Chemistry, Julia Burdge, 2st Ed McGraw Hill Chapter Chemical Quantities and Aqueous Reactions Mr Truong Minh Chien ; losedtales@yahoo.com http://tailieu.vn/losedtales http://mba-programming.blogspot.com 2011, NKMB Co., Ltd Overview Ch • Stoichiometry • Limiting Reagents Theoretical Yield Experimental Yield • • • • • • Molarity Molarity in calculations Electrolytes NIE’s Titrations REDOX Reaction Stoichiometry • the numerical relationships between chemical amounts • in a reaction is called stoichiometry the coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) molecules of C8H18 react with 25 molecules of O2 to form 16 molecules of CO2 and 18 molecules of H2O moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O Chemistry, Julia Burdge, 2nd e., McGraw Hill Predicting Amounts from Stoichiometry • 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? C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) moles C8H18 : 16 moles CO2 16 mol CO 22.0 moles C8 H18 × = 176 moles CO 2 mol C8 H18 Chemistry, Julia Burdge, 2nd e., McGraw Hill Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline • assuming that gasoline is octane, C8H18, the equation for the reaction is: C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) • the equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the Concept Plan will be: g C8H18 mol C8H18 Chemistry, Julia Burdge, 2nd e., McGraw Hill mol CO2 g CO2 Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline Given: Find: 3.4 x 1015 g C8H18 g CO2 Concept Plan: g C8H18 mol C8H18 mol 114.22 g Relationships: Solution: mol CO2 16 mol CO 2 mol C8 H18 44.01 g mol g CO2 mol C8H18 = 114.22g, mol CO2 = 44.01g, mol C8H18 = 16 mol CO2 mol C8 H18 16 mol CO 44.01 g CO 3.4 ×10 g C8 H18 × × × 114.22 g C8 H18 mol C8 H18 mol CO 15 = 1.0 ×1016 g CO Check: since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense Practice • According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C6H12O6(s) + O2(g) → CO2(g) + H2O(l) convert 9.0 g of glucose into moles (MM 180) convert moles of glucose into moles of water convert moles of water into grams (MM 18.02) convert grams of water into mL a) How? what is the relationship between mass and volume? density of water = 1.00 g/mL Chemistry, Julia Burdge, 2nd e., McGraw Hill Practice According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C6H12O6(s) + O2(g) → CO2(g) + H2O(l) 9.0 g C6 H12 O x mole C H12 O 6 mole H O 18.0 g H O mL H O x x x mole C H12 O mole H O 1.00 g H O 1.80 x 10 g = 5.4 mL H O Chemistry, Julia Burdge, 2nd e., McGraw Hill Limiting Reactant • for reactions with multiple reactants, it is likely that one • • of the reactants will be completely used before the others when this reactant is used up, the reaction stops and no more product is made the reactant that limits the amount of product is called the limiting reactant  sometimes called the limiting reagent  the limiting reactant gets completely consumed • reactants not completely consumed are called excess • reactants the amount of product that can be made from the limiting reactant is called the theoretical yield Chemistry, Julia Burdge, 2nd e., McGraw Hill Things Don’t Always Go as Planned! • many things can happen during the course of an experiment that cause the loss of product • the amount of product that is made in a reaction is called the actual yield generally less than the theoretical yield, never more! • the efficiency of product recovery is generally given as the percent yield actual yield Percent Yield = ×100% theoretical yield Chemistry, Julia Burdge, 2nd e., McGraw Hill 10 Electron Bookkeeping • for reactions that are not metal + nonmetal, or not • involve O2, we need a method for determining how the electrons are transferred chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction  even though they look like them, oxidation states are not ion charges! oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges Tro, Chemistry: A Molecular Approach 93 Rules for Assigning Oxidation States • rules are in order of priority free elements have an oxidation state =  Na = and Cl2 = in Na(s) + Cl2(g) monatomic ions have an oxidation state equal