1. CHEM 100 Fall 2013. chapter 4 -1. Instructor: Dr. Upali Siriwardane e-mail: upali@coes.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m Tu,Th,F 8:00 - 10:00 a.m. Or by appointment Test Dates: Chemistry 100(02) Fall 2013 September 30, 2013 (Test 1): Chapter 1 & 2 October 21, 2013 (Test 2): Chapter 3 & 4 November 13, 2013 (Test 3) Chapter 5 & 6 November 14, 2013 (Make-up test) comprehensive: Chapters 1-6 9:30-10:45:15 AM, CTH 328

2. CHEM 100 Fall 2013. chapter 4 -2. REQUIRED : Textbook: Principles of Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase the Mastering Chemistry Group Homework, Slides and Exam review guides and sample exam questions are available online: http://moodle.latech.edu/ and follow the course information links.http://moodle.latech.edu/ OPTIONAL : Study Guide: Chemistry: A Molecular Approach, 2nd Edition- Nivaldo J. Tro 2nd Edition Student Solutions Manual: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro 2nd Text Book & Resources

3. CHEM 100 Fall 2013. chapter 4 -3. 4.1 Global Warming and the Combustion of Fossil Fuels………………….127 4.2 Reaction Stoichiometry: How Much Carbon Dioxide?......................... 128 4.3 Limiting Reactant, Theoretical Yield, and Percent Yield………………. 133 4.4 Solution Concentration and Solution Stoichiometry………………….. 140 4.5 Types of Aqueous Solutions and Solubility…………………………….. 146 4.6 Precipitation Reactions………………………………………………….. 150 4.7 Representing Aqueous Reactions: Molecular, Ionic, and Complete Ionic Equations…………………………………………………………………........ 153 4.8 Acid–Base and Gas-Evolution Reactions……………………………..... 155 4.9 Oxidation–Reduction Reactions…………………………………………. 162 Chapter 4. Chemical Quantities and Aqueous Reactions

4. CHEM 100 Fall 2013. chapter 4 -4. Chapter 4. Chemical Quantities and Aqueous Reactions Global Warming Stoichiometry Reactions in One reactant in Limited Supply Limiting reactant Evaluating Success of Synthesis Theoretical Yield Actual Yield Percent Yield Solution Concentration Molecular, Ionic, and Complete Ionic Equations Precipitation Reactions Acid–Base and Gas-Evolution Reactions Oxidation–Reduction Reactions Stoichiometry in Solution Reactions

5. CHEM 100 Fall 2013. chapter 4 -5. What is Global Warming?

6. CHEM 100 Fall 2013. chapter 4 -6. Reaction Stoichiometry Law of Conversion of MatterAccording to the Law of Conversion of Matter: – Matter is neither created nor destroyed in a chemical reaction. –A balanced chemical equation illustrates the law of conversation of matter. In a balanced reaction: –Total mass of reactants = Total mass of products –A balanced reaction has the same type and quantity of atoms on both sides of the reaction. Stoichiometry is based on the law of conversion of matter. –Stoichiometry studies the quantitative aspects of chemical reactions.

7. CHEM 100 Fall 2013. chapter 4 -7. Reaction Stoichiometry: What it means 4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s) This equation means: 4 atoms Fe + 3 molecules O 2  2 molecules Fe 2 O 3 Fe 2 O 3 4 moles Fe + 3 moles O 2  2 moles Fe 2 O 3 4 moles Fe + 3 moles O 2  2 moles Fe 2 O 3 4 moles Fe atoms =Fe atoms and 6 moles O atoms and 6 mole O atoms 23.4 g Fe + 96.0 g O 2 = 415.4 g Fe 2 O 3 23.4 g Fe + 96.0 g O 2 = 415.4 g Fe 2 O 3 415.4 g = 415.4 g 415.4 g = 415.4 g of reactantsof products

8. CHEM 100 Fall 2013. chapter 4 -8. 1)Given the balanced reactions: 2H 2 (g) + O 2 (g) = 2 H 2 O(l), Write the mole conversion factors:

9. CHEM 100 Fall 2013. chapter 4 -9. 1.Need a balanced reaction to determine the stoichiometric relationship between: a.Reactants and reactants or b.Reactants and products or c.Products and products 2. Go to the mole: a.If mass is given, then divide by molecular mass: mass (g) /mol. mass (g/mole) = mole b. If volume and molarity (M) are given, then: M a x V a = moles A 3. Use the stoichiometric factor to convert from mole A to mole B to solve problem. Strategy Behind Solving STOICHIOMETRY Problems

10. CHEM 100 Fall 2013. chapter 4 -10. 10 Limiting Reactant For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others. When this reactant is used up, the reaction stops and no more product is made. The reactant that limits the amount of product is called the limiting reactant (limiting reagent). –The limiting reactant is completely consumed. Reactants not completely consumed are called excess reactants. The amount of product that can be made from the limiting reactant is called the theoretical yield.

11. CHEM 100 Fall 2013. chapter 4 -11. Limiting Reactant

12. CHEM 100 Fall 2013. chapter 4 -12. Analogy in Recipe : Making Cheese Sandwiches You were given 20 slices bread, 5 slices of cheese, 4 slices of ham If you want to make sandwiches containing two slices bread and one slice of cheese and one slice of ham How many sandwiches you could make? What is the limiting ingredient?

13. CHEM 100 Fall 2013. chapter 4 -13. Theoretical Yield and Limiting Reagent (Reactants) Percent Yield Using Stoichiometry to Predict

14. CHEM 100 Fall 2013. chapter 4 -14. : Problem: The following unbalanced equation is the chemical reaction associated with photosynthesis. CO 2 (g) + H 2 O(l)  C 6 H 6 O 6 (s) + O 2 (g) With adequate water, suppose a plant consumes 37.8 grams of CO 2 during the week. Determine how many grams of glucose (C 6 H 6 O 6 ) would be produced by the plant during this one-week period.

15. CHEM 100 Fall 2013. chapter 4 -15. Problem Strategy Problem Strategy: Determine how many grams of glucose (C 6 H 6 O 6 ) would be produced by the plant if 37.8 grams of CO 2 is consumed. 1. Need a balanced equation: 6 CO 2 (g) + 6 H 2 O(l)  C 6 H 6 O 6 (s) + 6 O 2 (g) 2. Determine moles of CO 2 consumed. 37.8 g CO 2 x (1 mol CO 2 /44.0 g) = 0.859 mol CO 2 6 mole CO 2 to 1 mole C 6 H 6 O 6 3. Determine how many moles of C 6 H 6 O 6 would be produced. Use the stoichiometric relationship between CO 2 and C 6 H 6 O 6. In this reaction the relationship is 6 mole CO 2 to 1 mole C 6 H 6 O 6. 1 mol C 6 H 6 O 6 /6 mol CO 2 0.859 mol CO 2 x ( 1 mol C 6 H 6 O 6 /6 mol CO 2 ) = 0.143 mol C 6 H 6 O 6 4. Determine the grams of C 6 H 6 O 6 produced. 0.143 mol C 6 H 6 O 6 x (1 mol / 180.0 g/1 mol C 6 H 6 O 6 ) = 25.8 g C 6 H 6 O 6

16. CHEM 100 Fall 2013. chapter 4 -16. 2) My recipe ratio for a bacon double cheeseburger is: 1 bun + 2 hamburger patties + 1 slices of cheese + 2 strips of bacon I f you start with: 2 bun, 6 patties, 4 slices of cheese, 6 strips of bacon a) How many bacon double cheeseburgers can you make? b) Which ingredient is limiting? c) What ingredients would be left over or in excess? hamburger bunhamburger pattiesslices of cheesestrips of bacon 1212 2646

17. CHEM 100 Fall 2013. chapter 4 -17. 3) Determine if the reaction is stoichiometric, or one reagent is in excess/one reagent is limiting for the reaction: 3H 2 (g) + 1N 2 (g) = 2 NH 3 (g) a)If 3 mole H 2 and 1.00 mole of N 2 are reacted according to the equation: Stoichiometric ratio (mole H 2 / mole of N 2 )= Actual mole ratio (mole H 2 / mole of N 2 ) = b) Is stoichiometric ratio equal, grate or less than actual mole ratio? c) Is the reaction stoichiometric, H 2 limiting or N 2 limiting?

