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CHEM 100 Fall 2013. chapter 4 -1. Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:30.

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Presentation on theme: "CHEM 100 Fall 2013. chapter 4 -1. Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:30."— Presentation transcript:

1 CHEM 100 Fall chapter Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m Tu,Th,F 8: :00 a.m. Or by appointment Test Dates: Chemistry 100(02) Fall 2013 September 30, 2013 (Test 1): Chapter 1 & 2 October 23, 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 chapter 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: 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 chapter Global Warming and the Combustion of Fossil Fuels………………… Reaction Stoichiometry: How Much Carbon Dioxide? Limiting Reactant, Theoretical Yield, and Percent Yield……………… Solution Concentration and Solution Stoichiometry………………… Types of Aqueous Solutions and Solubility…………………………… Precipitation Reactions………………………………………………… Representing Aqueous Reactions: Molecular, Ionic, and Complete Ionic Equations………………………………………………………………… Acid–Base and Gas-Evolution Reactions…………………………… Oxidation–Reduction Reactions…………………………………………. 162 Chapter 4. Chemical Quantities and Aqueous Reactions

4 CHEM 100 Fall chapter Chapter Chemical Quantities and Aqueous Reactions Solution Chemistry Solution Concentration Common types of chemical reactions Types of chemical equations Write molecular equation Write complete ionic equation Write a net ionic equation Solubilities of chemical compounds Precipitation Reactions Solubility Rules Acid–Base and Gas-Evolution Reactions Neutralization reactions Common acids and bases Oxidation–Reduction Reactions Stoichiometry in Solution Reactions

5 CHEM 100 Fall chapter Why Solution Chemistry? Collisions of reactants decides the rates of a reactions Why not gaseous Reactions? In Gases: Very fast Reactions (burning natural gas) Solution reactions are manageable! Liquids: Fast Reactions (precipitation of AgCl) Water is the most common solvent Why not solid Reactions? Solids: Very slow Reactions (making ceramics)

6 CHEM 100 Fall chapter ) Identify the gas phase, solution (liquid) and solid phase reactions. a) O 2 (g) + 2 H 2 (g)  2 H 2 O(l): b) AgNO 3 (aq) + NaCl(aq)  AgCl(s) insoluble salt + NaNO 3 (aq) c) 2C 4 H 10 (g) + 13O 2 (g)  8CO 2 (g) + 10H 2 O(g): d) 2 Al (s) + Fe 2 O 3 (s)  2 Fe(s) + Al 2 O 3 (s):

7 CHEM 100 Fall chapter Sugar in water Oxygen in water Air Dental fillings Saline What is a Solution? A solution:A solution: A homogeneous mixture of two or more components.

8 CHEM 100 Fall chapter Dissolution of (a) Ionic and (b) Molecular Compounds

9 CHEM 100 Fall chapter Components of a Solution Components of a Solution Solute Solvent substance that is present in smallest quantity dissolved substance(s) can be either a gas, a liquid, or a solid one or more present in a solution substance present in largest quantity only one per solution water in aqueous solutions

10 CHEM 100 Fall chapter Concentration Units a)Molarity (M) b)b) Molality (m) c)c) Mole fraction (a)(a) d) Mass percent (% weight) e) Volume percent (% volume) f) "Proof" g) ppm and ppb Proof = (w/w %) x 2

11 CHEM 100 Fall chapter Examples Calculate the molarity of a solution prepared by dissolving g of K 2 SO 4 in enough water to make mL solution. moles of solute Molarity(M) = Liters of solution

12 CHEM 100 Fall chapter Preparing 1.00 L of a 1.00 M NaCl Solution from a Solid Solute

13 CHEM 100 Fall chapter ) Molarity Calculations: Calculate the molarity of a solution prepared by dissolving g of K 2 SO 4 in enough water to make mL solution.

