3
Chemical Equations &
Reaction Stoichiometry
化學方程式
化學平衡及化學計量
CH4 燃燒
產生 H2O, CO2
Chapter Three Goals
1. Chemical Equations 化學方程式
2. Calculations Based on Chemical Equations
3. The Limiting Reactant Concept 限量反應物概念
4. Percent Yields from Chemical Reactions
5. Sequential Reactions 連續性反應
6. Concentrations of Solutions 溶液濃度
7. Dilution of solutions 溶液的稀釋
8. Using Solutions in Chemical Reactions
2
Chemical Equations 化學方程式
Chemical Equations
1. The substances that react, called reactants
reactants on left side of reaction
反應物
2. The substances formed, called products
products on right side of equation
產物
3. The relative amounts of the substances
involved
relative amounts of each using stoichiometric
coefficients係數
represent the number of molecule
3
Coefficients indicate the amount of each
Compound or element present and CAN be changed
to balance the equation
CH4 + 2O2
CO2+ 2H2O
Reactants
Products
Subscripts indicate the number of atoms of each
element present in the compound or element
present and CANNOT be changed when
balancing an equation
Chemical equations are based on experimental observation
4
Chemical Equations
Low of Conservation of Matter
• Attempt to show on paper what is happening at the
laboratory and molecular levels.
Ball-and-stick models
Chemical formula
Space-filling model
5
Chemical Equations
•Law of Conservation of Matter ：物質守恆定律
在任何物理與化學作用中,物質既不會被創造也不會被消
滅,而只是從一種狀態轉變到另一種狀態
– 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. the
father of modern chemistry
•Propane,C3H8, 丙烷burns in oxygen to give carbon dioxide
and water.
C3H8 + 5O2
3CO2+ 4H2O
6
Law of Conservation of Matter
reactants
C2H6O + O2
products
2CO2+ 3H2O
3 X 2 =6
atom count
2C, 6H, 3O
2C, 6H, 7O
Law of Conservation of Matter
4 oxygen atoms = 2O2
+
C2H6O + O2
2CO2+ 3H2O
C2H6O + 3 O2
2CO2+ 3H2O
Final atom count
2C, 6H, 7O
C2H6O + 3O2
2C, 6H, 7O
2CO2+ 3H2O
Law of Conservation of Matter
Al +3 HCl
atom count: Al, 3H, 3Cl
Al +6HCl
atom count:
Al, 6H, 6Cl
Al +6HCl
atom count:
Al, 6H, 6Cl
2Al +6HCl
Final atom count
2Al, 6H, 6Cl
2Al + 6HCl
AlCl3+ H2
Al, 2H, 3Cl
AlCl3+3 H2
Al, 6H, 3Cl
2AlCl3+ 3H2
2Al, 6H, 6Cl
2AlCl3+ 3H2
2Al, 6H, 6Cl
2AlCl3+ 3H2
Law of Conservation of Matter
• NH3 burns in oxygen to form NO & water
NH3 + O2
4NH3 + 5O2
NO2+ H2O
4NO2+ 6H2O
C7H16 + O2
C7H16 + 11O2
CO2+ H2O
7CO2+ 8H2O
• C7H16 burns in oxygen to form carbon dioxide
and water.
•Balancing equations is a skill acquired only with
lots of practice
•work many problems
10
Law of Conservation of Matter
Example 3-1 Balancing Chemical Equation
Balance the following chemical equations:
(a) P4 + Cl2
PCl3
(b) RbOH + SO2
Rb2SO3 + H2O
(b) P4O10 + Ca(OH)2
Ca3(PO4)2 + H2O
(a) P4 + 6Cl2
4 PCl3
(b) 2RbOH + SO2
Rb2SO3 + H2O
(b) P4O10+6Ca(OH)2
2Ca3(PO4)2+6 H2O
Exercise 8, 10
11
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).
