CHEM 100 Chapter 4 PDF

CHAPTER FOUR: CHEMICAL REACTIONS Part One: Ions in Aqueous Solution 4.1 Ionic Theory of Solutions and Solubility Rules: 1. Suppose you dissolve sodium chloride, NaCl in water. The ions in the solution begin to move, and these moving charges for...

CHAPTER FOUR: CHEMICAL REACTIONS Part One: Ions in Aqueous Solution 4.1 Ionic Theory of Solutions and Solubility Rules: 1. Suppose you dissolve sodium chloride, NaCl in water. The ions in the solution begin to move, and these moving charges form the electric current in the solution. 2. Pure water does not conduct electric current appreciably. It is the ions dissolved in the water that conduct charge. 3. In summary, although water is itself nonconducting, it has the ability to dissolve various substances, some of which go into solution as freely moving ions. An aqueous solution of ions is electrically conducting. 4. Electrolyte = substance that dissolves in water forming ions Chapter 4 Page 1  5.Classification of water-soluble substances. Wat er-soluble Substances Electrolytes Nonelectrolytes -dissociate to some extent -do not dissociate into in water to form ions ions in water -solution conducts electricity -solution is non-conductive Strong Weak Elect rolyt es Elect rolyt es -dissociate to large extent -dissociate to small extent -conduct well -conduct slightly HAc (acetic acid) st rong st rong most soluble acids bases salt HCl, HNO3, NaOH NaCl, CaCl 2 H 2 SO 4 a. Dissociation = process in which a solid ionic compound separates into ions in solution. NaCl ⎯⎯⎯→Na + + Cl − H O 2 b. Ionization = process in which a molecular compound separates to form ions. HCl ⎯⎯ ⎯→H + + Cl − HO 2 c. When one of the ions is H+, the substance is an acid. HCl ⎯⎯ ⎯→H + + Cl − HO 2 Chapter 4 Page 2  d. When OH- are produced, substance is a base. NaOH ⎯⎯ ⎯→Na + + OH − HO 2 e. When neither ion is H+ or OH-, substance is a salt. KBr ⎯⎯⎯→K + + Br − H O 2 Strong and weak acids. 1. Strong acids ionize completely in dilute aqueous solution. Example: HCl is a gas when pure, dissolves to large extent in water, and fully ionizes. HCl(g)⎯⎯⎯→HCl(aq)⎯⎯ H O 2 ⎯→H+ (aq) + Cl − (aq) H O 2 2. Know these: 3. H+ ions exist in aqueous solution attached to H O (i.e., hydrated). 2 H+ (aq) really means H 3O+ H + H O hydronium ion H Chapter 4 Page 3  4. Weak acids ionize only slightly. Example: acetic acid CH3COOH(aq) H+ (aq) + CH COO-(aq) 3 dominant form  99.5%  0.5% Abbreviation: HAc(aq) H+ (aq) + Ac- (aq) 5. Know these: 6. Note that we have introduced the idea of reversible reactions. ( ) Reaction is continually happening to appreciable extent in both directions to produce a balance in which both Reactants and Products are present. Bases = most commonly metal hydroxides. (Mx(OH)y) 1. Strong soluble bases: NaOH(s) ⎯⎯⎯→Na + (aq) + OH − (aq) H O 2 dissociate completely Chapter 4 Page 4  a. Insoluble bases: metal hydroxides which have low solubility in H2O, produce low amount of OH-(aq). Cu(OH)2, Zn(OH)2, Fe(OH)2, Fe(OH)3 b. Weak bases: soluble in H2O but ionize only slightly. Example: NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq) molecular compound Chapter 4 Page 5 Solubility rules for ionic compounds in aqueous solution. Let us summarize the main points in this section. Compounds that dissolve in water are soluble; those that dissolve little ,or not at all, are insoluble. Example 4.1 Using the Solubility Rules, determine whether the following compounds are soluble or insoluble in water: a) Hg2Cl2 b)KI c) lead(II) nitrate Answer: a)According to Rule 3 in Table 4.1,most compounds that contain chloride, Cl, are soluble. However,Hg2Cl2 is listed as one of the exceptions to this rule, so it does not dissolve in water. Therefore, Hg2Cl2 is not soluble in water. b)According to both Rule 1,Group IA compounds are soluble, and Rule 3,most iodides are soluble, KI is expected to be soluble. Therefore, KI is soluble in water. c) According to Rule 2,compounds containing nitrates, NO 3, are soluble. Since there are no exceptions to this rule, lead(II) nitrate, Pb(NO3)2, is soluble in water. Exercise 4.1 Determine whether the following compounds are soluble or insoluble in water. a) NaBr b) Ba(OH)2 c) calcium carbonate Chapter 4 Page 6  4.2 Molecular and Ionic Equations. 1. Three ways of writing equations. (illustrated by neutralization of Mg(OH)2 with HCl in aqueous solution.) a. Molecular equations = show complete formulas for all compounds. (best for stoichiometric calculations) 2HCl(aq) + Mg(OH)2(s) → MgCl2 (aq) + 2H2O(l) (insoluble) (soluble) b. Complete ionic equations = show predominant form of compounds as ions if they strongly dissociate. (the bridge to the net ionic equation) 2[H+(aq) + Cl-(aq)] + Mg(OH)2 (s) → [Mg2+ (aq) + 2Cl-(aq)] + 2H2O(l) c. Net ionic equations = complete ionic equation after “spectator ions” have been canceled out. 2H+(aq) + Mg(OH) (s) → Mg2+ (aq) + 2H O (l) 2 2 [This is the essence of the reaction.] 2. To get the net ionic equation: a. Write down the full molecular equation first. b. For each compound, ask: Is it a strong electrolyte? If not, keep full formula c. Cancel out ions that appear on both sides. These are “spectator ions.” Chapter 4 Page 7  Problem: Write down all 3 types of equations for the reaction of sodium sulfide with silver nitrate to produce silver sulfide and sodium nitrate. a. Molecular equation: Na2S(aq) + 2AgNO3(aq) → Ag2S(s) + 2NaNO3(aq)    all Na salts all nitrates insoluble soluble soluble sulfide b. Complete ionic equation: [2Na+(aq) + S2-(aq)] + 2[Ag+(aq) + NO3-(aq)] → Ag2S(s) + 2 [Na+(aq) + NO3-(aq)] c. Net ionic equation: S2-(aq) + 2Ag+(aq) → Ag2 S(s) (Simple formation of insoluble silver sulfide.) Solution a. According to the solubility rules presented in Table 4.1 and the problem statement,Ca(OH)2 and Ca(ClO4)2 are soluble ionic compounds, so they are strong. electrolytes. The problem statement notes that HClO4 is also a strong electrolyte. You write each strong electrolyte in the form of separate ions. Water,H2O, is a nonelec trolyte (or very weak electrolyte),so you retain its molecular formula. The complete ionic equation is: After canceling spectator ions and dividing by 2,you get the following net ionic equation: Chapter 4 Page 8  b. According to the solubility rules, NaOH and NaC2H3O2 are soluble ionic compounds, so they are strong electrolytes. The problem statement notes that HC 2H3O2 is a weak electrolyte, which you write by its molecular formula. Water,H2O, is a nonelectrolyte,so you retain its molecular formula also. The complete ionic equation is: and the net ionic equation is: Chapter 4 Page 9  Part Two: Types of Chemical Reactions There are millions. We will look at three important types: a) precipitation; b) acid-base; and c) oxidation-reduction. Three driving forces for ionic reaction to happen, all in which there is a removal of ions from solution: a. Formation of insoluble solid precipitate. (e.g. Ag2S) b. Formation of a very stable nonionized molecule. (e.g. H2O) c. Formation of a gas that escapes. (e.g., H2) A. Precipitation Reactions. (Metathesis Reaction I) (Section 4.3) 1. An insoluble solid (precipitate) forms and then settles out of solution. 2. Driving force is strong attraction between cations and anions, resulting in removal of ions from solution. 3. Example: addition of sodium phosphate solution to a calcium chloride solution. What happens? Pictorially: Chapter 4 Page 10  Na+ Ca2+ Na+ Ca2+ + PO 4 3- Cl- PO 4 3- Cl- Na+ Cl- solid Ca 3 (PO 4 ) 2 Ca3(PO4)2 ppts! 2Na3PO4 (aq) + 3CaCl2(aq) → Ca3(PO4)2(s) + 6NaCl(aq) Net ionic equation: 2PO 3- (aq) + 3Ca2+(aq) → Ca (PO ) (s) 4 3 4 2 4. Additional example: Potassium chromate solution added to lead nitrate solution. What happens? K+ Pb2+ soluble insoluble CrO 4 2- NO 3 - K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s) + 2KNO3(aq) Net ionic equation: CrO42-(aq) + Pb2+(aq) → PbCrO 4(s) yellow solid B. Acid/Base Reactions. (Metathesis Reaction II) (Section 4.4) 1. General form of neutralization between acid and base: acid + base → salt + H2O Chapter 4 Page 11  2. Driving force here is formation of H O from H+ and OH-. 2 3. Example - strong acid + strong base: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Total ionic reaction: [H+(aq) + Cl-(aq)] + [Na+(aq) + OH-(aq)] → [Na+(aq) + Cl-(aq)] + H O 2 Net ionic reaction for strong acid + strong base: H+(aq) + OH-(aq) → H 2O(l) 4. Example - weak acid + strong base: HAc(aq) + NaOH(aq) → NaAc(aq) + H2O(l) Total ionic reaction: - - HAc(aq) + [Na+(aq) + OH (aq)] → [Na+(aq) + Ac (aq)] + H O(l) 2 Net ionic reaction for weak acid + strong base: HAc(aq) + OH-(aq) → Ac-(aq) + H 2O(l) 5. Example of Acid-Base reaction with Formation of a Gas (a Driving Force for Reaction) reaction of insoluble carbonate salt with strong acid: 2HCl(aq) + CaCO3(s) → H2CO3(aq) + CaCl2(aq) limestone carbonic acid seashells (weak acid) chalk Chapter 4 Page 12  Net ionic reaction: 2H+(aq) + CaCO (s) → H CO (aq) + Ca2+(aq) 3 2 3 i.e. calcium carbonate dissolves in acid. driving force: further decomposition of H2CO3 to CO2(g) H2CO3(aq) → CO2(g) + H2O(l) escapes (origin of bubbles) C. Oxidation-Reduction (Redox) Reactions. (introduction) (Section 4.5) 1. Many reactions involve transfer of electrons between species, called oxidation- reduction reactions, abbreviated redox reactions. 2. To recognize these, we first need to introduce a concept called the oxidation number or oxidation state. 3. “Oxidation” originally referred to combining with Oxygen, resulting in increase in oxidation number. D. Oxidation Numbers. (keeping track of electrons) 1. Oxidation number (ox#), or oxidation state, of an element: a. In an ionic compound, oxidation number of an element = number of e- gained or lost by an atom of that element. Example: NaCl compound consists of Na+ and Cl- ion pair. ox# of Na = +1 Na+ said to be in “+1 oxidation state” ox# of Cl = -1 CaCO3 ox# of Ca = +2 Chapter 4 Page 13  b. In a molecular compound, ox# is assigned according to rules that aid in balancing redox equations. Meaning is different. Example: CO2 ox# of O is -2 ox# of C is +4 (+4)(-2) CO2 Example: CO ox# of O is -2 ox# of C is +2 (+2)(-2) CO C is said to be less fully oxidized to CO than in CO2. 2. Rules for assigning ox#’s: a. Ox#’s always assigned on a per atom basis. b. Ox#’s of any free uncombined element is zero. Examples: O2, S8, Cu(s) Chapter 4 Page 14  c. Ox#’s of an element in a simple monatomic ion equals the charge of the ion. Na+, Na is +1 d. Sum of ox#’s in a compound = 0. e. Sum of ox#’s in polyatomic ion = charge of ion. CO -2 3 C is +4; O is -2 sum = +4 + 3(-2) = -2 f. H is always +1 except when combined with a metal in a metal hydride: NaH, CaH2. (+1)(-1) NaH g. Group IA are always +1. h. Group IIA are always +2. i. Group IIIA are always +3, except in exotic compounds. j. Group VA are -3 in binary compound with a metal or with hydrogen. (+1)(-3) Na3N -sodium nitride (-3)(+1) NH3 -ammonium Otherwise, most often +3 or +5. k. Oxygen is -2 except in OF2 (+2) and in peroxides. (-1). (+1)(-1) H2O2 -example of exception (+1)(-1) Na2O2 -sodium peroxide l. Halogens are -1 in binary compounds with metals or H. Chapter 4 Page 15  3. Examples of assigning ox# in some compounds. (Hint: 1st do elements having hard and fast rules, then others by elimination.) a. Sodium hypochlorite (Clorox bleach): NaOCl (+1) -do Group IA first always; Na (-2) -do O next; O -Cl has to be what to sum to zero? +1 (+1) (-2)(+1) Na O Cl After bleach does its thing, Cl ends up as NaCl. Cl is -1 there. b. Potassium chromate: K2CrO4 (+1) -do Group IA first always; K (-2) -do O; O -Cr has to be what to sum to zero? +6 (+1) (+6)(-2) K2 Cr O4 c. Sulfuric acid: H2SO4 H is +1, except in hydrides. O is -2, except in peroxides and OF2 S is what to make sum = 0? +6 (+1)(+6)(-2) H2 S O4 d. Sulfurous acid: H2SO3 (+1)(+4)(-2) H2 S O3 Note: S is (+4) here and (+6) in sulfuric acid. Chapter 4 Page 16  4. “Oxidation” originally referred to combining with Oxygen, resulting in increase in ox#. Example - Combustion of methane is an oxidation of the Carbon: (-4)(+1) (0) (+4)(-2) CH4 + 2O2 → CO2 + 2H2O (-4) (+4) {C} → {C} oxidation of C 5. Oxidation = algebraic increase in ox# with corresp. loss of e-. 6. Reduction = algebraic decrease in ox# with corresp. gain of e-. 7. Oxidation and reduction always occur simultaneously. (0) (0) (+1)(-2) 2H2 + O2 → 2 H2O Oxygen (0) → (-2) reduced Hydrogen (0) → (+1) oxidized 8. Oxidizing agent: a. Oxidizes other species. b. Is itself reduced. 9. Example: Immerse copper wire in silver nitrate solution. Copper is dissolved as the Cupric ion and silver is deposited as metal. AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s) Net ionic equation: Ag+(aq) + Cu(s) → Cu2+(aq) + Ag(s) What species is oxidized? Cu (0) → (+2) What species is reduced? Ag (+1) → (0) What is the oxidizing agent? AgNO or Ag+(aq) 3 Chapter 4 Page 17 E. Many Oxidation-Reduction Reactions also fit under other categories, such as: 1. Combination reactions. 2. Decomposition reactions. 3. Displacement reactions. 4. Combustion reactions. F. Combination Reactions. 1. Combination reaction = reaction in which 2 or more substances combine to form a third substance. 2. Element + Element → Compound a. 2 Mg(s) + O2 → 2 MgO (metal + nonmetal → ionic cmpd) b. P4(s) + 6 Cl2(g) → 4 PCl3(s) (nonmetal+nonmetal → molec cmpd) 3. Compound + Element → Compound PCl3(s) + Cl2(g) → PCl5(s) 4. Compound + Compound → Compound CaO(s) + CO2(g) → CaCO3(s) (this one is NOT redox) G. Decomposition Reactions. 1. Example: 2 KClO3 → 2 KCl + 3O2 Chapter 4 Page 18 H. Displacement Reactions. 1. Reaction in which one element displaces another from a compound. M + AX → MX + A 2. The more readily a metal M displaces another element and forms a positive ion, the more active that metal is. 3. The metal doing the displacing is being oxidized. 4. Table 4.6 is Activity Series: Chapter 4 Page 19  5. [More Active Metal + → [Less Active Metal + Salt of Less Active Metal] Salt of More Active Metal] Example: Zinc metal is placed in copper(II) sulfate solution. Copper metal falls to bottom. Resulting solution contains zinc sulfate, ZnSO4. Formula unit equation: CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) Net ionic equation: Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+(aq) Displacement reaction: the more active metal, zinc, displaces ions of less active metal, copper, from aqueous solution. 