14 CHEM 100 Fall chapter solute = K 2 SO 4 ; F.W. = g/mol; mass= 200g moles of K 2 SO 4 = ? 200 g / g K 2 SO 4 = mol K 2 SO mL = ? Liters of solution = 0.5 L Molarity? mol K 2 SO 4 Molarity of K 2 SO 4 sol. = Liters of solution = 2.30 mole/Liter = 2.30 M (M = moles/liters)

15 CHEM 100 Fall chapter How do you calculate moles of substances in solutions Use concentration of solution to convert L or mL of solution in to moles What is concentration of a solution? The relative amounts of solute and solvent There are so many ways to show amount: g, mole, equivalents, volume

16 CHEM 100 Fall chapter ) Gram-mol-M conversions: How many grams of KNO 3 are contained in 500 mL of a M solution of potassium nitrate?

17 CHEM 100 Fall chapter Dilution Problems Why we dilute solutions? Preparing solutions by adding water to concentrated solutions moles before = moles after M i V i = MfVfMfVf M i = initial molarity V i = initial volume M f = final molarity V f = final molarity

18 CHEM 100 Fall chapter Solution Preparation by Dilution

19 CHEM 100 Fall chapter ) Dilution problems: How many mL of 2.00 M solution of nitric acid, HNO 3 are required with water to make a 250 mL of 1.50 M HNO 3 acid solution?

20 CHEM 100 Fall chapter ) What is concentration of Cl - in 0.4 M MgCl 2 aqueous solution if MgCl 2 a strong electrolyte? Ion Concentrations in Solutions

21 CHEM 100 Fall chapter Types of Chemical Reactions Based on driving force a) Precipitation Reactions (insoluble salt) b) Acid-base Reactions (neutralization) c) Gas-forming Reactions (escaping gas) d) Oxidation-reduction (REDOX)Reactions (electron transfer)

22 CHEM 100 Fall chapter ) Classify following solution reactions as precipitation, acid/base and redox reactions. a)KOH (aq)+ HNO 3 (aq)--> KNO 3 (aq) + H 2 O(l): b) HCl(aq) + NaHCO 3 (aq) --> NaCl(aq) + H 2 O(l) + CO 2 (g): c) Fe(s) + Cu(NO 3 ) 2 (aq) --> Fe(NO 3 ) 2 (aq) + Cu(s) d) AgNO 3 (aq) + NaCl(aq) --> AgCl(s) + NaNO 3 (aq) e) NaOH(aq) + HC 2 H 3 O 2 (aq) --> NaC 2 H 3 O 2 (aq) + H 2 O(l)

23 CHEM 100 Fall chapter Some acids, bases and their salts Acid Sodium salt Name Formula NameFormula Acetic acidHC 2 H 3 O 2 Sodium acetate NaC 2 H 3 O 2 Hydrogen chloride HClSodium chlorideNaCl Nitric acidHNO 3 Sodium nitrateNaNO 3 Phosphoric acidH 3 PO 4 Sodium phosphateNa 3 PO 4 Sulfuric acidH 2 SO 4 Sodium sulfateNa 2 SO 4 BaseChloride salt NameFormulaNameFormula Sodium hydroxide NaOH Sodium chlorideNaCl Barium oxide BaOBarium chlorideBaCl 2 Sodium oxideNa 2 OSodium chlorideNaCl AmmoniaNH 3 Ammonium chlorideNH 4 Cl

24 CHEM 100 Fall chapter ) Identify molecular compounds, acids, bases, and salts among the following: a) AgNO 3 b) NaCl c) C 6 H 12 O 6 d) H 3 PO 4 e) NaOH f) HCl, g) NaNO 3 h ) CH 3 OH i) CH 3 COOH j) H 2 SO 4 k) HC 2 H 3 O 2 l) KNO 3 m) HNO 3 n) MgO o) K 2 O p) SO 3

25 CHEM 100 Fall chapter How do find precursor Acid and base of a Salt Acid (A) + Base(B) = Salt + water (H 2 O) HA + BOH = BA + H2OH2O E.g. LiNO 3 B (Li)A (NO 3 ) OH H BOH(LiOH) HA(HNO 3 )

