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1、1Lecture 4 Chemical Reactions and QuantitiesChemical Reactions and EquationsCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第1頁，共82頁。Post-exam Error AnalysisWork in groups of 3 on examMaximum of 15 points possible15 (1/5)(theoretical actual)This is what I call PEA scoreExam

2、Correction Factor(wrong answer points / right answer points)Points you earned: Correction factor x PEA scoreExamples on the board2第2頁，共82頁。3Chemical ReactionIn a chemical reaction, A chemical change produces one or more new substances. There is a change in the composition of one or more substances.O

3、ld bonds are broken and new ones are formedCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第3頁，共82頁。4Chemical EquationsA chemical equation Gives the chemical formulas of the reactants on the left of the arrow and the products on the right. Reactants ProductC(s)O2 (g)CO2 (g)C

4、opyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第4頁，共82頁。5Symbols Used in EquationsSymbols are used inchemical equations to show The states of the reactants.The states of the products.The reaction conditions.TABLE 5.2Copyright 2019 by Pearson Education, Inc. Publishing as Ben

5、jamin Cummings第5頁，共82頁。6Chemical Equations are BalancedCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin CummingsChemical equations must be balanced!Atoms are not gained or lost.The number of atoms in the reactants is equal to the number of atoms in the products.第6頁，共82頁。7A Balanced C

6、hemical EquationIn a balanced chemical equation,There must be the same number of each type of atom on the reactant side and on the product side. Al + S Al2S3 Not Balanced 2Al + 3S Al2S3 Balanced 2Al = 2Al 3S = 3S第7頁，共82頁。8Learning CheckState the number of atoms of each element on thereactant side an

7、d the product side for each of thefollowing balanced equations:A. P4(s) + 6Br2(l) 4 PBr3(g)B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)第8頁，共82頁。9Learning CheckDetermine if each equation is balanced or not.A. Na(s) + N2(g) Na3N(s)B. C2H4(g) + H2O(l) C2H5OH(l)第9頁，共82頁。10Checking a Balanced EquationReactants

8、Products 1 C atom=1 C atom 4 H atoms=4 H atoms 4 O atoms = 4 O atomsCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第10頁，共82頁。11Guide to Balancing EquationsWrite out correct formulasCount the atoms of each element on both sides of the equationNever change subscriptsUse coeff

9、icients to balance each elementDo oxygen lastHydrogen is next to lastCheck final equation for balance第11頁，共82頁。12Balance the following chemical equation: NH3(g) + O2(g) NO(g) + H2O(g)Balancing Chemical Equations第12頁，共82頁。13Check the balance of atoms in the following: Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(

10、l) A. Number of H atoms in products. 1) 22) 43) 8B. Number of O atoms in reactants. 1) 22) 43) 8C. Number of Fe atoms in reactants. 1) 12) 33) 4Learning Check第13頁，共82頁。14Balance each equation and list the coefficients in the balanced equation going from reactants to products:A. _Mg(s) + _N2(g)_Mg3N2

11、(s) 1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1 B. _Al(s) + _Cl2(g) _AlCl3(s) 1) 3, 3, 22) 1, 3, 1 3) 2, 3, 2Learning Check第14頁，共82頁。15Equations with Polyatomic Ions Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第15頁，共82頁。16Learning CheckBarium nitrate reacts with sodium carbonate to

12、 form sodium nitrate and barium carbonate. What is the balanced chemical equation?第16頁，共82頁。17Balance and list the coefficients from reactants to products:A. _Fe2O3(s) + _C(s) _Fe(s) + _CO2(g) 1) 2, 3, 2,3 2) 2, 3, 4, 3 3) 1, 1, 2, 3 B. _Al(s) + _FeO(s) _Fe(s) + _Al2O3(s) 1) 2, 3, 3, 1 2) 2, 1, 1, 1

13、 3) 3, 3, 3, 1 C. _Al(s) + _H2SO4(aq) _Al2(SO4)3(aq) + _H2(g) 1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3Learning Check第17頁，共82頁。18Lecture 4.0: Chemical Reactions and QuantitiesTypes of ReactionsCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第18頁，共82頁。19Type of ReactionsChemi

14、cal reactions can be classified as Combination reactions.Decomposition reactions.Single Replacement reactions.Double Replacement reactions.第19頁，共82頁。20Combination In a combination reaction,Two or more elements (or simple compounds)combine to form one product +2Mg(s) + O2(g) 2MgO(s)2Na(s) + Cl2(g)2Na

