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Chapter 12 Stoichiometry
Composition Stoichiometry – mass relationships of elements in compounds
Reaction Stoichiometry – mass relationships between reactants and products in a chemical reaction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Chapter 9 Visual Concepts
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Stoichiometry
Visual Concept
Mole Ratio – conversion factor (unit factor) that relates one chemical to another – comes from the coefficients in a chemical equation
2 Al2O3 4 Al + 3 O2
2 mol Al2O3 yields 4 mol Al
2 mol Al2O3 yields 3 mol O2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 9
Mole Ratio
• A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction
Example: 2Al2O3(l) 4Al(s) + 3O2(g)
Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al
4 mol Al 3 mol O2 3 mol O2
, ,
Section 1 Introduction to Stoichiometry
In the standard stoichiometry calculations you should know, ALL ROADS LEAD TO MOLS. You can change any amount of any measurement of any material in the same equation with any other material in any measurement in the same equation. That is powerful. The setup is similar to Dimensional Analysis.
1. Start with what you know (GIVEN), expressing it as a fraction. 2. Use definitions or other information to change what you know to mols of that material. 3. Use the mol ratio to exchange mols of the material given to the mols of material you want to find. 4. Change the mols of material you are finding to whatever other measurement you need.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Chapter 9 Visual Concepts
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Molar Mass as a Conversion Factor
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Chapter 9 Visual Concepts
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Conversion of Quantities in Moles
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Chapter 9
Solving Stoichiometry Problems with Moles or
Grams
Section 2 Ideal Stoichiometric Calculations
How many grams of ammonia can you make with 25 grams of hydrogen?
N2 + 3H2 2NH3
You are given the mass of 25 grams of hydrogen. Start there.
141.67 g NH3
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Mass and Number of Moles of an Unknown
What mass of oxygen will react with 96.1 grams of propane?
First, write the balanced reaction:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
Second, convert the mass to moles. Note that the molecular weight of propane is 44.1 g/mole.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Chapter 9
Limiting Reactants• The limiting reactant is the reactant that limits the
amount of the other reactant that can combine and the amount of product that can form in a chemical reaction.
• The excess reactant is the substance that is not used up completely in a reaction.
Section 3 Limiting Reactants and Percentage Yield
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Limiting Reactants and Excess Reactants
Limiting Reactant – reactant that controls the amount of product that can be produced
For example, nitrogen gas is prepared by passing ammonia gas over solid copper(II) oxide at high temperatures. The other products are solid copper and water vapor.
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(g)
If 18.1 g of NH3 are reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N2 will be formed?
22
22
2
22
23
2
3
33
N g 06.1N molN g 28N mol 379.0
N less makesit b/creactant limiting is CuO
N mol 379.0CuO mol 3N mol 1
CuO g 79.5CuO molCuO g 4.90
N mol 532.0NH mol 2N mol 1
NH g 17NH molNH g 1.18
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Chapter 9
Percentage Yield
• The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant.
• The actual yield of a product is the measured amount of that product obtained from a reaction.
• The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.
percentage yield
actual yieldtheorectical yield
100
Section 3 Limiting Reactants and Percentage Yield
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Comparing Actual and Theoretical Yield
Actual Yield% Yield 100Theoretical Yield
What is the % yield of N2 in the previous problem if only 0.95 g were actually obtained from the reaction?
89.62% Yield %
100N g 1.06N g 0.95 Yield %
2
2