26 CHEM 100 Fall chapter ) Identify the precursor acid and base for the following salts: a) AgNO 3 b) NaCl c) NaNO 3 d) K 2 SO 4, e) NaC 2 H 3 O 2

27 CHEM 100 Fall chapter Types of Chemical Equations Molecular equation equation: Equation with formula, correct stoichiometric coefficients and physical form written within parenthesis. Ionic equation equation: All the ionic compounds soluble in water are separated into ions written with their ionic charge and (aq). Net Ionic equation equation: Ionic equation with all spectator ions removed from both sides.

28 CHEM 100 Fall chapter Chemical Reaction: NaCl(aq)+AgNO 3 (aq) -->AgCl(s)+ NaNO 3 (aq) Molecular equation: NaCl (aq) + AgNO 3 (aq) --> AgCl (s) + NaNO 3 (aq) Ionic Equation: Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq) --> AgCl(s) + Na + (aq) + NO 3 - (aq) Spectator Ions: Na + (aq) and NO - 3 (aq) Net Ionic Equation: Cl - (aq) + Ag + (aq) --> AgCl (s)

29 CHEM 100 Fall chapter Spectator Ions Ions appearing on both side of an ionic equation. We need to look at the ionic equation: Ionic Equation: Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq) --> AgCl(s) + Na + (aq) + NO 3 - (aq)

30 CHEM 100 Fall chapter Ionic equations When ionic substances dissolve in water, they dissociate into ions. AgNO 3 Ag + + NO 3 - KCl K + + Cl - When a reaction occurs, only some of the ions are actually involved in the reaction. Ag + Ag + + NO K + + Cl Cl - AgCl (s AgCl (s) + K + + NO 3 - H2OH2O H2OH2O

31 CHEM 100 Fall chapter Write molecular equation etc… 9) Write molecular, total ionic, and net ionic reactions for the following solution reactions a) Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) ----> BaSO 4 (s) + 2NaNO 3 (aq) Molecular equation: Total ionic equation: Spectator Ions: Net ionic equation:

32 CHEM 100 Fall chapter ) Write molecular, total ionic, and net ionic reactions for the following solution reactions b) HCl(aq) + NaOH(aq) ----> NaCl(aq) + H 2 O(l) Molecular equation: Total ionic equation: Spectator Ions: Net ionic equation:

33 CHEM 100 Fall chapter ) Write molecular, total ionic, and net ionic reactions for the following solution reactions c) NaOH(aq) + HC 2 H 3 O 2 (l) ----> NaC 2 H 3 O 2 (aq) + H 2 O(l) Molecular equation: Total ionic equation: Spectator Ions: Net ionic equation:

34 CHEM 100 Fall chapter Precipitation reactions They are double displacement reactions of ionic compounds where an insoluble salt is formed when two aqueous salt solutions are mixed. a)Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) ----> BaSO 4 (s) + 2NaNO 3 (aq) a)NaCl(aq)+AgNO 3 (aq) -->AgCl(s)+ NaNO 3 (aq)

35 CHEM 100 Fall chapter

36 CHEM 100 Fall chapter Precipitation of Barium Sulfate Double Displacement: Formation of insoluble salt is the diving force BaCl 2(aq) + Na 2 SO 4(aq)  2NaCl (aq) + BaSO 4(s) precipitat e

37 CHEM 100 Fall chapter Precipitation of Silver Chloride Formation of insoluble salt is the diving force AgNO 3 (aq) + NaCl  AgCl(s) + NaNO 3 (aq) precipitate