6. [Active Metal + Nonoxidizing Acid] → [Hydrogen + Salt of Acid] a. Example: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) strong acid soluble salt Net ionic equation: Zn(s) + H+(aq) + HSO -(aq) → Zn2+(aq) + SO 2-(aq) + H (g) 4 4 2 b. HNO3 is the common oxidizing acid. It reacts with active metals to produce oxides of nitrogen, but not hydrogen, H2. c. Very active metals can even displace hydrogen from water. Example: 2K(s) + 2H O(l) → 2[K+(aq) + OH-(aq)] + H (g) 2 2 7. [Active Nonmetal + → [Less Active Nonmetal + Salt of Less Active Nonmetal] Salt of More Active Nonmetal] a. Example: + - + - Cl2(g) + 2[Na (aq) + Br (aq)] → 2[Na (aq) + Cl (aq)] + Br2(l) chlorine sodium bromide sodium chloride bromine Chapter 4 Page 20  b. Each halogen will displace less active (heavier) halogens from their binary salts; i.e., the order of increasing activities is: (least active) I2 < Br2 < Cl2 < F2 (most active) I. Combustion Reactions. 1. Combustion reaction = a reaction in which a substance reacts with oxygen, sometimes with rapid release of heat to product a flame. 2. Oxygen goes from 0 to -2 oxidation state. 3. Of the combusted material, all Carbon ends up as CO or CO2, all Hydrogen ends up in H2O. 4. Example: Combustion of propane: C3H8(g) + O2(g) → J. Balancing Simple Redox Reactions. 1. The following redox reaction looks balanced, but what is wrong? Zn(s) + Ag+(aq) → Zn2+(aq) + Ag(s) The charge is not balanced. 2. Balance using the Half-Reaction method: (separate into two hypothetical half-reactions, one an oxidation, the other a reduction, by specifically putting electrons in the equations. Zn(s) → Zn2+(aq) + 2 e- oxidation 2[Ag (aq) + 1 e- → Ag(s) ] + reduction Zn(s) + 2Ag (aq) → Zn (aq) + 2Ag(s) + 2+ 3. More complicated cases are handled in Chapter 19, in connection with electrochemical cells. Chapter 4 Page 21  Part Three: Working with Solutions A. Molar Concentrations of Solution. (Section 4.7) 1. Many reactions carried out in solution for convenience. 2. Solution = homogeneous mixture of 2 or more substances. 3. Solvent = the dispersing medium. (substance usually present in greatest abundance) 4. Solute = substance dissolved in the solvent. (usually present in least amount) 5. Aqueous solutions most common. (water is solvent) 6. Concentrations expressed in various ways: a. Mass Percent of solute: mass of solute % solute = 100% mass of solution b. Molarity (M) or molar concentration: number of moles of solute molarity = number of liters of solution B. Mass Percent of Solute. 1. Problem involving percent solute: A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 g of solution. 5.00 g NaCl 5.00% means → 100 g solution 5.00 g NaCl  400g solution = 20.0g NaCl present 100 g solution (% solute)  mass solution = mass solute 100 Chapter 4 Page 22  2. Variation involving density: A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 mL of solution. The density of the solution is 1.10 g/mL. 400 mL solution x 1.10 g/mL = 440 g solution Then work like previous example: 5.00 g NaCl  440g solution = 22.0g NaCl present 100 g solution C. Molarity, or Molar Concentration (M). 1. Definition: moles of solute molarity = Liters of solution 2. Preparation of solutions of desired concentration: What mass of NaOH would you weigh out to prepare 2.00 L of 0.100 M NaOH? 0.100 M reads 0.100 molar meaning 0.100 moles NaOH per Liter of solution. 0.1 moles NaOH  2.00 L solution = 0.200 moles NaOH 1 Liter of solution 0.200 moles NaOH  40.0 gNaOH mol NaOH = 8.0 g NaOH Figure 4.18 Chapter 4 Page 23  3. Important relations: n moles solute M= molarity = V Volume of solution so: n = MV or: (moles of solute present) = (conc in molarity) x (Volume in Liters) 4. Problem: What mass of HCl is present in a 2.5 L jug of concentrated HCl solution (11.6 M)? a. First compute moles HCl present: moles = conc x volume = 11.6 mol/L x 2.5 L = 29.0 mol b. Now convert to grams HCl: 29.0 mol x 36.45 g HCl/mol = 1,057 g HCl Chapter 4 Page 24  5. Problem: Calculate the molarity of a solution in which 10.0 g NaOH are dissolved in 4.000 L solution.  