1. Limiting Reactants and Excess What is the Limiting Reagent (Reactant)? It is the substance in a chemical reaction that runs out first. The limiting reactant (reagent) determines how much product you can make. If you are given amounts of more than one reactant, determine how much product you can make with each of them. Whichever produces the LEAST amount of product is your limiting reagent.

2. Practice Problems Consider the reaction: 2 Al + 3 I 2  2 AlI 3 Determine the limiting reagent of the product, aluminum iodide, if one starts with: a). 1.20 moles of Al and 2.40 moles of iodine A. Al B. I 2

3. 1.20 moles of Al 2 moles AlI 3 1.20 moles AlI 3 2 moles Al 2.40 moles I 2 2 moles AlI 3 1.60 moles AlI 3 3 moles I 2 Al is limiting reactant

4. What is the limiting Reactant with 1.20 grams of Al and 2.40 grams of iodine? A 1.20g Al B 2.40 g I 2 1.20 g Al 1mol Al 2 mol AlI 3.0445 mols AlI 3 26.9815 g 2 mol Al 2.40 g I 2 1 mol I 2 2 mol AlI 3.00630 mols AlI 3 253.8089 g 3 mole I 2 I 2 is limiting Reactant! B

5. How many grams of excess reactant will remain? 2.40g I 2 1 mol I 2 2 mols Al 26.9815g 253.8089g 3 mole I 2 1 mol Al.0.170 g of Al reacted Started with 1.20g -0.170g reacted 1.03 g in excess

6. 15.00 g aluminum sulfide and 12.00 g of water react until the limiting reagent is used up. Here is the balanced equation for the reaction. Al 2 S 3 + 6 H 2 O  2 Al(OH) 3 + 3 H 2 S a.) What is the maximum mass of H 2 S which can be formed from these reagents ? A.10.21g B.11.35 g 15.00g Al 2 S 3 1 mol Al 2 S 3 3 mols H 2 S 34.08 g 150.16167g 1 mol Al 2 S 3 1 mol H 2 S 10.21g H2S 12.00g H2O 1 mol H2O 3 mols H 2 S 34.08 g 18.008g 6 mol H 2 O 1 mol H 2 S 11.35g H2S A. 10.21g

7. What is the limiting Reactant? A. H 2 O B. Al 2 S 3 B! How much of the excess reactant is used up? A, 1.200g B. 12.00 g C. 10.84 g 15.00g Al 2 S 3 1 mol Al 2 S 3 6 mols H 2 O 18.08 g 150.16167g 1 mol Al 2 S 3 1 mol H 2 0 10.84 g used up

8. Percent Yield Percent Yield = (Actual Yield / Theoretical Yield) * 100 Theoretical Yield is the amount that your stoichiometric calculations have predicted. Actual Yield is what you actually produced in an experiment.

9. Example: In the reaction between nitrogen gas and hydrogen gas producing ammonia: a.) If you start with 14 grams of N 2 and 6.0 grams of H 2, what will be the limiting reagent and the theoretical yield? b.) If 10. grams of the product was formed, what would be the percent yield?

10. Stoichiometry with Solutions Just like other problems but use Moles = Molarity X Liters to get moles, or Liters = Moles/Molarity to find the volume of a solution needed in a reaction.

11. How many grams of aluminum will be required to completely replace copper from a 208 mL of a 0.100M solution of copper(II) chloride? 2 Al + 3 CuCl 2  2 AlCl 3 + 3 Cu Answer = 0.374 g Al

12. How many milliliters of a 3.40M CuSO 4 solution will be needed to react with 510.0 g of AgCl in the following reaction? AgCl + CuSO 4  Ag 2 SO 4 + CuCl 2 Answer = 523 mL