3 Chemical Equations &
Reaction Stoichiometry
2
Chapter Three Goals 1. Chemical Equations 2. Calculations Based on Chemical Equations 3. The Limiting Reactant Concept 4. Percent Yields from Chemical Reactions 5. Sequential Reactions 6. Concentrations of Solutions 7. Dilution of solutions 8. Using Solutions in Chemical Reactions 9. Synthesis Question
3
Chemical Equations
• Symbolic representation of a chemical reaction that shows:
1. reactants on left side of reaction 2. products on right side of equation 3. relative amounts of each using
stoichiometric coefficients
4
Chemical Equations
• Attempt to show on paper what is happening at the laboratory and molecular levels.
5
Chemical Equations
• Look at the information an equation provides:
Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →
6
Chemical Equations
• Look at the information an equation provides:
reactants yields products
Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →
7
Chemical Equations
• Look at the information an equation provides:
reactants yields products
1 formula unit 3 molecules 2 atoms 3 molecules
Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →
8
Chemical Equations
• Look at the information an equation provides:
reactants yields products
1 formula unit 3 molecules 2 atoms 3 molecules
1 mole 3 moles 2 moles 3 moles
Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →
9
Chemical Equations
• Look at the information an equation provides:
reactants yields products 1 formula unit 3 molecules 2 atoms
3 molecules 1 mole 3 moles 2 moles
3 moles 159.7 g 84.0 g 111.7 g 132g
Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →
10
Chemical Equations
• Law of Conservation of Matter – There is no detectable change in quantity of matter in
an ordinary chemical reaction. – Balanced chemical equations must always include the
same number of each kind of atom on both sides of the equation.
– This law was determined by Antoine Lavoisier. • Propane,C3H8, burns in oxygen to give carbon
dioxide and water.
OH 4 CO 3 O 5 HC 22283 +→+ ∆
11
Law of Conservation of Matter
• NH3 burns in oxygen to form NO & water You do it!
12
Law of Conservation of Matter
• NH3 burns in oxygen to form NO & water
OH 6 + NO 4 O 5 + NH 4correctlyor
OH 3 + NO 2 O + NH 2
223
2225
3
→
→
∆
∆
13
Law of Conservation of Matter
• C7H16 burns in oxygen to form carbon dioxide and water.
You do it!
14
Law of Conservation of Matter
• C7H16 burns in oxygen to form carbon dioxide and water.
OH 8 + CO 7 O 11 + HC 222167 →∆
15
Law of Conservation of Matter
• C7H16 burns in oxygen to form carbon dioxide and water.
• Balancing equations is a skill acquired
only with lots of practice – work many problems
OH 8 + CO 7 O 11 + HC 222167 →∆
16
Calculations Based on Chemical Equations • Can work in moles, formula units, etc. • Frequently, we work in mass or weight
(grams or kg or pounds or tons).
Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →
17
Calculations Based on Chemical Equations • Example 3-1: How many CO molecules
are required to react with 25 formula units of Fe2O3?
CO of molecules 75unit formula OFe 1
molecules CO 3OFe units formula 25 = molecules CO ?32
32
=
×
18
Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?
325 OFe units formula 102.50=atoms Fe ? ×
19
Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?
=×
×
OFe units formula 1atoms Fe 2
OFe units formula 102.50=atoms Fe ?
32
325
20
Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?
atoms Fe 105.00 OFe units formula 1
atoms Fe 2OFe units formula 102.50=atoms Fe ?
5
32
325
×=×
×
21
Calculations Based on Chemical Equations • Example 3-3: What mass of CO is
required to react with 146 g of iron (III) oxide?
32
3232 OFe g 7.159
OFe mol 1OFe g 146 = CO g ? ×
22
Calculations Based on Chemical Equations • Example 3-3: What mass of CO is
required to react with 146 g of iron (III) oxide?
