2S (s) + 3O2 (g) 2SO3 (g) O < 1

Slides:

Advertisements

Similar presentations

Performing Calculations involving Limiting and Excess Reagents.

Advertisements

Chapter 5 Chemical Reactions

Chemical Quantities Chapter 9

Chemistry 12.3 “Limiting Reagent and Percent Yield”

Laboratory 08 LIMITING REACTANT LAB.

Limiting Reactants & Excess. Limiting Reactant Calculations  In many chemical reactions an excess of one reactant is added to ensure complete reaction.

Stoichiometry Introduction.

Limiting Reagent Coach Cox.

Percent Yield and Limiting Reactants

Section 3: Limiting Reactants

Limiting Reactants and Excess

Limiting Reactants & Percent Yield

Limiting reagent, Excess reactant, Theoretical or Percent yield

9.3 Notes Limiting reagents.

Limiting Reactant and Percent Yield Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how.

Stoichiometry of Chemical Equations and Formulas.

Limiting Reagents Determine the mass of P 4 O 10 formed if 25g P 4 and 50g O 2 are combined. P 4 + 5O 2  P 4 O 10 What is the given? We have to find the.

Limiting Reagent and Percent Yield

12.3 Limiting Reagent and Percent Yield

Limiting Reagent u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent.

2S (s) + 3O2 (g) 2SO3 (g) O < 1

Limiting Reagents and Percent Yield

Is this a balanced eq ? We only need ONE information during the reaction (reacted or produced)

12.3 Limiting Reagent and Percent Yield

Limiting Reagents. We have considered reactions with just the perfect amount of each reactant.

Stoichiometry Calculating Masses of Reactants and Products.

Percentage Yield.

11-3 Limiting Reactants Reactions to this point have had one ingredient in excess. When you are given two or both reactant amounts, you have a limiting.

Stoichiometry Chapters 7 and 9.

Stoichiometry Chapter 9. Step 1 Balance equations and calculate Formula Mass (FM) for each reactant and product. Example: Tin (II) fluoride, SnF 2, is.

Chapter #9 Stoichiometry. Chapter 9.1 Composition stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves.

Limiting Reactants and Percent Yield Definitions The Limiting Reactant is the reactant that limits the amounts of the other reactants that can combine.

Continuing Stoichiometry…. The idea.  In every chemical reaction, there is one reactant that will be run out (called the limiting reactant).  This will.

Limiting Reactions and Percent Yield Calculating by moles or mass ©2011 University of Illinois Board of Trustees

Theoretical yield vs. Actual yield. Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed,

1.3 Reacting Masses and Volumes Stoichiometry, Limiting & Excess Reactants, Theoretical & Percent Yields Mrs. Page - IB SL Chemistry

Reaction Stoichiometry. Objectives Understand the concept of stoichiometry. Be able to make mass-to-mass stoichiometric calculations.

Stoichiometry A chemical equation shows the reactants (left side) and products (right side) in a chemical reaction. A balanced equation shows, in terms.

Stoichiometry Warmup I have 1 mole of CO 2 gas at STP. How many grams of CO 2 do I have? How many Liters of CO 2 do I have? How many molecules of CO 2.

Limiting reagents In lab a reaction is rarely carried out with exactly the required amounts of each reactant. In lab a reaction is rarely carried out with.

Stoichiometry & Limiting Reactants. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometry. Warm-Up Example of stoich problem Magnesium metal burns in the air Mg (s) + O 2(g) MgO (s) 2 Mg (s) + O 2(g) 2 MgO (s) 3g? g.

8 H : 8 O Imagine that paperclips are atoms and that we’re starting with an equal number of hydrogen atoms and oxygen atoms. For example: Let’s assemble.

GOOD AFTERNOON! Prepare to take notes. You will be allowed to use these notes on the test next week! You will need: 1.Something to write on 2.Something.

Chapter 9 Stoichiometry Test REVIEW SHEET

Ch. 9-3 Limiting Reactants & Percent Yield. POINT > Define limiting reactant POINT > Identify which reactant is limiting in a reaction POINT > Define.

