2. Stoichiometry: the mass relationships between reactants and products. We will use the molar masses ( amount of grams in one mole of a element or compound) from the periodic table And the coefficients from the balanced equations to solve problems

3. When the following reaction occurs: Ba + Cl2  BaCl2 One atom of barium reacts with 1 molecule of chlorine to make one molecule of barium chloride. It does not mean that 1 gram of barium reacts with 1 gram of chlorine to make 1 gram of barium chloride! Remember- barium and chlorine are different sizes and have different masses! We use stoichiometry to relate the atoms to masses- something we can easily measure.

4. Not only signify the ratio of atoms that react- also the ratio of moles of that substance involved. Mole Ratio: a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. For example: 2Al 2 O 3  4Al + 3O 2 The mole ratio of Al 2 O 3 to Al is 2:4, the mole ratio of Al 2 O 3 to O 2 is 2:3

5. You can use the mole ratio to convert from moles of one substance to moles of another. Use the following reaction to answer the question: 4NH 3 + 6NO  5N 2 + 6H 2 O 1. If I have 20 moles of NH 3, how many moles of N 2 can I make? 2o mols NH 3 5 mols N 2 = 25 mols N 2 4 mols NH 3

6. Moles given Coefficient wanted Coefficient given Moles wanted

7. If I want to make 2 moles of NaCl, How many mols of NaI do I need? 2 mols If I have 10 moles of Cl 2, how many moles of I 2 can I make? 10 mols If I want 15 mols of I 2, how many mols of NaI do I need? 30 mols

8. You can use molar mass of the substance to convert from moles to grams. Remember, the molar mass on the periodic table is the number of grams in 1 mole. How many grams in 1 mole of Na? 23 g How many grams in 1 mole of NaCl? 58.5 g

9. 1 mol given Molar mass of given Grams of given Coefficient of given Coefficient of wanted Moles of wanted

10. If I have 100 g of NaI, how many moles of I 2 can I make? (molar mass of NaI is 150 g/mol) 100 g NaI 150 g NaI 1 mol NaI 2 mol NaI 1 mol I 2 0.033 mol I 2 If I want to make 150 g of NaCl, how many moles of NaI do I need? 2.6 mols

11. Moles of given 1 mol wanted Molar mass of wanted Coefficient of wanted grams of wanted Coefficient of given

12. If I have 10 moles of NaI, how many grams of I 2 can I make? 10 moles NaI 2 moles NaI 1 mol I 2 254 g I 2 = 1270 g I 2 If I need 15 moles of NaCl, how many grams of Cl 2 do I need? 532 grams

13. grams of given Molar mass of given 1 Mol given Coefficient of given Coefficient of wanted Molar mass of wanted 1 mole of wanted = Grams of wanted

14. If I have 500 g of NaI, how many grams of I 2 can I make? 2 mol NaI1 mol I 2 254 g I 2 = 423.3 g I 2 1 mol I 2 1 mol NaI 150g NaI 500 g NaI If I want 150 g of NaCl, how many grams of I 2 will I make? 325.6 g I 2

15. Remember 1 mol = 22.4 L of a gas.

16. Are you LOST???

18. 1. When you do a chemical experiment one of the reactants is going to be used up first. 2. The reactant that is used up first in a chemical experiment is called the limiting reactant 3. You can’t tell limiting reactants until you look at mole ratios

19. What is the limiting reactant in the picture?

20. 4. How to find the limiting reactant A. Step 1: Balance equation B. Step 2: Convert both reactants to moles C. Step 3: Start with one reactant and find how many grams of a product you will get (choose any product) D. Step 4: Do the same with the other reactant E. Step 5: whichever gives less product is the limiting reactant

21. 1. Use the following equation to answer the next question ____ MnO 2 + ____HCl  ____MnCl 2 + ____H 2 O + ____ Cl 2 You have 5.47 grams of MnO 2 and 9.861 grams of HCl. What is the limiting reactant?

22. A. Step 1: balance the chemical equation ____ MnO 2 + ____HCl  ____MnCl 2 + ____H 2 O + ____ Cl 2 1 41 22

23. B. Step 2: using the reactants, find the grams of a product stoichiometry. Essentially, you will be doing two stoichiometry problems. In this example I chose MnCl 2 as the product to find grams of 5.47 g of MnO 2 1 mol of MnO 2 87 g of MnO 2 1 mol of MnO 2 1 mol of MnCl 2 125.84 g of MnCl 2 7.91 g of MnCl 2 9.861 g of HCl1 mol of HCl 36.45 g of HCl4 mol of HCl 1 mol of MnCl 2 125.84 g of MnCl 2 8.511 g of MnCl 2

24. C. Step 3: Compare the two numbers you found for your products. The lower of the two numbers is the limiting reactant. In other words, it is the reactant that will run out first. MnO 2 is the limiting reactant

25. 1. Once the limiting reactant is found, we can use that value as the theoretical yield What this tells us is that we should get this amount of product after we do the experiment. 2. The amount of product that actually forms during an experiment is called actual yield It’s impossible to get 100% yields from our in class experiments.

26. 1. To find out how close to the maximum yield you have you calculate a percent yield 2. Percent error tells you how far off you are from producing the maximum yield. 3. Percent yield and percent error should add up to 100%

27. Percent yield equation Percent error equation

28. Use the equation for the following problems ____Al + ____O 2  ____Al 2 O 3 1. What is the theoretical yield of Al 2 O 3 if you start with 5.782 grams of Al?

29. A. Step 1: balance the chemical equation ____Al + ____O 2  ____Al 2 O 3 423

30. B. Convert 5.782 grams of Al to grams of Al 2 O 3 using stoichiometry. Doing this will give you the theoretical yield. In actuality, you have been finding theoretical yields the whole unit. 5.782 g of Al1 mol of Al 26.98 g of Al4 mol of Al 2 mol of Al 2 O 3 1 mol of Al 101.96 g of Al 2 O 3 10.93 g of Al 2 O 3

31. 2. If the actual yield is 8.64 grams of Al 2 O 3, what is the percent yield? = 79.9%

32. 3. What is the percent error? = 21%