Chemical Reactions and Stoichiometry - PowerPoint Presentation

1. Chemical Reactions and StoichiometryWriting and balancing Chemical EquationClassifying Chemical Reactions:Precipitation ReactionsAcid-base ReactionsDecomposition Reactions (nonredox)Oxidation-Reduction ReactionsReaction StoichiometryReaction YieldsQuantitative Chemical Analysis

2. Writing & Balancing Chemical EquationsRules for writing and balancing equations:Formulas of substances in the equation MUST BE CORRECTLY WRITTEN and not be altered during balancing.Use SMALLEST INTEGER coefficients to balance equations; coefficient “1” need not be indicated;Helpful steps in balancing equations:Begin with the compound that contains the most types of atoms.Balance atoms that appear only once on each side of the arrow.Then balance atoms that appear more than once on either side.Balance free elements last.Finally, check that smallest whole number coefficients are used; if a fraction occurs after all atoms are balanced, multiply the entire equation by a factor that would remove the fraction

3. The reaction between methane and oxygen gas that yields carbon dioxide and water (bottom) is given in the form of a chemical equation shown below (top):

4. Chemical Equation #1Describe the reaction: “Magnesium reacts with oxygen gas to form solid magnesium oxide.”Chemical identity of reactants and products (in correct symbols and formulas): Reactants = Mg(s) and O2(g); Product = MgO(s)Chemical equation: Mg(s) + O2(g)  MgO(s)Balanced equation: 2Mg(s) + O2(g)  2MgO(s)

5. Chemical Equation #2Describe the reaction: Lithium reacts with nitrogen gas to form solid lithium nitride:Chemical identity (in correct symbols and formulas): Reactants = Li(s) and N2(g); Product = Li3N(s)Chemical equation: Li(s) + N2(g)  Li3N(s)Balanced equation: 6Li(s) + N2(g)  2 Li3N(s)

6. Chemical Equation #3Describing the reaction:“Acetylene gas (C2H2) burns in air (reacts with oxygen gas) to form carbon dioxide gas and water vapor.”Equation: C2H2(g) + O2(g)  CO2(g) + H2O(g);Balancing: (balance the C, H, followed by O) C2H2(g) + O2(g)  2CO2(g) + H2O(g);Multiply equation by 2 to remove fraction and obtain the final equation: 2C2H2(g) + 5 O2(g)  4CO2(g) + 2H2O(g); 

7. Chemical Equation #4Describing the reaction:“Ammonia (NH3) reacts with oxygen gas to produce nitrogen monoxide gas and water vapor.”The equation: NH3(g) + O2(g)  NO(g) + H2O(g);Balancing the equation:2NH3(g) + 5/2 O2(g)  2NO(g) + 3H2O(g)Multiply equation by 2 to remove fraction and yield final equation:4NH3(g) + 5 O2(g)  4NO(g) + 6H2O(g)

8. StoichiometryStoichiometry – the quantitative relationship between reactants and products, as well as between one reactant to another in a chemical reaction.Start by understanding a balanced equation: Example: C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g);At molecular level this equation indicates that: one C3H8 molecule reacts with five O2 molecules to produce three CO2 molecules and four H2O molecules;At macroscopic level 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.

9. Interpreting Balanced EquationA balanced equation provides the mole-to-mole relationship between reactants and products, as well as between one reactant to another.An Example of a Stoichiometric calculation.Consider the following equation and the accompanying question: C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g) If 5.15 moles of C3H8 were completely reacted, how many moles of CO2 and H2O, respectively, will be formed? How many moles of O2 will be consumed? (Solutions on next slide)

10. Stoichiometric Calculation #1(from previous slide)Reaction: C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g);Stoichiometric calculations:(Calculating moles of CO2, H2O and O2)Moles of CO2 formed = 5.15 mol C3H8 x Moles of H2O formed = 5.15 mol C3H8 x = 20.6 molesMoles of O2 consumed = 5.15 mol C3H8 x = 25.8 moles  

11. Stoichiometric Calculation #2Consider the following reaction: 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)If 4.386 moles of octane (C8H18) were completely burned, how many moles of CO2 and H2O, respectively, will be produced? How many moles of O2 will be reacted?(solution on next slide)

12. Stoichiometric Calculation #2(Solution to problem from previous slide)Calculations:Calculate moles of CO2 and H2O formed and moles of O2 consumed when 4.386 moles of C8H18 are reacted:Moles of CO2 produced = 4.386 mol C8H18 x = 35.09 mol;Moles of H2O produced = 4.386 mol C8H18 x = 39.47 mol;Moles of O2 consumed = 4.386 mol C8H18 x = 54.82 mol 

