1. Ch. 12 Stoichiometry Objective: To learn how to use a complete chemical equation to calculate quantities of a substance

2. Chemical Equations / A balanced chemical equation provides the same kind of quantitative information that a recipe does. / If we were to make a bicycle:  Frame + Seat + 2 Wheels  Bike  F + S + 2 W  FSW 2 / Mr. Focht can make 345 bicycles in a week. How many wheels should I go out and buy on Sunday to be prepared for the work week / A balanced chemical equation provides the same kind of quantitative information that a recipe does. / If we were to make a bicycle:  Frame + Seat + 2 Wheels  Bike  F + S + 2 W  FSW 2 / Mr. Focht can make 345 bicycles in a week. How many wheels should I go out and buy on Sunday to be prepared for the work week

3. Balanced Equations / Balanced chemical equations tell what amounts of reactants to mix and what amounts of products to expect. The quantity is usually expressed in moles. / Can be expressed in liters, grams, or molecules / Why Would we need to know what amounts of reactants to mix or how much product is going to be produced? / Balanced chemical equations tell what amounts of reactants to mix and what amounts of products to expect. The quantity is usually expressed in moles. / Can be expressed in liters, grams, or molecules / Why Would we need to know what amounts of reactants to mix or how much product is going to be produced?

4. Stoichiometry / Calculations using balanced equations are called Stoichiometric calculations / Calc. chemical quantities within reaction / Form of Bookkeeping / Calculations using balanced equations are called Stoichiometric calculations / Calc. chemical quantities within reaction / Form of Bookkeeping

5. Interpreting Chemical Equations / A balanced chemical equation can be interpreted in terms of different quantities such as number of atoms, molecules or moles; mass; and volume. / Atoms: both number and types of atoms are not changed / Molecules: Indicates amt of molecules reacting/ being produced / Moles: Coefficients of balanced equation indicate relative # of moles of reactant and products / Most important info from equation / Mass: Obeys law of conservation of mass / Mass and atoms are conserved in every chemical reaction, however; molecules, formula units, moles, and volume may not be / A balanced chemical equation can be interpreted in terms of different quantities such as number of atoms, molecules or moles; mass; and volume. / Atoms: both number and types of atoms are not changed / Molecules: Indicates amt of molecules reacting/ being produced / Moles: Coefficients of balanced equation indicate relative # of moles of reactant and products / Most important info from equation / Mass: Obeys law of conservation of mass / Mass and atoms are conserved in every chemical reaction, however; molecules, formula units, moles, and volume may not be

6. N 2 (g) + 3H 2 (g)  2NH 3 (g) QuantityAmount of Reactants Amount of Products Quantity remains the same or is different Number of Atoms Number of Molecules moles Mass(g)

7. On your own: / Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:   2H 2 S (g) + 3O 2 (g)  2SO 2 (g) + 2H 2 O (g) / Determine the number of representative particles and moles for the reactants and products. / Determine the masses of the reactants and products. / Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is:   2H 2 S (g) + 3O 2 (g)  2SO 2 (g) + 2H 2 O (g) / Determine the number of representative particles and moles for the reactants and products. / Determine the masses of the reactants and products.

8. 8 Writing and using mole ratio  Writing and using mole ratios:  N 2 + 3H 2 --> 2NH 3  “1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol ammonia.”  Mole ratio: tells the ratio between 2 substances in a balanced chemical equation.  Writing and using mole ratios:  N 2 + 3H 2 --> 2NH 3  “1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol ammonia.”  Mole ratio: tells the ratio between 2 substances in a balanced chemical equation.