3232
3232 OFe mol 1
CO mol 3OFe g 7.159
OFe mol 1OFe g 146 = CO g ? ××
23
Calculations Based on Chemical Equations • Example 3-3: What mass of CO is
required to react with 146 g of iron (III) oxide?
CO g 8.76CO mol 1CO g 28.0
OFe mol 1CO mol 3
OFe g 7.159OFe mol 1OFe g 146 = CO g ?
3232
3232
=×
××
24
Calculations Based on Chemical Equations • Example 3-4: What mass of carbon
dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?
OFe mol 1CO mol 3OFe mol 540.0CO g ?
32
2322 ×=
25
Calculations Based on Chemical Equations • Example 3-4: What mass of carbon
dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?
2
2
32
2322 CO mol 1
CO g 0.44 OFe mol 1
CO mol 3OFe mol 540.0CO g ? ××=
26
Calculations Based on Chemical Equations • Example 3-4: What mass of carbon
dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?
? g CO mol Fe O3 mol CO
1 mol Fe O g CO
mol CO = 71.3 g CO
2 2 32
2 3
2
2
2
= × ×0 54044 01
..
27
Calculations Based on Chemical Equations • Example 3-5: What mass of iron (III) oxide
reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?
You do it!
28
Calculations Based on Chemical Equations
3232
32
2
32
2
2232
O Feg 5.10O Femol 1O Feg 7.159
CO mol 3 O Femol1
CO g 44.0 molCO 1CO g 8.65O Feg ?
=
×××=
• Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?
29
Calculations Based on Chemical Equations • Example 3-6: How many pounds of carbon
monoxide would react with 125 pounds of iron (III) oxide?
You do it!
30
Calculations Based on Chemical Equations
CO lb 7.65CO g 454
CO lb 1CO mol 1CO g 28
OFe mol 1CO mol 3
OFe g 7.159OFe mol 1
OFe lb 1OFe g 454OFe lb 125 = CO lb ?
3232
32
32
3232
=×
×××
×
YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!!
31
Limiting Reactant Concept
• Kitchen example of limiting reactant concept.
1 packet of muffin mix + 2 eggs + 1 cup of milk → 12 muffins
• How many muffins can we make with the following amounts of mix, eggs, and milk?
32
Limiting Reactant Concept • Mix Packets Eggs Milk 1 1 dozen 1 gallon
limiting reactant is the muffin mix 2 1 dozen 1 gallon 3 1 dozen 1 gallon 4 1 dozen 1 gallon 5 1 dozen 1 gallon 6 1 dozen 1 gallon 7 1 dozen 1 gallon
limiting reactant is the dozen eggs
33
Limiting Reactant Concept
• Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box?
( )( )( )
numbersmallest by the determinedsets 55 is make can number we maximum the
sets 99nut 1set 1nuts 99
sets 55 washers2set 1 washers110
sets 87bolt 1set 1bolts 87
=
=
=
34
Limiting Reactant Concept
• Look at a chemical limiting reactant situation.
Zn + 2 HCl→ ZnCl2 + H2
35
Limiting Reactant Concept
• Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?
CS2 + 3 O2 → CO2 + 2 SO2
36
Limiting Reactant Concept
• Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?
CS2 + 3 O2 → CO2 + 2 SO2
1 mol 3 mol 1 mol 2 mol
37
Limiting Reactant Concept
• Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?
CS O CO 2 SO1 mol 3 mol 1 mol 2 mol76.2 g 3(32.0 g) 44.0 g 2(64.1 g)
2 2 2 2+ → +3
38
Limiting Reactant Concept
• Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?
g 76.2CS mol 1CS g 6.95 SO mol ?
SO 2 CO O 3 CS
222
2222
×=
+→+
39
Limiting Reactant Concept
22
2
2
2
222
2222
SO g 161SO mol 1
SO g 1.64CS mol 1
SO mol 2g 76.2
CS mol 1CS g 6.95 SO g ?
SO 2 CO O 3 CS
=××
×=
+→+
What do we do next? You do it!
