Download presentation
Presentation is loading. Please wait.
Published byEmerald Johnston Modified over 8 years ago
1
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.1 The Mole 1
2
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Avogadro's Number Small particles such as atoms, molecules, and ions are counted using the mole. 1 mole = 6.02 x 10 23 items Avogadro’s number 602 000 000 000 000 000 000 000 = 6.02 x 10 23 2
3
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Mole of Atoms 1 mole of an element = 6.02 x 10 23 atoms of that element 1 mole of carbon= 6.02 x 10 23 atoms of carbon 1 mole of sodium= 6.02 x 10 23 atoms of sodium 3
4
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Avogadro's Number Avogadro’s number, 6.02 x 10 23, can be written as an equality and as two conversion factors. Equality: 1 mole = 6.02 x 10 23 particles Conversion Factors: 6.02 x 10 23 particles and 1 mole 1 mole 6.02 x 10 23 particles 4
5
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Guide to Calculating Atoms or Molecules 5
6
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check The number of atoms in 2.0 mole of Al atoms is: A. 2.0 Al atoms B. 3.0 x 10 23 Al atoms C. 1.2 x 10 24 Al atoms 6
7
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution The number of atoms in 2.0 moles of Al atoms is: Step 1 State the needed and given quantities: Given: 2.0 mole Al Needed: atoms of Al Step 2 Write a plan to convert moles to atoms or molecules: Avogadro’s number Moles of Al Atoms of Al 7
8
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution The number of atoms in 2.0 moles of Al atoms is: Step 3 Use Avogadro’s number to write conversion factors. 1 mole of Al = 6.02 x 10 23 atoms of Al 6.02 x 10 23 atoms Al and 1 mole A 1 mole Al 6.02 x 10 23 atoms Al 8
9
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution The number of atoms in 2.0 moles of Al atoms is: Step 4 Set up the problem to calculate the number of particles. C. 2.0 moles Al x 6.02 x 10 23 Al atoms 1 mole Al = 1.2 x 10 24 Al atoms 9
10
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.2 Molar Mass 10
11
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Molar Mass The molar mass is the mass of 1 mole of an element is the atomic mass expressed in grams 11
12
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Molar Mass from Periodic Table Molar mass is the atomic mass expressed in grams. 1 mole of Ag 1 mole of C 1 mole of S = 107.9 g of Ag = 12.0 g of C = 32.1 g of S 12
13
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Give the molar mass for each element (to the tenths place). A. 1 mole of K atoms =________ B. 1 mole of Sn atoms =________ C. 1 mole of Ca atoms=________ 13
14
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Give the molar mass for each element (to the tenths place). A. 1 mole of K atoms = 39.1 g of K B. 1 mole of Sn atoms =118.7 g of Sn C. 1 mole of Ca atoms= 40.0 g of Ca 14
15
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Guide to Calculating Molar Mass 15
16
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Molar Mass of the Compound C 2 H 6 O To calculate the molar mass of C 2 H 6 O: Step 1 Obtain the molar mass of each element. 1 mole of C = 12.0 g of C 1 mole of H = 1.01 g of H 1 mole of O = 16.0 g of O 16
17
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Molar Mass of the Compound C 2 H 6 O Step 2 Multiply each molar mass by the number of moles (subscripts) in the formula. 2 moles C x 12.0 g C = 24.0 g of C 1 mole C 6 moles H x 1.01 g H = 6.06 g of H 1 mole H 1 mole O x 16.0 g O = 16.0 g of O 1 mole O 17
18
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Molar Mass of the Compound C 2 H 6 O Step 3 Calculate the molar mass by adding the masses of the elements. 2 moles of C = 24.0 g of C 6 moles of H = 6.06 g of H 1 mole of O = 16.0 g of O Molar mass of C 2 H 6 O = 46.1 g of C 2 H 6 O 18
19
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check What is the molar mass of each compound? A. K 2 O B. Al(OH) 3 19
20
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Prozac, C 17 H 18 F 3 NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? A. 40.1 g/mole B. 262 g/mole C. 309 g/mole 20
21
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Calculations Using Molar Mass Molar mass conversion factors are fractions (ratios) written from the molar mass. relate grams and moles of an element or compound. for methane, CH 4, used in gas stoves and gas heaters, is 1 mole of CH 4 = 16.0 g of CH 4 (molar mass equality) Conversion factors: 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 21