9. 9 Using mole ratio  How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen?  N 2 + 3H 2 --> 2NH 3  Step 1: identify your known and unknown:  Known: 0.60 mol N 2  Unknown: ? mol NH 3  How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen?  N 2 + 3H 2 --> 2NH 3  Step 1: identify your known and unknown:  Known: 0.60 mol N 2  Unknown: ? mol NH 3

10. 10  N 2 + 3H 2 --> 2NH 3  Step 2: Use the unknown and known in the equation to develop a mole ratio:  2 mol NH 3 (unknown from equation)  1 mol N 2 (known from equation)  Step 3: multiply the known by the mole ratio:  0.60 mol N 2 x 2 mol NH 3= 1.2 mol NH 3  1 mol N 2  N 2 + 3H 2 --> 2NH 3  Step 2: Use the unknown and known in the equation to develop a mole ratio:  2 mol NH 3 (unknown from equation)  1 mol N 2 (known from equation)  Step 3: multiply the known by the mole ratio:  0.60 mol N 2 x 2 mol NH 3= 1.2 mol NH 3  1 mol N 2

11. 11  4Al + 3O 2 --> 2Al 2 O 3  How many mol of aluminum are needed to form 7.8 mol aluminum oxide?  Known: 7.8 mol Al 2 O 3  Unknown:? mol Al  Mole ratio: 4 mol Al (unknown from equation) * 2 mol Al 2 O 3 (known from equation)   7.8 mol Al 2 O 3 x 4 mol Al=16 mol Al  2 mol Al 2 O 3  4Al + 3O 2 --> 2Al 2 O 3  How many mol of aluminum are needed to form 7.8 mol aluminum oxide?  Known: 7.8 mol Al 2 O 3  Unknown:? mol Al  Mole ratio: 4 mol Al (unknown from equation) * 2 mol Al 2 O 3 (known from equation)   7.8 mol Al 2 O 3 x 4 mol Al=16 mol Al  2 mol Al 2 O 3

12. 12 Mass to Mole/ Mole to Mass  4Al + 3O 2 --> 2Al 2 O 3  When 5.00g of Al reacts with Oxygen, how many moles of Aluminum Oxide is formed?  Step 1: Identify known and unknown  Known: Al= 5.00g  Unknown: Al 2 O 3= ? moles  4Al + 3O 2 --> 2Al 2 O 3  When 5.00g of Al reacts with Oxygen, how many moles of Aluminum Oxide is formed?  Step 1: Identify known and unknown  Known: Al= 5.00g  Unknown: Al 2 O 3= ? moles

13. 13  4Al + 3O2 --> 2Al 2 O 3  Step 2: Set up the problem  Known x molar mass x mole ratio  5.00g Al x 1 mole Al x 2 mol Al 2 O 3  27.0g Al 4 mol Al  = 0.0926 mol Al 2 O 3  4Al + 3O2 --> 2Al 2 O 3  Step 2: Set up the problem  Known x molar mass x mole ratio  5.00g Al x 1 mole Al x 2 mol Al 2 O 3  27.0g Al 4 mol Al  = 0.0926 mol Al 2 O 3

14. 14  4Al + 3O 2 --> 2Al 2 O 3  How many grams of Oxygen are needed to produce 1.23 moles of Aluminum Oxide  Step 1: Identify known and unknown  Know: 1.23 mole aluminum oxide  Unknown: ? grams Oxygen  4Al + 3O 2 --> 2Al 2 O 3  How many grams of Oxygen are needed to produce 1.23 moles of Aluminum Oxide  Step 1: Identify known and unknown  Know: 1.23 mole aluminum oxide  Unknown: ? grams Oxygen

15. 15 1.23  4Al + 3O 2 --> 2Al 2 O 3  Step 2: Set up the equation:  Known x Mole ratio x Molar mass  1.23 mol Al 2 O 3 x 3 mole O2 x 32.0 g O 2  2 mole Al 2 O 3 1 mole O 2  = 59.0g O 2  4Al + 3O 2 --> 2Al 2 O 3  Step 2: Set up the equation:  Known x Mole ratio x Molar mass  1.23 mol Al 2 O 3 x 3 mole O2 x 32.0 g O 2  2 mole Al 2 O 3 1 mole O 2  = 59.0g O 2

16. 16 Mass to Mass Problems  Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00g of lithium hydroxide?  How do we solve this problem?  Step 1: Write a balanced chemical equation:  2 LiOH (s) + CO 2 (g) --> Li 2 CO 3 (s) + H 2 O (l)  Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00g of lithium hydroxide?  How do we solve this problem?  Step 1: Write a balanced chemical equation:  2 LiOH (s) + CO 2 (g) --> Li 2 CO 3 (s) + H 2 O (l)

