3 chemical equations & reaction stoichiometry 化學方程式化學平衡及化學計量 ch 4...

33 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

2

ChapterChapter Three Goals Three Goals1.Chemical Equations 化學方程式2.Calculations Based on Chemical

Equations3.The Limiting Reactant Concept 限量反應

物概念4.Percent Yields from Chemical

Reactions5.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

3

Chemical EquationsChemical Equations 化學化學方程式方程式

Chemical Equations1. The substances that react, called

reactantsreactants on left side of reaction 反

應物 2. The substances formed, called

productsproducts 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

4

CH4 + 2O2 CO2+ 2H2OReactants Products

Coefficients indicate the amount of each

Compound or element present and CAN be changed to balance the equation

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 observationChemical equations are based on experimental observation

5

Chemical EquationsChemical Equations

• Attempt to show on paper what is happening at the laboratory and molecular levels.

Ball-and-stick models

Chemical formula

Space-filling model

Low of Conservation of MatterLow of Conservation of Matter

6

Chemical EquationsChemical 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

Law of Conservation of Matter

C2H6O + O2 CO2+ H2O

reactants products

3 X 2 =6

2C, 6H, 3O 2C, 6H, 7Oatom count

2 3

C2H6O + O2 2CO2+ 3H2O2C, 6H, 7O 2C, 6H, 7O

Law of Conservation of Law of Conservation of MatterMatter

C2H6O + O2 2CO2+ 3H2O4 oxygen atoms = 2O2

+

3

C2H6O + 3O2 2CO2+ 3H2O

Final atom count

Law of Conservation of MatterLaw of Conservation of MatterAl + HCl AlCl3+ H2

2Al + 6HCl 2AlCl3+ 3H2

Final atom count

3atom count: Al, 3H, 3Cl Al, 2H, 3Cl

Al + HCl AlCl3+ H26

Al, 6H, 6Cl Al, 6H, 3Cl3

Al +6HCl AlCl3+ 3H2Al, 6H, 6Cl 2Al, 6H, 6Cl

2

Al +6HCl 2AlCl3+ 3H22Al, 6H, 6Cl 2Al, 6H, 6Cl

2

atom count:

atom count:

10

Law of Conservation of MatterLaw of Conservation of Matter• NH3 burns in oxygen to form NO & water

NH3 + O2 NO2+ H2O4NH3 + 5O2 4NO2+ 6H2O

• C7H16 burns in oxygen to form carbon dioxide and water.

C7H16 + O2 CO2+ H2OC7H16 + 11O2 7CO2+ 8H2O

•Balancing equations is a skill acquired only with lots of practice

•work many problems

Example 3-1 Balancing Chemical EquationBalance the following chemical equations:Example 3-1 Balancing Chemical EquationBalance the following chemical equations:

11

Law of Conservation of MatterLaw of Conservation of Matter

(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

12

Calculations Based on Chemical Calculations Based on Chemical EquationsEquations

reactants yields products

1 formula unit 3 molecules 2 atoms 3 molecules1 mole 3 moles 2 moles 3 moles159.7 g 84.0 g 111.7 g 132g

Fe2O3 + 3CO 2Fe2+ 3CO2

•Can work in moles, formula units, etc.•Frequently, we work in mass or weight (grams or kg or pounds or tons).

13


Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?

?CO molecules=25 formula units Fe2O3 x3CO molecules

1Fe2O3 formula unit=75 molecule of CO

Fe2O3 +3CO 2Fe+3CO2

14


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?