15、Cl(s)SO3(g) + H2O(l)H2SO4(aq) ABAB第20頁，共82頁。21DecompositionIn a decomposition reaction, One substance splits into two or more simpler substances.2HgO(s)2Hg(l) + O2(g)2KClO3(s)2KCl(s) + 3O2(g) Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第21頁，共82頁。22Single ReplacementIn a

16、single replacement reaction, One element takes the place of a different element in a reacting compound. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第22頁，共82頁。23Zn and HCl is a Single Replacement Reacti

17、onCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第23頁，共82頁。24Double ReplacementIn a double replacement, Two elements in the reactants exchange places. AgNO3(aq) + NaCl(aq)AgCl(s) + NaNO3(aq) ZnS(s) + 2HCl(aq)ZnCl2(aq) + H2S(g)Copyright 2019 by Pearson Education, Inc. Publis

18、hing as Benjamin Cummings第24頁，共82頁。25Example of a Double ReplacementCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第25頁，共82頁。26Chemical Reactions and QuantitiesOxidation-Reduction ReactionsCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第26頁，共82頁。27

19、Oxidation and ReductionAn oxidation-reduction reaction Provides us with energy from food.Provides electrical energy in batteries.Occurs when iron rusts.4Fe(s) + 3O2(g) 2Fe2O3(s)Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第27頁，共82頁。28An oxidation-reduction reactionTransfe

20、rs electrons from one reactant to another.A Loss of Electrons is Oxidation (LEO) Zn(s) Zn2+(aq) + 2e-A Gain of Electrons is Reduction(GER)Cu2+(aq) + 2e- Cu(s)Electron Loss and Gain第28頁，共82頁。29Oxidation and ReductionCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第29頁，共82頁。30

21、Zn and Cu2+oxidationZn(s) Zn2+(aq) + 2e-Silvery metalreduction Cu2+(aq) + 2e- Cu(s) Blue orange Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第30頁，共82頁。31Electron Transfer from Zn to Cu2+ Oxidation: electron loss Reduction: electron gainCopyright 2019 by Pearson Education,

22、 Inc. Publishing as Benjamin Cummings第31頁，共82頁。32Write the separate oxidation and reduction reactions for the following equation.2Cs(s) + F2(g) 2CsF(s) Each cesium atom loses an electron to form cesium ion.2Cs(s) 2Cs+(s) + 2e oxidationFluorine atoms gain electrons to form fluoride ions.F2(s) + 2e- 2

23、F(s) reductionWriting Oxidation and Reduction Reactions第32頁，共82頁。33Pause for ALE exerciseDo, in groups, problems from the active learning exercise sheet第33頁，共82頁。34Lecture 4.0: Chemical Reactions and QuantitiesThe MoleCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第34頁，共82頁

24、。35Collection TermsA collection term states a specific number of items.1 dozen donuts = 12 donuts1 ream of paper = 500 sheets1 case = 24 cansCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第35頁，共82頁。36A mole is a collection that containsThe same number of particles as there

25、are carbon atoms in 12.0 g of carbon 12C.6.02 x 1023 atoms of an element (Avogadros number). 1 mole element Number of Atoms1 mole C = 6.02 x 1023 C atoms1 mole Na = 6.02 x 1023 Na atoms1 mole Au= 6.02 x 1023 Au atomsA Mole of Atoms 第36頁，共82頁。37A mole Of a covalent compound has Avogadros number of mo

26、lecules.1 mole CO2 = 6.02 x 1023 CO2 molecules1 mole H2O = 6.02 x 1023 H2O moleculesOf an ionic compound contains Avogadros number of formula units.1 mole NaCl = 6.02 x 1023 NaCl formula units1 mole K2SO4 = 6.02 x 1023 K2SO4 formula unitsA Mole of a Compound第37頁，共82頁。38Particle in One-Mole SamplesTA

27、BLE 5.3Copyright 2019 by Pearson Education, Inc Publishing as Benjamin Cummings第38頁，共82頁。39Avogadros number 6.02 x 1023 can be written as anequality and two conversion factors.Equality:1 mole = 6.02 x 1023 particlesConversion Factors:6.02 x 1023 particles and 1 mole 1 mole6.02 x 1023 particlesAvogad

