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Stoichiometry Calculations

Unit 7 - Stoichiometry. Stoichiometry Calculations. Proportional Relationships. Ratio of eggs to cookies. I have 5 eggs. How many cookies can I make?. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar. 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs

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Stoichiometry Calculations

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  1. Unit 7 - Stoichiometry Stoichiometry Calculations

  2. Proportional Relationships Ratio of eggs to cookies • I have 5 eggs. How many cookies can I make? 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 5 eggs 5 dozen 2 eggs = 12.5 dozen cookies

  3. Proportional Relationships • Stoichiometry • mass relationships between substances in a chemical reaction • based on conversion factors and a chemical equation

  4. Proportional Relationships 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O (g) • In this equation there are 2 molecules of benzene (C6H6) reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water. • This equation could also be read as 2 moles of benzene (C6H6) reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water.

  5. Proportional Relationships Since the relationship between the number of molecules and the number of moles present is 6.02 x 1023, a common factor between all species involved in the equation, we can establish the MOLE RATIO between the species involved in a chemical reaction.

  6. Proportional Relationships • Mole Ratio • indicated by coefficients in a balanced equation

  7. Proportional Relationships 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O (g) The MOLE RATIO for oxygen and carbon dioxide is 15 : 12. It can be written as: 12 moles CO2or as 15 moles of O2 15 moles O2 12 moles of CO2 • NOTE:The MOLE RATIO is used for converting moles of one substance into moles of another substance. Without the balanced equation there is no other relationship between two different compounds.

  8. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio moles  moles • Molar mass moles  grams • Avogadro’s # moles  particles • Molar volume moles  liters gas • Mole ratio moles  moles Core step in all stoichiometry problems!! 4. Calculate and check answer.

  9. Mole Road Map - REVIEW MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF GAS AT STP Molar Volume (22.4 L/mol) molar mass 6.02  1023 (particles/mol) (g/mol)

  10. Stoichiometry Island Diagram Known Unknown Substance A Substance B Mass Mass 1 mole = molar mass (g) 1 mole = molar mass (g) Use coefficients from balanced chemical equation Volume Mole Mole Volume 1 mole = 22.4 L @ STP 1 mole = 22.4 L @ STP (gases) (gases) MOLE RATIO! 1 mole = 6.02 x 1023 particles (atoms, molecules, or formula units) 1 mole = 6.02 x 1023 particles (atoms, molecules, or formula units) Particles Particles

  11. Mole – Mole Practice Problem (1-Step) How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3 2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3 2 mol KClO3 6 mol x mol KClO3 = 9 mol O2 = 6 mol KClO3 3 mol O2 O2 KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  12. Mole – Mass Practice Problem (2-Steps) • How many grams of nitrogen must react with 2.3 moles of hydrogen? N2(g) + H2(g)  NH3(g) 3 2 ? g 2.3 mol 2.3 mol H2 1 mol N2 3 mol H2 28.02 g N2 1 mol N2 = 21 g N2

  13. PRACTICE! • Stoichiometry WS #1 (1-Step & 2-Step Problems)

  14. Mass – Mass Practice Problem (3-Steps) Cu + 2 AgNO3 2 Ag + Cu(NO3)2 How many grams of silver will be formed from 12.0 g copper? 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag 40.7 g 2 mol Ag 107.87 g Ag 1 mol Cu x g Ag = 12.0 g Cu = 40.7 g Ag 1 mol Cu 1 mol Ag 63.55 g Cu Cu Ag Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  15. Mass – Volume Practice Problem (3-Steps) 2KClO3  2KCl + 3O2 How many grams of KClO3 are required to produce 9.00 L of O2 at STP? ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L O2 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3 32.8 g 2 mol KClO3 122.55 g KClO3 1 mol O2 x g KClO3 = 9.00 L O2 = 32.8 g KClO3 3 mol O2 1 mol KClO3 22.4 L O2 O2 KClO3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  16. Stoichiometry Practice 2 KClO3 2 KCl + O2 3 500. g x g x L (196 g) 3 mol O2 22.4 L O2 1 mol KClO3 x L O2 = 500. g KClO3 = 137 L O2 137 L 2 mol KClO3 1 mol O2 122.55 g KClO3 KClO3 O2 32 g O2 1 mol O2 x g O2 = 137 L O2 = 196 g O2 1 mol O2 22.4 L O2 2 mol KCl 74.55 g KCl 1 mol KClO3 x g KCl = 500. g KClO3 = 304 g KCl (304 g) 2 mol KClO3 1 mol KCl 122.55 g KClO3 KClO3 KCl

