II. Stoichiometry in the Real World (p )

Slides:

Advertisements

Similar presentations

II. Stoichiometry in the Real World (p. 379 – 388) Stoichiometry – Ch. 11.

Advertisements

Percent Yield and Limiting Reactants

Mole Review 1.) Calculate the number of moles in 60.4L of O2. 2.) How many moles are there in 63.2g of Cl2? 1 mol O2 60.4L O2 = 2.7 mol O2 22.4L O2 1mol.

Stoichiometry Continued…

II. Stoichiometry in the Real World (p ) Stoichiometry – Ch. 9.

Mole Review 1.) Calculate the number of moles in 60.4L of O2. 2.) How many moles are there in 63.2g of Cl2? 1 mol O2 60.4L O2 = 2.7 mol O2 22.4L O2 1mol.

Limiting Reactants and Percent Yield

II. Stoichiometry in the Real World Limiting Reagents and % yield (p ) Stoichiometry – Ch. 12.

Chapter 9 - Section 3 Suggested Reading: Pages

II. Stoichiometry in the Real World * Limiting Reagents

Unit 08 – Moles and Stoichiometry I. Molar Conversions.

Chapter 9 Pages Proportional Relationships u I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2.

Stoichiometric Calculations (p )

Lecturer: Amal Abu- Mostafa.  Available Ingredients ◦ 4 slices of bread ◦ 1 jar of peanut butter ◦ 1/2 jar of jelly Limiting Reactant Limiting Reactant.

Limiting/Excess Reactants and Percent Yield

II. Gas Stoichiometry. 1 mol of a gas=___ L at STP A. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm.

Chemical Quantities – Ch. 9.

Limiting Reagents and Percent Yield

Stoichiometric Calculations

Unit 8: Percent Yield Calculations

Chapter 12 Stoichiometry The study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.Stoichiometry.

When copper (II) reacts with silver nitrate, the number of grams of copper required to produce 432 grams of silver is:Warm-Up CuAgNO 3 Ag22+Cu(NO 3 ) 2.

Chapter 12 Cookies? u When baking cookies, a recipe is usually used, telling the exact amount of each ingredient If you need more, you can double or.

Chapter 9 Stoichiometry. Definition of “Stoichiometry”: the mathematics of chemical equations Important Concepts: 1. You MUST have a balanced equation!

Stoichiometric Calculations Stoichiometry – Ch. 9.

Stoichiometric Calculations Start Your Book Problems NOW!! Stoichiometry.

Stoichiometric Calculations Stoichiometry – Ch. 8.

The Study of Stoichiometry I. Stoichiometric Calculations.

I. I.Stoichiometric Calculations Topic 9 Stoichiometry Topic 9 Stoichiometry.

C. Johannesson I. I.Stoichiometric Calculations (p ) Stoichiometry – Ch. 9.

Limiting Reagents & Percent Yield Chapter 9 Notes Part III.

C. Johannesson II. Stoichiometry in the Real World (p ) Stoichiometry – Ch. 9.

II. Limiting Reactants Stoichiometry – 3.7. A. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b.

II. Stoichiometry in the Real World Stoichiometry – Ch. 11.

II. Stoichiometry in the Real World Stoichiometry – Unit. 10.

II. Stoichiometry in the Real World Stoichiometry.

I. I.Stoichiometric Calculations Topic 6 Stoichiometry Topic 6 Stoichiometry.

I. I.Stoichiometric Calculations Stoichiometry – Ch. 10.

I. I.Stoichiometric Calculations Stoichiometry. A. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla.

Stoichiometry: Limiting Reactants Chapter 9 Lesson 3.

Jeopardy $100 Stoich Vocab Stoich Calculations Stoich True/False Percent Yield Limiting Reactants $200 $300 $400 $500 $400 $300 $200 $100 $500 $400 $300.

Stoichiometry – Ch What would be produced if two pieces of bread and a slice of salami reacted together? + ?

Stoichiometry in the Real World Stoichiometry – Ch. 11.

II. Stoichiometry in the Real World (p )

Percent Yield in a Chemical Reaction.

Percent Yield actual yield % yield = x 100 theoretical yield

Limiting Reagents and Percent Yield

II. Stoichiometry in the Real World

Stoichiometric Calculations

Unit 8: Stoichiometry: Part 1

Stoichiometric Calculations (p )

Ch. 9: Calculations from Chemical Equations

Chapter 12 Review.

Chapter 12 Review.

Agenda: 1/13/2017 Go over the procedure for the Molarity Lab

Limiting and Excess Reactants

Stoichiometry in the Real World

II. Stoichiometry in the Real World (p )

Formation of Ammonia.

Stoichiometric Calculations (p )

II. Stoichiometry in the Real World (p. 368 – 375)

II. Stoichiometry in the Real World

14.5 NOTES Stoichiometry.

Limiting Reactants and Percent Yield

STOICHIOMETRY!!!!.

Limiting/Excess Reactants and Percent Yield

Stoichiometric Calculations (p )

Presentation transcript:

II. Stoichiometry in the Real World (p. 288-294) Stoichiometry – Ch. 9 II. Stoichiometry in the Real World (p. 288-294)

A. Limiting Reactants Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly Limiting Reactant bread Excess Reactants peanut butter and jelly

A. Limiting Reactants Limiting Reactant used up in a reaction determines the amount of product Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? g 37 g 79.1 g of zinc react with 37 g of 2.5M HCl. Identify the limiting and excess reactants. How many grams of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 79.1 g 37 g ? g

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 37g ? g 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 2.02 g H2 1 mol = g H2

A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 37g ? g 37g HCl 1 mol HCl 36.46g HCl 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 = g H2

A. Limiting Reactants Zn: gH2 HCl: g H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: g H2 left over zinc

B. Percent Yield measured in lab calculated on paper

B. Percent Yield When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g Theoretical Yield: 45.8 g K2CO3 1 mol K2CO3 138.21 g 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g actual: 46.3 g Theoretical Yield = 49.4 g KCl 46.3 g 49.4 g % Yield =  100 = 93.7%