1 Lecture 4 Chemical Reactions and Quantities Chemical Reactions and Equations Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings.

Post-exam Error Analysis Work in groups of 3 on exam Maximum of 15 points possible 15 – (1/5)(theoretical – actual) This is what I call PEA score Exam Correction Factor (wrong answer points / right answer points) Points you earned: Correction factor x PEA score Examples on the board 2

3 Chemical Reaction In a chemical reaction, A chemical change produces one or more new substances. There is a change in the composition of one or more substances. Old bonds are broken and new ones are formed Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

4 Chemical Equations A chemical equation  Gives the chemical formulas of the reactants on the left of the arrow and the products on the right. Reactants Product C(s) O 2 (g) CO 2 (g) Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

5 Symbols Used in Equations Symbols are used in chemical equations to show  The states of the reactants.  The states of the products.  The reaction conditions. TABLE 5.2 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

6 Chemical Equations are Balanced Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings Chemical 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.

7 A Balanced Chemical Equation In 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 Al 2 S 3 Not Balanced 2Al + 3S Al 2 S 3 Balanced 2Al = 2Al 3S = 3S

8 Learning Check State the number of atoms of each element on the reactant side and the product side for each of the following balanced equations: A. P 4 (s) + 6Br 2 (l) → 4 PBr 3 (g) B. 2Al(s) + Fe 2 O 3 (s) → 2Fe(s) + Al 2 O 3 (s)

9 Learning Check Determine if each equation is balanced or not. A. Na(s) + N 2 (g) → Na 3 N(s) B. C 2 H 4 (g) + H 2 O(l) → C 2 H 5 OH(l)

11 Guide to Balancing Equations Write out correct formulas Count the atoms of each element on both sides of the equation Never change subscripts Use coefficients to balance each element Do oxygen last Hydrogen is next to last Check final equation for balance

12 Balance the following chemical equation: NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) Balancing Chemical Equations

13 Check the balance of atoms in the following: Fe 3 O 4 (s) + 4H 2 (g) 3Fe (s) + 4H 2 O (l) A. Number of H atoms in products. 1) 22) 43) 8 B. Number of O atoms in reactants. 1) 22) 43) 8 C. Number of Fe atoms in reactants. 1) 12) 33) 4 Learning Check

14 Balance each equation and list the coefficients in the balanced equation going from reactants to products: A. __Mg(s) + _N 2 (g)__Mg 3 N 2 (s) 1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1 B. __Al(s) + __Cl 2 (g) __AlCl 3 (s) 1) 3, 3, 22) 1, 3, 1 3) 2, 3, 2 Learning Check

16 Learning Check Barium nitrate reacts with sodium carbonate to form sodium nitrate and barium carbonate. What is the balanced chemical equation?

17 Balance and list the coefficients from reactants to products: A. __Fe 2 O 3 (s) + __C(s) __Fe(s) + __CO 2 (g) 1) 2, 3, 2,3 2) 2, 3, 4, 3 3) 1, 1, 2, 3 B. __Al(s) + __FeO(s) __Fe(s) + __Al 2 O 3 (s) 1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1 C. __Al(s) + __H 2 SO 4 (aq) __Al 2 (SO 4 ) 3 (aq) + __H 2 (g) 1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3 Learning Check

19 Type of Reactions Chemical reactions can be classified as  Combination reactions.  Decomposition reactions.  Single Replacement reactions.  Double Replacement reactions.

20 Combination In a combination reaction,  Two or more elements (or simple compounds) combine to form one product + 2Mg(s) + O 2 (g) 2MgO(s) 2Na(s) + Cl 2 (g)2NaCl(s) SO 3 (g) + H 2 O(l)H 2 SO 4 (aq) AB AB

21 Decomposition In a decomposition reaction,  One substance splits into two or more simpler substances. 2HgO(s)2Hg(l) + O 2 (g) 2KClO 3 (s)2KCl(s) + 3O 2 (g) Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

22 Single Replacement In a single replacement reaction,  One element takes the place of a different element in a reacting compound. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Fe(s) + CuSO 4 (aq) FeSO 4 (aq) + Cu(s) Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

24 Double Replacement In a double replacement,  Two elements in the reactants exchange places. AgNO 3 (aq) + NaCl(aq)AgCl(s) + NaNO 3 (aq) ZnS(s) + 2HCl(aq)ZnCl 2 (aq) + H 2 S(g) Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

27 Oxidation and Reduction An oxidation-reduction reaction  Provides us with energy from food.  Provides electrical energy in batteries.  Occurs when iron rusts. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

28 An oxidation-reduction reaction  Transfers electrons from one reactant to another. A Loss of Electrons is Oxidation (LEO) Zn(s) Zn 2+ (aq) + 2e - A Gain of Electrons is Reduction(GER) Cu 2+ (aq) + 2e - Cu(s) Electron Loss and Gain

