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Chemistry-140 Lecture 9

Chemistry-140 Lecture 9. Chapter Highlights balance simple chemical equations mass relationships limiting reagents percent yield chemical analysis. Chapter 4: Chemical Equations. 2 Al( s ) + 3 Br 2 ( l ) Al 2 Br 6 ( s ). Chemistry-140 Lecture 9. Chemical Equations.

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Chemistry-140 Lecture 9

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  1. Chemistry-140 Lecture 9 • Chapter Highlights • balance simple chemical equations • mass relationships • limiting reagents • percent yield • chemical analysis Chapter 4: Chemical Equations

  2. 2 Al(s) + 3 Br2(l) Al2Br6(s) Chemistry-140 Lecture 9 Chemical Equations reactants products

  3. 2 Al(s) + 3 Br2(l) Al2Br6(s) Chemistry-140 Lecture 9 Balanced Chemical Equations • a balanced chemical equation shows: • relative amounts of reactants • relative amounts of products • the physical states of all species • a balanced chemical equation does NOT show: • experimental conditions • gain or loss of energy (heat, light, etc..) • the rate of the reaction

  4. 2 Al(s) + 3 Br2(l) 1 Al2Br6(s) + Chemistry-140 Lecture 9 Reaction Stoichiometry the stoichiometric coefficients are 2, 3 & 1

  5. C4H10(g) + O2(g) CO2(g) + H2O(g) Chemistry-140 Lecture 9 • a chemical equation must be balanced before useful quantitative information can be obtained Balancing Chemical Equations • Example: • a combustion reaction occurs when a hydrocarbon compound (such as butane C4H10) burns in air to give carbon dioxide and water

  6. Step 1: Write down the correct formulas of the reactants • and products C4H10(g) + O2(g) CO2(g) + H2O(l) C4H10(g) + O2(g) 4 CO2(g) + H2O(l) Chemistry-140 Lecture 9 Basic Steps in Balancing a Chemical Equation • Step 2: Balance the number of C atoms

  7. C4H10(g) + O2(g) 4 CO2(g) + 5H2O(l) Chemistry-140 Lecture 9 Basic Steps in Balancing a Chemical Equation • Step 3: Balance the number of H atoms • Step 4: Balance the number of O atoms 2 C4H10(g) +13 O2(g) 8 CO2(g) + 10 H2O(l) • Step 5: Check that the numbers of atoms of each element are balanced

  8. Chemistry-140 Lecture 9 Weight Relations in Stoichiometric Reactions Question: The reaction of elemental phosphorus, P4(s), with chlorine gas, Cl2(g) gives phosphorus trichloride, PCl3(l). What mass of chlorine is required to react completely with 1.45 g of phosphorus and what mass of product will result?

  9. Chemistry-140 Lecture 9 Method: direct calculation not possible grams reactant P4 grams product PCl3 x x (MW) moles reactant P4 moles reactant PCl3 stoichiometric factor

  10. P4(s) + 6 Cl2(g) 4 PCl3(l) Chemistry-140 Lecture 9 Answer: Step 1: Write the balanced equation for the reaction. This determines the stoichiometric factors to be used in the calculations. That is one mole of P4 requires six moles of Cl2 and four moles of PCl3 will be produced.

  11. Chemistry-140 Lecture 9 Answer: Step 2: Calculate the number of moles of P4 in 1.45 g. (1.45 g P4) = Number of moles of P4 in 1.45 g is 1.17 x 10-2 mol. 1.17 x 10-2 mol P4

  12. Chemistry-140 Lecture 9 Answer: Step 3: Use the stoichiometric factors to calculate the number of moles of Cl2 required. (1.17 x 10-2 mol P4) = Number of moles of Cl2 required for complete reaction is 7.02 x 10-2 mol. 7.02 x 10-2 mol Cl2

  13. Chemistry-140 Lecture 9 Answer: Step 4: Calculate the number of grams of Cl2 in 7.02 x 10-2 mol. (7.02 x 10-2 mol Cl2) = Number of grams of Cl2 required for complete reaction is 4.98g. 4.98 g Cl2