?Fe atoms=2.5x105 formula units Fe2O3 x 2 Fe atoms1 Fe2O3 formula unit

=5.0x105 Fe atoms

Fe2O3 +3CO 2Fe+3CO2

15


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 3 mol CO1 mol Fe2O3

=76.8g CO

1mol Fe2O3

159.7g Fe2O3

x x 28.0g CO1 mol CO

16


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 CO21 mol Fe2O3

=71.3g CO2

x 44.0g CO2

1 mol CO2

17


• 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 x1 mol Fe2O3

3 mol CO2

=10.5g Fe2O3

1mol CO244.0g CO2

x x 1 mol Fe2O3

159.7g Fe2O3

18


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

3 mol CO159.7 Fe2O3

=65.7lb CO

x

x1 mol CO

28g CO

1lb Fe2O3

454g Fe2O3

x454g CO

1lb CO

YOU MUST BE PROFICIENT WITH THESE

TYPES OF PROBLEMS!!!

19

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?

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?

Exercise 12, 14,18

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 x2 mol H2 O 1 mol CH4

=7.0 mol H2O

20

Example 3-4, 3-5, 3-6, 3-7 Mass of a Reactant Required, Mass of a Product Formed1.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.

Example 3-4, 3-5, 3-6, 3-7 Mass of a Reactant Required, Mass of a Product Formed1.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.

Exercise 22, 24, 26,28

CH4 +2O2 CO2+2H2O? O2 g =1.2 mol CH4 x 2 mol O2

1 mol CH4=76.8g O2

32.0g O2 1 mol O2

x

? O2 mol =24g CH4 x 2 mol O2 1 mol CH4

=96.0g O2 32.0g O2

1 mol O2

x1mol CH4 16.0g CH4

x

? CH4 mol =96g O2 x 1 mol CH4 2 mol O2

=24.0g CH4 16g CH4

1 mol CH4

x1mol O2 32g O2

x

? CO2 g =6.0 mol CH4 x 1 mol CO2 1 mol CH4

=264.0g CO2 44.0g CO2

1 mol CO2

x

2121

Example 3-8 Mass of a Reactant RequiredPhosphorus, P4, burns with excess oxygen to form tetraphosphorus decoxide, P4

O10. In this reaction, what mass of P4 reacts with 1.5mol of O2?

Example 3-8 Mass of a Reactant RequiredPhosphorus, P4, burns with excess oxygen to form tetraphosphorus decoxide, P4

O10. In this reaction, what mass of P4 reacts with 1.5mol of O2?

Exercise 28

P4 + O2 P4O10

? P4 g =1.5 mol O2 x1 mol P4 5 mol O2

=37.2g P4 124.0g P4 1 mol P4

x

5

Calculations Based on Chemical Equations

22

Limiting Reactant ConceptLimiting Reactant Concept 限量反應物限量反應物•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 Eggs Milk 1 1 dozen 1 gallon

limiting reactant is the muffin mix2 1 dozen 1 gallon3 1 dozen 1 gallon4 1 dozen 1 gallon5 1 dozen 1 gallon6 1 dozen 1 gallon7 1 dozen 1 gallon

limiting reactant is the dozen eggs

23

Limiting Reactant ConceptLimiting Reactant Concept

CO + 2H2 CH3OH C O H

H2 molecule: excess

That is:Not enough CO molecules to react with all H2 moleculeLimiting reactant (limiting reagent)

2424

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?

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?

Exercise 28

? mol CH4 =16.0g CH4 x =1.0mol CH4 1mol CH4 16.0g CH4

CH4 +2O2 CO2+2H2O

? mol O2 =48.0g O2 x =1.5mol O2 1mol O2 32.0g O2

1mol 2mol 1mol 2mol CH4 excess

? mol CH4 =1.5mol O2 x =0.75mol CH4 1mol CH4 2mol O2

? g CO2 =1.5mol O2 x

=33.0g CO2

1.0mol CO2 2.0mol O2

44.0g CO2 1.0mol CO2

x

2525

Example 3-10 Limiting ReactantWhat 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?

Example 3-10 Limiting ReactantWhat 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?