10.0 g    = moles NaOH 40.0 g/mol  molarity = Vol solution (L) 4.000 L = 0.0625 mol/L = 0.0625 M D. Dilution of Solutions. (Section 4.8) 1. Dilution is simply the addition of solvent to an existing solution. Moles of solute remains fixed. 2. Therefore: moles solute before = moles solute after n1 = n2 M1 V1= M2 V2 molarity x volume = molarity x volume before before after after Chapter 4 Page 25  3. Since V is on both sides, can use V in any units as long as consistent. 4. Dilution problem: If 10.0 mL of 3.0 M NaCl is diluted to 250.0 mL, what is the new molarity of NaCl solution? V1M1 = V2M2 (10.0 mL) x (3.0 M) = (250.0 mL) x M2 10.0 mL  3.0 M M2 = = 0.12 M 250 mL E. Volumetric Analysis (Section 4.10) (Solution Stoichiometry) 1. Carrying out reactions in solution: Problem: What volume of 0.100 M HCl(aq) would exactly neutralize 400 mL of 0.200 M NaOH(aq)? HCl+ NaOH→NaCl +H2O moles HCl = moles NaOH (at neutralization) nHCl = nNaOH MHCl VHCl = MNaOH VNaOH 0.100M VHCl = 0.200M  0.400 L 0.200 M  0.400 L VHCl = = 0.800 L = 800 mL 0.100 M Chapter 4 Page 26 F. Method of Titration: Volumetric analysis is a method of analysis based on titration. F.Titration = process in which a solution of one reactant (the titrant) is carefully added to a solution of another reactant. Volume of titrant required for complete reaction is measured. 2. Complete reaction is determined by a color change in the indicator. a. Example: the indicator phenolphthalein is colorless in acidic solution and pink in basic solution. (An indicator is a substance that undergoes a color change when a reaction approaches completion.) b. The indicator is placed in acidic reactant solution. The base is then added until the first drop produces a pink color that persists. c. When stoichiometric amounts have reacted, we call this the end point or equivalence point of the titration. d. Example: reacting H2SO4 with NaOH, the equivalence point is when: H2SO4+ 2NaOH→ Na2SO4 + 2H2O n NaOH = 2 x n H2SO4 e. Example: reacting HCl with NaOH, equivalence point is when: HCl+ NaOH→NaCl +H2O nHCl=nNaOH Chapter 4 Page 27  3. Problem: 20.0 mL of HCl solution are required to neutralize 10.0 g of NaOH. What is the molarity of the HCl solution? HCl+ NaOH→NaCl +H2O nHCl=nNaOH=10.0g/40g/mol=0.25 mol CHCl = nHCl / vHCl= 0.25 mol/0.02 L=12.5M 4. In the neutralization of phosphoric acid with sodium hydroxide, what volume of 0.200 M NaOH is required to neutralize 400 mL of 0.50 M H3PO4? H3PO4+ 3NaOH→ Na3PO4 + 3H2O n NaOH = 3 x n H3PO4 3 × CH3PO4 × VH3PO4=CNaOH ×VNaOH 3 × 0.5 M × 0.4 L= 0.2M ×VNaOH VNaOH = 3.00L Chapter 4 Page 28  5. Problem: 50.0 mL of H2SO4 of unknown concentration is titrated with a 0.100 M NaOH titrant solution. The end point is reached after 40.0 mL of titrant are added. Calculate molarity of H2SO4 solution. H2SO4 + 2 NaOH → Na2SO4 + 2H2O end point when: (moles of NaOH) = 2 x (moles of H2SO4) VbMb = 2 x (VaMa) 40.0 mL x 0.100 M = 2 x (50.0 mL x Ma) Ma = 0.0400 M H2SO4 6. In the neutralization of 2.00 g of MgO with HCl, 25.0 mL of an HCl solution are required. What is the molarity of the HCl solution? 2 HCl + MgO → MgCl2 + H2O Chapter 4 Page 29  Balancing Redox Reactions The Half-Reaction Method  Balancing Redox Reactions The Half-Reaction Method  Balancing Redox Reactions The Half-Reaction Method Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions Balancing Redox Reactions 1 Balancing Redox Reactions Balancing Redox Reactions