40
Limiting Reactant Concept
22
2
2
2
2
222
22
2
2
2222
2222
SO g 147SO mol 1
SO g 1.64O mol 3
SO mol 2O g 32.0O mol 1O g 110SO g ?
SO g 161SO mol 1
SO g 1.64CS mol 1
SO mol 2g 76.2
CS mol 1CS g 6.95 SO g ?
SO 2 CO O 3 CS
=×××=
=×××=
+→+
• Which is limiting reactant? • Limiting reactant is O2. • What is maximum mass of sulfur dioxide? • Maximum mass is 147 g.
41
Percent Yields from Reactions • Theoretical yield is calculated by assuming that
the reaction goes to completion. – Determined from the limiting reactant calculation.
• Actual yield is the amount of a specified pure product made in a given reaction. – In the laboratory, this is the amount of product that is
formed in your beaker, after it is purified and dried. • Percent yield indicates how much of the product
is obtained from a reaction.
% yield = actual yieldtheoretical yield
× 100%
42
Percent Yields from Reactions
• Example 3-9: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?
43
Percent Yields from Reactions
523
52
52352523
2523523
HCOOCCH g 1.19 OHHC g 0.46
HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?
yield al theoretic theCalculate 1.OH HCOOCCH OHHC + COOHCH
=
×
+→
44
Percent Yields from Reactions
523
52
52352523
2523523
HCOOCCH g 1.19 OHHC g 0.46
HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?
yield al theoretic theCalculate 1.OH HCOOCCH OHHC + COOHCH
=
×
+→
45
Percent Yields from Reactions
yield.percent theCalculate .2HCOOCCH g 1.19
OHHC g 0.46 HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?
yield al theoretic theCalculate 1.OH HCOOCCH OHHC + COOHCH
523
52
52352523
2523523
=
×
+→
46
Percent Yields from Reactions
%5.77%100HCOOCCH g 19.1
HCOOCCH g 14.8= yield %
yield.percent theCalculate .2HCOOCCH g 1.19
OHHC g 0.46 HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?
yield al theoretic theCalculate 1.OH HCOOCCH OHHC + COOHCH
523
523
523
52
52352523
2523523
=×
=
×
+→
47
Sequential Reactions N O2 NH2
HNO3
H2SO4
Sn
Conc. HCl
benzene nitrobenzene aniline
• Example 3-10: Starting with 10.0 g of benzene (C6H6), calculate the theoretical yield of nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).
48
Sequential Reactions
××benzene g 0.78benzene mol 1benzene g 10.0 = nenitrobenze g ?
49
Sequential Reactions
nenitrobenze g 8.15nenitrobenze mol 1nenitrobenze g 0.123
benzene mol 1nenitrobenze mol 1
benzene g 0.78benzene mol 1benzene g 10.0 = nenitrobenze g ?
=×
××
• Next calculate the mass of aniline produced. You do it!
50
Sequential Reactions N O2 NH2
HNO3
H2SO4
Sn
Conc. HCl
benzene nitrobenzene aniline
××nenitrobenze g 123.0
nenitrobenze mol 1nenitrobenze g 15.8 = aniline g ?
51
Sequential Reactions N O2 NH2
HNO3
H2SO4
Sn
Conc. HCl
benzene nitrobenzene aniline
? g aniline = 15.8 g nitrobenzene 1 mol nitrobenzene123.0 g nitrobenzene
1 mol aniline1 mol nitrobenzene
93.0 g aniline1 mol aniline
g aniline
× ×
× = 119.
52
Sequential Reactions
• If 6.7 g of aniline is prepared from 10.0 g of benzene, what is the percentage yield?
You do it!
%56%100aniline g 11.9aniline g 6.7 = yield % =×
53
Concentration of Solutions • Solution is a mixture of two or more substances
dissolved in another. – Solute is the substance present in the smaller amount. – Solvent is the substance present in the larger amount. – In aqueous solutions, the solvent is water.
