3. Grilled Cheese Sandwich Idea….. Bread + Cheese  ‘Cheese Melt’ 2 B + C  B 2 C If you have 100 pieces of bread and 30 slices of cheese, how many sandwiches could you make?

4. Grilled Cheese Sandwich Idea….. 30 sandwiches…… but why? Because you can only make as many sandwiches as you have of the ingredient with the least amount. So you could say the cheese limits the amount of sandwiches that you can make!

5. In chemical equations…… There is often a situation very much like the grilled cheese sandwiches There is often more of one compound then another in a chemical reaction and the amount of products that are produces are limited by one of the reactants

6. Container 1 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269

7. Before and After Reaction 1 All the hydrogen and nitrogen atoms combine. Before the reaction After the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269

8. Container 2 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

9. Before and After Reaction 2 Before the reactionAfter the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

10. Real-World Stoichiometry: Limiting Reactants LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 366 Ideal Stoichiometry Limiting Reactants

11. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl 2 (g)  AlCl 3 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g ? g A. 200 gB. 125 gC. 667 gD. ???

12. Another example: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the exactly amount of ingredients listed above, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

13. Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction stops. This is called the limiting reactant.

14. Steps to Solving limiting reagent problems: 1. Write a balanced equation if its not given in the question. 2. Calculate the number of moles for each reactant (convert from g to mol) 3. Pick one of the products. 4. Using mol to mol conversions, calculate the number of mol of product for each reactant 5. The lower number of mol is the limiting reagent!

15. Example: 1)2 Al + 3 Cl 2  2 AlCl 3 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2) X mol Al = 10.0 gx 1 mol Al 26.98g = 0.371 mol Al X mol Cl 2 = 35.0 gx 1 mol Cl 2 70.90g = 0.494 mol Cl 2

16. 2 Al + 3 Cl 2  2 AlCl 3 0.371 mol Al and 0.494 mol Cl 2 3) Product = AlCl 3 4) X mol AlCl 3 = 0.371 mol Al Example: X 2 mol AlCl 3 2 mol Al = 0.371 mol AlCl 3 X mol AlCl 3 = 0.494 mol Cl 2 X 2 mol AlCl 3 3 mol Cl 2 = 0.329 mol AlCl 3 Pick the lower one, this is the limiting reagent

17. Example: Therefore, Cl 2 is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a stop!

18. Limiting Reactant Practice: 1) HF gas is produced by the double displacement reaction of CaF 2 with H 2 SO 4. Determine the limiting reagent if 10.0 g of CaF 2 reacts with 15.5 g of H 2 SO 4. 2) 2 Al + 3 CuCl 2  3 Cu + 2 AlCl 3. If 0.25 g of Al reacts with 0.51 g of CuCl 2, determine the limiting reactant. 3) 15.0 g of potassium reacts with 15.0 g of iodine. Determine which reactant is limiting and how much product (in grams) is made.

19. Day 2 Finding amount of excess Often limiting reagent questions will ask you to find the excess of the non limiting reagent reactant

20. Finding the Amount of Excess Find the amount actually used, using the limiting reagent to calculate moles then convert to grams Then you can subtract that amount from the starting (given amount) to find the amount of excess.

21. Example: 1)2 Al + 3 Cl 2  2 AlCl 3 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. How much excess reactant is left after the reaction? 2) X mol Al = 10.0 gx 1 mol Al 26.98g = 0.371 mol Al X mol Cl 2 = 35.0 gx 1 mol Cl 2 70.90g = 0.494 mol Cl 2

22. 2 Al + 3 Cl 2  2 AlCl 3 0.371 mol Al and 0.494 mol Cl 2 3) Product = AlCl 3 4) X mol AlCl 3 = 0.371 mol Al Example: X 2 mol AlCl 3 2 mol Al = 0.371 mol AlCl 3 X mol AlCl 3 = 0.494 mol Cl 2 X 2 mol AlCl 3 3 mol Cl 2 = 0.329 mol AlCl 3 Pick the lower one, this is the limiting reagent

23. The question said there was 10.0g of Al to start with: X mol Al = 0.494 mol Cl 2 X 2 mol Al 3 mol Cl 2 = 0.329 mol Al = 8.88 g Al after reaction X 26.98 g 1 mol Al X g Al = 0.329 mol Al 10.0g – 8.88g = 1.22g of excess Al after the reaction