13. Stoichiometric Calculation #3Mass-to-mole-to-mole-to-mass relationship:If the mass of reactant is given, instead of mole, this must be converted to mole, because the balanced equation only provides a mole-to-mole relationship. Example-1: In the following reaction, if 653.0 g of octane, C8H18, are completely combusted, how many grams of CO2 are formed? 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)Calculation process: (1) convert mass of C8H18 to moles; (2) calculate moles of CO2 using stoichiometric relationship of C8H18 to CO2 given in the balanced equation; (3) convert moles of CO2 to grams. (next slide)

14. Stoichiometric Calculation #3Mass-to-mole-to-mole-to-mass relationship:Reaction: (from previous slide)2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g);Calculation-1: Moles C8H18 reacted = 653.0 g C8H18 x = 5.718 molesMoles CO2 formed = 5.718 mol C8H18 x = 45.74 moles Mass of CO2 formed = 45.74 mol CO2 x = 2013 g CO2 

15. Stoichiometric Calculation #4Mass-to-mole-to-mole-to-mass relationship:Example-2: In the following reaction, if 653.0 g of octane, C8H18, are completely combusted, how many grams of H2O are produced? 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)Calculation process: (1) convert grams of C8H18 to mol; (2) calculate moles of H2O using stoichiometric relationship of C8H18 to H2O H2O from balanced equation; (3) convert moles of H2O to grams.(see next slide)

16. Stoichiometric Calculation #4Mass-to-mole-to-mole-to-mass relationship:Reaction: (from previous slide) 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g);Calculation-2: Moles C8H18 reacted = 653.0 g C8H18 x = 5.718 molesMoles H2O formed = 5.718 mol C8H18 x = 51.46 moles Mass of H2O formed = 51.46 mol H2O x = 927.4 g H2O 

17. Stoichiometric Calculation #5Mass-to-mole-to-mole-to-mass relationship:Example-3: In the following reaction, if 653.0 g of octane, C8H18, are completely combusted, how many grams of O2 are consumed? 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)Calculation process: (1) convert grams of C8H18 to mol; (2) calculate moles of O2 using the stoichiometric relationship of C8H18 to O2 given in balanced equation; (3) convert moles of O2 to grams. (next slide)

18. Stoichiometric Calculation #5Mass-to-mole-to-mole-to-mass relationship:Reaction: (from previous slide)2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g);Calculation-3:Moles C8H18 reacted = 653.0 g C8H18 x = 5.718 molesMoles O2 formed = 5.718 mol C8H18 x = 71.48 moles Mass of O2 formed = 71.48 mol H2O x = = 2287 g O2 

19. Stoichiometry Involving Limiting ReactantLimiting reactant One that got completely consumed in a chemical reaction before the other reactant(s).The product yields depend on the amount of the limiting reactant.

20. Making cheese burgers to illustrate the concepts of limiting and excess reactants. (Ingredient: 1 hamburger bun + 1 hamburger patty + 1 slice of cheese + 2 tomato slices = 1 cheese burger)

21. The Concept of Limiting ReactantFor example, in the synthesis of water according to the following equation: 2H2(g) + O2(g)  2H2O(l), The equation indicates that 2 H2 molecules react with 1 O2 molecule to produce 2 H2O molecules. If a reaction is carried out with 10 molecules of H2 and 7 molecules of O2, O2 will be in excess, since 10 H2 will require only 5 O2 molecules. H2 will be the limiting reactant; the reaction will produce 10 molecules of H2O.

22. Limiting and Excess Reactants in the Synthesis of H2O: 2H2 + O2  2H2O

23. Synthesis of NH3

24. Stoichiometry Calculations involving limiting reactant Exercise-#1:Ammonia is produced by the reaction of N2 with H2 according to the following equation: N2(g) + 3H2(g)  2NH3(g) If 5.0 moles of N2 are reacted with 12.0 moles of H2, which is the limiting reactant? How many moles of NH3 are formed? How many moles of the excess reactant remains after the reaction? (Answer: (a) H2; (b) 8.0 moles; (c) 1.0 mol)

