1. TOPIC 17: INTRO TO STOICHIOMETRY EQ: EQ: How does a balanced chemical equation help you predict the number of moles and masses of reactants and products?

2. MOLE RATIOS 2H 2 + O 2 → 2H 2 O The mole ratio is the relationship of the number of moles of the chemicals that participate in a chemical equation. For ex: For every 2 moles of hydrogen we can get 2 moles of water. For every one mole of oxygen we can get 2 moles of water. For every two moles of hydrogen we need one mole of oxygen.

3. MOLE RATIOS 2H 2 + O 2 → 2H 2 O We can also say: 2 moles of H 2 = 2 moles of H 2 O 1 mole of O 2 = 2 moles of H 2 O 2 moles of H 2 = 1 mole of O 2 What can we do with things equal to each other? CONVERSION FACTORS!!!

4. STEPS TO CALCULATE STOICHIOMETRIC PROBLEMS 1. Correctly balance the equation. 2. Start with given number and units. 3. Convert to moles. 4. Use mole ratios to SWITCH to moles of desired chemical. 5. Convert moles back into final unit.

5. MOLE TO MOLE CONVERSIONS 2Al 2 O 3  Al + 3O 2 2Al 2 O 3  Al + 3O 2 - each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 - 2 moles of Al 2 O 3 = 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are the two possible conversion factors to use in the solution of the problem.

6. MOLE TO MOLE CONVERSIONS How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 2 O 3  Al + 3O 2 2Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 5.01 mol O 2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

7. PRACTICE: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O 1)If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.60 mol) 2)How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) 3)If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol)

8. HOW DO YOU GET GOOD AT THIS?

9. YOUR TURN!! 1)How many moles of copper are needed to produce 13.0 moles of copper (I) oxide? 2)How many moles of copper (I) oxide would be produced if only 0.25 moles of oxygen were available? 3)If you want to produce 3.25 moles of copper (I) oxide, how many moles of copper would you need to start the reaction? 4)If you would have 5.5 moles of oxygen to start the reaction, how many moles of copper would you need? 5)How many moles of oxygen are needed to produce 2.5 moles of copper (I) oxide? 4Cu(s) + O 2 (g)  2Cu 2 O(s)

10. MASS-MASS PROBLEM: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 (6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

11. ANOTHER EXAMPLE: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

12. HOW DO YOU GET GOOD AT THIS?

13. YOUR TURN!! 1)You produced 11.7 grams of copper (I) oxide. How many grams of oxygen did you need? 2)If you want to produce 10.5 grams of copper (I) oxide, how many grams of copper do you need at the beginning of the reaction? 3)When you have 15.7 grams of oxygen available, how many grams of copper (I) oxide can you produce? 4)You only have 6.50 grams of copper available. How many grams of oxygen do you need to produce the copper (I) oxide? 5)If there is only 12.5 grams of copper available, what mass (in grams) of copper (I) oxide can be produced? 4Cu(s) + 1O 2 (g)  2Cu 2 O(s)

14. VOLUME-VOLUME CALCULATIONS: How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O 2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4 Notice anything relating these two steps?

15. AVOGADRO TOLD US: Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Moles are numbers of particles Moles are numbers of particles You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. 1 mole = 22.4 L @ STP

16. SHORTCUT FOR VOLUME-VOLUME? How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O2O2 2 L O 2 1 L CH 4 = 8.75 L CH 4 Note: This only works for Volume-Volume problems.

17. SECTION 12.3 LIMITING REAGENT & PERCENT YIELD OBJECTIVES: OBJECTIVES: Identify the limiting reagent in a reaction. Identify the limiting reagent in a reaction.

18. TOPIC #18: LIMITING REAGENT EQ: EQ: Given the amount of reactants in a chemical reaction, how are the amounts of products calculated (in moles and grams)?

19. “LIMITING” REAGENT If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? The limiting reagent is the reactant you run out of first. The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make The limiting reagent determines how much product you can make

20. HOW DO YOU FIND OUT WHICH IS LIMITED? The chemical that makes the least amount of product is the “limiting reagent”. The chemical that makes the least amount of product is the “limiting reagent”. You can recognize limiting reagent problems because they will give you 2 amounts of chemical You can recognize limiting reagent problems because they will give you 2 amounts of chemical Do two stoichiometry problems, one for each reagent you are given. Do two stoichiometry problems, one for each reagent you are given.

21. If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? 2Cu + S  Cu 2 S 2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is the Limiting Reagent, since it produced less product.

22. ANOTHER EXAMPLE: If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 the CuSO 4 is limited, so Cu = 20.6 g the CuSO 4 is limited, so Cu = 20.6 g How much excess reagent will remain? How much excess reagent will remain? Excess = 4.47 grams

23. YOUR TURN!! 3 NH 4 NO 3 + Na 3 PO 4  (NH 4 ) 3 PO 4 + 3 NaNO 3 If 45.0 grams of ammonium nitrate react with 25.0 grams of sodium phosphate, how many grams of sodium nitrate can be formed? 1.Which substance is the limiting reagent? 2.How much sodium nitrate can be formed? 3.How much ammonium phosphate can be formed?

24. The Concept of: A little different type of yield than you had in Driver’s Education class.

25. WHAT IS YIELD? Yield is the amount of product made in a chemical reaction. Yield is the amount of product made in a chemical reaction. There are three types: There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield = Actual Theoretical x 100

26. EXAMPLE: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu What is the actual yield? What is the actual yield? What is the theoretical yield? What is the theoretical yield? What is the percent yield? What is the percent yield? = 6.78 g Cu = 13.8 g Cu = 49.1 %

27. DETAILS ON YIELD Percent yield tells us how “efficient” a reaction is. Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! Theoretical yield will always be larger than actual yield! Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring