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II. Stoichiometry in the Real World * Limiting Reagents

More Stoichiometry!. II. Stoichiometry in the Real World * Limiting Reagents. A. Limiting Reactants. X. Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly. Limiting Reactant bread. Excess Reactants peanut butter and jelly. Jam Sandwich Equation. X.

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II. Stoichiometry in the Real World * Limiting Reagents

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  1. More Stoichiometry! II. Stoichiometry in the Real World * Limiting Reagents

  2. A. Limiting Reactants X • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly • Limiting Reactant • bread • Excess Reactants • peanut butter and jelly

  3. Jam Sandwich Equation X 2 loaves bread + 1 jar of jam  20 sandwiches • Mole Ratio 2 : 1 : 20 • What would happen if we had 1 loaf of bread and 1 jar of jam? • Bread = limiting reactant • Jam = excess reactant (1/2 jar left) • Produce 10 sandwiches

  4. Limiting Reactant • used up in a reaction • determines the amount of product • Excess Reactant • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  5. Example 1 • Iron burns in air to form a solid black oxide (FeO). If you had 50.0 g of iron and 60.0 g of oxygen, what is the limiting and excess reactant? How much iron oxide could be produced? 2 Fe (s) + O2(g) 2 FeO (s)

  6. 2 Fe (s) + O2(g) 2 FeO (s) mass = 50 g mass = 50 g mass = 60 g mass = ?

  7. Example 2 • 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many grams of zinc chloride would be produced? Zn + 2HCl  ZnCl2 + H2 mass = ? 79.1 g 0.90 L 2.5M

  8. Zn + 2 HCl  ZnCl2 + H2 V = 0.90 L of 2.5M mass = ? mass = 79.1 g

  9. Example 2 Results left over zinc Limiting reactant: HCl Excess reactant: Zn Product Formed: x g ZnCl2

  10. Student Example • Sodium carbonate is needed in the manufacturing of glass, but very little occurs naturally. It can be made from the double replacement reaction between calcium carbonate and sodium chloride. If you had 5.00 g of each what is the limiting and excess reactant? How much sodium carbonate would be formed? CaCO3 + 2 NaCl  CaCl2 + Na2CO3

  11. CaCO3 + 2 NaCl  CaCl2 + Na2CO3 mass = 5.0 g mass = 5.0 g mass = ?

  12. Assignment Stoichiometry: Limiting Reagent Worksheet

  13. B. Percent Yield measured in lab calculated on paper

  14. Ivan Buz

  15. When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  16. Example 45.8 g sm # moles = # moles = mm 138.2 g/mol K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: mass = ? mass = 45.8 g actual: 46.3 g # moles = 0.662 mol mass = (# mol) (mm) mass = (0.662 mol) (74.6 g/mol) # moles = 0.331 mol mass KCl = 49.4 g

  17. Example Continued  100 % 46.3 g 49.4 g Actual Yield Theoretical Yield  100 % = Theoretical Yield = 49.4 g KCl % Yield = = 93.7 %

  18. Assignment • Pg 295 # 2, 5, 7, 11, 12, 15, 18, 20, 25 - 30

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