18. CHEM 100 Fall 2013. chapter 4 -18. 4) Determine if the reaction is stoichiometric, or one reagent is in excess/one reagent is limiting for the reaction: reaction: 2H 2 (g) + 1O 2 (g) = 2 H 2 O (l) a) If 1 mole H 2 and 2.00 mole of O 2 are reacted according to the equation: Stoichiometric ratio (mole H 2 / mole of O 2 )= Actual mole ratio (mole H 2 / mole of O 2 ) = b) Is stoichiometric equal, grate or less than actual mole ratio? c) Is the reaction stochiometric, H 2 limiting or O 2 limiting?

19. CHEM 100 Fall 2013. chapter 4 -19. COMPARE THEORETICAL YIELDS TO DETERMINE THE LIMITING REACTANT: If all 8.10 g O 2 were used, then 17.2 g of Al 2 O 3 would be produced. If all 5.40 g Al were used, then 10.2 g of Al 2 O 3 would be produced. 10.2 g < 17.2 g The limiting reactant is Al. Theoretical yield is 10.2 g Al 2 O 3. To determine the percent yield of the reaction: (4.50 g/10.2) x 100 = 44.1% 44.1% is the percent yield for this reaction.

20. CHEM 100 Fall 2013. chapter 4 -20. Which reactant will remain when the reaction is complete? Al was the limiting reactant. O 2 was in excess –Therefore, O 2 was in excess. But by how much? – First find how much oxygen gas was required. –Then find how much oxygen gas is in excess.

21. CHEM 100 Fall 2013. chapter 4 -21. 4 Al + 3 O 2 products 0.200 mol = LR 0.253 mol O 2 available – O 2 required = excess O 2 O 2 available – O 2 required = excess O 2 0.253 mol O 2 - 0.150 mol O 2 = 0.103 mol O 2 left over = 0.103 mol O 2 in excess, or 3.30 grams O 2 0.200 mol Al x (3 mol O 2 /4 mol Al) = 0.15 mol O 2 0.150 mol of O 2 is required to react with all 0.200 mol of Al. How to Determine the Amount of Excess Reagent Left Over

22. CHEM 100 Fall 2013. chapter 4 -22. 5) 2 Al (s) + Fe 2 O 3 (s) = 2 Fe(l) + Al 2 O 3 takes place in the thermite mixture when it is ignited by a magnesium ribbon. A thermite mixture contains a mass ratio of 1 to 2 for Al and Fe 2 O 3. Which one is the limiting reagent?

23. CHEM 100 Fall 2013. chapter 4 -23. 6) Consider the reaction: 2H 2 (g) + O 2 (g) = 2 H 2 O(l) Equal weights of H 2 and O 2 are placed in a balloon and then ignited. Assume reaction goes to completion, which gas is the excess reagent? a) How many moles of H 2 O will be produced by 0.80 mole of O 2 and excess H 2, according to the equation? b) How many moles of H 2 O will be produced by 25.6g of O 2 and excess H 2, according to the equation? c) How many grams moles of H 2 O will be produced by 25.6g of O 2 and excess H 2, according to the equation?

24. CHEM 100 Fall 2013. chapter 4 -24. 7) Two moles of Mg and five moles of O 2 are placed in a reaction vessel, and then the Mg is ignited to produce MgO(F.W.=40.31 g/mole) a)The balanced chemical reaction: b) The limiting reactant? c) Actual mole ratio: d) Calculated mole ratio: e) How many moles of MgO are formed? f) What is the weight of MgO formed?