38 CHEM 100 Fall chapter Solubility rules for ionic compounds a)All acids are soluble. b)All b)All Na +, Na +, K+ K+ K+ K+ and NH 4 + NH 4 + salts are soluble soluble. c)All nitrate and acetate salts are soluble soluble. d)All d)All chlorides except AgCl and Hg 2 Cl 2 PbCl 2 are soluble soluble. e)All sulfates are soluble except except PbSO 4, Hg 2 SO 4, SrSO 4 and BaSO 4. f)All sulfides are insoluble except except those of the Group IA (1), IIA (2) and ammonium sulfide. g)All hydroxides are insoluble except except those of the group IA(1) and IIA Ba(OH) 2. Sr(OH) 2 and Ca(OH) 2

39 CHEM 100 Fall chapter Illustration of Some Solubility Rules

40 CHEM 100 Fall chapter ) Which of the following salts (ionic compounds) is soluble/insoluble in water? a)NaCl b) Li 2 CO 3 c) AgCl d) PbBr 2 e)NH 4 NO 3 f) Ca(NO 3 ) 2 g) CaSO 4 h) CaCO 3 i) Mg 3 (PO 4 ) 2 j) MnO 2 k) Al(OH) 3 l) BaSO 4 m) CH 3 CO 2 Na n) Cu(OH) 2 o) Fe(ClO 4 ) 2

41 CHEM 100 Fall chapter Precipitation or Not MgI 2 + NaNO 3 = 2 NaI + Mg(NO 3 ) 2 Ba(NO 3 ) 2 +Na 2 SO 4 = BaSO NaNO 3 AgCl +NaNO 3 = AgNO 3 + NaCl

42 CHEM 100 Fall chapter Acid/base Reactions substance that donates H + ions to solution sour-tasting substances substances whose aqueous solutions are capable of turning blue litmus indicators red dissolves certain metals to form salts react with bases or alkalis to form salts substance that donates a OH -1 ion to solution hydroxides and oxides of metals bitter tasting, slippery solutions turn litmus blue react with acids to form salts Acid Base

43 CHEM 100 Fall chapter Neutralization Reactions Formation of water is the diving force acid + base  “salt” + water HCl + NaOH  NaCl + H 2 O H 2 SO 4 + 2KOH  K 2 SO 4 + 2H 2 O Salt Substances produced by the reaction of an acid with a base Characterized by ionic bonds and high melting points Electrical conductivity when melted or when in solution Has a crystalline structure when in the solid state

44 CHEM 100 Fall chapter Ionization of Acids in Water

45 CHEM 100 Fall chapter Titrations

46 CHEM 100 Fall chapter Ionic Equations Strong Acid/base Molecular Equation: HCl(aq) + NaOH(aq) ----> NaCl(aq) + H 2 O(l) Total Ionic Equation: H + + Cl -1 + Na + + OH -1  Na + +Cl -1 + H 2 O Net Ionic Equation: H + + OH -1  H 2 O NaOH(aq) + HC 2 H 3 O 2 (aq) -----> NaC 2 H 3 O 2 (aq) + H 2 O(l) Na + (aq) + OH - (aq) + HC 2 H 3 O 2 (aq) -----> Na + (aq) +C 2 H 3 O 2 - (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH -1  C 2 H 3 O 2 - (aq) + H 2 O

47 CHEM 100 Fall chapter Common Acids and Bases

48 CHEM 100 Fall chapter G as-Forming Exchange Reaction: CO 2, SO 2, H 2 S Escape of a gas is the diving force

49 CHEM 100 Fall chapter Gas-Forming Reactions Metal carbonates + acid  CO 2 (g) + salt + water Na 2 CO 3 (aq) + 2HCl(aq)  H 2 O(l) + CO 2 (g) + 2 NaCl(aq) Net ionic:CO H +  H 2 O(l) + CO 2 (g) CaCO 3 (s) + 2 HCl(aq)  CO 2 (g) + H 2 O(l) + CaCl 2 (aq) Net ionic: CaCO 3 (s) + 2 H +  CO 2 (g) + H 2 O(l) + Ca 2+ Metal (s) + acid  Gas + salt Mg(s) + 2 HCl(aq)  H 2 (g) + MgCl 2 (aq) Net ionic: Mg(s) + 2 H +  H 2 (g) + Mg 2+

