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Chapter 9 Chemical Quantities in Reactions

Chapter 9 Chemical Quantities in Reactions. 9.1 Mole Relationships in Chemical Equations. 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 9 Chemical Quantities in Reactions

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  1. Chapter 9 Chemical Quantities in Reactions 9.1 Mole Relationships in Chemical Equations

  2. Conservation of Mass The law of conservation of massindicates 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. Antoine Lavoisier • Lavoisier (1743-1794) • He was so called the "Father of modern chemistry", because he thought “the Law of Conservation of Mass.” • He also helped construct the metric system. • In 1771, at 28-yr-old, he married 13-yr-old Marie-Anne Pierrettewho assisted in his lab. • This scientist sadly lost his head during the French Revolution.

  4. Conservation of Mass Reactants Products 2 mol of Ag + 1 mol of S = 1 mol of Ag2S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 = 247.9 g

  5. Reading an Equation in Moles Consider the following equation: 2Fe(s) + 3S(s) Fe2S3(s) An equation can be read in moles by placing the words “mol of” between each coefficient and formula. 2 mol of Fe + 3 mol of S 1 mol of Fe2S3

  6. Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for two substances in an equation. 2Fe(s) + 3S(s) Fe2S3(s) Fe and S2 mol Fe and 3 mol S 3 mol S 2 mol Fe Fe and Fe2S3 2 mol Fe and 1 mol Fe2S3 1 mol Fe2S3 2 mol Fe S and Fe2S3 3 mol S and 1 mol Fe2S3 1 mol Fe2S3 3 mol S

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

  8. Solution 3H2(g) + N2(g) 2NH3(g) A. A mole factor for H2 and N2 is 2) 1 mol N2 3 mol H2 B. A mole factor for NH3 and H2 is 2) 2 mol NH3 3 mol H2

  9. Using Mole-Mole Factors

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

  11. Calculations with Mole-Mole Factors STEP 1 Write the given and needed number of moles. Given 12.0 mol of O2Need moles of Fe STEP 2 Write a plan to convert the given to the needed moles. mol of O2 mol of Fe

  12. Calculations with Mole-Mole Factors (continued) STEP 3 Use coefficients to write relationship and mole-mole factors. 4 mol of Fe = 3 mol of O2 4 mol Fe and 3 mol O2 3 mol O2 4 mol Fe STEP 4Set up problem using the molemole factor that cancels given moles. 12.0 mol O2 x 4 mol Fe = 16.0 mol Fe (C) 3 mol O2

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