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Chapter 6 Chemical Reactions and Quantities

Chapter 6 Chemical Reactions and Quantities. 6.6 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 or destroyed no change in total mass occurs in a reaction

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Chapter 6 Chemical Reactions and Quantities

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  1. Chapter 6 Chemical Reactions and Quantities 6.6Mole Relationships inChemical Equations

  2. Law of Conservation of Mass The Law of Conservation of Mass indicates that in an ordinary chemical reaction, • matter cannot be created or destroyed • no change in total mass occurs in a reaction • mass of products is equal to mass of reactants

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

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

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

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

  7. Solution Consider the following equation: 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. Guide to Using Mole–Mole Factors

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

  10. Solution 3) 16.0 moles of Fe STEP 1 Given: 12 moles of O2 4Fe(s) + 3O2(g) 2Fe2O3(s) Need: moles of Fe STEP 2 Plan: moles of O2 moles of Fe STEP 3 Write mole–mole factors from coefficients: 4 moles of Fe = 3 moles of O2 4 moles Fe and 3 moles O2 3 moles O2 4 moles Fe

  11. Solution (continued) STEP 4Set up problem to cancel moles of O2: 12.0 moles O2 x 4 moles Fe = 16.0 moles of Fe 3 moles O2

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