25. CHEM 100 Fall 2013. chapter 4 -25. 9) Write the balanced chemical equation for the reaction where zinc is producing silver in a single displacement reaction with silver chloride.

26. CHEM 100 Fall 2013. chapter 4 -26. Theoretical and Percent Yield Theoretical Yield –Predicting what could/should/would be produced from a given amount of reactant(s) Limiting reagent (reactant) –The determining reactant –Using stoichiometric relationship(s) from balanced reaction Percent Yield –Actual amount produced –% yield = (actual yield/theoretical yield) x 100

27. CHEM 100 Fall 2013. chapter 4 -27. Strategy: 1. Check to see if reaction given is balanced. 2. Determine limiting reactant and theoretical yield. Solution: 1. Balance reaction: 2 Mg(s) + O 2 (g)  2 MgO(s) 2. Determine limiting reactant * theoretical yield.

28. CHEM 100 Fall 2013. chapter 4 -28. Predicting Theoretical Yield Problem: Determine the theoretical yield for the following reaction: Mg(s) + O 2 (g)  MgO(s) when 42.5 g Mg(s) and 33.8 g O 2 (g) are reacted.

29. CHEM 100 Fall 2013. chapter 4 -29. 10) If 15.00 g of Zn (A.M.= 65.39 g/mole) reacts with silver with 25.00 g AgCl (F.M.= 143.35 g/mole)? a)Actual moles of Zn= 0.229 b) Actual moles of AgCl= 0.174 c) Stoichiometric ratio (mole Zn/ mole of AgCl)= ½ = 0.5 d) Actual mole ratio (mole Zn/ mole of AgCl)= 1.32 e) What is limiting reactant? b) Mole of Ag produced? c) Grams of Ag produced (theoretical yield)? d) If the actual yield of Ag was found to be 16.1 g Ag, the percent yield =

30. CHEM 100 Fall 2013. chapter 4 -30. 8) When 10.0 g pure solid calcium carbonate, CaCO 3 is heated and converted to solid calcium oxide CaO and CO 2 gas: a)The balanced chemical reaction: b)How much calcium oxide should be theoretically obtained? c)If 4.20 g of CaO is actually produced what is the percent yield of CaO?

31. CHEM 100 Fall 2013. chapter 4 -31. Ammonium nitrate (NH 4 NO 3 ) decomposes Problem: Ammonium nitrate (NH 4 NO 3 ) decomposes to N 2 O(g) and H 2 O(g). 1. Predict how many grams of H 2 O will be produced from the decomposition of 0.454 kg of NH 4 NO 3. 2. If 131 grams of water is produced, what is the % yield for this reaction? Strategy:1. Write a balanced reaction. 2. Determine the theoretical yield. 3. Calculate the percent (%) yield.

32. CHEM 100 Fall 2013. chapter 4 -32. Ammonium nitrate (NH 4 NO 3 ) decomposes Problem Solution: Ammonium nitrate (NH 4 NO 3 ) decomposes to N 2 O(g) and H 2 O(g). to N 2 O(g) and H 2 O(g). Balance equation: NH 4 NO 3 (s)  N 2 O(g) + 2 H 2 O(g) 1. Predict how many grams of H 2 O will be produced from the decomposition of 0.454 kg of NH 4 NO 3. 0.454 kg NH 4 NO 3 x (1000 g/1 kg) x (1 mol NH 4 NO 3 /80.04 g) = 5.68 mol NH 4 NO 3 5.68 mol NH 4 NO 3 x (2 mol H 2 O/1mol NH 4 NO 3 ) = 11.4 mol H 2 O 11.4 mol H 2 O x (18.0 g/1 mol H 2 O) = 204 grams H 2 O 204 grams of H 2 O is the predicted or theoretical yield.

33. CHEM 100 Fall 2013. chapter 4 -33. Ammonium nitrate (NH 4 NO 3 ) decomposes Problem: Ammonium nitrate (NH 4 NO 3 ) decomposes to N 2 O(g) and H 2 O(g). Balance equation: NH 4 NO 3 (s)  N 2 O(g) + 2 H 2 O(g) If 131 grams of water is produced, what is the % yield for this reaction? If 131 grams of water is produced, what is the % yield for this reaction? % yield = (131 g/204 g) x 100 = 64.2% 62.4% is the percent yield for the reaction.

