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Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 4 Chemical quantities and Aqueous reactions Chapt 4 Chemical reactions 4.2- Reaction Stoichiometry Stoichiometry is the study of the numerical relationship between chemical quantities (reactants and products) in a chemical reaction is based on the balanced chemical equation and on the relationship between mass and moles Such calculations are fundamental to most quantitative work in chemistry The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction Chapt 4 Chemical reactions Reaction Stoichiometry Interpretation of the balanced chemical equation; 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 2 molecules C8H18 + 25 molecules O2 2 moles C8H18 + 25 moles O2 16 molecules CO2 + 18 molecules of H2O 16 moles CO2 +18 moles H2O Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation Chapt 4 Chemical reactions 1 Stoichiometry: Mole –to-Mole conversion The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18? 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 2 moles C8H18 : 16 moles CO2 Chapt 4 Chemical reactions Mass Relationships in Chemical Equations(Mass –to – Mass conversion ) Estimate the mass of CO2 produced in 2006 by the combustion of 3.5 × 1015 g petroleum(octane C8H18). Solution: The equation for the reaction is: 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) g C8H18 mol C8H18 mol CO2 g CO2 Given: 3.4 × 1015 g C8H18 Find: g CO2 Chapt 4 Chemical reactions Example cont’d Molar mass: Mole ratio : 1mol C8H18 = 114.22g 1 mol CO2 = 44.01 g 2 mol C8H18:16 mol CO2 Chapt 4 Chemical reactions 2 4.3 –Limiting reactant, Theoretical and percent Yield Limiting reactant or reagent(L.R)- The reactant that makes the least amount of the product and is completely consumed in the reaction that limits the amount of the product in a chemical reaction. Excess Reactant- Any reactant that occurs in a quantity greater than is required to completely react with L.R. Theoretical Yield-The amount of product that can be made in a chemical reaction based on the amount of L.R. Actual Yield-The amount of product actually produced by a reaction. Chapt 4 Chemical reactions Limiting reactant, Theoretical and percent Yield The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). %Yield actual yield 100% theoretical yield Chapt 4 Chemical reactions Limiting reactant, Theoretical and percent Yield: Making Pizzas •If we had 4 crusts,15 oz. tomato sauce, and 10 cu cheese how many pizzas can we make? We know that: 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza tomato sauce- limiting reactant theoretical yield (3 pizzas) Chapt 4 Chemical reactions 3 Limiting and Excess Reactants in the Combustion of Methane If we have 5 molecules of CH4 and 8 molecules of O2, how many molecules of CO2 can be produced? which is the limiting reactant? CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) H H H H C C H H H H + H H C H H H H H H C C H H H H O O O O O O O O O O O O O O O O Chapt 4 ?co2 Chemical reactions Finding Limiting Reactant, Theoretical Yield, and Percent Yield How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? Identify the limiting reactant and determine the theoretical yield of N2(g). 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are produced, what is the percent yield? Chapt 4 Chemical reactions Solutions Mostly chemical reaction occur in Solutions Solutions are homogeneous mixtures Solute + solvent Solute: The component of the solution in small quantity Solvent: The component that is in large quantity the composition can vary from one sample to another Chapt 4 Chemical reactions 4 Solution concentration and solution stoichiometry To describe solutions accurately, we must describe how much of each component is present Qualitative description Dilute solutions have a small amount of solute compared to solvent Concentrated solutions have a large amount of solute compared to solvent Chapt 4 Chemical reactions Solution Concentration: Molarity