Fe2O3 + 3CO
reactants
1 formula unit 3 molecules
1 mole
3 moles
159.7 g
84.0 g
2Fe2+ 3CO2
yields
products
2 atoms
2 moles
111.7 g
3 molecules
3 moles
132g
12
Calculations Based on Chemical
Equations
Example 3-1: How many CO molecules are
required to react with 25 formula units of Fe2O3?
Fe2O3 +3CO
2Fe+3CO2
3CO molecules
?CO molecules=25 formula units Fe2O3 x
1Fe2O3 formula unit
=75 molecule of CO
13
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?
Fe2O3 +3CO
2Fe+3CO2
?Fe atoms=2.5x105 formula units Fe2O3 x
2 Fe atoms
1 Fe2O3 formula unit
=5.0x105 Fe atoms
14
Calculations Based on Chemical
Equations
Example 3-3: What mass of CO is required to react
with 146 g of iron (III) oxide?
Fe2O3 +3CO
2Fe+3CO2
?g CO=146g Fe2O3 x 1mol Fe2O3 x 3 mol CO x 28.0g CO
159.7g Fe2O3 1 mol Fe2O3 1 mol CO
=76.8g CO
15
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?
Fe2O3 +3CO
2Fe+3CO2
?g CO2=0.54mol Fe2O3 x 3 mol CO2 x 44.0g CO2
1 mol Fe2O3 1 mol CO2
=71.3g CO2
16
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?
Fe2O3 +3CO
2Fe+3CO2
?g Fe2O3=8.65g CO2 x 1mol CO2 x 1 mol Fe2O3 x 159.7g Fe2O3
1 mol Fe2O3
44.0g CO2
3 mol CO2
=10.5g Fe2O3
17
Calculations Based on Chemical
Equations
Example 3-6: How many pounds of carbon
monoxide would react with 125 pounds of iron
(III) oxide?
Fe2O3 +3CO
2Fe+3CO2
?lb CO =125lb Fe2O3 x 454g Fe2O3 x 3 mol CO
1lb Fe2O3
159.7 Fe2O3
28g CO
1lb CO
x
x
=65.7lb
1 mol CO
454g CO
CO
YOU MUST BE PROFICIENT WITH THESE
TYPES OF PROBLEMS!!!
18
Calculations Based on Chemical
Equations
Example 3-2, 3-3 Number of Molecules, Number of Moles Formed
1.How many O2 molecule react with 47 CH4 molecules
according to the preceding equation?
2.How many moles of water could be produced by the
reaction of 3.5mol of methane with excess oxygen?
CH4 +2O2
CO2+2H2O
? O2 molecules =47 CH4 molecules x 2 O2 molecule
1 CH4 molecule
=94 O2 molecule
? mol H2 O =3.5mol CH4 x 2 mol H2 O
1 mol CH4
=7.0 mol H2O
Exercise 12, 14,18
19
Example 3-4, 3-5, 3-6, 3-7 Mass of a Reactant Required, Mass of a
Product Formed
1. What mass of O2 is required to react completely with 1.2mol of CH4?
2. What mass of O2 is required to react completely with 24.0g of CH4?
3. What mass of CH4, is required to react with 96.0g of O2?
4. Calculated the mass of CO2, in grams, that can be produced by
burning 6.0mol of CH4 in excess O2.
CH4 +2O2
CO2+2H2O
? O2 g =1.2 mol CH4 x 2 mol O2 x 32.0g O2 =76.8g O2
1 mol CH4 1 mol O2
? O2 mol =24g CH4 x 1mol CH4 x 2 mol O2 x 32.0g O2 =96.0g O2
16.0g CH4 1 mol CH4
1 mol O2
? CH4 mol =96g O2 x 1mol O2 x 1 mol CH4 x 16g CH4 =24.0g CH4
2 mol O2 1 mol CH4
32g O2
? CO2 g =6.0 mol CH4 x 1 mol CO2 x 44.0g CO2 =264.0g CO2
1 mol CH4
1 mol CO2
Exercise 22, 24, 26,28
20
Calculations Based on Chemical
Equations
Example 3-8 Mass of a Reactant Required
Phosphorus, P4, burns with excess oxygen to form tetraphosphorus decoxide, P4
O10. In this reaction, what mass of P4 reacts with 1.5mol of O2?