22
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Guide to Calculating Moles from Mass 22
23
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? 23
24
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Converting Mass to Moles of Compound NaCl A box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box? Step 1 State the given and needed quantities. Given: 737 g of NaCl Need:moles of NaCl Step 2 Write a plan to convert grams to moles. molar mass Grams of NaCl Moles of NaCl 24
25
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Converting Mass to Moles of Compound NaCl Step 3 Determine the molar mass and write conversion factors. Molar Mass 1 mole Na x 23.0 g Na = 23.0 g of Na 1 mole Na 1 mole Cl x 35.5 g Cl = 35.5 g of Cl 1 mole Cl 58.5 g of NaCl Conversion Factors 1 mole of NaCl = 58.5 g of NaCl 58.5 g NaCl and 1 mole NaCl 1 mole NaCl 58.5 g NaCl 25
26
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Converting Mass to Moles of Compound NaCl Step 4 Set up the problem to convert grams to moles. 737 g NaCl x 1 mole NaCl = 12.6 moles of NaCl 58.5 g NaCl 26
27
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Map: Mass – Moles – Particles 27
28
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.3 Chemical Reactions and Equations 28
29
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chemical Changes A chemical change occurs when a substance is converted into one or more substances with different formulas and different properties may be observed by the formation of bubbles, a change in color, or production of a solid 29
30
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Visible Evidence of a Chemical Reaction 30
31
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Which of the following represent a chemical change? A. burning a candle B. ice melting C. toast burns D. water turns to steam 31
32
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Writing a Chemical Equation 32
33
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Symbols Used in Chemical Equations To write a chemical equation an arrow indicates reactants form products reactants are written on the left side of the arrow products are written on the right side of the arrow multiple reactants or products are separated by a + sign the delta {Δ} sign indicates that heat is used to start the reaction 33
34
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Identifying a Balanced Equation In a balanced chemical equation, no atoms are lost or gained the number of reacting atoms is equal to the number of product atoms 34
35
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Balancing a Chemical Equation 35
36
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Balance the following equation, Fe 3 O 4 (s) + H 2 (g) Fe(s) + H 2 O(l) 36
37
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Balancing Equations with Polyatomic Ions 37
38
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Balancing Equations with Polyatomic Ions When balancing equations with polyatomic ions, balance each polyatomic ion as a unit. 2Na 3 PO 4 (aq) + 3MgCl 2 (aq) Mg 3 (PO 4 ) 2 (s) + 6NaCl(aq) 38
39
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check 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 39
40
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution A. 2) 2, 3, 4, 3 2Fe 2 O 3 (s) + 3C(s) 4Fe(s) + 3CO 2 (g) B. 1) 2, 3, 3, 1 2Al(s) + 3FeO(s) 3Fe(s) + 1Al 2 O 3 (s) C. 2) 2, 3, 1, 3 2Al(s) + 3H 2 SO 4 (aq) 1Al 2 (SO 4 ) 3 (aq) + 3H 2 (g) 40
41
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.4 Types of Reactions 41
42
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Types of Reactions Chemical reactions can be classified as combination reactions decomposition reactions single replacement reactions double replacement reactions combustion reactions 42
43
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Combination Reactions In a combination reaction, two or more elements form one product 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) 43
44
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Decomposition Reaction 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) 44
45
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Single Replacement Reaction In a single replacement reaction, one element takes the place of a different element in another reacting a compound. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Fe(s) + CuSO 4 (aq)FeSO 4 (aq) + Cu(s) 45