• The concentration of a solution defines the amount of solute dissolved in the solvent. – The amount of sugar in sweet tea can be defined by its
concentration. • One common unit of concentration is:
w/w% symbol thehas solute of massby %solvent of mass + solute of mass =solution of mass
%100solution of masssolute of mass = solute of massby % ×
54
Concentration of Solutions
• Example 3-11: What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH?
NaOH g 0.20solution g 100.0NaOH g 8.00solution g 0.250 =
55
Concentration of Solutions
• Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.
nsol' g .400NaOH g 8.00
solution g 100.0NaOH g 32.0 =solution g ?
=
×
56
Concentration of Solutions
• Example 3-13: Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL.
You do it!
NaOH g 2.26nsol' g 100
NaOH g 8.00
nsol' mL 1nsol' g 1.09nsol' mL 300.0 = NaOH g ?
=
××
57
Concentrations of Solutions
• Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.
You do it!
58
Concentrations of Solutions
• Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.
solution mL .300solution g 1.11solution mL 1
KOH g 12.0solution g 100.0KOH g 40.0 =solution mL ?
=
××
59
Concentrations of Solutions
• Second common unit of concentration:
mLmmol
Lmoles
solution of liters ofnumber solute of moles ofnumber molarity
=
=
=
M
M
60
Concentrations of Solutions
• Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.
You do it!
61
Concentrations of Solutions
• Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.
42
424242
SOH g 98.1SOH mol 1
nsol' L 75.1 SOH g 12.5
nsol' LSOH mol ?
×=
62
Concentrations of Solutions
• Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.
42
42
42
424242
SOH 0728.0L
SOH mol 0728.0SOH g 98.1SOH mol 1
nsol' L 75.1 SOH g 12.5
nsol' LSOH mol ?
M=
=
×=
63
Concentrations of Solutions
• Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 .
You do it!
64
Concentrations of Solutions
• Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 .
? g Ca(NO L 0.800 mol Ca(NO
L164 g Ca(NO mol Ca(NO
g Ca(NO
33
3
33
) .)
))
)
22
2
22
3 50
1459
= × ×
=
65
Concentrations of Solutions
• One of the reasons that molarity is commonly used is because:
M x L = moles solute and
M x mL = mmol solute
66
Concentrations of Solutions
• Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?
g/L 1185or g/mL 1.185=density us tells1.185 =gravity specific
67
Concentrations of Solutions
• Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?
××nsol' g 100
HCl g 31.36solution L
solution g 1185 = HCl/L mol ?
1185g/Lor g/mL 1.185=density us tells1.185 =gravity specific
68
Concentrations of Solutions
• Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?
HCl 11.80HCl g 36.46
HCl mol 1nsol' g 100
HCl g 31.36solution L
solution g 1185 = HCl/L mol ?
1185g/Lor g/mL 1.185=density us tells1.185 =gravity specific
M=
××
69
Dilution of Solutions
• To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet.” – How fountain drinks are made from syrup.
• The number of moles of solute in the two solutions remains constant.
• The relationship M1V1 = M2V2 is appropriate for dilutions, but not for chemical reactions.
70
Dilution of Solutions
• Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.
71
Dilution of Solutions
• Example 3-18: If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution?
M
MM
MMMM
20.1mL 100.0
mL 0.100.12mL 100.0mL 0.10 0.12
VV
2
2
2211
=
×=
×=×=
72
Dilution of Solutions
• Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?
You do it!
73
Dilution of Solutions
• Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?
mL 333or L 0.333 18.0
2.40 L 2.50V
V V
V V
1
1
221
2211
=
×=
×=
=
MM
MMMM
74
Using Solutions in Chemical Reactions
• Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.
75
Using Solutions in Chemical Reactions • Example 3-20: What volume of 0.500 M BaCl2
is required to completely react with 4.32 g of Na2SO4?