P4 +5 O2
P4O10
? P4 g =1.5 mol O2 x 1 mol P4 x 124.0g P4 =37.2g P4
1 mol P4
5 mol O2
Exercise 28
21
Limiting Reactant Concept限量反應物
• 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?
•
Mix Packets
1
2
3
4
5
6
7
Eggs
Milk
1 dozen
1 gallon
limiting reactant is the muffin mix
1 dozen
1 gallon
1 dozen
1 gallon
1 dozen
1 gallon
1 dozen
1 gallon
1 dozen
1 gallon
1 dozen
1 gallon
limiting reactant is the dozen eggs
22
Limiting Reactant Concept
CO + 2H2
CH3OH
C
O
H
H2 molecule: excess
That is:
Not enough CO molecules to
react with all H2 molecule
Limiting reactant
(limiting reagent)
23
Example 3-9 Limiting Reactant
What mass of CO2 could be formed by the reaction of
16.0g of CH4 with 48.0g of O2?
CH4 +2O2
CO2+2H2O
CH4 excess
1mol 2mol
1mol 2mol
? mol CH4 =16.0g CH4 x 1mol CH4 =1.0mol CH4
16.0g CH4
? mol O2 =48.0g O2 x 1mol O2 =1.5mol O2
32.0g O2
? mol CH4 =1.5mol O2 x 1mol CH4 =0.75mol CH4
2mol O2
? g CO2 =1.5mol O2 x 1.0mol CO2 x 44.0g CO2
2.0mol O2
1.0mol CO2
=33.0g CO2
Exercise 28
24
Example 3-10 Limiting Reactant
What is the maximum mass of Ni(OH)2 that could be
prepared by mixing two solutions that contain 25.9g of
NiCl2 and 10.0g of NaOH, respectively?
NiCl2 +2NaOH
Ni(OH)2+2NaCl
NiCl2 excess
? mol NiCl2 =25.9g NiCl2 x 1mol NiCl2 =0.2mol NiCl2
129.6g NiCl
1mol
2mol
1mol
2mol
2
? mol NaOH =10.0g NaOH x 1mol NaOH =0.25mol NaOH
40.0gNaOH
? g Ni(OH)2 =0.25mol NaOH x 1.0mol Ni(OH)2 x 92.7g Ni(OH)2
2.0mol NaOH
1.0mol Ni(OH)2
=11.6g Ni(OH)2
Exercise 34,36
25
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 +3O2
CO2+2SO2
CS2 excess
1mol 2mol
1mol CS2
=1.25mol CS2
? mol CS2 =95.6g CS2 x
76.2g CS2
? mol O2 =110.0g O2 x 1mol O2 =3.44mol O2
32.0g O2
1mol
3mol
? g SO2 =3.44mol O2 x 2.0mol SO2 x 64.0g SO2
3.0mol O2
1.0mol SO2
=146.6g SO2
26
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 yield
x100%
Theoretical yield
27
Percent Yields from Reactions
Example 3-11 percent Yield
A 15.6g sample of C6H6 is mixed with excess HNO3. We isolate
18.0g of C6H5NO2. What is the percent yield of C6H5NO2 in
this reaction?
C6H6 +HNO3
C6H5NO2+H2O
1mol 1mol
1mol
1mol
Theoretical yield
? g C6H5NO2=15.6g C6H6 x 123.0g 1C6H5NO2
78.0g C6H6
=24.6g C6H5NO2
% yield = 18.0g x100%
24.6g
=73.2%
Exercise 44
28
Percent
Yields
from
Reactions
Example 3-9: A 10.0 g sample of ethanol, C H OH, was boiled with
2 5
excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate,
CH3COOC2H5. What is the percent yield?
