Chemicals reactions, Limit reactant, Yield.

1. Chapter 9
Stoichiometry

2. Introduction to Stoichiometry
Section 1

3.  Much knowledge of chemistry based on
quantitative analysis of substances in
chemical reactions
 Composition stoichiometry  deals
with mass relationships of elements in
compounds
 Reaction stoichiometry  deals with
mass relationships between reactants
and products in a chemical reaction

4.  Based on chemical equations and law of
conservation of matter
 All calculations start with balanced
chemical equation
 This gives numbers of moles of
reactants and products

5. Reaction-Stoichiometry Problems
 Can be classified according to
information given in problem and info
you are expected to find, the unknown
 May both be reactants, may both be
products, or both
 Masses usually expressed in grams
 Solved by using ratios to convert given
quantities by following methods:

6. Problem Type 1:
 Given and unknown quantities are
amounts in moles
 General plan:
Amt of given (mol)  amt of unknown
(mol)

7. Problem Type 2:
 Given is in moles and unknown is a
mass in grams
Amt given (mol)  amt unknown (mol) 
amt unknown (g)

8. Problem Type 3:
 Given is mass (g) and unknown is in
moles
Mass given (g)  amt given (mol) 
amount unknown (mol)

9. Problem Type 4:
 Given is mass (g) and unknown is
mass (g)
Mass given  amt given (mol)  amt
unknown (mol)  mass unknown

10. Mole Ratio
 Solving any reaction-stoichiometry
problem requires use of mole ratio to
convert from moles to grams
 Mole ratio  conversion factor that
relates amounts in moles of any two
substances involved in a chemical
reaction
 Get directly from balanced equation

11. Example: 2Al2O3(l)  4Al(s) + 3O2(g)
 Moles ratios:

12. 2Al2O3(l)  4Al(s) + 3O2(g)
 Use mole ratios to convert from amount
in moles of one substance to amount in
moles of another
 Ex. 13.0 mol Al2O3

13. Molar Mass
 Mass of one mole of a substance
 It is the conversion factor that relates
mass of substance to amount in moles
 To solve stoichiometry problems, you
need to determine molar mass of
substances

14.  Molar masses of substances:
 Al2O3 = 101.96 g/mol
 O2 = 32.00 g/mol
 Al = 26.98 g/mol
 Can use these as conversion factors

16.  Find number of grams of Al equal to 26.0 mol of
Al

17. Ideal Stoichiometric Calculations
Section 2

18. Ideal Stoichiometric Calculations
 Equations are very important because
you get the mole ratio directly from it
 First thing to solving these types of
problems is balancing the equation
 Chemical equations help make
predictions about reactions without
having to run experiments (and waste
resources) in lab

19.  Calculations in this book are theoretical
 They tell amounts of reactants and
products under ideal conditions (where
all reactants completely converted into
products)
 Hardly ever happens in real life

20. Conversions of Quantities in
Moles
 If asked for moles of product made from specific
number of moles of reactant:

21.  Plan requires 1 conversion factor
 Mole ratio

22. Sample Problem

23. 1. Analyze
 Given:
 Amount of CO2 = 20 mol
 Unknown:
 Amount of LiOH in moles

24. 2. Plan
 Amount CO2 (mol)  amount LiOH (mol)
 Requires a mole to mole ratio between CO2 and
LiOH

25. 3. Compute

26.  How many moles of sodium will react with
water to produce 4.0 mol of hydrogen in the
following reaction?
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
 8.0 mol Na

27.  How many moles of lithium chloride will be
formed by the reaction of chlorine with 0.046
mol of lithium bromide in the following
reaction?
2LiBr(aq) + Cl2(g)  2LiCl(aq) + Br2(l)
 0.046 mol LiCl

28. Conversions of Amounts in Moles to
Mass
 Usually asked to find mass in grams of product
formed

29. Practice Problem
 In photosynthesis, plants use energy from the
sun to produce glucose, C6H12O6, and oxygen
from the reaction of carbon dioxide and water.
What mass, in grams, is produced when 3.00
mol of water react with carbon dioxide?

30. 1. Analyze
 Given:
 Amount H2O = 3.00 mol
 Unknown:
 Mass of glucose produced (in g)

31. 2. Plan
 Chemical equation is
 Need 2 conversion factors:
 Mole ratio of CO2 to H2O
 Molar mass of CO2

33. 3. Compute

34. Practice Problem 1
 Phosphorous burns in air to produce a
phosphorous oxide in the following reaction.
4P(s) + 5O2(g)  P4O10(s)
What mass of phosphorous will be needed to
produce 3.25 mol P4O10?
 403 g

35. Practice Problem 2
 Hydrogen peroxide breaks down, releasing
oxygen, in the following reaction.
2H2O2(aq)  2H2O(l) + O2(g)
What mass of oxygen is produced when 1.840
mol of H2O2 decompose?
 29.44 g

36. Conversions of Mass to Amounts in
Moles
 In this type of problem you are starting with mass
of some substance
 Plan:

37. Sample Problem

38. 1. Analyze
 Given:
 Mass of NH3 = 824 g
 Unknown:
 a. amount of NO produced (in mol)
 b. amount of H2O produced (in mol)

