Chapter 12 Stoichiometry 12.2 Chemical Calculations

Slides:



Advertisements
Similar presentations
Chemical Quantities Chapter 9
Advertisements

Starter S moles NaC 2 H 3 O 2 are used in a reaction. How many grams is that?
April 3, 2014 Stoichiometry. Stoichiometry is the study of quantities of materials consumed and produced in chemical reactions Stoikheion (Greek, “element”)
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Stoichiometry Chapter 8. Stoichiometry Chemical equations Limiting reagent Problem types Percent yield Mass-mass Mole - mole other.
Aim: Using mole ratios in balanced chemical equations.
Stoichiometry Calculating Masses of Reactants and Products.
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Chapter 9 Chemical Quantities. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 2 Information Given by the Chemical Equation Balanced equations.
The Mole & Stoichiometry!
Stoichiometry Chemical Quantities Chapter 9. What is stoichiometry? stoichiometry- method of determining the amounts of reactants needed to create a certain.
10.3 Percent Composition and Chemical Formulas 1 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Chapter 10 Chemical Quantities.
Stoichiometry and cooking with chemicals.  Interpret a balanced equation in terms of moles, mass, and volume of gases.  Solve mole-mole problems given.
Ch. 12 Stoichiometry 12.1 The Arithmetic of Equations.
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.
12.3 Limiting Reagent and Percent Yield > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry 12.1.
Chemistry Chapter 9 - Stoichiometry South Lake High School Ms. Sanders.
12.1 The Arithmetic of Equations > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry 12.1 The Arithmetic.
Ch. 9.1 & 9.2 Chemical Calculations. POINT > Define the mole ratio POINT > Use the mole ratio as a conversion factor POINT > Solve for unknown quantities.
12.2 Chemical Calculations > 12.2 Chemical Calculations > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry.
Chapter 9 Chemical Quantities.
Chapter 12 Review “Stoichiometry”
Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield
Solving a Stoichiometry Problem
Stoichiometry.
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Intro to Ch 9 Pg 267 #2= work w/partner (a-f)=10 min
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Chapter 9 Chemical Quantities in Reactions
Stoichiometry “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy.
Stoichiometry: Chapter 9.
MASS - MASS STOICHIOMETRY
Unit 4: Stoichiometry Stoichiometry.
Calculations with Equations
Calculations with Equations
Chemical Reactions Unit
Chapter 9 Stoichiometry
Gas Stoichiometry At STP
Chapter 5 Chemical Reactions and Quantities
Limiting Reactant/Reagent Problems
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
12.2 Chemical Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chapter 8 Chemical Quantities in Reactions
Chapter 9 Stoichiometry
Chapter 8 Chemical Quantities in Reactions
Ch. 9 Notes -- Stoichiometry
Chapter 6 Chemical Reactions and Quantities
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chemical Calculations
Created by C. Ippolito June 2007
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chapter 9.2 Ideal Stoichiometric Calculations
Chemical Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Stoichiometry Chapter 12.
Stoichiometry Chemistry II Chapter 9.
DRILL Besides mass what is conserved during a chemical reaction?
Chapter 9 Stoichiometry
Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield
Chapter 9 Chemical Quantities in Reactions
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Stoichiometry.
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Happy Tuesday! Please get you’re S’mores Lab and your Stoichiometry Worksheet Packet from Thursday and Friday ready to turn in for a grade. Copyright ©
Stoichiometry.
Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
Calculation of Chemical Quantities
Chapter 5 Chemical Quantities and Reactions
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 mass (in grams) of 5.4 moles of H2O? Do Now What is the mass (in grams) of 5.4 moles of H2O? How many molecules of CH4 are in 6 moles of CH4? How many moles of O2 molecules are in 44.8 L of O2 at STP? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Writing and Using Mole Ratios In chemical calculations, mole ratios are used to convert between a given number of moles of a reactant or product to moles of a different reactant or product. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Combustion of Gasoline Chemical Equations Combustion of Gasoline 2C8H18 + 25O2  16CO2 + 18H2O ∆H = -10,224 kJ How many moles of CO2 will be produced from the combustion of 30 moles of C8H18? (note: 30 moles C8H18 is ~ 1 gallon of gasoline) http://www3.epa.gov/climatechange/kids/ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2H2(g) + O2(g) ↔ 2H2O(l) ∆H = -482 kJ Chemical Equations Combustion of Hydrogen / Electrolysis 2H2(g) + O2(g) ↔ 2H2O(l) ∆H = -482 kJ How many moles of O2 are required to react with 10 moles of H2? http://energy.gov/eere/videos/energy-101-fuel-cell-technology Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Writing and Using Mole Ratios Mass-Mass Calculations N2(g) + 3H2(g)  2NH3(g) If I want to produce 150 grams of NH3, how many grams of hydrogen will I need? 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.

Calculating the Mass of a Product Sample Problem 12.4 Calculating the Mass of a Product Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is: N2(g) + 3H2(g)  2NH3(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Change given unit to moles Sample Problem 12.4  2 mol NH3 3 mol H2 5.40 g H2  1 mol H2 2.0 g H2 17.0 g NH3 1 mol NH3 Given quantity Change given unit to moles Mole ratio Change moles to grams = 31 g NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.