CH3COOH+C2H5OH
CH3COOC2H5+H2O
Theoretical yield
? g CH3COOC2H5 =10.0g C2H5OH x 88.0g CH3COOC2H5
46.0g C2H5OH
=19.1g CH3COOC2H5
Actual yield
x100%
Theoretical yield
14.8g
x100%
% yield =
19.1g
% yield =
=77.5%
29
Sequential Reactions
Example 3-12 Sequential Reactions
At high temperatures, carbon reacts with water to produce a mixture of carbon
monoxide, CO and hydrogen, H2.
Carbon monoxide is separated from H2 and then used to separate nickel from
cobalt by forming a gaseous compound, nickel tetracarbonyl, Ni(CO)4.
What mass of Ni(CO)4 could be obtained from the CO produced by the reaction of
75.0g of carbon? Assume 100% yield.
C + H2O
1mol 1mol
CO+H2
1mol
Ni + 4CO
1mol 4mol
Ni(CO)4
1mol
? g CO=75.0g C x 1mol C X 1mol CO =6.25mol CO
12.0g C
1mol C
? g Ni(CO)4=6.25mol CO x 1mol Ni(CO)4 X 171g Ni(CO)4
4mol CO
1mol Ni(CO)4
=267.2g Ni(CO)4
Exercise 50
30
Example 3-13 Sequential Reactions
Phosphoricacud,H3PO4, is a very important compound used to make
fertilizers. It is also present in cola drink.
H3PO4 can be prepared in a two-step process.
Reaction1:
Reaction2:
P4 +5O2
1mol
5mol
P4O10
1mol
P4O10+6H2O
1mol
6mol
4H3PO4
4mol
We allow 272g of phosphorus to react with excess oxygen, which forms
tetraphosphorus decoxide, P4O10, in 89.5% yield. In the second step
reaction, a 96.8% yield of H3PO4 is obtained. What mass of H3PO4 is
obtained?
? g P4O10=272g P4x 1mol P4 x 1mol P4O10 x 284g P4O10 X 89.5%
1mol P4O10
1mol P4
124g P4
=558g P4O10
? g H3PO4= 558g P4O10x 1mol P4O10x 4mol H3PO4 x 98.0g H3PO4
1mol H3PO4
284g P4O10 1mol P4O10
x 96.8% =746g H PO
3
4
Exercise 52,54
31
Concentration of Solutions
•Solution is a mixture of two or more substances
dissolved in another.
– Solute 溶質 the dissovled phase of a solution
– Solvent 溶劑the dispersed medium of a solution
– In aqueous solutions 水溶液, 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:
Mass of solute x100%
% by mass of solute =
Mass of solution
Mass of solution= mass of solute + mass of solvent
% by mass of solute has the symbol % w/w
32
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?
Mass of solute x100%
Mass of solution
250.0g solution x 8.0g NaOH
100.0g solution =20.0g NaOH
% by mass of solute =
Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that
contains 32.0 g of NaOH.
? g solution= 32.0g NaOH x
100.0g solution
8.0g NaOH
=400.0g solution
33
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.
sol’n X
? g solution= 300.0ml sol’n x 1.09g
1ml sol’n
8.0g NaOH
100.0g sol’n
=26.2g NaOH
Sol’n weight
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.
? ml solution= 40.0g KOH x
100.0g sol’n X 1mL sol’n
12.0g KOH
1.1g sol’n
=300ml solution
Sol’n weight
34
Concentration of Solutions
Example 3-14 Percent of Solute
Calculated the mass of nickel(II) sulfide, NiSO4, contained in 200.0g
of a 6% solution of NiSO4.
6.0g NiSO4
? g NiSO4= 200.0g sol’n x 100.0g solution =12.0g NiSO4
Example 3-15 Mass of Solution
A 6% NiSO4 solution contained 40.0g NiSO4. Calculate the mass of
the solution.
