1. Stoichiometry Using the Balanced Equation

2. What does the balanced equation really mean? Particles: Moles: 2H 2 + O 2  2H 2 O 2 molecules H 2 + 1 molecule O 2  2 molecules H 2 O 2 moles H 2 + 1 mole O 2  2 moles H 2 O Which is more useful? moles

3. 2H 2 + O 2  2H 2 O Like a recipe, the balanced equation tells you in what ratios the ingredients must be mixed. Therefore, for every 2 moles of H 2 you will need 1 mole of O 2. Mathematically, this can be written 2 moles H 2 = 1 mole O 2 What can you do with equalities? Dimensional Analysis

4. 2H 2 + O 2  2H 2 O How many conversion factors (called mole ratios) can be written from this 1 equation? 2 moles H 2 = 1 mole O 2 2 moles H 2 = 2 mole H 2 O 1 mole O 2 = 2 moles H 2 O

5. 2H 2 + O 2  2H 2 O How many moles of O 2 are needed to react with 6.4 mol H 2 ? 6.4 mol H 2 = mol O 2 mol H 2 mol O 2 2 1 3.2

6. 2H 2 + O 2  2H 2 O How many moles of H 2 O are produced from 6.4 mol H 2 ? 6.4 mol H 2 = mol H 2 O mol H 2 mol H 2 O 2 2 6.4

7. Multi-step Stoichiometry EVERY stoiciometry problem involves moving from 1 substance to another! 2H 2 + O 2  2H 2 O How many molecules of H 2 does it take to make 647 g of H 2 O? 647 g ? molecules The only way to move from 1 substance to another is the mole ratio!

8. Multi-step Stoichiometry Tools of Stoichiometry 1. Mole Ratiobalanced equation coefficients moles A  moles B 2. Molar Massperiodic table mass = 1 mol mass A (g)  1 mole A 3.Avogadro's # 6.02  10 23 particles = 1 mol particles particles A  1 mole A 4.Volume 22.4 L at STP 1 mole A  22.4 L of A Every stoichiometry problem uses the mole ratio!!! You have three tools for stoichiometry problems!

9. Multi-step Stoichiometry To answer the original question, you need to use all 3 tools. How many molecules of H 2 does it take to make 647 g of H 2 O? 2H 2 + O 2  2H 2 O 647 g ? molecules This problem not only involves changing substances (mole ratio), but also mass (molar mass) and molecules (Avogadro’s number).

10. First, you want to set up a dimensional analysis problem. Put your given and target in the equation. 2H 2 + O 2  2H 2 O 647 g ? molecules 647 g H 2 O = ? molecules H 2

11. Next, convert g of H 2 O to moles of H 2 O using molar mass. 2H 2 + O 2  2H 2 O 647 g ? molecules 647 g H 2 O = ? molecules H 2 g H 2 O mol H 2 O 18.0153 1

12. You don’t want H 2 O, so put moles H 2 O in the bottom of the middle step (remember you have to be in moles to move from 1 substance to another). You want to change to H 2, so put moles of H 2 on top (mole ratio is moles to moles). The numbers come from the balanced equation! Next, EVERY problem must have a mole ratio step, so put that next! mol H 2 O mol H 2 2H 2 + O 2  2H 2 O 647 g ? molecules 2 2647 g H 2 O = ? molecules H 2 18.0153gH2OgH2O mol H 2 O1

13. Since, we need molecules and the middle step gets us to moles H 2, we need to use Avogadro’s number. 2H 2 + O 2  2H 2 O 647 g ? molecules 647 g H 2 O = ? molecules H 2 mol H 2 O mol H 2 2 2 gH2OgH2O mol H 2 O 18.0153 1 mol H 2 molecule H 2 1 6.02  10 23 Answer = 2.16  10 25 molecules H 2

14. Calculating Stoichiometric Problems 1.Balance the equation 2.Convert given to moles 1.Set up mole ratios. 2.Use mole ratios to calculate moles of desired chemical. 3.Convert moles to target.