25. Limiting Reactants and Reaction YieldsExercise-#2:Ammonia is produced in the following reaction: N2(g) + 3H2(g)  2NH3(g) If 118 g of nitrogen gas is reacted with 31.5 g of hydrogen gas, which reactant will be completely consumed at the end of the reaction? How many grams ammonia will be produced when the limiting reactant is completely reacted and the yield is 100%? How many grams of the excess reactant will be unreacted? Answer: (a) N2; (b) 143 g NH3; (c) 6.0 g of H2

26. Theoretical, Actual and Percent YieldsTheoretical Yield: Yield of product(s) calculated calculated based on the stoichiometry of balanced equation and the amount of limiting reactant (assuming the reaction goes to completion and there is no competing reaction).Actual Yield (aka experimental yield)The amount of product actually obtained from laboratory experimentPercent Yield = x 100% 

27. Limiting Reactant & YieldsExercise-#3:The production of ammonia is carried out according to the following equation: N2(g) + 3H2(g)  2NH3(g)In one reaction at 227oC, the reactor is charged with 805.0 g of N2 and 167.5 g of H2. Determine the limiting reactant.What is the expected yield of ammonia?If 831.3 g of NH3 are actually produced, calculate the percentage yield. (Answer: (a) H2; (b) 943.3 g NH3; (b) Percent yield = 88.13%)

28. Reactions in Aqueous SolutionMany reactions taking place in the environments and in our body occur in aqueous solution in which water is the solvent.

29. Water as a Universal SolventWater molecule is very polar and has a strong dipole property;Water interacts strongly with ionic and polar molecules;Strong interactions enable water to dissolve many substances – both ionic and nonionic;The solubility of ionic compounds depends on the relative strength of ion-dipole interactions between the ions and water molecules;Many ionic compounds dissolve in water because of the strong ion-dipole interactions;Polar nonionic compounds, such as alcohol and sugar, dissolve in water due to strong dipole-dipole interactions and hydrogen bonding;

30. Dissolution of Ionic Compounds: Ionic compounds dissociate into separate cations and anions when dissolved in water.

31. Particles present in Aqueous Solutions containing Ionic and Molecular Compounds

32. Free ions in solution conduct electric current; such substances are called Electrolytes:such as ionic compounds and strong acids and bases

33. Electrolytes and NonelectrolytesElectrolytes – solutions capable of conducting electric current - contain ions that move freely;Nonelectrolytes – solutions not capable of conducting electric current - contains neutral molecules only;Strong electrolytes – ionic compounds, strong acids and strong bases; they dissociate completely when dissolved in water, producing a lot of free ions;Weak electrolytes – weak acids or weak bases: they only ionize partially when dissolved in water; solutions contain mostly neutral molecules and very little free ions.

34. Strong and Weak ElectrolytesExamples of strong electrolytes – they ionize completely: NaCl(aq)  Na+(aq) + Cl-(aq) H2SO4(aq)  H+(aq) + HSO4-(aq) Ca(NO3)2(aq)  Ca2+(aq) + 2NO3-(aq);Examples of weak electrolytes – they do not ionize completely: HC2H3O2(aq) ⇄ H+(aq) + C2H3O2-(aq) NH4OH(aq) ⇄ NH4+(aq) + OH-(aq); Mg(OH)2(s) ⇄ Mg2+(aq) + 2 OH-(aq)

35. NonelectrolytesSubstances that do not ionize in aqueous solution are nonelectrolytes;Most organic compounds are nonelectrolytes;Solutions cannot conduct electricity – no charged particles.Examples of nonelectrolytes: C6H12O6, C12H22O11, CH3OH, C2H5OH, C3H7OH, HOC2H4OH, etc.

36. Types of Reactions in Aqueous Solution1. Ion-Exchange ReactionsPrecipitation reactions;Acid-Base (or Neutralization) reactions;Gas-producing reactions2. Oxidation-Reduction (Redox) ReactionsCombination reactions involving free elements;Decomposition reactions that produces free elements;Combustion reactionsSingle-Replacement reactionsReactions involving strong oxidizing reagents

37. Precipitation ReactionsReactions that produce insoluble products (or precipitates) when two aqueous solutions are mixed.Examples:(1) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) (2) Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq) (3) BaCl2(aq) + H2SO4(aq)  BaSO4(s) + 2HCl(aq) (4) 3AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + 3NaNO3(aq)

38. Precipitation Reactions Formation of a precipitate (water-insoluble product) when two aqueous solutions are mixed:

39. Precipitates may be white or colored Examples:

40. Predicting Precipitation ReactionsCan we predict whether a precipitate will form when two solutions are mixed?How do we identify which of the products will form the precipitate?The answer: we need to know the solubility of ionic compounds in water;We use the “Solubility Rules” to predict and identify the precipitates in ion-exchange reaction.