32 Write the separate oxidation and reduction reactions for the following equation. 2Cs(s) + F 2 (g) 2CsF(s) Each cesium atom loses an electron to form cesium ion. 2Cs(s) 2Cs + (s) + 2e − oxidation Fluorine atoms gain electrons to form fluoride ions. F 2 (s) + 2e - 2F − (s) reduction Writing Oxidation and Reduction Reactions

33 Pause for ALE exercise Do, in groups, problems from the active learning exercise sheet

35 Collection Terms A collection term states a specific number of items.  1 dozen donuts = 12 donuts  1 ream of paper = 500 sheets  1 case = 24 cans Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

36 A mole is a collection that contains  The same number of particles as there are carbon atoms in 12.0 g of carbon 12 C.  6.02 x 10 23 atoms of an element (Avogadro’s number). 1 mole element Number of Atoms 1 mole C = 6.02 x 10 23 C atoms 1 mole Na = 6.02 x 10 23 Na atoms 1 mole Au= 6.02 x 10 23 Au atoms A Mole of Atoms

37 A mole  Of a covalent compound has Avogadro’s number of molecules. 1 mole CO 2 = 6.02 x 10 23 CO 2 molecules 1 mole H 2 O = 6.02 x 10 23 H 2 O molecules  Of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x 10 23 NaCl formula units 1 mole K 2 SO 4 = 6.02 x 10 23 K 2 SO 4 formula units A Mole of a Compound

39 Avogadro’s number 6.02 x 10 23 can be written as an equality and two conversion factors. Equality: 1 mole = 6.02 x 10 23 particles Conversion Factors: 6.02 x 10 23 particles and 1 mole 1 mole6.02 x 10 23 particles Avogadro’s Number

40 Using Avogadro’s Number Avogadro’s number is used to convert moles of a substance to particles. How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 10 23 Cu atoms 1 mole Cu = 3.0 x 10 23 Cu atoms Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

41 Using Avogadro’s Number Avogadro’s number is used to convert particles of a substance to moles. How many moles of CO 2 are in 2.50 x 10 24 molecules CO 2 ? 2.50 x 10 24 molecules CO 2 x 1 mole CO 2 6.02 x 10 23 molecules CO 2 = 4.15 moles CO 2

42 Subscripts and Moles The subscripts in a formula give  The relationship of atoms in the formula.  The moles of each element in 1 mole of compound. Glucose C 6 H 12 O 6 In 1 molecule: 6 atoms C 12 atoms H6 atoms O In 1 mole: 6 moles C 12 moles H 6 moles O

44 Factors from Subscripts Subscripts used for conversion factors  Relate moles of each element in 1 mole compound.  For aspirin C 9 H 8 O 4 can be written as: 9 moles C 8 moles H 4 moles O 1 mole C 9 H 8 O 4 1 mole C 9 H 8 O 4 1 mole C 9 H 8 O 4 and 1 mole C 9 H 8 O 4 1 mole C 9 H 8 O 4 1 mole C 9 H 8 O 4 9 moles C 8 moles H 4 moles O

47 Molar Mass of CaCl 2  For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl 2 to the nearest 0.1 g as follows. ElementNumber of Moles Atomic MassTotal Mass Ca140.1 g/mole40.1 g Cl235.5 g/mole71.0 g CaCl 2 111.1 g

48 Molar Mass of K 3 PO 4 Determine the molar mass of K 3 PO 4 to 0.1 g. ElementNumber of Moles Atomic MassTotal Mass in K 3 PO 4 K339.1 g/mole 117.3 g P131.0 g/mole 31.0 g O416.0 g/mole 64.0 g K 3 PO 4 212.3 g

50 Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH 4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 Conversion Factors from Molar Mass

51  Mole factors are used to convert between the grams of a substance and the number of moles. Calculations Using Molar Mass Grams Mole factor Moles Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

52 Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al Calculating Grams from Moles

54 Law of Conservation of Mass The Law of Conservation of Mass indicates that in an ordinary chemical reaction,  Matter cannot be created or destroyed.  No change in total mass occurs in a reaction.  Mass of products is equal to mass of reactants.

55 Conservation of Mass + ReactantsProducts 2 moles Ag + 1 moles S = 1 mole Ag 2 S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 = 247.9 g Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

56 Consider the following equation: 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) This equation can be read in “moles” by placing the word “moles” between each coefficient and formula. 4 moles Fe + 3 moles O 2 2 moles Fe 2 O 3 Reading Equations with Moles

57 A mole-mole factor is a ratio of the moles for any two substances in an equation. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Fe and O 2 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe Fe and Fe 2 O 3 4 moles Fe and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 4 moles Fe O 2 and Fe 2 O 3 3 moles O 2 and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 3 moles O 2 Writing Mole-Mole Factors

58 How many moles of Fe 2 O 3 can be produced from 6.0 moles O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship:3 mole O 2 = 2 mole Fe 2 O 3 Write a mole-mole factor to determine the moles of Fe 2 O 3. 6.0 mole O 2 x 2 mole Fe 2 O 3 = 4.0 moles Fe 2 O 3 3 mole O 2 Calculations with Mole Factors