Concentration - Amount of solute in a given amount of solution. Molarity (M)- is the amount of solute in moles divided by volume of solution (Liters) molar concentration (mol L-1) Chapt 4 Chemical reactions Preparing 1 L of a 1.00 M NaCl Solution Chapt 4 Chemical reactions 5 Example:Calculating molarity What is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution? g NH3 mol NH3 mL sol’n L sol’n Chapt 4 M Chemical reactions Using Molarity in Calculations Molarity can be used as conversion factor between the moles of solute and liters of solution. If a sugar solution concentration is 2.0 M, 1 L solution : 2 moles sugar Chapt 4 Chemical reactions Example 4.6 a) How many liters of 0.125 M NaOH contains 0.255 mol NaOH? b) Determine the mass of CaCl2 (MM = 110.98) in 1.75 L of 1.50 M solution. Chapt 4 Chemical reactions 6 Dilution Stock solutions (concentrated) To make solutions of lower concentrations from these stock solutions, more solvent is added The amount of solute doesn’t change, just the volume of solution moles solute in solution 1 = moles solute in solution 2 Dilution equation M 1 ·V1 = M2 ·V2 The concentrations and volumes of the stock and new solutions are inversely proportional. Chapt 4 Chemical reactions Preparing 3.00 L of 0.500 M CaCl2 from a 10.0 M stock solution(dilution) Chapt 4 Chemical reactions Sample Problems Diluting a solution How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? M1V1 = M2V2 Chapt 4 Chemical reactions 7 Solution Stoichiometry Since 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. Chapt 4 Chemical reactions What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3 ) 2 in the reaction 2 KCl( aq) + Pb(NO3 ) 2 ( aq) PbCl2 ( s) + 2 KNO3 ( aq) 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl Chapt 4 Chemical reactions Quantitative Analysis:Volumetric Analysis Volumetric analysis is a method of analysis based on titration Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete End point- The point of completion of the reaction indicated by the colour change of a chemical called indicator Chapt 4 Chemical reactions 8 Volumetric Analysis Titration of an unknown amount of HCl with NaOH HCl (aq) End point : colorless to pink Chapt 4 Chemical reactions Volumetric Analysis Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH: H 2SO4 (aq) 2NaOH(aq) 2 H 2O ( l ) Na2SO4 (aq) Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many milliliters of 0.250 M NaOH must be added to completely react with the sulfuric acid? Solution: Convert 0.0350L of H2SO4(35.0 mL) to moles of H2SO4 (using the molarity of the acid). Then, convert to moles of NaOH (from the balanced chemical equation) Finally, convert to volume of NaOH solution (using the molarity of NaOH Chapt 4 Chemical reactions Finally, convert to volume of NaOH solution (using the molarity of NaOH ( 0.0350 L ) 0.175 mole H 2SO 4 2 mol NaOH 1 L NaOHsoln. 1 L H 2SO 4 solution 1 mol H2SO4 0.250mol NaOH 0 . 0490 L NaOH solution ( 49.0 mL of NaOH solution) Chapt 4 Chemical reactions 9 Practice problem To neutralize 37.6 mL of Ba(OH)2 solution, 43.8 mL of 0.107 M HCl is needed. What is the molarity of the base. 2 HCl + Ba(OH)2 Chapt 4 BaCl2 + 2 H2O Chemical reactions What happens when a solute dissolves? There are attractive forces between the solute particles holding them together attractive forces between the solvent molecules on mixing the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules If the attractions between solute and solvent are strong enough, the solute will dissolve Chapt 4 Chemical reactions Electrolytes and Nonelectrolytes Electrolytes- Materials that dissolve in water to form a solution that will conduct electricity Nonelectrolytes- Materials that dissolve in water to form a solution that will not conduct electricity Chapt 4 Chemical reactions 10 Electrolyte and non electrolyte solution Ionic compounds dissociate into ions when they dissolve. Chapt 4 Molecular compounds do not dissociate when they dissolve. Chemical reactions Strong and Weak Electrolytes