41. Solubility Rules for Predicting PrecipitatesSoluble salts:Compounds of alkali metals (esp. Na+ and K+) and NH4+ are soluble;Compounds containing nitrate, NO3-, and acetate. C2H3O2-, are soluble; AgC2H3O2 is moderately soluble;Most chlorides, bromides, and iodides are soluble; exception: AgX, Hg2X2, PbX2, and HgI2; (where X = Cl-, Br-, I-)Most sulfates are soluble; exception: CaSO4, SrSO4, BaSO4, PbSO4 and Hg2SO4.Insoluble or slightly soluble salts:Most hydroxides (OH-); sulfides (S2-); carbonates (CO32-); chromates (CrO42-), and phosphate (PO43-) are only slightly soluble, except those associated with Rule #1 above.

42. Predicting Precipitation ReactionsExercise#4: Complete and balance the following equations; predict and identify the precipitates (if formed).CaCl2(aq) + Na2CO3(aq)  ? NH4NO3(aq) + MgSO4(aq)  ? Pb(NO3)2(aq) + KI(aq)  ? AgNO3(aq) + Na2CrO4(aq)  ?

43. Equations for Precipitation ReactionsMolecular equation: Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)Total Ionic equation: Pb2+ + 2NO3- + 2K+ + CrO42-  PbCrO4(s) + 2K+ + 2NO3- (K+ and NO3- are spectator ions)Net ionic equation: Pb2+(aq) + CrO42-(aq)  PbCrO4(s)

44. Acid-Base (or Neutralization) ReactionsAcids – compounds that produce hydrogen ions (H+) when dissolved in aqueous solution;Bases – compounds that produce hydroxide ions (OH-) in aqueous solutions.Some examples of acids and strong bases:Acids: HCl, HClO4, HNO3, H2SO4, H3PO4, and HC2H3O2;Bases: NaOH, KOH, Ba(OH)2, and NH3.

45. Acid-Base ReactionsSome examples of acid-base reactions: HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) H2SO4(aq) + 2KOH(aq)  H2O(l) + K2SO4(aq)HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq) 2HNO3(aq) + Ba(OH)2(aq)  2 H2O(l) + Ba(NO3)2(aq)

46. Equations for Acid-Base Reaction #1Reaction between a strong acid and a strong base: Molecular equation: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)Total ionic equation:H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l)Net ionic equation: H+(aq) + OH-(aq)  H2O(l)

47. Equations for Acid-Base Reaction #2Reaction between a weak acid and a strong base: Molecular equation: HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)Total ionic equation:HC2H3O2(aq) + Na+ + OH-  Na+ + C2H3O2-(aq) + H2O(l)Net ionic equation: HC2H3O2(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)

48. Equations for Acid-Base Reaction #3Reaction between a strong acid and a weak base: Molecular equation: HCl(aq) + NH3(aq)  NH4Cl(aq)Total ionic equation: H+(aq) + Cl-(aq) + NH3(aq)  NH4+(aq) + Cl-(aq)Net ionic equation: H+(aq) + NH3(aq)  NH4+(aq)

49. Gas-Producing ReactionsReactions that produce CO2 gas: CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g); NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g);Reaction that produces SO2 gas: Na2SO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + SO2(g);Reaction that produces H2S gas: Na2S(aq) + 2HCl(aq)  2NaCl(aq) + H2S(g);

50. Equations for Reaction involving Solid Reactant #1Reaction between calcium carbonate and hydrochloric acid: Molecular equation: CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g);Total ionic equation:CaCO3(s) + 2H+ + 2Cl-  Ca2+ + 2Cl- + H2O(l) + CO2(g)Net ionic equation: CaCO3(s) + 2H+(aq)  Ca2+(aq) + H2O(l) + CO2(g)

51. Equations for Reaction involving Solid Reactant #2Reaction between sodium bicarbonate with hydrochloric acid:Molecular equation: NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g);Total ionic equation:NaHCO3(s) + H+ + Cl-  Na+ + Cl- + H2O(l) + CO2(g)Net ionic equation: NaHCO3(s) + H+(aq)  Na+(aq) + H2O(l) + CO2(g)

52. Quantitative Analysis:The determination of the amount or concentration of a substance in a sample.We will discuss three types of quantitative analysis: Volumetric Analysis (Titration) Gravimetric Analysis Combustion Analysis Quantitative Chemical Analysis

53. TitrationsTitration analysis allows for very exact quantitative analysis of reactions in solution. Titration is commonly used for analysis of acid-base reactions but works well for other reactions like redox. Titrations involve reactions of two solutions: Titrant – Solution added from a buret; contains a reactant of known concentration; Analyte – Solution containing a reactant of unknown concentration; placed in a beaker or an Erlenmeyer flask.

