1. II. Stoichiometry in the Real World Stoichiometry

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

3. A. Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b 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 b 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  ZnCl 2 + H 2 79.1 g ? L 2.25 mol

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

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

8. A. Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Theoretical Yield: 25 L H 2 left over zinc

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

10. B. Percent Yield b When 45.8 g of K 2 CO 3 was reacted with excess HCl, 46.3 g of KCl was formed. a)Calculate the theoretical yield of KCl b)Calculte the % yield K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g ? g actual: 46.3 g

11. B. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

12. B. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g

13. Finding the Amount of Excess Reactant Left Over 1. calculate the amount of the excess reactant used Set up another stoich problem. Start with the LIMITING REACTANT and solve for the EXCESS REACTANT 2. subtract the amount of excess reactant used from the amount given to find the amount left over. b Can we find the amount of excess reactant in the next problem?

14. 4 Na (s) + O 2(g)  2 Na 2 O (s) b 5.00 g of sodium reacted with 5.00 g of oxygen a. How many grams of product can form? b. What is the limiting reactant? c. How much excess reactant is left over? (hint: first find the amount of excess reactant used in the reaction, and subtract from the amount given)

15. a. 6.74 g sodium oxide b. Na c. 3.26 g of oxygen was left over

16. b 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?  4 Na (s) + O 2(g)  2 Na 2 O (s) b Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 6.74 g of Na 2 O b 5.00g Na ( 1 mole Na ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 6.74 g of Na 2 O 23 g Na 4 mole Na 1 mol Na 2 O 19.38 g of Na 2 O b 5.00g O 2 ( 1 mole O 2 ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 19.38 g of Na 2 O 32 g O 2 1 mole O 2 1 mol Na 2 O b 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.

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

18. Finding Excess Practice b 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI b Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant 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!