17. 17  Understand that each coefficient represents the number of moles of each substance.  2 LiOH (s) + CO 2 (g) --> Li 2 CO 3 (s) + H 2 O (l)  Pay attention to the substance and units asked for in the problem......we need grams of CO 2.  Step 2: Find the ratio between the unknown substance and the known substance in the problem:  The mole ratio of unknown = 1 mole CO 2  known 2 mole LiOH  Understand that each coefficient represents the number of moles of each substance.  2 LiOH (s) + CO 2 (g) --> Li 2 CO 3 (s) + H 2 O (l)  Pay attention to the substance and units asked for in the problem......we need grams of CO 2.  Step 2: Find the ratio between the unknown substance and the known substance in the problem:  The mole ratio of unknown = 1 mole CO 2  known 2 mole LiOH

18. 18  Step 3: Write down the quantity and substance you are starting with in the problem and convert to moles, and multiply by the mole ratio.  1.00 g LiOH x 1 mol LiOH x 1 mol CO 2 =  23.9 g LiOH 2 mol LiOH  Step 3: Write down the quantity and substance you are starting with in the problem and convert to moles, and multiply by the mole ratio.  1.00 g LiOH x 1 mol LiOH x 1 mol CO 2 =  23.9 g LiOH 2 mol LiOH

19. 19  Step 4: But we need grams of CO 2, so we must convert moles of CO 2 to grams:  1.00 g LiOH x 1 mol LiOH x 1 mol CO 2 x 44.0g CO 2 =  23.9 g LiOH 2 mol LiOH1 mol CO 2  =0.921 g CO 2  Mathematically, the problem is:  1.00/23.9/2 x 44.0 = 0.92  Step 4: But we need grams of CO 2, so we must convert moles of CO 2 to grams:  1.00 g LiOH x 1 mol LiOH x 1 mol CO 2 x 44.0g CO 2 =  23.9 g LiOH 2 mol LiOH1 mol CO 2  =0.921 g CO 2  Mathematically, the problem is:  1.00/23.9/2 x 44.0 = 0.92

20. 20  Chocolate chip cookie recipe:--makes 4 dozen.  4 c. flour  2 eggs  2 c. butter  2 tsp. vanilla  2 tsp. baking powder  1 c. milk  1 tsp. salt  2 bags chocolate chips  Chocolate chip cookie recipe:--makes 4 dozen.  4 c. flour  2 eggs  2 c. butter  2 tsp. vanilla  2 tsp. baking powder  1 c. milk  1 tsp. salt  2 bags chocolate chips I only have 2/3 bag of chocolate chips. What do I do? How does this affect my cookie-making? I don’t have enough chips to make 4 dozen cookies. So, it’s my limiting ingredient. It will control how many cookies I can make.

21. 21  Original Recipe:  4 c. flour  2 eggs  2 c. butter  2 tsp. vanilla  2 tsp. baking powder  1 c. milk  1 tsp. salt  2 bags chocolate chips  Original Recipe:  4 c. flour  2 eggs  2 c. butter  2 tsp. vanilla  2 tsp. baking powder  1 c. milk  1 tsp. salt  2 bags chocolate chips So, how do I modify this recipe for 2/3 bag of chips? How much flour do I need? 0.667 bag x 4 c. flour = 1.33 c. flour 2 bags How much butter do I need? 0.667 bag x 2 c. butter = 0.667 c. butter 2 bags How to modify a recipe: Find your limiting ingredient. Multiply the given amount of that ingredient by a ratio of the needed ingredient over the given ingredient from the original recipe.

22. 22  Mg + 2HCl --> MgCl2 + H2  Identify the limiting reagent when 6.00g HCl reacts with 5.00g Mg  1. Convert given quantities to moles:  6.00g HCl x 1 mol = 0.164 mol HCl Given  36.5g  5.00g Mg x 1 mol = 0.206 mol Mg Given  24.3g  2. Use mole ratios to find needed quantities:  0.164 mol HCl x 1 mol Mg = 0.082 mol Mg  2 mol HCl  HCl is limiting reagent.  Mg + 2HCl --> MgCl2 + H2  Identify the limiting reagent when 6.00g HCl reacts with 5.00g Mg  1. Convert given quantities to moles:  6.00g HCl x 1 mol = 0.164 mol HCl Given  36.5g  5.00g Mg x 1 mol = 0.206 mol Mg Given  24.3g  2. Use mole ratios to find needed quantities:  0.164 mol HCl x 1 mol Mg = 0.082 mol Mg  2 mol HCl  HCl is limiting reagent. 22