54. Acid-Base Titration Equivalence Point – point in the titration where enough titrant has been added to react with exactly all of the analyte. End-point - the point in which the indicator color changes color that signals that the equivalent point is reached.Indicator – Special dye that changes color.The indicator is added to the analyte so that it changes color at or very close to the equivalence point.

55. Acid-Base Titration

56. Stoichiometry Calculation in Aqueous SolutionExample-#1: A 10.00-mL sample of vinegar 10.00-mL is titrated with 0.2560 M NaOH solution, which requires 32.40 mL of the NaOH to completely neutralize the acetic acid in the sample. (a) Calculate the molar concentration of acetic acid in the vinegar. (b) How many gram of acetic acid are present in 473 mL (~1 pint) of vinegar? If the density of vinegar is 1.0 g/mL, calculate the mass percent of acetic acid in vinegar.Reaction: HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq)

57. Stoichiometry Calculation in Aqueous SolutionSolving Example-1 problem:Mole of NaOH used = 0.3240 L x = 0.008294 molMole HC2H3O2 in vinegar sample = 0.008294 mol;Molarity of HC2H3O2 in vinegar = Mol HC2H3O2 in 473 mL vinegar = 0.473 L x = 0.392 molMass of HC2H3O2 in 473 mL vinegar = 0.392 mol x = 23.6 gMass percent of acetic acid in vinegar = x 100 = 5.0% 

58. Stoichiometry Calculation in Aqueous SolutionSolving Example-1 problem:Mole of NaOH used = 0.3240 L x (0.2560 mol/L) = 0.008294 molMole HC2H3O2 in vinegar sample = 0.008294 mol;Molarity of HC2H3O2 in vinegar = (0.008294 mol/0.01000 L) = 0.8294 M Mole of HC2H3O2 in 473 mL of vinegar = 0.473 L x (0.8294 mol/L) = 0.392 molMass of HC2H3O2 in 473 mL of vinegar = 0.392 mol x (60.05 g/mol) = 23.6 gMass percent of acetic acid in vinegar = (23.6 g/473 g) x 100 = 5.0%

59. Stoichiometry Calculation in Aqueous SolutionExercise-#1: A 5.00-mL sample of sulfuric acid of unknown concentration is diluted to 100.0 mL. A 20.0 mL portion of the dilute acid solution is then titrated with 0.265 M of NaOH. (a) If 33.0 mL of the NaOH solution were required to neutralize the acid solution, what is the molar concentration of H2SO4 in the diluted acid? (b) Calculate the molarity of H2SO4 in the original solution.(Answer: (a) 0.219 M; (b) 4.37 M;

60. Stoichiometry Calculation in Aqueous SolutionSolving Exercise-#1 problem:(1) Write the balanced equation; (2) Calculate mole of NaOH used; (3) Use stoichiometry given in balanced equation to calculate mole of H2SO4 reacted; (4) Divide mole of H2SO4 by volume (in L) of solution titrated to get molarity of H2SO4 in dilute solution; (5) Multiply molarity of diluted H2SO4 by dilution factor to obtain molarity in original solution.

61. Uses of Reactions in Aqueous Solutions1. Dissolving Insoluble Compounds: Fe2O3(s) + 6HNO3(aq)  2Fe(NO3)3(aq) + 3H2O(l); Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)2. Syntheses of Inorganic Compounds: AgNO3(aq) + NaBr(aq)  AgBr(s) + NaNO3(aq); Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)3. Extraction of Metals form Solution: Mg2+(aq) + Ca(OH)2(aq)  Mg(OH)2(s) + Ca2+(aq);

62. Gravimetric Analysis The most common form of gravimetric analysis involves reactions in which the product of interest is isolated as a precipitate. The product is removed by filtration from the reaction mixture, dried, and weighed. The mass of the product is related back to the analyte and the percent yield is calculated.Gravimetric analysis is suitable for products that are not soluble or only slightly soluble in water.