46
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Double Replacement Reaction 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) 46
47
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Double Replacement Reaction: BaSO 4 47
48
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Combustion Reaction In a combustion reaction, a carbon-containing compound burns in oxygen gas to form carbon dioxide (CO 2 ) and water (H 2 O) energy is released as a product in the form of heat CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + energy 48
49
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Classify each of the following reactions as combination, decomposition, single replacement, double replacement, or combustion. A. 2Al(s) + 3H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (s) + 3H 2 (g) B. Na 2 SO 4 (aq) + 2AgNO 3 (aq) Ag 2 SO 4 (s) + 2NaNO 3 (aq) C. 3C(s) + Fe 2 O 3 (s) 2Fe(s) + 3CO(g) D. C 2 H 4 (g) + 2O 2 (g) 2CO 2 (g) + 2H 2 O(g) 49
50
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution A. 2Al(s) + 3H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (s) + 3H 2 (g) Single Replacement B. Na 2 SO 4 (aq) + 2AgNO 3 (aq) Ag 2 SO 4 (s) + 2NaNO 3 (aq) Double Replacement C. N 2 (g) + O 2 (g) 2NO(g) Combination D. C 2 H 4 (g) + 2O 2 (g) 2CO 2 (g) + 2H 2 O(g) Combustion 50
51
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.5 Oxidation−Reduction Reactions 51
52
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Oxidation−Reduction Reactions 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) 52
53
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Oxidation−Reduction In an oxidation–reduction reaction, electrons are transferred from one substance to another. 53
54
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Oxidation and Reduction 2Cu(s) + O 2 (g) 2CuO OIL RIG Oxidation is loss of electrons. 2Cu(s) 2Cu 2+ (s) + 4e − Reduction is gain of electrons. O 2 (g) + 4e − 2O 2− (s) 54 The green patina on copper is due to oxidation
55
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Identify each of the following as oxidation or reduction: __A. Sn(s) Sn 4+ (aq) + 4e − __B. Fe 3+ (aq) + 1e − Fe 2+ (aq) __C. Cl 2 (g) + 2e − 2Cl − (aq) 55
56
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Identify each of the following as oxidation or reduction: Oxidation__A. Sn(s) Sn 4+ (aq) + 4e − Reduction__B. Fe 3+ (aq) + 1e − Fe 2+ (aq) Reduction__C. Cl 2 (g) + 2e − 2Cl − (aq) 56
57
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Oxidation – Reduction in Biological Systems The oxidation of a typical biochemical molecule can involve the transfer of two hydrogen atoms (or 2H + and 2e − ) to a proton acceptor such as the coenzyme FAD (flavin adenine dinucleotide). 57
58
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Characteristics of Oxidation and Reduction 58
59
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.6 Mole Relationships in Chemical Equations 59
60
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. 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 the mass of products is equal to mass of reactants 60
61
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Reading Equations in Moles Consider the following equation: 2Fe(s) + 3S(s) Fe 2 S 3 (s) This equation can be read in “moles” by placing the word “moles of” between each coefficient and formula. 2 moles of Fe + 3 moles of S 1 mole of Fe 2 S 3 61
62
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Writing Conversion Factors from Equations A mole–mole factor is a ratio of the moles for any two substances in an equation. 2Fe(s) + 3S(s) Fe 2 S 3 (s) Fe and S 2 moles Fe and 3 moles S 3 moles S 2 moles Fe Fe and Fe 2 S 3 2 moles Fe and 2 moles Fe 2 S 3 1 mole Fe 2 S 3 2 moles Fe S and Fe 2 S 3 3 moles S and 1 mole Fe 2 S 3 1 mole Fe 2 S 3 3 moles S 62
63
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Calculations with Mole Factors How many moles of Fe 2 O 3 can form from 6.0 moles of O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship: 3 moles of O 2 = 2 moles of Fe 2 O 3 Use a mole–mole factor to determine the moles of Fe 2 O 3. 6.0 moles O 2 x 2 moles Fe 2 O 3 = 4.0 moles of Fe 2 O 3 3 moles O 2 63
64
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Guide to Using Mole–Mole Factors 64
65
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check How many moles of Fe are needed for the reaction of 12.0 moles of O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) A. 3.00 moles of Fe B. 9.00 moles of Fe C. 16.0 moles of Fe 65