50 CHEM 100 Fall chapter Reaction of Metal Carbonates with Acids Molecular Equation: CaCO 3(s) + 2HCl (aq)  CaCl 2(aq) + H 2 CO 3(aq) H 2 CO 3(aq)  H 2 O + CO 2(g) Total Ionic Equation: CaCO 3(s) + 2H + + 2Cl -1  Ca Cl -1 + H 2 O + CO 2(g) Net Ionic Equation: CaCO 3(s) + 2H +  Ca +2 + H 2 O + CO 2(g)

51 CHEM 100 Fall chapter Reaction of Metal Carbonates with Acids Alka-Seltzer NaHCO 3(aq) + HCl (aq)  NaCl (aq) + H 2 O + CO 2(g) Net Ionic Equation: HCO H +  H 2 O + CO 2(g) Tums CaCO 3(s) + 2HCl (aq)  CaCl 2(aq) + H 2 O + CO 2(g) Net Ionic Equation: CO H +  H 2 O + CO 2(g)

52 CHEM 100 Fall chapter Reaction of Metal Sulfites with Acids Molecular Equation: CaSO 3(s) + 2HCl (aq)  CaCl 2(aq) + H 2 SO 3(aq) H 2 SO 3(aq)  H 2 O + SO 2(g) Total Ionic Equation: CaSO 3(s) + 2H + + 2Cl -1  Ca Cl -1 + H 2 O + SO 2(g) Net Ionic Equation: CaSO 3(s) + 2H +  Ca +2 + H 2 O + SO 2(g)

53 CHEM 100 Fall chapter Reaction of Metal Sulfides with Acids Molecular Equation: Na 2 S (aq) + 2HCl (aq)  2NaCl (aq) + H 2 S (g) Total Ionic Equation: 2Na + + S H + + 2Cl -1  2Na + + 2Cl -1 + H 2 S (g) Net Ionic Equation: S H +  H 2 S (g)

54 CHEM 100 Fall chapter Oxidation–Reduction Reactions Electronstransferred Electrons are transferred from one compound to the other resulting in a chemical change. E.g. Zn(s) + 2HCl(aq) ---> ZnCl 2 (aq) + H 2 (g) Ox # 0 +1, -1 2+, includes Single Replacement Reactions Oxidation –loss of electrons Reduction – gain of electrons oxidizing agent – substance that causes oxidation reducing agent – substance that cause reduction

55 CHEM 100 Fall chapter Ionic Equations Weak Acid/base Molecular Equation: NaOH(aq) + HC 2 H 3 O 2 (aq) -----> NaC 2 H 3 O 2 (aq) + H 2 O(l) Total Ionic Equation: Na+(aq) + OH - (aq) + HC 2 H 3 O 2 (aq) ----> Na + (aq) + C 2 H 3 O 2 - (aq) + H 2 O(l) Net Ionic Equation: HC 2 H 3 O 2 (aq) + OH -  C 2 H 3 O 2 - (aq) + H 2 O

56 CHEM 100 Fall chapter Acid/base Ionic Equations Molecular Equation: H 2 SO 4 + 2KOH  K 2 SO 4 + 2H 2 O Total Ionic Equation: 2H + + SO Na + + 2OH -1  2Na + +2Cl H 2 O Net Ionic Equation: 2H + + 2OH -1  2H 2 O

57 CHEM 100 Fall chapter Oxidation-Reduction (Redox) Reactions Electronstransferred Electrons are transferred from one compound to the other resulting in a chemical change. E.g. Zn(s) + 2HCl(aq) ---> ZnCl 2 (aq) + H 2 (g) Ox # 0 +1, -1 2+, includes Single Replacement Reactions Oxidation –loss of electrons Reduction – gain of electrons oxidizing agent – substance that causes oxidation reducing agent – substance that cause reduction