63. In gravimetric analysis the product of interest is isolated at a precipitate.Precipitate is removed from a reaction mixture by filtration (as shown in the diagram).

64. Stoichiometry in Aqueous SolutionExample-#2: According to the following reaction: BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)How many grams of BaSO4 will be formed when 100.0 mL of 0.100 M BaCl2 is reacted with 150.0 mL of 0.100 M Na2SO4? If the actual yield of barium sulfate is 2.05 g, what is the percent yield?Solving the problem: To solve this problem, find the limiting reactant - calculate the mole of each reactant by multiplying the molarity by the volume (in liters) of each solution.

65. Stoichiometry Calculation in Aqueous SolutionSolving Example-#2:Mole of BaCl2 = 0.1000 L x 0.100 mol/L = 0.0100 mole;Mole of Na2SO4 = 0.1500 L x 0.100 mol/L = 0.0150 mole; BaCl2 is the limiting reactant; Mole of BaSO4 expected = 0.0100 mole; Mass of BaSO4 expected = 0.0100 mol x 233.39 g/mol = 2.33 g ( b) Percent yield = (2.05 g/2.33 g) x 100% = 88.0%

66. Stoichiometry Calculation in Aqueous SolutionExercise-#1:(a) How many grams of AgCl can theoretically be produced when 150.0 mL of 0.100 M AgNO3 solution is reacted with 50.0 mL of 0.200 M CaCl2 solution? (b) Calculate the percent yield if 1.92 g of AgCl is obtained.Reaction: 2AgNO3(aq) + CaCl2(aq)  2AgCl(s) + Ca(NO3)2(aq)How to solving the problem:Determine the limiting reactant; (2) calculate the mole of limiting reactant; (3) Use reaction stoichiometry from balanced equation to determine mole of AgCl; (4) calculate mass of AgCl.(Answer: (a) 2.15 g AgCl; (b) 89.3%)

67. Combustion Analysis The elemental composition of hydrocarbon compounds can be determined by combustion analysis.An exact known mass of the compound is combusted in the following reaction: Example: CxHyOz(?) + O2(g)  CO2(g) + H2O(g)Masses of carbon and hydrogen in compound are calculated from mass of CO2 and H2O collected, respectively;Mass of oxygen in compound is obtained by subtracting mass of compound combusted with masses of carbon + hydrogen;Percent composition of each element is calculated from the mass, and the empirical formula of the compound can be determined.

68. Combustion Analysis This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.

69. Combustion Analysis of HydrocarbonExample-#1:A 1.372-g sample of compound composed of carbon, hydrogen and oxygen, is completely combusted in excess oxygen, which yields 3.460 g of CO2 and 0.605 g of water. (a) Calculate the mass percent composition of carbon, hydrogen and oxygen, respectively, in the compound.(b) Determine the empirical formula of the compound.Reaction: CxHyOz + O2(g)  xCO2(g) + y/2 H2O(l)

70. Combustion Analysis of HydrocarbonSolving example-#1 problem:Calculate mass of carbon and hydrogen from mass of CO2 and H2O collected, respectively;Calculate mass percent of carbon and hydrogen from their respective masses and mass of sample; Subtract mass percent of carbon and hydrogen from 100 to obtain mass percent of oxygen in the compound;Use mass percent of each element to determine the empirical formula of the compound in the form of CxHyOz

71. Combustion Analysis of HydrocarbonCalculation-a: (calculating mass percent of C, H & O)Mass of carbon = 3.460 g CO2 x = 0.9442 g Mass of hydrogen = 0.605 g H2O x = 0.0677 g Mass percent of carbon = x 100 = 68.82% CMass percent of hydrogen = x 100 = 4.93% HMass percent of oxygen = 100 – (68.82 + 4.93) = 26.25% O 

72. Combustion Analysis of HydrocarbonCalculation-b: determine empirical formula using mass percent(Calculate mole of each element using mass percent and obtain a simple mole ratio of C:H:O)Mole of carbon in compound = = 5.73 mol C;Mole of hydrogen in compound = = 4.89 mol H;Mole of oxygen in compound = = 1.64 mol ODividing each mole by 1.64 = : : = 3.5 : 3.0 : 1.0;Multiplying by 2 yields simple ratios = 7 : 6 : 2Empirical formula = C7H6O2  