23. 23  Analyze:  Given: Needed:  0.164 mol HCl  0.206 mol Mg 0.082 mol Mg  We have more Mg than is needed, so Mg is in excess. Mg is the excess reagent. Therefore, HCl is the limiting reagent.  Analyze:  Given: Needed:  0.164 mol HCl  0.206 mol Mg 0.082 mol Mg  We have more Mg than is needed, so Mg is in excess. Mg is the excess reagent. Therefore, HCl is the limiting reagent.

24. 24  2C 2 H 2 + 5O 2 --> 4CO 2 + 2H 2 O  How many grams of water can be produced by the reaction of 2.40 mol C 2 H 2 with 7.40 mol O 2 ?  First, identify the limiting reagent:  2.40 mol C 2 H 2 x 5 mol O 2 = 6.00 mol O 2  2 mol C 2 H 2  7.40 mol O 2 x 2 mol C 2 H 2 = 2.96 mol C 2 H 2 given 5 mol O 2 needed  C 2 H 2 is limiting reagent.  2C 2 H 2 + 5O 2 --> 4CO 2 + 2H 2 O  How many grams of water can be produced by the reaction of 2.40 mol C 2 H 2 with 7.40 mol O 2 ?  First, identify the limiting reagent:  2.40 mol C 2 H 2 x 5 mol O 2 = 6.00 mol O 2  2 mol C 2 H 2  7.40 mol O 2 x 2 mol C 2 H 2 = 2.96 mol C 2 H 2 given 5 mol O 2 needed  C 2 H 2 is limiting reagent.

25. 25  C 2 H 2 is limiting reagent, so use that to find grams of water produced.  2C 2 H 2 + 5O 2 --> 4CO 2 + 2H 2 O  2.40 mol C 2 H 2 x 2 mol H 2 O x 18.0g H20 = 43.2g H 2 O  2 mol C2H2 1 mole H2O  C 2 H 2 is limiting reagent, so use that to find grams of water produced.  2C 2 H 2 + 5O 2 --> 4CO 2 + 2H 2 O  2.40 mol C 2 H 2 x 2 mol H 2 O x 18.0g H20 = 43.2g H 2 O  2 mol C2H2 1 mole H2O

26. 26 Percent Yield / Theoretical Yield - the maximum amount of product that could be formed from given amounts of reactants / Actual Yield - The amount of product that will actually form when the experiment is done / Percent Yield = actual yield x 100% / theoretical yield / Theoretical Yield - the maximum amount of product that could be formed from given amounts of reactants / Actual Yield - The amount of product that will actually form when the experiment is done / Percent Yield = actual yield x 100% / theoretical yield

27. 27 / CaCO 3 CaO + CO 2 / What is the percent yield if 13.1g of CaO is produced when 24.8 g CaCO 3 is heated? / Step 1- Find theoretical yield / 24. 8g CaCO3 x 1 mole CaCO3 x 1 mol CaO x 56.1 g CaO = / 100.1g CaCO3 1 mol CaCO3 1 mol CaO / =13.9g CaO -------Theoretical Yield / CaCO 3 CaO + CO 2 / What is the percent yield if 13.1g of CaO is produced when 24.8 g CaCO 3 is heated? / Step 1- Find theoretical yield / 24. 8g CaCO3 x 1 mole CaCO3 x 1 mol CaO x 56.1 g CaO = / 100.1g CaCO3 1 mol CaCO3 1 mol CaO / =13.9g CaO -------Theoretical Yield

28. 28 / Step 2- Plug theoretical yield calculated and actual yield into percent yield equation. / 13.1g CaO x 100% = 94.2% ----percent yield / 13.9g CaO / Step 2- Plug theoretical yield calculated and actual yield into percent yield equation. / 13.1g CaO x 100% = 94.2% ----percent yield / 13.9g CaO