SOL Review 6 Stoichiometry. Consider: 4NH 3 + 5O 2  6H 2 O + 4NO Many conversion factors exist: 4 NH 3 6 H 2 04NO 5O 2 (and others) 5 O 2 4 NO4 NH 3.

Unit 6: Limiting Reactants and Percent Yield Chapter 11.3 and 11.4.

LIMITING REACTIONS INB PAGE 43. ESSENTIAL QUESTION: Why is the limiting reactant not always the reactant with fewer moles?

Stoichiometry. Stoichiometry Stoichiometry – the process of using a balanced chemical equation to calculate the relative amounts of reactants and products.

Stoichiometry From the Greek words “stoicheion” meaning “element”

Chapter 9: Stoichiometry

Chapter 9 STOICHIOMETRY

Quantity Relationships in Chemical Reactions

Warm Up WRITE DOWN three questions that you still have about limiting reagents or stoichiometry. Today is a review of stoichiometry and limiting reagents.

Chapter 12 CHEMICAL STOICHIOMETRY

PROBLEM 4 OF LIMITING REAGENT SET

2 KOH (s) + H2SO4 (g) K2SO4 (aq) + 2 H2O (g)

LIMITING REAGENT EXAMPLES 9/27/11

Limiting Reagents.

How many moles of water are made by

Limiting Reagent and Percent Yield

PROBLEM 4 OF LIMITING REAGENT SET

CHAPTER 12.2 BASIC STOICHIOMETRY

#5 of 11-6 LIMITING REAGENT SET

Other way to identify the Limiting Reactant:

1.50 g. NH mol NH ( ) 1 mol = 17.00g STEP 1-Convert to moles

LIMITING REAGENT (LR) gets completely used up in the reaction

Reaction Stoichiometry

Presentation transcript:

2S (s) + 3O2 (g) 2SO3 (g) O2 3 0.03125 .666 < 1 5)Sulfur and oxygen react in a combination reaction to produce sulfur trioxide, an environmental 5) pollutant: 2S (s) + 3O2 (g) 2SO3 (g) In a particular experiment, the reaction of 1.0 g S with 1.0 g O2 produced 0.80 g of SO3. The % yield in this experiment is __________. AND calculate the moles of unreacted excess reagent. Calculate the moles of the excess reagent that is unconsumed at the end ofreaction. MOLES = MASS/GFM S MOLES = 1.0 g s/ 32.04g/mol MOLES = 0.031191 mol S MOLES = MASS/GFM O2 MOLES = 1.0 g O2 / 32.00g/mol MOLES = 0.03125mol O2 S = 2 = 0.031191 , O2 3 0.03125 .666 < 1 The actual ratio is larger than the theoretical, therefore the actual ratio fraction is too large, the denominator is to small, O2 limiting.

INTRODUCING ICE CHART SOLUTION SKILL THE ICE CHART WILL ALLOW YOU TO CALCULATE MOLES OF ALL SPECIES AFTER THE REACTION IS COMPLETED. MOLES OF ALL PRODUCTS. MOLES OF ALL EXCESS REACTANTS REMAINNING. THE FASTEST WAY TO APPROACH A REACTION IN WHICH YOU ARE CALCULATING MANY SUBSTANCES AT ONCE. 2S + 3O2  2SO3 INITIAL 0.031191 mol 0.03125mol LIMITING 0.0 CHANGE -2X -3X END 0.031191 – 2X 3X

DETERMINING THE VALUE OF “X” X IS DETERMINED BY THE LIMITING REAGENT. IN THIS PROBLEM 0.03125mol = 3X FROM THE LIMITING REAGENT OXYGEN. X = 0.0104111 MOL. THE EXCESS REAGENT IS SULFER, WHICH IS 0.031191 – 2X 0.031191 – 2(0.0104111) 0.01036 MOLE OF S REMAINS 4. THE PRODUCT SO3 = 2X = 2(0.0104111) = 0.0208222 MOL ( 1.667 GRAMS)