73. Combustion Analysis of HydrocarbonExercise-#1: A 1.352-g sample of compound composed of carbon, hydrogen and oxygen, is completely combusted in excess oxygen, which yields 3.850 g of CO2 and 0.902 g of water. (a) Calculate the mass percent composition of carbon, hydrogen and oxygen, respectively, in the compound.(b) Determine the empirical formula of the compound.(Answer: (a) 77.71% C; 7.46%; 14.83% O; (b) C7H8O)

74. Oxidation-Reduction ReactionsOxidation  loss of electrons and increase in oxidation number;Reduction  gain of electrons and decrease in oxidation number;Oxidation-reduction (or Redox) reaction  one that involves transfer of electrons from one reactant to the other;Oxidizing agent  the reactant that gains electrons and got reduced;Reducing agent  the reactant that loses electrons and got oxidized.

75. Oxidation-Reduction ReactionsIn the following reactions, identify all reactions that are redox reactions:1. 2KMnO4(aq) + 16HCl(aq)  2MnCl2(aq) + 2KCl(aq) + 5Cl2(aq) + 8H2O(l);2. 2KClO3(s)  2KCl(s) + 3 O2(g);3. CaCO3(s)  CaO(s) + CO2(g);4. Mg(OH)2(s)  MgO(s) + H2O(g);5. Mg(s) + ZnSO4(aq)  MgSO4(aq) + Zn(s);6. Cr2O72-(aq) + 3C2H5OH(l) + 2H+(aq)  2Cr3+(aq) + 3CH3COOH(aq) + 4H2O(l)

76. Guidelines for Determining Oxidation Numbers of ElementsAtoms in the free elemental form are assigned oxidation number zero;The sum of oxidation number (o.n.) in neutral molecules or formula units is zero; the sum of oxidation number of atoms in a polyatomic ion is equal to the net charge of the ion (in magnitude and sign); In their compounds, each Group IA metal is assigned o.n. = +1; each Group IIA metal o.n. = +2; boron and aluminum each o.n. = +3, and fluorine o.n. = –1; Hydrogen is assigned o.n. = +1 in compounds or polyatomic ions with nonmetals, but o.n. = –1 in metal hydrides; In compounds and polyatomic ions, oxygen is assigned o.n. = -2, except in peroxides, in which its o.n. = –1; In binary compounds with metals, chlorine, bromine, and iodine each has o.n. = -1; sulfur, selenium, and tellurium each has o.n. = -2.

77. Types of Redox ReactionsReactions between metals and nonmetals;Combustion reactions (reactions with molecular oxygen);Single replacement reactions;Decomposition reactions that form free elements;Reactions involving strong oxidizing agents;Disproportionation reactions.

78. Types of Redox ReactionsReactions between metals and nonmetals:4Al(s) + 3 O2(g)  2Al2O3(s); 3Mg(s) + N2(g)  Mg3N2(s);

79. Type of Redox ReactionsCombustion reactions:CH4(g) + 2 O2(g)  CO2(g) + 2H2O(g); 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g); C2H5OH(l) + 3 O2(g)  2CO2(g) + 3H2O(g);

80. Types of Redox ReactionsSingle-Replacement Reactions: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g); Cu(s) + 2AgNO(aq)  Cu(NO3)2(aq) + 2Ag(s); Cl2(aq) + 2KBr(aq)  2KCl(aq) + Br2(aq);

81. Types of Redox ReactionsDecomposition that produces free elements:2HgO(s)  2Hg(l) + O2(g); (NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + 4H2O(g)

82. Types of Redox ReactionsReactions involving strong oxidizing reagents, such as permanganate, dichromate and nitrate: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq)  2Cr3+(aq) + 7H2O(l) + 3 O2(g) 2Cr(OH)4-(aq) + 3H2O2(aq) + 2 OH-(aq)  2CrO42-(aq) + 8H2O(l) 2NO3-(aq) + Cu(s) + 8H+(aq)  Cu2+(aq) + 2NO(g) + 4H2O(l)

83. Types of Redox ReactionsDisproportionation reaction: Cl2 (g) + 2NaOH (aq)  NaOCl (aq) + NaCl (aq) + H2O (l); 3Br2 (aq) + 6NaOH (aq)  NaBrO3 (aq) + NaBr (aq) + H2O(l);

84. Balancing Redox Reactions by Half-Equation MethodExample-1: Balance the following oxidation-reduction reaction in acidic solution: MnO4-(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq); Solution-1: Note: the above equation is both not balanced and not complete; it only shows components (reactants) that undergoes changes is oxidation numbers;Redox reactions in acidic solution means that H+ ion must be added to the net ion equation in order to balance it; water is one of the products in most redox reactions that involves this type of oxidizing agents.