  14. Chemistry-140 Lecture 9 Answer: Step 5: Use the stoichiometric factors to calculate the number of moles of PCl3 that will be produced. (1.17 x 10-2 mol P4) = Number of moles of PCl3 produced by complete reaction is 4.68 x 10-2 mol . 4.68 x 10-2 mol PCl3

  15. Chemistry-140 Lecture 9 Answer: Step 6: Calculate the number of grams of PCl3 in 4.68 x 10-2 mol. (4.68 x 10-2 mol PCl3) = Number of grams of PCl3 produced in the reaction of 1.45 g of P4 with is 4.98g of Cl2 was 6.43 g PCl3. 6.43 g PCl3

  16. C3H8(g) + O2(g) 3 CO2(g) + 4 H2O(l) Chemistry-140 Lecture 9 • In practice: Reagents are rarely combined in the exact stoichiometric ratio described by the balanced equation Stoichiometric Reactions & Limiting Reagents • For example: In this combustion reaction, excess oxygen is used. The amount of CO2 and H2O produced is dependent on the amount of propane ONLY. Propane is the limiting reagent.

  17. Chemistry-140 Lecture 9 A Reaction Involving a Limiting Reagent Example 4.3: Methanol, CH3OH, is an excellent fuel and can be made from the direct reaction of carbon monoxide, CO, and hydrogen, H2. Suppose 356 g of CO are mixed with 65.0 g of H2. What mass of methanol will be formed ?

  18. CO(g) + 2 H2(g) CH3OH(l) Chemistry-140 Lecture 9 Answer: Step 1: Write the balanced equation for the reaction.

  19. Chemistry-140 Lecture 9 Answer: Step 2: Calculate the number of moles of each reagent. (356 g CO) = (65.0 g H2) = 12.7 mol CO 32.2 mol H2

  20. Answer: Step 3: What is the ratio of moles of the reagents. Are the reagents present in the stoichiometric ratio. = = BUT stoichiometry is 2 moles of H2 to 1 mole of CO, therefore, H2 is in excess and CO must be a limiting reagent! Chemistry-140 Lecture 9

  21. Chemistry-140 Lecture 9 Answer: Step 4: Calculate the number of grams of methanol that can be produced. (12.7 mol CO) = Number of grams of methanol produced in the reaction was 407 g CH3OH. 407 g CH3OH

  22. Chemistry-140 Lecture 9 • In practice: Reactions rarely (if ever!!) produce all the product predicted by the balanced equation. The ratio of the actual yield to the theoretical yield expressed as a percentage is known as the percent yield Percent Yield

  23. Percent Yield = x 100% Chemistry-140 Lecture 9 Percent Yield • Actual yield: the quantity of material actually produced in an experiment • Theoretical yield: the quantity of material that would be produced IF the reaction produced the maximum product allowable by the stoichiometry and limiting reagents

  24. C7H6O3(s) + C4H6O3(l) C9H8O4(s) + CH3CO2H(l) salicylic acid acetic anhydride acetylsalicylic acid (ASA) acetic acid Chemistry-140 Lecture 9 Percent Yield Section 4.5: Suppose you made aspirin (ASA) in the laboratory according to the following reaction: Suppose you begin with 14.4 g of salicylic acid, use an excess of acetic anhydride and obtain 6.26 g of ASA. What was the percent yield of ASA?

  25. Chemistry-140 Lecture 9 Answer: Step 1: Calculate the moles of the limiting reagent, salicylic acid. (14.4 g C7H6O3) = 1.04 x 10-1 mol C7H6O3

  26. Chemistry-140 Lecture 9 Answer: Step 2: Use the stoichiometric ratios to determine the expected number of moles of ASA . (1.04 x 10-1 mol C7H6O3) = = 1.04 x 10-1 mol ASA

  27. Chemistry-140 Lecture 9 Answer: Step 3: Calculate the number of grams of ASA to be produced. (1.04 x 10-1 mol ASA) = The theoretical yield is 18.8 g of ASA. 18.8 g ASA

  28. Percent Yield = x 100% Answer: Step 3: Calculate the percent yield of ASA in this particular experiment. = x 100% = The percent yield of ASA 33.3 %. Chemistry-140 Lecture 9 33.3 %