100.0g soln
? g sol’n= 40.0g NiSO4 x 6.0g NiSO =667g soln
4
35
Concentration of Solutions
Example 3-16 Mass of Solute
Calculated the mass of NiSO4 present in 200.0ml of a 6% solution of NiSO4.
The density of the solution is 1.06g/ml at 25oC.
? g NiSO4= 200.0ml sol’n x 1.06g sol X 6.0g NiSO4
1.0ml sol
100.0g solution
=12.70g NiSO4
Exercise 58
Example 3-17 Percent solute and Density
What volume of a solution that is 15% iron(III) nitrate contains 30.0g of Fe(NO3) 3? The
density of the solution id 1.16g/ml at 25oC.
100.0g sol’n x 1.16g sol
30.0g
Fe(NO
)
x
? ml sol’n=
3 3 15.0g Fe(NO )
1.0ml sol
3 3
=172ml soln
Exercise 60
36
Concentrations of Solutions
•Second common unit of concentration:
Molarity is defined as the number of moles of
solute per literof solution.
Molarity =
M=
Number of moles of solute
Number of liters of solution
moles
L
mmoles
M = mL
37
0.010M KMnO4 sol.
0.395g KMnO4 (MW158) in 250ml distilled H2O
38
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.
? mol H2SO4 =
L sol’n
12.5g H2SO4
98.1g H2SO4
= 0.0728 M H2SO4
1.75L sol’n
Example 3-16: Determine the mass of calcium nitrate required to
prepare 3.50 L of 0.800M Ca(NO3)2 .
? g Ca(NO4) 2= 3.50L x 0.8 mol Ca(NO4) 2
1.0L sol
x
164.0g Ca(NO4) 2
1mol Ca(NO4) 2
=459.0g Ca(NO4) 2
39
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?
Specific gravity =1.185
 density=1.185g/mL of 1185g/L
? M HCl= 1185g sol
1.0L sol
x 36.31g HCl x 1mol HCl
100g sol
36.46g HCl
=11.80M HCl
40
Concentration of Solutions
Example 3-18 Molarity
Calculated the Molarity (M) a solution that contains 3.65g of HCl in 2.00L
of solution.
? mol HCl
L sol’n
=
3.65g HCl
36.5g HCl
2.00L sol’n
= 0.05 M HCl
Exercise 62
Example 3-19 Mass of solute
Calculate the mass of Ba(OH)2 required to prepare 2.50L of 0.0600M
solution of barium hydroxide.
0.06 mol Ba(OH) 2 X 171.3g Ba(OH) 2
? g Ba(OH) 2= 2.50L x
1.0L sol
1mol Ba(OH) 2
=25.7g Ba(OH) 2
Exercise 64
41
Concentration of Solutions
Example 3-20 Molarity
A sample of commercial sulfuric acid is 96.4% H2SO4 by mass, and its specific
gravity is 1.84. Calculate the molarity of this sulfuric acid solution.
Specific gravity =1.84
 density=1.84g/mL of 1840g/L
? M H2SO4= 1840g sol
1.0L sol
x
96.4g H2SO4 1mol H2SO4
x
98.1g H2SO4
100g sol
=18.1 M H2SO4
Exercise 70
42
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
43
Dilution of Solutions 溶液的稀釋
• 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.
44
Dilution of Solutions
• Common method to dilute a solution involves
the use of volumetric flask定量瓶, pipet, and
suction bulb.
100ml
0.1M K2CrO4
1L
0.1M K2CrO4
0.01M K2CrO4
45
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?
M1V1 = M2V2
12.0M X 10.0 ml = M2 x 100.0 ml
M2 = 1.2 M
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?
18.0 M x V1 L = 2.40M x 2.5 L
V1 = 2.40M x 2.5L
V1 = 0.333L
18.0M
46
Dilution of Solutions
Example 3-21 Dilution
How many milliliters of 18.0M H2SO4 are required to prepare 1.00L of a 0.9M
solution of H2SO4.