85. Balancing Redox Reactions by Half-Equation MethodBalancing process:the first step in balancing a redox reaction by half-equation method is to break the net ionic equation into two half equations: oxidation and reduction half-equations: MnO4-(aq)  Mn2+(aq) (1)H+ ion will be added on the left side and H2O on the right; all 4 oxygen in MnO4- will become H2O in reactions in acidic solution. Therefore, to balance this half-equation, So, 8H+ is added on the left side and 4H2O on the right. The half-equation with all atoms balanced will look as follows: MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l); (1)

86. Balancing Redox Reactions by Half-Equation MethodBalancing process (continued):Next step is to count the total charges on both side of the half-equation; note that they are not equal. Add enough electrons on the side that has more positive charges or less negative charges, so that the total charges on both sides become equal (in magnitude and sign); for this equation, you need to add 5e- to the left side of the equation: MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l); (1)Now we have a balanced reduction half-equation.(How do you know it is “reduction” half-equation?)

87. Balancing Redox Reactions by Half-Equation MethodBalancing process (continued):Next, write the other half-equation; for this half-equation, you only need to add an electron on the right to balance the charge. (This is the oxidation half-equation) Fe2+(aq)  Fe3+(aq) + e-; (2)To obtain the overall equation, we will add the two balanced half-equations, but first we have to multiply the half-equation (2) by 5 so that the number of electron is equal to that in half-equation (1), as shown below. When you add the two half-equations, electrons must cancel out; electron cannot appear in the final net ionic equation. The overall equation should not contain any electron. 5Fe2+(aq)  5Fe3+(aq) + 5e-; (3)

88. Balancing Redox Reactions by Half-Equation MethodBalancing process (final step):Now combine the two half-equations (1) and (3) to obtain the following balanced net ionic equation:MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l);

89. Balancing Redox Reactions by Half-Equation MethodExample-2: Balance the following oxidation-reduction reaction in acidic solution: Cr2O72-(aq) + H2O2(aq)  Cr3+(aq) + O2(g) + H2O(l);Balancing process: Write the two half-equations and balance them.Reduction half-equation:Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l); (1)Oxidation half-equation: H2O2(aq)  O2(g) + 2H+(aq) + 2e-; (2)

90. Balancing Redox Reactions by Half-Equation MethodBalancing process (continued):Multiply the oxidation half-equation by 3 to make the number of electrons equal with that of the reduction-half equation. 3H2O2(aq)  3 O2(g) + 6H+(aq) + 6e-; (2)Add the two half-equations, canceling all the electrons on both side of the equation; all the H+ ions on the right-hand side, and the same number of H+ ions on the left-side. The final (balanced) net ionic equation will be as follows:Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq)  2Cr3+(aq) + 3O2(g) + 7H2O(l);

91. Balancing Redox Reactions by Half-Equation MethodExample-3: Balance the following oxidation-reduction reaction in basic solution: Cr(OH)4-(aq) + H2O2(aq)  CrO42-(aq) + H2O(l); Balancing process: Write the two half-equations and balance them, adding electrons on either side as needed to balance the charges. Note that this is a reaction in basic solution; the final equation will have OH- ion instead of H+ ion.

92. Balancing Redox Reactions by Half-Equation MethodBalancing process (continued):Oxidation half-equation:Cr(OH)4-(aq) + 4OH-(aq)  CrO42-(aq) + 4H2O(l) + 3e-; (1)Reduction half-equation: H2O2(aq) + 2e-  4OH-(aq); (2)

93. Balancing Redox Reactions by Half-Equation MethodBalancing process (continued):Multiply the oxidation half-equation by 2 and the reduction half-equation by 3 to make the number of electrons in both half-equations equal. 2Cr(OH)4-(aq) + 8OH-(aq)  2CrO42-(aq) + 8H2O(l) + 6e-; (1) 3H2O2(aq) + 6e-  6OH-(aq); (2)

94. Balancing Redox Reactions by Half-Equation MethodBalancing process (continued):Now add the two half-equations, canceling all the electrons on both sides of the equation, all the 6OH- ions on the right-hand side, and the same number of OH- ions on the left-hand side. The overall balanced net ionic equation will appear as follows:2Cr(OH)4-(aq) + 3H2O2(aq) + 2OH-(aq)  2CrO42-(aq) + 8H2O(l);