  29. Chemistry-140 Lecture 10 • Chapter Highlights • definitions of electrolytes & nonelectrolytes • recognize acids & bases • predict solubility of ionic compoundsin water • determine net ionic equations • learn four basic reactions types & predict products • acid base • precipitation • gas-forming • oxidation -reduction Chapter 5: Reactions in Aqueous Solution

  30. NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) Chemistry-140 Lecture 10 Definition: a substance whose aqueous solution conducts electricity is called an electrolyte Electrolytes • a substance can be a strong electrolyte, a weak electrolyte or a nonelectrolyte depending on the degree of dissociation (ionization) in solution • Example: • For sodium chloride, the ionic solid dissociates 100 % in water forming exclusively Na+ and Cl- ions in solution • 100 % dissociation = strong electrolyte

  31. Chemistry-140 Lecture 10 Electrolytes pure water acetic acid solution potassium dichromate solution

  32. Chemistry-140 Lecture 10 • Strong electrolytes : Substances that dissociate completely inwater. Simple salts like NaCl that are combination of a metal and a nonmetal • Weak electrolytes : Substances that do not dissociate fully inwater but do form some ions. Usually molecular compounds like acetic acid (CH3COOH) with ionizable groups (H+) • Nonelectrolytes : Substances that do not dissociate inwater to form ions. Molecular compounds which are soluble but which remain intact as the molecule in solution Identifying Electrolytes

  33. Chemistry-140 Lecture 10 Question: What type of electrolytes are these compounds? a) Epsom’s salt: MgSO4 . 7 H2O b) Methanol: CH3OH c) Acetic acid: CH3COOH

  34. MgSO4 . 7 H2O(s) + H2O(l) Mg2+(aq) + SO42-(aq) CH3OH(l) + H2O(l) CH3OH(aq) CH3COOH(l) + H2O(l) H+(aq) + CH3COO-(aq) Chemistry-140 Lecture 10 Answer: a) b) c) strong electrolyte nonelectrolyte weak electrolyte

  35. NaCl(aq) + KNO3(aq) Na+(aq) + Cl-(aq) + K+(aq) + NO3-(aq) NaCl(aq) + AgNO3(aq) Na+(aq) + NO3-(aq) + AgCl(s) Chemistry-140 Lecture 10 Understanding & Predicting Reactions in Solution • Driving Force: a property of the reaction that can be identified as the reason for product formation • Examples: solid formation

  36. Chemistry-140 Lecture 10 Types of Reactions • the reaction type depends on the driving force of the reaction. There are four basic types • Formation of an insoluble compound • Formation of a nonelectrolyte • Formation of a gas • Transfer of electrons

  37. NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s) Chemistry-140 Lecture 10 • certain combinations of cations and anions are soluble; that is they dissolve in water….. • if a compound will not dissolve in water it is insoluble • if a combination of anion and cation results in the formation of an insoluble solid, this is a precipitate Solubility • Example: precipitate

  38. All salts of Cl-, Br- & I- Halides of Ag+, Hg22+ & Pb2+ Fluorides of Mg2+, Ca2+, Sr2+ & Pb2+ Salts containing F- Salts of SO4- Sulfates of Sr2+, Ba2+ & Pb2+ Chemistry-140 Lecture 10 Soluble Compounds Exceptions Almost all salts of Na+, K+ & NH4+ Salts of NO3-, ClO3-, ClO4-, CH3CO2-

  39. All salts of CO32-, PO43-, C2O42-, CrO42-, S2- Most metal hydroxides (OH-) & oxides (O2-) Salts of NH4+, and alkali metal cations Chemistry-140 Lecture 10 Exceptions Insoluble Compounds

  40. Chemistry-140 Lecture 10 • the balanced equation that results from the omission of all spectator ions is the net ionic equation • spectator ions are the ions which do not participate in the reaction Net Ionic Equations Example: Write a balanced net ionic equation for the reaction of AgNO3 with CaCl2 to produce AgCl and Ca(NO3)2.

  41. 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2 2 AgNO3(aq) + CaCl2(aq) 2 AgCl(s) + Ca(NO3)2(aq) Chemistry-140 Lecture 10 Step 1: Write the complete balanced equation with appropriate stoichiometry Step 2: Decide on the physical state (eg solubility) of each compound.