  17. Baking Soda Lab NaHCO3 + HCl NaCl + H2O + CO2 Na NaHCO3 HCl + HCO3 + H Cl H2CO3 H2CO3 sodium bicarbonate hydrochloric acid sodium chloride baking soda table salt H2O + CO2 (g) (l) (g) D heat actual yield ? g gas gas D excess x g 5 g theoretical yield actual yield % yield = x 100 % theoretical yield

  18. PRACTICE! • Stoichiometry WS #2 (3-Step Problems)

  19. Unit 7 - Stoichiometry II. Stoichiometry in the Real World

  20. Limiting Reactants + plus 16 tires excess 48 tires 8 cars 8 car bodies C + 4 T CT4

  21. The Limiting Reactant 3 W + 2 P + S + H + F W3P2SHF A balanced equation for making a tricycle might be: excess of all other reactants 50 P 25 W3P2SHF excess of all other reactants 50 S 50 W3P2SHF 50 S and excess of all other reactants 50 P 25 W3P2SHF

  22. Limiting Reactants • Limiting Reactant • used up in a reaction • determines the amount of product • Excess Reactant • what is left over after reaction • added to ensure that the other reactant is completely used up

  23. Visualizing Limiting Reactant H2 + O2 H2O 2 2 2.5 1.5 4.5 3.5 1 7 1 5 5 8 0 3 4 0 8 3 2 6 4 2 4 2 3 6 5 7 1 ___ mole H2 ___ mole O2 ___ mole H2O Limiting Reactant Excess Limiting reactant determines amount of product.

  24. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: • limiting reactant AND • amount of product formed

  25. Limiting Reactants • 81.0 g of copper reacts with 25.0 g of sulfur to produce solid copper (I) sulfide. • Identify the limiting and excess reactants. • How many grams of product are formed? 2 Cu(s) + S(s) Cu2S(s) ? g 81.0 g 25.0 g

  26. Limiting Reactants 2 Cu(s) + S(s) Cu2S(s) 81.0 g ? g 25.0 g 81.0 g Cu 1 mol Cu2S 2 mol Cu 159.17 g Cu2S 1 mol Cu2S 1 mol Cu 63.55 g Cu = 101 g Cu2S

  27. Limiting Reactants 2 Cu(s) + S(s) Cu2S(s) 81.0 g ? g 25.0 g 25.0 g S 1 mol Cu2S 1 mol S 1 mol S 32.07 g S 159.17 g Cu2S 1 mol Cu2S = 124 g Cu2S

  28. Limiting Reactants Cu:101 g Cu2S S: 124 g Cu2S Limiting reactant: Cu Excess reactant: S Product Formed: 101 g Cu2S

  29. Percent Yield measured in lab calculated on paper

  30. Actual Yield Theoretical Yield % Yield = When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. actual yield 46.3 g K2CO3 + 2 HCl KCl + H2O + CO2 2 + H2CO3 ? g excess 45.8 g theoretical yield Theoretical yield 1 mol K2CO3 2 mol KCl 74.55 g KCl = 49.4 g KCl x g KCl = 45.8 g K2CO3 49.4 g 49.4 g KCl 1 mol K2CO3 1 mol KCl 138.21 g K2CO3 46.3 g KCl = x 100 % Yield % Yield = 93.7% efficient

  31. PRACTICE • S’more Lab • Stoichiometry WS #4 (#1 & 2)

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