60 Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can form from 2.50 moles N 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) The plan needed would be moles N 2 moles NH 3 grams NH 3 The factors needed would be: mole factor NH 3 /N 2 and the molar mass NH 3

61 Moles to Grams The setup for the solution would be: 2.50 mole N 2 x 2 moles NH 3 x 17.0 g NH 3 1 mole N 2 1 mole NH 3 given mole-mole factor molar mass = 85.0 g NH 3

62 The reaction between H 2 and O 2 produces 13.1 g water. How many grams of O 2 reacted? 2H 2 (g) + O 2 (g) 2H 2 O(g) ? g 13.1 g The plan and factors would be g H 2 O mole H 2 O mole O 2 g O 2 molar mole-mole molar mass H 2 O factor mass O 2 Calculating the Mass of a Reactant

63 The setup would be: 13.1 g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O 2 18.0 g H 2 O 2 moles H 2 O 1 mole O 2 molar mole-mole molar mass H 2 O factor mass O 2 = 11.6 g O 2 Calculating the Mass of a Reactant

64 Calculating the Mass of Product When 18.6 g ethane gas C 2 H 6 burns in oxygen, how many grams of CO 2 are produced? 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) 18.6 g ? g The plan and factors would be g C 2 H 6 mole C 2 H 6 mole CO 2 g CO 2 molar mole-mole molar mass C 2 H 6 factor mass CO 2

65 Calculating the Mass of Product The setup would be 18.6 g C 2 H 6 x 1 mole C 2 H 6 x 4 moles CO 2 x 44.0 g CO 2 30.1 g C 2 H 6 2 moles C 2 H 6 1 mole CO 2 molar mole-mole molar mass C 2 H 6 factor mass CO 2 = 54.4 g CO 2

66 Active learning exercise

68 Theoretical, Actual, and Percent Yield Theoretical yield  The maximum amount of product, which is calculated using the balanced equation. Actual yield  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)

69 You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burn and you throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies Calculating Percent Yield

70 Limiting Reactant A limiting reactant in a chemical reaction is the substance that  Is used up first.  Limits the amount of product that can form and stops the reaction.

71 Reacting Amounts In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item? Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

72 Reacting Amounts Only 4 place settings are possible. Initially Used Left over Plates 5 41 Forks 6 42 Spoons 4 4 0 Knives 7 4 3 The limiting item is the spoon. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

73 Example of An Everyday Limiting Reactant How 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. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

74 Example of An Everyday Limiting Reactant How 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 © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

75 Limiting Reactant When 4.00 moles H 2 is mixed with 2.00 moles Cl 2,how many moles of HCl can form? H 2 (g) + Cl(g)  2HCl (g) 4.00 moles 2.00 moles ??? moles  Calculate the moles of product that each reactant, H 2 and Cl 2, could produce.  The limiting reactant is the one that produces the smallest amount of product.

76 Limiting Reactant HCl from H 2 4.00 moles H 2 x 2 moles HCl = 8.00 moles HCl 1 moles H 2 (not possible) HCl from Cl 2 2.00 moles Cl 2 x 2 moles HCl = 4.00 moles HCl 1 mole Cl 2 (smaller number of moles, Cl 2 will be used up first) The limiting reactant is Cl 2 because it is used up first. Thus Cl 2 produces the smaller number of moles of HCl.

77 Check Calculations InitiallyH 2 4.00 moles Cl 2 2.00 moles 2HCl 0 mole Reacted/ Formed -2.00 moles +4.00 moles Left after reaction 2.00 moles Excess 0 mole Limiting 4.00 moles

78 Limiting Reactants Using Mass If 4.80 moles Ca are mixed with 2.00 moles N 2, which is The limiting reactant? 3Ca(s) + N 2 (g)  Ca 3 N 2 (s) moles of Ca 3 N 2 from Ca 4.80 moles Ca x 1 mole Ca 3 N 2 = 1.60 moles Ca 3 N 2 3 moles Ca (Ca is used up) moles of Ca 3 N 2 from N 2 2.00 moles N 2 x 1 mole Ca 3 N 2 = 2.00 moles Ca 3 N 2 1 mole N 2 (not possible) Ca is used up when 1.60 mole Ca 3 N 2 forms. Thus, Ca is the limiting reactant.

79 Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l)

80 Limiting Reactants Using Mass Moles H 2 O from H 2 : 8.00 g H 2 x 1 mole H 2 x 2 moles H 2 O = 3.97 moles H 2 O 2.02 g H 2 2 moles H 2 (not possible) Moles H 2 O from O 2 : 24.0 g O 2 x 1 mole O 2 x 2 moles H 2 O = 1.50 moles H 2 O 32.0 g O 2 1 mole O 2 O 2 is limiting The maximum amount of product is 1.50 moles H 2 O, which is converted to grams. 1.50 moles H 2 O x 18.0 g H 2 O = 27.0 g H 2 O 1 mole H 2 O

81 More Active Learning Exercises

1 Lecture 4 Chemical Reactions and Quantities Chemical Reactions and Equations Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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1 Lecture 4 Chemical Reactions and Quantities Chemical Reactions and Equations Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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