2. Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

3. Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Be sure you have a balanced reaction before you start! Example: 2 Na + Cl 2  2 NaCl Example: 2 Na + Cl 2  2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles? What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

4. 2H 2 + O 2   2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Two molecules of hydrogen and one molecule of oxygen form two molecules of water. 2 Al 2 O 3  Al + 3O 2 2 Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 Now try this: 2Na + 2H 2 O  2NaOH + H 2

5. Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O 2 H 2 + O 2  2 H 2 O How many moles of reactants are needed? How many moles of reactants are needed? What if we wanted 4 moles of water? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

6. Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2  2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

7. Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

8. Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl 5.00 moles Na 1 mol Cl 2 70.90g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

9. Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

10. Mass-Mole We can also start with mass and convert to moles of product or another reactant We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 10.0 g H 2 O 1 mol H 2 O 2 mol C 2 H 6 18.0 g H 2 O 6 mol H 2 0 = 0.185 mol C 2 H 6

11. Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

12. Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

13. Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N 2 + 3 H 2  2 NH 3 N 2 + 3 H 2  2 NH 3 2.00g N 2 1 mol N 2 2 mol NH 3 17.06g NH 3 28.02g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

14. Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

15. Mole to Mole conversions 2 Al 2 O 3  Al + 3O 2 2 Al 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 each time we use 2 moles of Al 2 O 3 we will also make 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

16. 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? 2 Al 2 O 3  Al + 3O 2 2 Al 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

17. 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 If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed?If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.6 mol) How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol)

18. Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 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 2 x 101.96) ÷ (26.98 x 4) = 12.3 g Al 2 O 3

19. Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much 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

20. 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 concerning these two steps?

21. 50.0 mL 6.0 M ? g 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal = g NaHCO 3 H 2 SO 4 50.0 mL 1 mol H 2 SO 4 NaHCO 3 2 mol NaHCO 3 84.0 g mol NaHCO 3 50.4

22. When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

23. When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

24. Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

25. What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 2HCl(aq) + Ba(OH) 2 (aq)  2H 2 O(l) + BaCl 2 0.40 M 47.1 mL 0.75 M ? mL = mL HCl Ba(OH) 2 47.1 mL 1 mol Ba(OH) 2 HCl 2 mol 0.40 mol HCl HCl 1000 mL 176 Units match Solution Stoichiometry

26. Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L = mol Ba(OH) 2 L Ba(OH) 2 25.00 x 10 -3 L Ba(OH) 2 Units Already Match on Bottom! 0.0629 Units match on top!

27. Limiting Reactant: 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 specified amount of all ingredients listed, 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?

28. Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other 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). 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 s. This is called the limiting reactant. The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

29. Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

30. Limiting Reactant: Example 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? 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 Al + 3 Cl 2  2 AlCl 3 Start with Al: Start with Al: Now Cl 2 : Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

31. LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete. We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

32. Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

33. If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product (copper (I) sulfide) will be formed? If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of 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 Limiting Reagent

34. Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem? Can we find the amount of excess potassium in the previous problem?

35. Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Hide Based on: H 2 O = mol O 2 0.10 mol H2OH2O 0.150 Limiting/Excess/ Reactant and Theoretical Yield Problems :

36. Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Based on: H 2 O = mol O 2 0.10 mol H 2 O 0.150 What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant? It was limited by the amount of KO 2. H 2 O = excess (XS) reactant!

37. Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

38. Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Convert ALL of the reactants to the SAME product (pick any product you choose.) 3. The lowest answer is the correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!