  42. AgNO3(aq) Ag+(aq) + NO3-(aq) CaCl2(aq) Ca2+(aq) + 2 Cl-(aq) Ca(NO3)2(aq) Ca2+(aq) + 2 NO3-(aq) 2 Ag+(aq) + 2 NO3-(aq) + Ca2+(aq) + 2 Cl-(aq) 2 AgCl(s) + Ca2+(aq) + 2 NO3-(aq) Chemistry-140 Lecture 10 Step 3: Recognize that all soluble ionic compounds dissociate to form ions in aqueous solution

  43. complete ionic equation 2 Ag+(aq) + 2 NO3-(aq) + Ca2+(aq) + 2 Cl-(aq) 2 AgCl(s) + Ca2+(aq) + 2 NO3-(aq) net ionic equation Ag+(aq) + Cl-(aq) AgCl(s) Chemistry-140 Lecture 10 Step 4: Identify the spectatorions and remove them from the complete ionic equation to give the net ionic equation. Simplify the resulting equation in terms of stoichiometric coefficients. • The sum of ion charges is the same on • both sides of the net ionic equation

  44. Overall Reaction Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq) Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq) PbI2(s) + 2 K+(aq) + 2 NO3-(aq) Net Ionic Equation Pb2+(aq) + 2 I-(aq) PbI2(s) Chemistry-140 Lecture 10 Precipitation Reactions • Write the net ionic equation for • the reaction of Pb(NO3)2 with KI.

  45. HCl(g) + H2O(l) H+(aq) + Cl-(aq) NaOH(s) + H2O(l) Na+(aq) + OH-(aq) Chemistry-140 Lecture 10 • Acid: any substance that , when dissolved in water, increases the concentration of hydrogen ions, H+, in the water • Base: any substance that, when dissolved in water, increases the concentration of hydroxide ions, OH-, in the water Acids & Bases

  46. HCl(g) + H2O(l) H+(aq) + Cl-(aq) CH3COOH(l) + H2O(l) H+(aq) + CH3COO-(aq) Chemistry-140 Lecture 10 • A strong acid or strong base: an acid or base which ionizes completely in water; a strong electrolyte • A weak acid or base: an acid or base which does not ionize completely in water; a weak electrolyte Strong Vs. Weak

  47. HNO3(aq) + KOH(aq) KNO3(aq) + HOH(l) H+(aq) + NO3-(aq) + K+(aq) + OH-(aq) K+(aq) + NO3-(aq) + H2O(l) Net Ionic Equation H+(aq) + OH-(aq) H2O(l) Chemistry-140 Lecture 10 Acid-Base Reactions I • Write the net ionic equation for • the reaction of HNO3 with KOH. Overall Reaction

  48. Net Ionic Equation 2 CH3CO2H(aq) + Ca(OH)2(s) Ca2+(aq) + 2 (CH3CO2)- (aq) + 2 H2O(l) Chemistry-140 Lecture 10 Acid-Base Reactions II • Write the net ionic equation for • the reaction of CH3CO2H with Ca(OH)2. Overall Reaction 2 CH3CO2H(aq) + Ca(OH)2(s) Ca(CH3CO2)2(aq) + 2 HOH(l)

  49. Chemistry-140 Lecture 10 Some Common Acids & Bases Strong Acids Strong Bases HCl, HBr, HI, HNO3, H2SO4, HClO4 NaOH, KOH, Ca(OH)2 Weak Acids Weak Bases CH3CO2H, H3PO4, HF, H2CO3 NH3 H2SO4(l) H+(aq) + HSO4-(aq) Note: HSO4-(aq) H+(aq) + SO42-(aq)

  50. 2 H+(aq) + S2-(aq) H2S(g) H+(aq) + CN-(aq) HCN(g) H2CO3(aq) H2O(l) + CO2(g) H2SO3(aq) H2O(l) + SO2(g) Chemistry-140 Lecture 10 Gas-Forming Reactions • The acids of some nonmetal ions are gases and a small number of aqueous acids easily decompose to form a gaseous product. Examples

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