28、ros Number第39頁，共82頁。40Using Avogadros NumberAvogadros number is used to convertmoles of a substance to particles. How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 1023 Cu atoms 1 mole Cu = 3.0 x 1023 Cu atomsCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第40頁，共8

29、2頁。41Using Avogadros NumberAvogadros number is used to convertparticles of a substance to moles.How many moles of CO2 are in 2.50 x 1024 molecules CO2? 2.50 x 1024 molecules CO2 x 1 mole CO2 6.02 x 1023 molecules CO2= 4.15 moles CO2第41頁，共82頁。42Subscripts and MolesThe subscripts in a formula giveThe

30、relationship of atoms in the formula.The moles of each element in 1 mole of compound.GlucoseC6H12O6In 1 molecule: 6 atoms C 12 atoms H6 atoms OIn 1 mole: 6 moles C 12 moles H 6 moles O第42頁，共82頁。43Subscripts State Atoms and Moles1 mole C9H8O4 = 9 moles C 8 moles H 4 moles O Copyright 2019 by Pearson

31、Education, Inc. Publishing as Benjamin Cummings第43頁，共82頁。44Factors from Subscripts Subscripts used for conversion factorsRelate moles of each element in 1 mole compound. For aspirin C9H8O4 can be written as:9 moles C 8 moles H 4 moles O 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4and 1 mole C9H8O4 1 mo

32、le C9H8O4 1 mole C9H8O4 9 moles C 8 moles H 4 moles O 第44頁，共82頁。45Lecture 4.0: Chemical Reactions and QuantitiesMolar MassCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第45頁，共82頁。46The molar mass is The mass of one mole of a substance.The atomic mass of an element expressed

33、 in grams.Molar MassCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第46頁，共82頁。47Molar Mass of CaCl2For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.ElementNumber

34、 of Moles Atomic MassTotal MassCa140.1 g/mole40.1 gCl235.5 g/mole71.0 gCaCl2 111.1 g第47頁，共82頁。48Molar Mass of K3PO4Determine the molar mass of K3PO4 to 0.1 g.ElementNumber of Moles Atomic MassTotal Mass in K3PO4K339.1 g/mole 117.3 gP131.0 g/mole 31.0 gO416.0 g/mole 64.0 gK3PO4 212.3 g第48頁，共82頁。49One

35、-Mole Quantities 32.1 g 55.9 g 58.5 g 294.2 g 342.3 gCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第49頁，共82頁。50Methane CH4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.

36、0 g CH4 and 1 mole CH4 1 mole CH4 16.0 g CH4Conversion Factors from Molar Mass第50頁，共82頁。51 Mole factors are used to convert between the grams of a substance and the number of moles. Calculations Using Molar Mass GramsMole factorMolesCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cu

37、mmings第51頁，共82頁。52Aluminum is often used for the structure oflightweight bicycle frames. How many gramsof Al are in 3.00 moles of Al?3.00 moles Al x 27.0 g Al = 81.0 g Al1 mole Al mole factor for AlCalculating Grams from Moles第52頁，共82頁。53Lecture 4.0 Chemical Reactions and QuantitiesMole Relationship

38、s in Chemical EquationsCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第53頁，共82頁。54Law of Conservation of MassThe Law of Conservation of Mass indicates that in anordinary chemical reaction, Matter cannot be created or destroyed.No change in total mass occurs in a reaction.Ma

39、ss of products is equal to mass of reactants.第54頁，共82頁。55Conservation of Mass + ReactantsProducts2 moles Ag + 1 moles S = 1 mole Ag2S2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 = 247.9 gCopyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第55頁，共82頁。56Consider the following equati

40、on:4Fe(s) + 3O2(g) 2Fe2O3(s)This equation can be read in “moles” by placing theword “moles” between each coefficient and formula.4 moles Fe + 3 moles O2 2 moles Fe2O3 Reading Equations with Moles第56頁，共82頁。57A mole-mole factor is a ratio of the moles for any twosubstances in an equation. 4Fe(s) + 3O2

41、(g) 2Fe2O3(s)Fe and O2 4 moles Fe and 3 moles O2 3 moles O2 4 moles FeFe and Fe2O3 4 moles Fe and 2 moles Fe2O3 2 moles Fe2O34 moles FeO2 and Fe2O3 3 moles O2 and 2 moles Fe2O3 2 moles Fe2O33 moles O2Writing Mole-Mole Factors第57頁，共82頁。58How many moles of Fe2O3 can be produced from 6.0 moles O2?4Fe(s