M1V1 = M2V2
18.0M x V1 ml = 0.9 M x 1000.0 ml
V1 = 50.0ml
Example 3-22 Amount of solute
Calculate (a) the number of moles of H2SO4 and (b) the number of grams of H2SO4 in
500ml of 0.324M H2SO4 solution.
moles
? mol H2SO4 = 0.5 L x 0.324M =0.162mole
V
? g H2SO4 =0.162 mole x 98.1 =15.9g
M=
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.
48
Using Solutions in Chemical
Reactions
Example 3-20: What volume of 0.500M BaCl is
2
required to completely react with 4.32 g of Na2SO4?
Na2SO4+BaCl2
BaSO4+ 2 NaCl
? mol BaCl2=4.32g Na2SO4 x 1mol Na2SO4 x 1mol BaCl2
142g Na2SO4
1mol Na2SO4
=0.03 mol BaCl2
moles
M=
V
moles
0.03mol BaCl2
V=
=0.06 L BaCl2
=
0.5M
BaCl
M
2
49
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? (b)What mass of Al(OH)3 precipitates in (a)?
Al(NO3)3+3NaOH
Al(OH)3+ 3 NaNO3
? mol Al(NO3) 3 = 50ml/1000mlx0.2 =0.01 mol Al(NO3)3
? mol NaOH =0.01 mole Al(NO3) 3x 3mol NaOH =0.03 mol NaOH
1mol Al(NO3)2
moles
= 0.03 mol NaOH =0.15 L NaOH
L=
M
0.2 M NaOH
? g Al(OH)3 =0.01 mole Al(NO3)3x 1mol Al(OH)3 x 78g Al(OH)3
1mol Al(NO3)2 1mol Al(OH)2
=0.78g Al(OH)3
50
Using Solutions in Chemical
Reactions
• Titrations 滴定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
51
Example 3-23 Solution Stoichiometry
Calculate the volume in liters and in milliliters of a 0.324 M
solution of surfuric acid required to react completely with
2.792g of Na2CO3 according to the equation.
H2SO4+Na2CO3
Na2SO4+ CO2+H2O
? mol Na2CO3 =2.792g x 1.0 mol Na2CO3 =0.026mole
106.0g Na2CO3
? mol H2SO4 =0.026 mol Na2CO3 x
=0.026mole H2SO4
M=
moles
V
? L H2SO4 =
1mol H2SO4
1mol Na2CO3
0.026mole
0.324M
=0.08L
=80.0ml
Example 3-24 Volume of Solution Required
Find the volume in liters and in milliliters of a 0.505 M NaOH
solution required to react 40.0ml of 0.505M H2SO4 solution
according to the reaction.
H2SO4+2NaOH
Na2SO4+ 2H2O
? mol H2SO4 =0.505g x 40.0x10-3 L =20.2x10-3mole
? mol NaOH =20.2x10-3 mol H2SO4 x
2mol NaOH
1mol H2SO4
=40.2x10-3mole NaOH
moles
M=
V
-3mole
40.2x10
? L NaOH =
=0.08L
0.505 M
=80.0ml
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+ HCl
KCl + H2O
? mol HCl = (43.2ml/1000ml) x 0.223M =9.63x10-3mol HCl
? mol KOH = 9.63x10-3 mol HCl x 1mol KOH
1mol HCl
=9.63x10-3 mol KOH
3 mol KOH
moles
9.63x10=
M=
=0.249 M KOH
-3
V
38.7 x10 L KOH
54
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?
Ba(OH)2 + 2HCl
BaCl2+ 2 H2O
? mol HCl = (44.1ml/1000ml) x 0.103M = 4.54x10-3mol HCl
1mol Ba(OH)2
? mol HCl =4.54x10-3 mol HCl x
2mol HCl
= 2.27x10-3 mol Ba(OH)2
moles
M=
=
V
2.27x10-3 mol Ba(OH)2
-2 M KOH
=
5.93x10
38.3 x10-3L HCl
55