NaCl 2 + BaSO BaCl + SONa 4242 →
76
Using Solutions in Chemical Reactions • Example 3-20: What volume of 0.500 M BaCl2
is required to completely react with 4.32 g of Na2SO4?
××=
→
42
42422
4242
SONa g 142SONa mol 1 SOgNa 4.32 BaCl L ?
NaCl 2 + BaSO BaCl + SONa
77
Using Solutions in Chemical Reactions • Example 3-20: What volume of 0.500 M BaCl2
is required to completely react with 4.32 g of Na2SO4?
L 0.0608 BaCl mol 0.500
BaCl L 1SONa mol 1
BaCl mol 1SONa g 142SONa mol 1 SOgNa 4.32 BaCl L ?
NaCl 2 + BaSO BaCl + SONa
2
2
42
2
42
42422
4242
=×
××=
→
78
Using Solutions in Chemical Reactions • Example 3-21: (a)What volume of 0.200
M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO3)3?
( ) ( )You do it!
3333 NaNO 3OHAlNaOH 3NOAl +→+
79
Using Solutions in Chemical Reactions • Example 3-20: (a)What volume of 0.200 M
NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?
( )
nsol' NaOH mL 150or L 0.150 NaOH mol 0.200
NaOH L 1)Al(NO mol 1
NaOH mol 3 nsol' )Al(NO L 1
n sol' )Al(NO mol 0.200mL 1000
L 1nsol' )Al(NO mL 50.0 = NaOH mL ?
NaNO 3Al(OH)NaOH 3NOAl
3333
33
33
3333
=
××
×
+→+
80
Using Solutions in Chemical Reactions • (b)What mass of Al(OH)3 precipitates in
(a)? You do it!
81
Using Solutions in Chemical Reactions • (b) What mass of Al(OH)3 precipitates in
(a)?
3
3
3
33
3
33
33
333
Al(OH) g 780.0Al(OH) mol 1Al(OH) g 0.78
)Al(NO mol 1Al(OH) mol 1
nsol' )Al(NO L1 )Al(NO mol 0.200
mL 1000 L1nsol' )Al(NO mL 50.0 Al(OH) g ?
=
××
×=
82
Using Solutions in Chemical Reactions • Titrations are a method of determining the
concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. – Requires special lab glassware
• Buret, pipet, and flasks – Must have an indicator also
83
Using Solutions in Chemical Reactions • Example 3-22: What is the molarity of a KOH
solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?
OH + KCl HCl + KOH 2→
84
Using Solutions in Chemical Reactions • Example 3-22: What is the molarity of a KOH
solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?
HCl mmol 9.63 = HCl 0.223 mL 43.2OH + KCl HCl + KOH 2
M×→
85
Using Solutions in Chemical Reactions • Example 3-22: What is the molarity of a KOH
solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?
KOH mmol 63.9HCl mmol 1
KOH mmol 1HCl mmol 9.63
HCl mmol 9.63 = HCl 0.223 mL 43.2OH + KCl HCl + KOH 2
=×
×→
M
86
Using Solutions in Chemical Reactions • Example 3-22: What is the molarity of a KOH
solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?
KOH 249.0KOH mL 38.7KOH mmol 9.63
KOH mmol 63.9HCl mmol 1
KOH mmol 1HCl mmol 9.63
HCl mmol 9.63 = HCl 0.223 mL 43.2OH + KCl HCl + KOH 2
M
M
=
=×
×→
87
Using Solutions in Chemical Reactions • Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?
HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1OH 2 + BaCl HCl 2 + Ba(OH) 222
M→
88
Using Solutions in Chemical Reactions • Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?
HCl mmol 2 Ba(OH) mmol 1HCl mmol 54.4
HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1OH 2 + BaCl HCl 2 + Ba(OH)
2
222
×
→M
89
Using Solutions in Chemical Reactions • Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?