39. 2. Plan
 First write balanced equation

40.  Mole ratios needed:

41. 3. Compute

42. Mass-Mass Calculations
 Since you can never measure moles
directly, mass-mass calculations are more
common

43. Sample Problem

44. 1. Analyze
 Given:
 Amount of HF = 30.00 g
 Unknown:
 Mass of SnF2 produced in grams

45. 2. Plan

46. 3. Compute

47. Practice Problem 1
 Calculate the mass of silver bromide
produced from 22.5 g of silver nitrate in the
following reaction:
2AgNO3 + MgBr2  2AgBr + Mg(NO3)2
 24.9 g AgBr

48. Practice Problem 2
 What mass of acetylene, C2H2, will be
produced from the reaction of 90. g of calcium
carbide, CaC2, with water in the following
reaction?
CaC2(s) + 2H2O(l)  C2H2(s) + Ca(OH)2(aq)
 37 g

49. Limiting Reactants and Percent
Yield
Section 3

50.  In experiments, a reaction is rarely ever done with
exact amounts of reactants
 Usually one or more reactants is in excess (too
much)
 Once one of the reactants is used up, the
reaction stops
 Substance used up first is the limiting reactant

51.  Limiting reactant  the reactant that limits the
amounts of the other reactants that can combine and
the amount of product that can form in a chemical
reaction
 Excess reactant  the substance that is not used up
completely in reaction
 Consider the following reaction
C(s) + O2(g)  CO2(g)
 According to equation, 1 mol C reacts with 1 mol
oxygen to form 1 mol carbon dioxide
 Supposed you could mix 5 mol C with 10 mol O2

52.  Consider the following reaction
C(s) + O2(g)  CO2(g)
 According to equation, 1 mol C reacts with 1 mol
oxygen to form 1 mol carbon dioxide
 Supposed you could mix 5 mol C with 10 mol O2

53. C(s) + O2(g)  CO2(g)


54. Problem Limit Reactant
Problem LR :Aluminum oxidizes according
to the following equation: 4Al + 3O2 
2Al2O3
Powdered Al (0.048 mol) is placed into a
container containing 0.030 mol O2. What is
the limiting reactant?
O2

55. 1. Analyze
 Given:
 Amount of HF = 2.0 mol
 Amount of SiO2 = 4.5 mol
 Unknown:
 Limiting reactant

56. 2. Plan
 The given amount of either reactant is used to
calculate required amount of other reactant
 Calculated amount compared with amount you
actually have
 Limiting reactant can be identified

58.  Under ideal conditions, 2.0 mol HF requires 0.50
mol SiO2 for complete reaction
 Because 4.5 mol SiO2 is available, that is more
than is required so….
 HF is limiting reactant

59. Practice Problem 1
 Aluminum oxidizes according to the following
equation: 4Al + 3O2  2Al2O3
 Powdered Al (0.048 mol) is placed into a
container containing 0.030 mol O2. What is
the limiting reactant?
 O2

60. Practice Problem 2
 Heating zinc sulfide in the presence of oxygen
yields the following:
ZnS + O2  ZnO + SO2
 If 1.72 mol ZnS is heated in the presence of 3.04
mol O2, which reactant will be used up?
 ZnS

61. Practice Problem 3
 Use the following equation for the oxidation of
aluminum in the following problems:
4Al + 3O2  2Al2O3

62. 4Al + 3O2  2Al2O3
 Which reactant is limiting if 0.32 mol Al and
0.26 mol O2 are available?
 Al
 How many moles of Al2O3 are formed from the
reaction of 6.38 x 10-3 mol of O2 and 9.15 x 10-3
mol of Al?
 4.25 x 10-3 mol Al2O3
 If 3.17 g Al and 2.55 g of O2 are
available, which reactant is limiting?
 O2

63. Practice Problem 4
ZrSiO4 + 2Cl2  ZrCl4 + SiO2 + O2
 What mass of ZrCl4 can be produced if 862 g
of ZrSiO4 and 950 g of Cl2 are available?
 1.10 x 103 g

64. Percent Yield
 Amounts of products calculated in problems
represent theoretical yields
 Theoretical yield  maximum amount of product
that can be made from a given amount of
reactant
 In lab, amount of product is usually less than
theoretical yield

65. Why?
 Some reactant may be used competing in side
reactions that reduce amount of product
 Once product is formed, usually collected in
impure form
 Some product lost during purification
 Actual yield  measured amount of product
gotten from reaction

66.  Chemists usually interested in efficiency of
reaction
 Expressed by comparing actual and theoretical
yields
 Percent yield  ratio of actual yield to theoretical
yield multiplied by 100

67. Sample Problem

68. 1. Analyze
 Given:
 Mass C6H6 = 36.8 g
 Mass of Cl2 = excess
 Actual yield of C6H5Cl = 38.8 g
 Unknown:
 Percent yield of C6H5Cl

69. 2. Plan
 First do mass-mass calculation to find theoretical
yield

70.  Then percent yield can be found

71. 3. Compute

72. Practice Problem
 Calculate the percent yield in each of the
following cases
 T Yield = 50.0 g, A Yield = 41.9 g
 83.8%
 T yield = 290 kg, A Yield = 270 kg
 93%
 T Yield = 0.00192 g, A yield = 0.00089 g
 46%