58 CHEM 100 Fall chapter Oxidation-Reduction Reactions

59 CHEM 100 Fall chapter

60 CHEM 100 Fall chapter

61 CHEM 100 Fall chapter Single replacement reactions If various metals are in water, we observe that some are more reactive than others. 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g) (fast) Ca (s) + 2H 2 O (l) Ca(OH) 2 (s) + H 2 (g) (slow) Mg (s) + H 2 O (l) no reaction This indicates that the order of reactivity of these metals towards water is Na > Ca > Mg activity series. We can show the reactivity of metals towards water and acids using an activity series.

62 CHEM 100 Fall chapter Recognizing Redox Reactions

63 CHEM 100 Fall chapter Oxidation number or State A number assigned to a atom in compounds, ions and polyatomic ions to show the number of electrons relative to an atom in the element.

64 CHEM 100 Fall chapter Copper Oxide and Hydrogen Gas

65 CHEM 100 Fall chapter Activation Series of Metals

66 CHEM 100 Fall chapter Rules for Assigning Ox # a) Oxidation number of atoms in an element is zero (0). e.g. O 2 b) Monoatomic ions: Ox # equal to charge. E.g. Na +, Ox # = +1 c) Sum of the oxidation numbers in an element, compound is equal to zero. Sum of the oxidation numbers in an ion, cation or anion is equal to the ionic charge d) As a rule ONs of H =+1, and O=-2 almost most of the time. The group number in the periodic table could be used for main group elements (p and s blocks). d and f block elements show variable ONs E.g. Fe shows either +3 or +2.

67 CHEM 100 Fall chapter Oxidation State What is the oxidation state of Cl in HClO 4 ? H  +1 O  -2 neutral compound, thus sum equals zero 4O  4  -2 = -8 H  1  +1 = +1 0 = +1 + (y) + (-8) y = +7

68 CHEM 100 Fall chapter Oxidation State What is the oxidation state of S in H 2 SO 4 ? H  +1 O  -2 neutral compound, thus sum equals zero 4O  4  -2 = -8 2H  2  +1 = +2 0 = +2 + (x) + (-8) x = +6

69 CHEM 100 Fall chapter Assigning the Oxidation State 11) Assign the oxidation states to each atom in a)NaCl: b)O 2 : c)CBr 4 : d)S 8 : e)MnO 2 : f)KMnO 4 : g)K 2 Cr 2 O 7 :

70 CHEM 100 Fall chapter a) NaCl + AgNO > AgCl + NaNO 3 b) NaOH + HCl ----> NaCl + H 2 O c) Zn + 2HCl ----> ZnCl 2 + H 2 d) 2Cr + 6HCl ----> 2CrCl 3 + 3H 2 e) MnO 2 + 4HBr ----> Br 2 + MnBr 2 + 2H 2 O Which of the following reactions are redox?

71 CHEM 100 Fall chapter Redox Half-Reactions Redox reactions MUST be charge and mass balanced! Balancing redox reactions: –Note the environment that the reaction is occurring in (neutral, acidic, basic). –Split the redox reaction into two separate half-reactions. Determine which species is being oxidized and which is being reduced using oxidation states. »The oxidation half-reaction has electrons as products. »The reduction half-reaction has electrons as reactants. –Mass and charge balance each half-reaction. Mass balancing Charge balancing

72 CHEM 100 Fall chapter Half Redox Reactions two half-reactions in a redox reaction: one where the oxidation is talking place and the other where reduction is taking place. E.g.2 Na + Cl > 2NaCl ON OxidationNa ----> Na + + e - ; Na increase ON, > +1 Reduction Reduction Cl 2 + 2e > 2Cl - ; Cl decrease ON, > -1