2
2
222
Ba(OH) mmol 2.27 HCl mmol 2
Ba(OH) mmol 1HCl mmol 54.4
HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1OH 2 + BaCl HCl 2 + Ba(OH)
=
×
→M
90
Using Solutions in Chemical Reactions • Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?
22
2
2
2
222
Ba(OH) 0593.0Ba(OH) mL 3.38Ba(OH) mmol 27.2
Ba(OH) mmol 2.27 HCl mmol 2
Ba(OH) mmol 1HCl mmol 54.4
HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1OH 2 + BaCl HCl 2 + Ba(OH)
M
M
=
=
×
→
91
Synthesis Question
• Nylon is made by the reaction of hexamethylene diamine
CH2CH2CH2
CH2CH2CH2NH2
NH2
C
OH
O CH2
CH2
CH2
CH2
C
OH
O
with adipic acid.
92
Synthesis Question
in a 1 to 1 mole ratio. The structure of nylon is:
CNH
O
CH2
CH2
CH2
CH2
CH2
CH2
C
O
CH2
CH2
CH2
NH ** n
where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day?
93
Synthesis Question
moleculesnylon g/mol 10 1.02 450,000 g/mol] [226
000,450 ])162()142()221( 12) (12[ moleculenylon 1 of massMolar
8
units of #atoms Oatoms Natoms Hatoms C
×=
=
×+×+×+×=
94
Synthesis Question
g 106.81 lb
g 454lb) 10 (1.5 poundsmillion 1.5
moleculesnylon g/mol 10 1.02 450,000 g/mol] [226
000,450 ])162()142()221( 12) (12[ moleculenylon 1 of massMolar
8
6
8
units of #atoms Oatoms Natoms Hatoms C
×=
×=
×=
=
×+×+×+×=
95
Synthesis Question
( )nylon of mol 6.68
g 101.02nylon mol 1 g 106.81 moleculesnylon of mol #
g 106.81 lb
g 454lb) 10 (1.5 poundsmillion 1.5
moleculesnylon g/mol 10 1.02 450,000 g/mol] [226
000,450 ])162()142()221( 12) (12[ moleculenylon 1 of massMolar
88
8
6
8
units of #atoms Oatoms Natoms Hatoms C
=
×
×=
×=
×=
×=
=
×+×+×+×=
96
Synthesis Question
:requiresnylon of mol 6.68 make toformed,nylon of moleper 450,000 diamine enehexamethyl of mole 1 plus
450,000 acid adipic of mole 1 usesreaction formation nylon theBecause×
×
97
Synthesis Question
( ) lb 1066.9g 454
lb 1g 104.39 g/mol 146450,000 6.68 - acid adipic
:requiresnylon of mol 6.68 make toformed,nylon of moleper 450,000 diamine enehexamethyl of mole 1 plus
450,000 acid adipic of mole 1 usesreaction formation nylon theBecause
58 ×=
×=××
××
98
Synthesis Question
( )
( ) lb 1068.7g 454
lb 1g 1049.3 g/mol 116450,000 6.68 - diamine enehexamethyl
lb 1066.9g 454
lb 1g 104.39 g/mol 146450,000 6.68 - acid adipic
:requiresnylon of mol 6.68 make toformed,nylon of moleper 450,000 diamine enehexamethyl of mole 1 plus
450,000 acid adipic of mole 1 usesreaction formation nylon theBecause
58
58
×=
×=××
×=
×=××
××
99
Group Activity
Manganese dioxide, potassium hydroxide and oxygen react in the following fashion:
OH 2 KMnO 4 O 3 + KOH 4 + MnO 4 2422 +→ A mixture of 272.9 g of MnO2, 26.6 L of 0.250
M KOH, and 41.92 g of O2 is allowed to react as shown above. After the reaction is finished, 234.6 g of KMnO4 is separated from the reaction mixture. What is the per cent yield of this reaction?
3 Chemical Equations &
Reaction Stoichimoetry