Strong electrolytes are materials that dissolve completely as ions and their solutions are good conductor of electricity ionic compounds and strong acids ionize virtually 100% HCl(aq) H+ (aq) + Cl-(aq) Weak electrolytes are materials that dissolve mostly as molecules (only partially ionize) This situation is denoted by writing the equation with a double arrow HC2H3O2(aq) H+ (aq) + C2H3O2-(aq) Most soluble molecular compounds are either nonelectrolytes or weak electrolytes their solutions are poor conductor of electricity Chapt 4 Chemical reactions Practice Problem Name and classify each of the following compounds as strong electrolyte,weak electrolyte or nonelectrolyte. Write down the dissociation equation where appropriate. BaCl2 C6H6 HBr HNO3 (NH4) 2CO3 Chapt 4 Chemical reactions 11 Solubility of Ionic Compounds Solubility-The amount of a substance dissolved in a given amount of solvent at a given temperature defined as greater than 0.1g sloute/100g of the solvent If the solubility is low- precipitate will form The degree of solubility depends on the temperature Even insoluble compounds dissolve, just not enough to be meaningful Solubility rules Chapt 4 Chemical reactions Chapt 4 Chemical reactions Practice Problem Determine if each of the following is soluble in water KOH AgBr CaCl2 Pb(NO3) 2 PbSO4 Chapt 4 Chemical reactions 12 Types of the chemical reaction Most of the reactions we will study fall into one of the following categories: Precipitation Reactions Formation of an insoluble ionic substance (precipitate) Acid-Base Reactions Reaction between an acid and base Oxidation-Reduction Reactions Involve transfer of electrons Chapt 4 Chemical reactions Precipitation reactions Precipitation reactions- Reactions between aqueous solutions of ionic compounds that produce an insoluble ionic compound in water the insoluble product is called a precipitate For example, reaction between potassium iodide and lead(II) nitrate solution 2 KI(aq) + Pb(NO3) 2(aq) PbI2(s) + 2 KNO3(aq) Chapt 4 Chemical reactions Chapt 4 Chemical reactions 13 Process for Predicting the Products of a Precipitation Reaction 1. 2. 3. 4. Determine what ions each aqueous reactant has Determine formulas of possible products Exchange ions (+) ion from one reactant with (–) ion from other Balance charges of combined ions to get formula of each product Determine solubility of each product in water Use the solubility rules If product is insoluble or slightly soluble, it will precipitate If neither product will precipitate, write no reaction after the arrow. Chapt 4 Chemical reactions Process for Predicting the Products of a Precipitation Reaction 5. 6. If either product is insoluble, write the formulas for the products after the arrow—writing (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate Balance the equation. Chapt 4 Chemical reactions Practice Problem Predict the products and balance the equation. KCl(aq) + AgNO3(aq) Na2S(aq) + CaCl2(aq) Write an equation for the reaction that takes place when an aqueous solution of (NH4) 2SO4 is mixed with an aqueous solution of Pb(C2H3O2) 2 (NH4) 2SO4(aq) + Pb(C2H3O2) 2(aq) Chapt 4 Chemical reactions 14 Quantitative analysis: Gravimetric analysis Gravimetric Analysis- is a type of quantitative analysis in which the amount of a species in a material is determined by converting the species into a product that can be isolated and weighed. Precipitation reactions are often used in gravimetric analysis Chapt 4 Chemical reactions Gravimetric analysis Consider the problem of determining the amount of lead in a sample of drinking water. Adding sodium sulfate (Na2SO4) to the sample will precipitate lead(II) sulfate. Na2 SO4 (aq) Pb2 (aq) 2 Na (aq) PbSO4 ( s) The PbSO4 can then be filtered, dried, and weighed Chapt 4 Chemical reactions Gravimetric analysis Suppose a 1.00 L sample of polluted water was analyzed for lead(II) ion, Pb2+ , by adding an excess of sodium sulfate to it. The mass of lead(II) sulfate that precipitated was 229.8 mg. What is the mass