42、) + 3O2(g) 2Fe2O3(s)Relationship:3 mole O2 = 2 mole Fe2O3Write a mole-mole factor to determine the moles of Fe2O3.6.0 mole O2 x 2 mole Fe2O3 = 4.0 moles Fe2O3 3 mole O2 Calculations with Mole Factors第58頁，共82頁。59Lecture 4.0: Chemical Reactions and QuantitiesMass Calculations for Reactions Copyright 2

43、019 by Pearson Education, Inc. Publishing as Benjamin Cummings第59頁，共82頁。60Moles to GramsSuppose we want to determine the mass (g) of NH3that can form from 2.50 moles N2. N2(g) + 3H2(g) 2NH3(g)The plan needed would be moles N2 moles NH3 grams NH3The factors needed would be: mole factor NH3/N2 and the

44、 molar mass NH3第60頁，共82頁。61Moles to GramsThe setup for the solution would be:2.50 mole N2 x 2 moles NH3 x 17.0 g NH3 1 mole N2 1 mole NH3 given mole-mole factor molar mass = 85.0 g NH3第61頁，共82頁。62The reaction between H2 and O2 produces 13.1 g water. How many grams of O2 reacted?2H2(g) + O2(g) 2H2O(g

45、) ? g 13.1 gThe plan and factors would be g H2O mole H2O mole O2 g O2 molar mole-mole molar mass H2O factor mass O2Calculating the Mass of a Reactant第62頁，共82頁。63The setup would be:13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2 18.0 g H2O 2 moles H2O 1 mole O2 molar mole-mole molar mass H2O factor m

46、ass O2 = 11.6 g O2Calculating the Mass of a Reactant第63頁，共82頁。64Calculating the Mass of ProductWhen 18.6 g ethane gas C2H6 burns in oxygen, howmany grams of CO2 are produced? 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 18.6 g ? g The plan and factors would be g C2H6 mole C2H6 mole CO2 g CO2 molar mole-mole

47、molar mass C2H6 factor mass CO2第64頁，共82頁。65Calculating the Mass of ProductThe setup would be18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2 30.1 g C2H6 2 moles C2H6 1 mole CO2 molar mole-mole molar mass C2H6 factor mass CO2= 54.4 g CO2第65頁，共82頁。66Active learning exercise第66頁，共82頁。67Lecture 4.0:

48、 Chemical Reactions and QuantitiesPercent Yield and Limiting Reactants Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第67頁，共82頁。68Theoretical, Actual, and Percent YieldTheoretical yield The maximum amount of product, which is calculated using the balanced equation.Actual yi

49、eld The amount of product obtained when the reaction takes place.Percent yield The ratio of actual yield to theoretical yield.percent yield = actual yield (g) x 100 theoretical yield (g)第68頁，共82頁。69You prepared cookie dough to make 5 dozen cookies.The phone rings and you answer. While you talk, a sh

50、eetof 12 cookies burn and you throw them out. The restof the cookies are okay. What is the percent yield ofedible cookies?Theoretical yield 60 cookies possibleActual yield 48 cookies to eatPercent yield 48 cookies x 100 = 80% yield 60 cookies Calculating Percent Yield第69頁，共82頁。70Limiting ReactantA l

51、imiting reactant in a chemical reaction is thesubstance that Is used up first.Limits the amount of product that can form and stops the reaction.第70頁，共82頁。71Reacting AmountsIn a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon.How many table settings arepossible from 5 plates, 6 forks,4 s

52、poons, and 7 knives?What is the limiting item?Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第71頁，共82頁。72Reacting AmountsOnly 4 place settings are possible. Initially Used Left overPlates 5 41Forks 6 42Spoons 4 4 0Knives 7 4 3The limiting item is the spoon.Copyright 2019 by

53、 Pearson Education, Inc. Publishing as Benjamin Cummings第72頁，共82頁。73Example of An Everyday Limiting ReactantHow many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.Copyri

54、ght 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第73頁，共82頁。74Example of An Everyday Limiting ReactantHow many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter?With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.Copyright 2019 by Pearson Education, Inc. Publishing as Benjamin Cummings第74頁，共82頁。75Limiting ReactantWhen 4.00 moles H2 is mixed with