66
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.7 Mass Calculations for Reactions 66
67
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Mass Calculations in Equations 67
68
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can be produced from 32 grams of N 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) Step 1 Use molar mass to convert grams of given to moles. 1 mole of N 2 = 28.0 g of N 2 1 mole N 2 and 28.0 g N 2 28.0 g N 2 1 mole N 2 32 g N 2 x 1 mole N 2 = 1.1 mole of N 2 28.0 g N 2 68
69
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Moles to Grams Step 2 Write a mole–mole factor from the coefficients in the equation. 1 mole of N 2 = 2 moles of NH 3 1 mole N 2 and 2 moles NH 3 2 moles NH 3 1 mole N 2 69
70
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Moles to Grams Step 3 Convert moles of given to moles of needed using the mole–mole factor. 1.1 mole N 2 x 2 moles NH 3 = 2.2 moles of NH 3 1 mole N 2 70
71
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Moles to Grams Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of NH 3 = 17.0 g of NH 3 1 mole NH 3 and 17.0 g NH 3 17.0 g NH 3 1 mole NH 3 2.2 moles NH 3 x 17.0 g NH 3 = 37 g of NH 3 1 mole NH 3 71
72
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check How many grams of O 2 are needed to produce 45.8 grams of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) A. 38.4 g of O 2 B. 13.8 g of O 2 C. 1.38 g of O 2 72
73
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution How many grams of O 2 are needed to produce 45.8 grams of Fe 2 O 3 in the following reaction? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Step 1 Use molar mass to convert grams of given to moles. 1 mole of Fe 2 O 3 = 159.7 g of Fe 2 O 3 1 mole Fe 2 O 3 and 159.7 g Fe 2 O 3 159.7 g Fe 2 O 3 1 mole Fe 2 O 3 45.8 g Fe 2 O 3 x 1 mole Fe 2 O 3 = 0.287 mole of Fe 2 O 3 159.7 g Fe 2 O 3 73
74
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Step 2 Write a mole–mole factor from the coefficients in the equation. 3 moles of O 2 = 2 moles of 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 Step 3 Convert moles of given to moles of needed using the mole–mole factor. 0.287 mole Fe 2 O 3 x 3 moles O 2 = 0.430 mole of O 2 2 moles Fe 2 O 3 74
75
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Step 4 Convert moles of needed substance to grams using molar mass. 1 mole of O 2 = 32.0 g of O 2 1 mole O 2 and 32.0 g O 2 32.0 gO 2 1 mole O 2 0.430 mole O 2 x 32.0 g O 2 = 13.8 g of O 2 1 mole O 2 The answer is B, 13.8 g of O 2. 75
76
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 5 Chemical Quantities and Reactions 5.8 Energy in Chemical Reactions 76
77
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Molecules Must Collide for Reaction Three conditions for a reaction to occur are: 1. collision: The reactants must collide. 2. orientation: The reactants must align properly to break and form bonds. 3. energy: The collision must provide the energy of activation. 77
78
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Exothermic Reaction In an exothermic reaction, heat is released the energy of the products is less than the energy of the reactants heat is a product C(s) + 2H 2 (g) CH 4 (g) + 18 kcal 78
79
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Endothermic Reactions In an endothermic reaction, heat is absorbed the energy of the products is greater than the energy of the reactants heat is a reactant (added) 79 N 2 (g) + O 2 (g) + 43.3 kcal 2NO(g)
80
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Identify each reaction as exothermic or endothermic. A. N 2 (g) + 3H 2 (g) 2NH 3 (g) + 22 kcal B. CaCO 3 (s) + 133 kcal CaO(s) + CO 2 (g) C. 2SO 2 (g) + O 2 (g) 2SO 3 (g) + heat 80
81
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Reaction Rate The reaction rate is the speed at which reactant is used up is the speed at which product forms increases when temperature rises because reacting molecules move faster, providing more colliding molecules with energy of activation increases with increase in concentration of reactants 81
82
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Catalyst A catalyst increases the rate of a reaction lowers the energy of activation is not used up during the reaction 82
83
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Factors That Increase Reaction Rate 83
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.