of lead in a liter of the water? Express the answer as mg of lead per liter of solution. Na 2SO 4 ( aq ) Pb 2 ( aq ) 2 Na ( aq ) PbSO 4 ( s ) First we must obtain the mass percentage of lead in lead(II) sulfate, by dividing the molar mass of lead by the molar mass of PbSO4, then multiplying by 100. % Pb 207.2 g/mol 303.3 g/mol 100 68 .32 % Use the % of Pb to calculate the mass of lead in 229.8 mg of lead(II) sulfate Chapt 4 Chemical reactions 15 4.7 Representing aqueous reactions Molecular equation: in which reactants and products are written as molecules, even though they may actually exist in solution as ions 2 KOH(aq) + Mg(NO3) 2(aq) 2 KNO3(aq) + Mg(OH)2(s) (strong) (strong) (strong) (insoluble) Complete ionic equation: Equation that describe the actual dissolved species in solution(ions) Aqueous strong electrolytes are written as ions Insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form 2 K+ (aq) + 2 OH-(aq) + Mg2+(aq) + 2 NO3-(aq) Chapt 4 2 K+ (aq) + 2 NO3-(aq) + Mg(OH)2(s) Chemical reactions Net Ionic equation Spectator ions- Ions that appear on both side of the equation and they can be cancelled e.g. K+ and NO3 2 K+ (aq) + 2 OH-(aq) + Mg2+(aq) + 2 NO3-(aq) Mg(OH)2(s) 2 K+ (aq) + 2 NO3-(aq) + Net Ionic equation- An equation showing only the species that actually change(react) during the reaction Eliminating spectator ions 2 OH-(aq) + Mg2+(aq) Chapt 4 Mg(OH)2(s) Chemical reactions Examples 1. Write a balanced equation for the reaction that takes place when an aqueous solution of NiCl2 is mixed with an aqueous solution of Na3PO4. Molecular equation (to be balanced) NiCl 2 Na 3 PO 4 Ni 3 ( PO 4 ) 2 NaCl Ionic equation Net Ionic equation Chapt 4 Chemical reactions 16 4.8- Acid–Base Reactions neutralization reactions Acid-Base reactions also called neutralization reactions because the acid and base neutralize each other’s properties Arrhenius Acids ionize in water to form H+ ions. More precisely, the H+ from the acid molecule is donated to a H2O molecule to form hydronium ion, H3 O+ . Most chemists use H+ and H3O+ interchangeably. H ( aq ) H 2 O ( l ) Chapt 4 H 3 O ( aq ) Chemical reactions Polyprotic Acids Polyprotic Acids-contain more than one ionizable proton and release them equentially. e.g. sulfuric acid(H2SO4)or phosphoric acid Dissociation of H3 PO4 (phosphoric acid) All three dissociations are incomplete H3PO4 (aq) + H2O (l) H2PO4- (aq) + H3O+ (aq) H2PO4 (aq) + H2O (l) HPO42- (aq) + H3O+ (aq) HPO42- (aq) + H2O (l) PO43- (aq) + H3O+ (aq) Chapt 4 Acids are found in lemons, limes, and vinegar. Vitamin C and aspirin are also acids Chemical reactions Many common household products, such as bleach and ammonia, are bases. Chapt 4 Chemical reactions 17 Acid-Base reactions Arrhenius Bases- dissociate in water to form OH ions. NaOH (s ) H 2O Na ( aq ) OH (aq ) Acid-Base neutralization reactions- H+ ion from acid combine with OH- from the base to form water Net ionic equation: H+ (aq) + OH (aq) Chapt 4 H2O(l) Chemical reactions Common Acids and bases Chapt 4 Chemical reactions Bronsted- Lowery Concept The Brønsted-Lowry acid is defined as the species (molecule or ion) that donates a proton(H+ ) to another species in a proton-transfer reaction HNO 3 ( aq ) acid H 2O ( l ) NO 3 ( aq ) H 3 O ( aq ) base base acid A Brønsted-Lowry base is defined as the species (molecule or ion) that accepts the proton (H+ ) in a proton-transfer reaction NH 3 (aq ) H 2O( l ) base NH 4 (aq ) OH (aq ) acid Chapt 4 Chemical reactions 18 Summary of Acid-Base concepts In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution. The Arrhenius Concept: acid: proton (H+ ) donor base: hydroxide ion (OH-) donor The Brønsted-Lowry Concept: acid: proton (H+ ) donor base: proton (H+ ) acceptor Chapt 4 Chemical reactions Practice Example 1. Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide Chapt 4 Chemical reactions Acid-Base Neutralization reaction I. Neutralization reaction of the strong acid, HCl(aq) and a strong base, KOH(aq). Molecular equation. HCl ( aq ) KOH ( aq ) KCl ( aq ) H 2 O ( l ) Net ionic equation: H+ (aq) + OH (aq) H2O(l) II. Neutralization Reaction of Weak acid and strong base In a reaction involving HCN(aq), a weak acid, and KOH(aq), a strong base, the product is KCN, a strong electrolyte. The net ionic equation HCN ( aq ) OH ( aq ) Chapt 4 CN ( aq ) H 2O ( l ) Chemical reactions 19 Gas-Evolution reactions Gas-Evolution reactions -Aqueous reactions that form a gas when two solutions are mixed K2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g) Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq) H2SO3 H2O(l) + SO2(g) Chapt 4 Chemical reactions Chapt 4 Chemical reactions Gas-Evolution reactions Carbonates react with acids to form CO2, carbon dioxide gas Sulfites react with acids to form SO2, sulfur dioxide gas Sulfides react with acids to form H2S, hydrogen sulfide gas Chapt 4 Chemical reactions 20 Example 4.14 Write a molecular equation for the reaction that occurs when an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves Write the formulas of the reactants and possible products Na2CO3(aq) + HNO3(aq) H2CO3(l) + NaNO3(aq) Check to see if either product decomposes Na2CO3(aq) + 2 HNO3(aq) Chapt 4 2 NaNO3(aq) + CO2(g) + H2O(l) Chemical reactions Oxidation-Reduction reactions(Redox) Oxidation–reduction reactions in which electrons are transferred from one reactant to another many involve the reaction of a substance with O2(g) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) Oxidation is defined as the loss of electrons OIL (oxidation is loss of electrons ) Reduction is defined as the gain of electrons. RIG (Reduction Is Gain of electrons) Ger Leo Chapt 4 Chemical reactions Oxidation-Reduction Reaction Oxidation and reduction always occur simultaneously. The # of e- lost by the substance being oxidized = the # of e - gained by the substance being reduced Atoms that lose electrons are being oxidized Atoms that gain electrons are being reduced 2 Na(s) + Cl2(g) 2 Na+Cl–(s) Na Na+ + 1 e– oxidation Cl2 + 2 e– 2 Cl– reduction Na is oxidized and Cl2 is reduced Chapt 4 Chemical reactions 21 Combustion as Redox 2 H2 ( g) + O2 ( g) 2 H 2 O(g) Chapt 4 Chemical reactions Oxidation numbers or state Oxidation number is a simple way of keeping track of electrons in a reaction The oxidation number (or oxidation state) of an atom in a substance is the actual charge of the atom if it exists as a monatomic ion oxidation states are not ion charges! Oxidation states are imaginary charges assigned based on a set of rules. Ion charges are real, measurable charges Rules for assigning oxidation states(page 148) Chapt 4 Chemical reactions Rules for assigning oxidation states Rule Applies to Statement 1 Elements The oxidation number of an atom in an element is zero. 2 Monatomic ions The oxidation number of an atom in a monatomic ion equals the charge of the ion. 3 Oxygen The oxidation number of oxygen is –2 in most of its compounds. (An exception is O in H2O2 and other peroxides, where the oxidation number is –1.) 4 Hydrogen The oxidation number of hydrogen is +1 in most of its compounds except in metal hydrides. 5 Halogens Fluorine is –1 in all of its compounds. The other halogens are –1 unless the other element is another halogen or oxygen. 6 Compounds and ions The sum of the oxidation numbers of the atoms in a compound is zero. The sum in a polyatomic ion equals the charge on the ion. Chapt 4 Chemical reactions 22 Sample problems Oxidation no. of Cr in Cr2O7 2- (dichromate ions) ? 