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Chapter 5 Chemical Reactions and Quantities

Chapter 5 Chemical Reactions and Quantities. 5.7 Mole Relationships in Chemical Equations. Law of Conservation of Mass. The Law of Conservation of Mass indicates that in an ordinary chemical reaction, Matter cannot be created nor destroyed.

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Chapter 5 Chemical Reactions and Quantities

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  1. Chapter 5 Chemical Reactions and Quantities 5.7 Mole Relationships in Chemical Equations

  2. Law of Conservation of Mass The Law of Conservation of Massindicates that in an ordinary chemical reaction, • Matter cannot be created nor destroyed. • No change in total mass occurs in a reaction. • Mass of products is equal to mass of reactants.

  3. Conservation of Mass 2 moles Ag + 1 moles S = 1 mole Ag2S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product

  4. Reading Equations In Moles Consider the following equation: 4 Fe(s) + 3 O2(g) 2Fe2O3(s) This equation can be read in “moles” by placing the word “moles” between each coefficient and formula. 4 moles Fe + 3 moles O22 moles Fe2O3

  5. Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for any two substances in an equation. 4Fe(s) + 3O2(g) 2Fe2O3(s) Fe and O24 moles Fe and 3 moles O2 3 moles O2 4 moles Fe Fe and Fe2O34 moles Fe and 2 moles Fe2O3 2 moles Fe2O3 4 moles Fe O2 and Fe2O33 moles O2 and 2 moles Fe2O3 2 moles Fe2O3 3 moles O2

  6. Learning Check Consider the following equation: 3 H2(g) + N2(g) 2 NH3(g) A. A mole-mole factor for H2 and N2 is 1) 3 moles N22) 1 mole N2 3) 1 mole N2 1 mole H2 3 moles H2 2 moles H2 B. A mole-mole factor for NH3 and H2 is 1) 1 mole H22) 2 moles NH3 3) 3 moles N2 2 moles NH3 3 moles H2 2 moles NH3

  7. Solution 3H2(g) + N2(g) 2NH3(g) A. A mole-mole factor for H2 and N2 is 2) 1 mole N2 3 moles H2 B. A mole-mole factor for NH3 and H2 is 2) 2 moles NH3 3 moles H2

  8. Calculations with Mole Factors How many moles of Fe2O3 can form from 6.0 mole O2? 4Fe(s) + 3O2(g) 2Fe2O3(s) Relationship: 3 mole O2 = 2 mole Fe2O3 Write a mole-mole factor to determine the moles of Fe2O3. 6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3 3 mole O2

  9. Guide to Using Mole Factors

  10. Learning Check How many moles of Fe are needed for the reaction of 12.0 moles O2? 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe

  11. Solution 3) 16.0 moles Fe 12.0 moles O2 x 4 moles Fe = 16.0 moles Fe 3 moles O2

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