15. Another Example How many grams of O 2 are needed to use up 13.6 moles of H 2 ? 13.6 mol H 2 = g O 2 mol H 2 mol O 2 2 1 g O 2 mol O 2 31.9988 1 2H 2 + O 2  2H 2 O 13.6 molg ? 218

16. 12.3 Limiting Reactants Cake Recipe 2 cups flour 1 cup butter 1tsp. salt 1/2 cup sugar If I have 6 cups of flour, 4 cups of butter, 112 tsp. of salt, and 429 cups of sugar, how many cakes can I make?

17. 12.3 Limiting Reactants Cake RecipeYou Have in the Cupboard 2 cups flour6 cups flour 1 cup butter4 cup butter 1tsp. Salt112 tsp. salt 1/2 cup sugar429 cups sugar 6 cups flour = cakes cups flour cakes 2 1 3 4 cups butter = cakes cups butter cakes 1 1 4

18. 12.3 Limiting Reactants Cake RecipeYou Have in the Cupboard 2 cups flour6 cups flour 1 cup butter4 cup butter 1tsp. Salt112 tsp. salt 1/2 cup sugar429 cups sugar 112 tsp. salt = cakes tsp. salt cakes 1 1 112 429 cups sugar = cakes cups sugar cakes.5 1 858

19. 12.3 Limiting Reactants Flour controls or limits how many cakes we can make because after we make 3 cakes we run out of flour. It also controls how much of the other ingredients we need!

20. 12.3 Limiting Reactants How much butter is left over? First find how much was used and then subtract how much is left. FLOUR controls how much we use. 6 cups flour = cups butter cups flour cup butter 2 1 3 4 cups butter – 3 cups butter = 1 cup butter left

21. 12.3 Limiting Reactants Limiting reactant - the reactant that makes the least. It controls everything about the reaction because when it is used up the reaction stops. Chemical reactions work the same way!!!!

22. 12.3 Limiting Reactants 2H 2 + O 2  2H 2 O 2.50 g 3.40 g How much water can be made? You must find which reactant controls (limiting reactant) the reaction. 2.50 g H 2 = g H 2 O mol H 2 mol H 2 O 2 2 g H 2 mol H 2 2.01588 1 mol H 2 O g H 2 O 1 18.0153 22.3 3.40 g O 2 = g H 2 O mol O 2 mol H 2 O 1 2 g O 2 mol O 2 31.9988 1 mol H 2 O g H 2 O 1 18.0153 3.83 O 2 is therefore the limiting reactant and 3.83 g of water is produced!

23. 12.3 Limiting Reactants 2H 2 + O 2  2H 2 O 2.50 g 3.40 g How much H 2 is left over? 1.Find out how much H 2 was used up by the 3.40 g O 2 2.Then subtract that amount from 2.50 g to find out how much H 2 is left

24. 12.3 Limiting Reactants 2H 2 + O 2  2H 2 O 2.50 g 3.40 g How much H 2 is left over? 3.40 g O 2 = g H 2 mol O 2 mol H 2 1 2 g O 2 mol O 2 31.9988 1 mol H 2 g H 2 1 2.01588.428 2.50 g.428 g - 2.07 g H 2 left over

25. 12.4 Percent Yield When we calculated that 3.83 g of H 2 O could be made in the previous sample problem, we didn’t actually perform the reaction and measure the resulting water. We made theoretical prediction of how much could be made. Actual Yield – do the reaction to find the yield Theoretical Yield – do the math to predict the yield

26. 12.4 Percent Yield When doing stoichiometric calculations, you are calculating what is known as a theoretical yield - the maximum amount that could be produced during the reaction. Under ideal conditions, this is exactly what will happen. In reality, however, the actual yield will be less than what is predicted.

27. 12.4 Percent Yield