2(Cr) + 7(-2) = -2 2Cr = -2 -7(-2) 2Cr = -2 +14 = +12 and Cr = +6 Oxidation no. of Cl in ClO4- (perchlorate ion) Cl + 4(-2) = -1 Cl = -1-4 (-2) = -1 + 8 = + 7 Oxidation no. of N in NH4+ ion? N + 4(+1) = +1 N = +1 - 4(+1) = -3 1. Assign Oxidation numbers to all atoms in the following: CO2 , K2Cr2O7 ,PCl5, HNO2 Chapt 4 Chemical reactions Identifying Redox Reactions Oxidation occurs when an atom’s oxidation state increases during a reaction Reduction occurs when an atom’s oxidation state decreases during a reaction. CH4 + 2 O2 –4 +1 0 CO2 + 2 H2O +4 –2 +1 –2 oxidation reduction Chapt 4 Chemical reactions Redox reaction An oxidizing agent is a species that oxidizes another species; it is itself reduced. A reducing agent is a species that reduces another species; it is itself oxidized. Loss of 2 e- oxidation reducing agent Fe(s ) Cu 2 (aq ) Fe 2 (aq ) Cu(s ) oxidizing agent Gain of 2 e- reduction Chapt 4 Chemical reactions 23 Redox reaction Activity Series - ranks the elements (neutral atom) in order of their reducing ability of cation (Mn+ ) in aqueous solution 1. Any element higher in the activity series will react with the ion of any element(in sol’n) lower in the activity series 2. Position of Hydrogen- indicates which metal will react with aq. acid H+ (aq) to produce H2 gas 3. Most reactive metals are at the top of the activity series (strong reducing agent) 4. Least reactive metal - bottom of the activity series ( weak reducing agent) Chapt 4 Chemical reactions Activity Series Zn0 (s) + 2Ag+ (aq) 2Ag(s) + Zn2+(aq) Chapt 4 Chemical reactions Practice Problem Assign oxidation states, determine the elements oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reaction. Sn4+ + Ca F2 + S Sn2+ + Ca2+ SF4 Chapt 4 Chemical reactions 24 Balancing Simple OxidationReduction Reactions Oxidation half – one-half of the reaction in which oxidation occurs Reduction half - one-half of the reaction in which reduction occurs a balanced equation must have a charge balance as well as a mass balance. 2 E.g. Zn ( s ) Ag ( aq ) Zn ( aq ) Ag ( s ) The number of electrons lost in the oxidation half-reaction must equal the number gained in the reduction half-reaction, Ag ( aq ) Zn ( s ) Zn 2 e ( aq ) Chapt 4 Ag ( s ) 2e reduction half-reaction oxidation half-reaction Chemical reactions Adding the two half-reactions together, the electrons cancel, Zn(s) Zn2 (aq) 2e 2Ag (aq) 2e 2Ag(s) Zn(s) 2Ag (aq) oxidation half-reaction reduction half-reaction Zn 2 (aq) 2Ag(s) the balanced oxidation-reduction reaction Chapt 4 Chemical reactions Redox reaction:Combustion Reactions Some Common Oxidation-Reduction Reactions Most of the oxidation-reduction reactions fall into one of the following simple categories: Combustion Reactions Combination Reactions Decomposition Reactions Displacement Reactions Combustion reactions-Reactions in which O2(g) is a reactant Combustion reactions release lots of energy 2 C8H18(g) + 25 O2(g) Chapt 4 16 CO2(g) + 18 H2O(g) Chemical reactions 25 Combination Reactions A combination reaction is a reaction in which two substances combine to form a third substance. Combination reaction of sodium and chlorine Decomposition Reaction: A decomposition reaction is a reaction in which a single compound reacts to give two or more substances. Decomposition reaction of mercury(II) oxide Displacement Reactions A displacement reaction (also called a single- replacement reaction) is a reaction in which an element reacts with a compound, displacing an element from it. – Displacement reaction of zinc and hydrochloric acid. Chapt 4 Chemical reactions Combustion Products 1. 3. Write the equation for the complete combustion of CH3OH(l). Write the skeleton equation CH3OH(l) + O2(g) CO2(g) + H2O(g) Balance the equation Chapt 4 Chemical reactions Key Skills Relating quantities in a chemical equation. Determining limiting reactant, calculating theoretical and % yield Calculating and using molarity as a conversion factor Determining solution dilutions; using solution stoichiometry to find volume and amounts Predicting the solubility of a compound Writing equations for precipitation reaction; molecular, ionic and net ionic equations. Writing equations for Acid-base reactions, Gas-evolution reactions Assigning oxidation states and Identifying Redox reactions Writing equations for combustion reactions Chapt 4 Chemical reactions 26 This document was created with Win2PDF available at http://www.win2pdf.com. The unregistered version of Win2PDF is for evaluation or non-commercial use only. This page will not be added after purchasing Win2PDF.