34. CHEM 100 Fall 2013. chapter 4 -34. Problem: When 28.6 kg of carbon reacted with 88.2 kg of TiO 2, 42.8 kg of Ti was obtained. Reaction: TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Strategy:1. Determine limiting reactant. 2. Calculate theoretical yield. 3. Calculate percent yield.

35. CHEM 100 Fall 2013. chapter 4 -35. Problem Solution Problem Solution: When 28.6 kg of carbon reacted with 88.2 kg of TiO 2, 42.8 kg of Ti was obtained. Reaction: TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) It is balanced! 1. Determine limiting reactant. 28.6 kg carbon:28.6 kg C x (1000 g/1 kg) = 2.86 x 10 4 g C 88.2 kg TiO 2 88.2 kg TiO 2 x (1000 g/1 kg) = 8.82 x 10 4 g TiO 2 2.86 x 10 4 g C x (1 mol C/12.0 g) x (1 mol Ti/2 mol C) = 1.19 x 10 3 mol TiTHIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL CARBON IS USED. 8.82 x 10 4 g TiO 2 x (1 mol TiO 2 /79.9 g) x (1 mol Ti/1 mol TiO 2 ) = 1.10 x 10 3 mol TiTHIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL TiO 2 IS USED.

36. CHEM 100 Fall 2013. chapter 4 -36. Problem Solution Problem Solution: When 28.6 kg of carbon reacted with 88.2 kg of TiO 2, 42.8 kg of Ti was obtained. Reaction: TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) 2. Determining the theoretical yield from limiting reactant. = 1.19 x 10 3 mol TiTHIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL CARBON IS USED. = 1.10 x 10 3 mol TiTHIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL TiO 2 IS USED. 1.10 x 10 3 < 1.19 x 10 3 ; THEREFORE, the limiting reagent is TIO 2 and the theoretical yield in moles of Ti is 1.10 x 10 3.

37. CHEM 100 Fall 2013. chapter 4 -37. Problem Solution: When 28.6 kg of carbon reacted with 88.2 kg of TiO 2, 42.8 kg of Ti was obtained. Reaction: TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) 3. Determining percent yield: The theoretical yield in moles of Ti is 1.10 x 10 3. 1.10 x 10 3 mol Ti x (47.9 g/1 mol Ti) = 5.29 x 10 4 g Ti 42.8 kg is actual yield of Ti. 42.8 kg x (1000 g/1 kg) = 4.28 x 10 4 g Ti was produced. (4.28 x 10 4 g Ti/5.29 x 10 4 g Ti) x 100 = 80.9% 80.9% is the percent yield for this reaction.

38. CHEM 100 Fall 2013. chapter 4 -38. PROBLEM: Given the following chemical reaction: Al(s) + O 2 (g)  Al 2 O 3 (s) 1.Determine how many grams of Al 2 O 3 can form when 5.40 grams of Al and 8.10 grams of O 2 are reacted. 2.If 4.50 grams of Al 2 O 3 was produced, what is the percent yield for this reaction? 3.Which reactant was in excess and how much (grams) of this reactant remained after the reaction came to completion?

39. CHEM 100 Fall 2013. chapter 4 -39. Step 1: Balance the reaction and calculate theoretical yield for each reactant. Balanced reaction:4 Al(s) + 3 O 2 (g)  2 Al 2 O 3 (s) 5.40 g Al x (1 mol/27.0 g Al) = 0.200 mol Al Al 2 O 3 0.200 mol Al x 2 mol Al 2 O 3 = 0.100 mol Al 2 O 3 4 mol Al stoichiometric factor 0.100 mol Al 2 O 3 x 101.96 g = 10.2 g Al 2 O 3 1 mol Al 2 O 3 all 5.40 g Alonly 10.2 grams of Al 2 O 3 COULD This means that if all 5.40 g Al were consumed, then only 10.2 grams of Al 2 O 3 COULD be produced. NOW DETERMINE THE THEORITICAL YIELD IF ALL 8.10 grams of O 2 WERE USED.

