol H2O 29.2 g H2O  1 mol H2O 18.0 g H2O Given quantity Change to moles Mole ratio Change to molecules = 4.88  1023 molecules O2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Volume-Volume Stoichiometric Calculations Sample Problem 12.6 Volume-Volume Stoichiometric Calculations Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen react with an excess of nitrogen monoxide? Assume conditions are at STP. 2NO(g) + O2(g)  2NO2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

34 L O2    1 mol O2 22.4 L O2 1 mol NO2 22.4 L NO2 2 mol NO2 Sample Problem 12.6 34 L O2    1 mol O2 22.4 L O2 1 mol NO2 22.4 L NO2 2 mol NO2 Given quantity Change to moles Mole ratio Change to liters = 68 L NO2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Volume of a Gas Needed for a Reaction Sample Problem 12.7 Finding the Volume of a Gas Needed for a Reaction Assuming STP, how many liters of oxygen are needed to produce 20.4 L SO3 according to this balanced equation? 2SO2(g) + O2(g)  2SO3(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Volume of a Gas Needed for a Reaction Sample Problem 12.7 Finding the Volume of a Gas Needed for a Reaction Assuming STP, how many liters of oxygen are needed to produce 20.4 L SO3 according to this balanced equation? 2SO2(g) + O2(g)  2SO3(g) 10.2 L O2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? 22.4 L H2O 1 mol H2O 501 g CH4  16.05 g CH4 1 mol CH4  2 mol H2O = 1.40  103 L H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.