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Chapter 12 Stoichiometry 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.
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Take a worksheet and get started on it!
Chemical Equations Do Now: Take a worksheet and get started on it! Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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Using Equations Stoichiometry Calculation of quantities of reactants and products in chemical reactions Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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__N2(g) + __H2(g) __NH3(g)
Using Equations Balance the equation for producing ammonia. __N2(g) + __H2(g) __NH3(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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Writing and Using Mole Ratios
N2(g) + 3H2(g) 2NH3(g) The three mole ratios derived from the balanced equation above are: 2 mol NH3 1 mol N2 3 mol H2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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Calculating Moles of a Product
Sample Problem 12.3 Calculating Moles of a Product N2 + 3H2 2NH3 How many moles of NH3 are produced when 0.60 mol of nitrogen reacts with hydrogen? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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2 mol NH3 0.60 mol N2 = 1.2 mol NH3 1 mol N2 Sample Problem 12.3
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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Combustion of Gasoline
Chemical Equations Combustion of Gasoline 2C8H O2 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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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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? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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END OF 12.1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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