40. CHEM 100 Fall 2013. chapter 4 -40. Step 1: Balance the reaction and calculate theoretical yield for each reactant. Balanced reaction:4 Al(s) + 3 O 2 (g)  2 Al 2 O 3 (s) 8.10 g O 2 x ( 1 mol/32.00 g) = 0.253 mol O 2 Al 2 O 3 0.253 mol O 2 x 2 mol Al 2 O 3 = 0.169 mol Al 2 O 3 3 mol O 2 stoichiometric factor Al 2 O 3 Al 2 O 3 0.169 mol Al 2 O 3 x 101.96 g = 17.2 g Al 2 O 3 Al 2 O 3 1 mol Al 2 O 3 8.10 grams of O 2 17.2 grams MOSTCOULD This means that if all 8.10 grams of O 2 were consumed, then 17.2 grams is the MOST that COULD be produced.

41. CHEM 100 Fall 2013. chapter 4 -41. PROBLEM: Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn metal, what volume of 2.50 M HCl is needed to covert the Zn metal completely to Zn 2+ ions? Step 1: Write the balanced equation. Zn(s) + 2 HCl(aq)  ZnCl 2 (aq) + H 2 (g) Step 2: Calculate amount of Zn. 10.0 g Zn x (1 mol Zn/65.39 g) = 0.153 mol Zn Step 3: Use the Step 3: Use the stoichiometric factor. 0.153 mol Zn x (2 mol HCl/1 mol Zn) = 0.306 mol HCl Step 4: Calculate volume of HCl required. 0.306 mol HCl x (1.00 L/2.50 mol) = 0.122 L HCl

42. CHEM 100 Fall 2013. chapter 4 -42. Problem:76.80 grams of apple juice (malic acid) requires 34.56 mL of 0.664 M NaOH to reach the endpoint in a titration. What is the (w/w)% of malic acid in this sample? Reaction: Reaction:C 4 H 6 O 5 (aq) + 2 NaOH(aq) --> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(l) (malic acid) Step 1: Calculate amount of NaOH used. M x V = (0.664 M)(0.03456 L) = 0.0229 mol NaOH Step 2: Calculate amount of malic acid titrated. 0.0229 mol NaOH x (1 mol malic acid/2 mol NaOH) = 0.0115 mol malic acid Step 3:Calculate the grams of malic acid in 0.0115 moles. 0.0115 mol malic acid x (134 g/1 mol malic acid) = 1.54 g Step 4: Calculate the weight (w/w)% malic acid in apple sample. (1.54 g/76.80) x 100 = 2.01 (w/w)%

43. CHEM 100 Fall 2013. chapter 4 -43. Chemical Analysis Problem: An impure sample of the mineral contains Na 2 SO 4. The mass of mineral sample = 0.123 g All of the Na 2 SO 4 in the sample is converted to insoluble BaSO 4. The mass of BaSO 4 is 0.177 g. Determine the mass percent of Na 2 SO 4 in the mineral.

44. CHEM 100 Fall 2013. chapter 4 -44. Answer: Reaction equations: Na 2 SO 4 (aq) + BaCl 2 (aq)  2 NaCl(aq) + BaSO 4 (s) 0.177 g BaSO 4 x (1 mol/233.4 g) = 7.58 x 10 -4 mol BaSO 4 7.58 x 10 -4 mol BaSO 4 x (1 mol Na 2 SO 4 /1 mol BaSO 4 ) = 7.58 x 10 -4 mol Na 2 SO 4 7.58 x 10 -4 mol Na 2 SO 4 x (142.0 g/1 mol) = 0.108 g Na 2 SO 4 (0.108 g Na 2 SO 4 /0.123 g sample)100% = 87.6% Na 2 SO 4