to their charge  Na = +1 and Cl = -1 in NaCl (a) the sum of the oxidation states of all the atoms in a compound is  Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = Tro, Chemistry: A Molecular Approach 94 Rules for Assigning Oxidation States (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion  N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1 (a) Group I metals have an oxidation state of +1 in all their compounds  Na = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds  Mg = +2 in MgCl2 Tro, Chemistry: A Molecular Approach 95 Rules for Assigning Oxidation States in their compounds, nonmetals have oxidation states according to the table below  nonmetals higher on the table take priority Nonmetal Oxidation State Example F -1 CF4 H +1 CH4 O -2 CO2 Group 7A -1 CCl4 Group 6A -2 CS2 Group 5A -3 NH3 Tro, Chemistry: A Molecular Approach 96 Practice – Assign an Oxidation State to Each Element in the following • • • • Br2 K+ LiF CO2 • SO42• Na2O2 Tro, Chemistry: A Molecular Approach 97 Practice – Assign an Oxidation State to Each Element in the following • • • • Br2 Br = 0, (Rule 1) K+ K = +1, (Rule 2) LiF Li = +1, (Rule 4a) & F = -1, (Rule 5) CO2 O = -2, (Rule 5) & C = +4, (Rule 3a) • SO42- O = -2, (Rule 5) & S = +6, (Rule 3b) • Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a) Tro, Chemistry: A Molecular Approach 98 Oxidation and Reduction Another Definition • 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 + O2 → CO2 + H2O -4 +1 +4 –2 +1 -2 oxidation reduction Tro, Chemistry: A Molecular Approach 99 Oxidation–Reduction • oxidation and reduction must occur simultaneously  if an atom loses electrons another atom must take them • the reactant that reduces an element in another reactant is called the reducing agent  the reducing agent contains the element that is oxidized • the reactant that oxidizes an element in another reactant is called the oxidizing agent  the oxidizing agent contains the element that is reduced Na(s) + Cl2(g) → Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent Tro, Chemistry: A Molecular Approach 100 Identify the Oxidizing and Reducing Agents in Each of the Following H2S + NO3– + H+ → S + NO + H2O MnO2 + HBr → MnBr2 + Br2 + H2O Tro, Chemistry: A Molecular Approach 101 Identify the Oxidizing and Reducing Agents in Each of the Following red ag ox ag +1 -2 +5 -2 H2S + NO3– + H+ → S + NO + H2O +1 oxidation ox ag +1 -1 +2 -2 +1 -2 reduction red ag +4 -2 MnO2 + HBr → MnBr2 + Br2 + H2O +2 -1 +1 -2 oxidation reduction Tro, Chemistry: A Molecular Approach 102 Go To Chapter 18: Balancing Redox Reactions in Acidic/Basic solutions Tro, Chemistry: A Molecular Approach 103 Combustion Reactions • Reactions in which O2(g) is a • • reactant are called combustion reactions Combustion reactions release lots of energy Combustion reactions are a subclass of oxidation-reduction reactions C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) Tro, Chemistry: A Molecular Approach 104 Combustion Products • to predict the products of a combustion reaction, combine each element in the other reactant with oxygen Reactant Combustion Product contains C CO2(g) contains H H2O(g) contains S SO2(g) contains N NO(g) or NO2(g) contains metal M2On(s) Tro, Chemistry: A Molecular Approach 105 Practice – Complete the Reactions • combustion of C3H7OH(l) • combustion of CH3NH2(g) Tro, Chemistry: A Molecular Approach 106 Practice – Complete the Reactions C3H7OH(l) + O2(g) → CO2(g) + H2O(g) CH3NH2(g) + O2(g) → CO2(g) + H2O(g) + NO2(g) Tro, Chemistry: A Molecular Approach 107 ... mol C8 H18 44 .01 g mol g CO2 mol C8H18 = 1 14. 22g, mol CO2 = 44 .01g, mol C8H18 = 16 mol CO2 mol C8 H18 16 mol CO 44 .01 g CO 3 .4 ×10 g C8 H18 × × × 1 14. 22 g C8 H18 mol C8 H18 mol CO 15 = 1.0 ×1016... 20 04 by the combustion of 3 .4 x 1015 g gasoline Given: Find: 3 .4 x 1015 g C8H18 g CO2 Concept Plan: g C8H18 mol C8H18 mol 1 14. 22 g Relationships: Solution: mol CO2 16 mol CO 2 mol C8 H18 44 .01... McGraw Hill O + H H O + O H H 11 Limiting and Excess Reactants in the Combustion of Methane CH4(g) + O2(g) → CO2(g) + H2O(g) • If we have molecules of CH4 and molecules of O2, which is the limiting