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II. Stoichiometry in the Real World

Stoichiometry. II. Stoichiometry in the Real World. A. Limiting Reactants. 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. A. Limiting Reactants. Limiting Reactant used up in a reaction

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II. Stoichiometry in the Real World

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  1. Stoichiometry II. Stoichiometry in the Real World

  2. A. Limiting Reactants • 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. A. Limiting Reactants • 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 • The cheaper and more common chemical is used in excess

  4. A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. (this means you will set up 2 stoich problems) 3. Smaller answer indicates: • limiting reactant • amount of product that can be made

  5. A. Limiting Reactants • 79.1 g of zinc react with 2.25 mol of HCl. • Identify the limiting and excess reactants. • How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 2.25 mol

  6. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 2.25 mol 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 1 mol Zn 22.4 L H2 1 mol H2 = 27.1 L H2

  7. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 2.25 mol 2.25 mol HCl 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2

  8. A. Limiting Reactants left over zinc Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2

  9. Finding the Amount of Excess Reactant Left Over • calculate the amount of the excess reactant needed to react with the limiting reactant • subtract that amount from the given amount to find the amount left over. • Can we find the amount of excess potassium in the next problem?

  10. Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? • 4Na (s)+ O2(g)2 Na2O(s) • Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): • 5.00g Na (1 mole Na) ( 2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O 23 g Na 4 mole Na 1 mol Na2O • 5.00g O2 (1 mole O2) ( 2 mole Na2O)( 62 g Na2O) = 19.38 g of Na2O 32 g O2 1 mole O2 1 mol Na2O • Notice you can not have two different masses produced for the same product! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide.

  11. Finding the Excess Left Over • ) How much excess reactant (O2) was leftover ? • 4Na (s)+ O2(g)2 Na2O(s) • To answer that question we first need to determine how much O2 (g) was used. • Set up a stoichiometry problem starting with the limiting reactant and solving for the excess reactant • The amount of O2 used is calculated by: • 5.00g Na (1 mole Na)( 1 mole O2)( 32 g O2) = 1.74 g of O2 was used 23 g Na 4 mole Na 1 molO2 • The amount of oxygen gas left over can be calculated by subtracting the used mass of oxygen from the starting mass of oxygen. • 5.00 g O2 – 1.74 g O2 = 3.26 g of oxygen left over

  12. Finding Excess Practice • 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI • Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I2 1 mol I2 2 mol K 39.1 g K 254 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it! Given amount of excess reactant

  13. B. Percent Yield measured in lab • calculated on paper • Determined by limiting reactant

  14. B. Percent 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. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  15. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

  16. B. Percent Yield 46.3 g 49.4 g K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.4 g KCl 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield =

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