73 CHEM 100 Fall chapter Mass and charge balance each half-reaction. Mass balancing Balance your metal then nonmetal elements. Balance your H next. Balance you O last. –Environment concerns, specifically acidic and basic environments Acidic –Balance H as H +. –Balance O as water, H 2 O, Basic –Balance H as H +. –Balance O as OH -. –Counterbalance H + with OH - to make H 2 O. –Charge balancing Use oxidation states to determine number of electrons transferred. »The oxidation half-reaction has electrons as products. »The reduction half-reaction has electrons as reactants. Add the two half-reactions and simplify. Redox Half-Reactions

74 CHEM 100 Fall chapter Separating Half Reactions a) Zn + 2HCl ----> ZnCl 2 + H 2 b) MnO 2 +4HBr----> Br 2 + MnBr 2 + 2H 2 O c) 10K + 2KNO > N 2 + 6K 2 O

75 CHEM 100 Fall chapter Reaction:Cu(s) + AgNO 3 (aq) ---> Ag(s) + Cu(NO 3 ) 2 (aq) Oxidation number:Cu = Ag = N= O = Ag = Cu= N= O= Reducing agent: Oxidizing agent: 12) Identify the reducing and oxidizing agent (reactants) in the following reactions a) b) c Reaction:Fe(s) + Cu(NO 3 ) 2 (aq ) ---> Fe(NO 3 ) 2 (aq) + Cu(s) Oxidation number:Fe = Cu = N= O = Fe= N= O= Cu= Reducing agent: Oxidizing agent: Reaction: Ca(s) + H 2 O(l) ---> Ca(OH) 2 (aq) + H 2 (g) Oxidation number:Ca = H = O = Ca = O= H= H= Reducing agent: Oxidizing agent:

76 CHEM 100 Fall chapter Activation Series of Metals metals higher in series react with compounds of those below metals become less reactive to water top to bottom metals become less able to displace H 2 from acids top to bottom

77 CHEM 100 Fall chapter Activity series of metals - various metals in HCl Iron Zinc Magnesium

78 CHEM 100 Fall chapter Activation Series of Metals Zn (s) + CuSO 4(aq)  ZnSO 4(aq) + Cu (s) Cu (s) + 2AgNO 3(aq)  Cu(NO 3 ) 2(aq) + Ag (s) Fe (s) + 2HCl (aq)  FeCl 2(aq) + H 2(g) Zn (s) + 2HBr (aq)  ZnBr 2(aq) + H 2(g)

79 CHEM 100 Fall chapter Stoichiometric Relationships

80 CHEM 100 Fall chapter Solution Stoichiometry Molarity relates the moles of solute to the liters of solution. – It can be used to convert between amount of reactants and/or products in a chemical reaction.

81 CHEM 100 Fall chapter Stoichiometric calculations of solutions reactions Check whether chemical equation is balanced from volume and “M” of solutionsget the moles from volume and “M” of solutions find the limiting reactant calculate moles of products from the limiting reactant convert moles of the products to grams find the actual yield of the reaction calculate % yield of the reaction

82 CHEM 100 Fall chapter ) How many mLs of M BaCl 2 are required to react completely with 25 mL of M Fe 2 (SO 4 ) 3 ? 3 BaCl 2 (aq) + Fe 2 (SO 4 ) 3 (aq)---> 3 BaSO 4 (s) + 2 Fe Cl 3 (aq) Stoichiometric conversion factors: 3 BaCl 2 = 1 Fe 2 (SO 4 ) 3

83 CHEM 100 Fall chapter Calculate the moles of Fe 2 (SO 4 ) 3 : moles = Molarity x Liters of solution M x L = mole Fe 2 (SO 4 ) 3 Then convert Fe 2 (SO 4 ) 3 to BaCl 2 mole, BaCl 2 moles to liters and liters to mL mole Fe 2 (SO 4 ) > BaCl 2 moles mol Fe 2 (SO 4 ) 3 x 3 = BaCl 2 moles = Molarity x Liters of solution = x Liters BaCl 2 = 0.15L = 150 mL of BaCl 2


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