Exercise 34,36

? mol NiCl2 =25.9g NiCl2 x =0.2mol NiCl2 1mol NiCl2

129.6g NiCl2

NiCl2 +2NaOH Ni(OH)2+2NaCl

? mol NaOH =10.0g NaOH x =0.25mol NaOH1mol NaOH40.0gNaOH

1mol 2mol 1mol 2mol NiCl2 excess

? g Ni(OH)2 =0.25mol NaOH x

=11.6g Ni(OH)2

1.0mol Ni(OH)2 2.0mol NaOH

92.7g Ni(OH)2 1.0mol Ni(OH)2

x

26

Limiting Reactant ConceptLimiting 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?

? mol CS2 =95.6g CS2 x =1.25mol CS2 1mol CS2

76.2g CS2

CS2 +3O2 CO2+2SO2

? mol O2 =110.0g O2 x =3.44mol O21mol O2 32.0g O2

1mol 3mol 1mol 2mol CS2 excess

? g SO2 =3.44mol O2 x

=146.6g SO2

2.0mol SO2 3.0mol O2

64.0g SO2 1.0mol SO2

x

27

Percent Yields from ReactionsPercent 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.

x100%% yield =Actual yield

Theoretical yield

2828

Example 3-11 percent YieldA 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?

Example 3-11 percent YieldA 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?

Exercise 44

? g C6H5NO2=15.6g C6H6 x

=24.6g C6H5NO2

123.0g 1C6H5NO2 78.0g C6H6

Percent Yields from Reactions

C6H6 +HNO3 C6H5NO2+H2O1mol 1mol 1mol 1mol

Theoretical yield

x100%% yield =18.0g 24.6g

=73.2%

29

Percent Yields from ReactionsPercent Yields from ReactionsExample 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?

CH3COOH+C2H5OH CH3COOC2H5+H2O

? g CH3COOC2H5 =10.0g C2H5OH x

=19.1g CH3COOC2H5

88.0g CH3COOC2H546.0g C2H5OH

Theoretical yield

x100%% yield =14.8g 19.1g

=77.5%

x100%% yield =Actual yield

Theoretical yield

30

Sequential ReactionsSequential ReactionsExample 3-12 Sequential ReactionsAt 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.

Example 3-12 Sequential ReactionsAt 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.

Exercise 50

? g CO=75.0g C x =6.25mol CO 1mol C12.0g C

C + H2O CO+H21mol 1mol

Ni + 4CO Ni(CO)41mol 1mol 4mol 1mol

1mol CO1mol C

X

? g Ni(CO)4=6.25mol CO x

=267.2g Ni(CO)4

1mol Ni(CO)44mol CO

171g Ni(CO)41mol Ni(CO)4

X

31

Example 3-13 Sequential ReactionsPhosphoricacud,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:

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?

Example 3-13 Sequential ReactionsPhosphoricacud,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:

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?

Exercise 52,54

? g P4O10=272g P4x

=558g P4O10

1mol P4

124g P4

P4 +5O2 P4O101mol 5mol 1mol 1mol 6mol 4mol

1mol P4O101mol P4

x

P4O10+6H2O 4H3PO4

284g P4O101mol P4O10

x 89.5%X

? g H3PO4= 558g P4O10x

=746g H3PO4

1mol P4O10

284g P4O10

4mol H3PO41mol P4O10

x 98.0g H3PO41mol H3PO4

x

96.8%x

32

Concentration of SolutionsConcentration 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:

x100%% by mass of solute =Mass of solute

Mass of solutionMass of solution= mass of solute + mass of solvent

% by mass of solute has the symbol % w/w

33

Concentration of SolutionsConcentration of SolutionsExample 3-11: What mass of NaOH is required to prepare 250.0

g of solution that is 8.00% w/w NaOH?

x100%% by mass of solute =Mass of solute

Mass of solution

=20.0g NaOH250.0g solution x 8.0g NaOH100.0g solution

Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.

? g solution= 100.0g solution8.0g NaOH

32.0g NaOH x

=400.0g solution

34

Concentration of SolutionsConcentration of SolutionsExample 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.

=26.2g NaOH

? g solution= 300.0ml sol’n x 1.09g sol’n1ml sol’n

8.0g NaOH100.0g sol’n

X

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.

=300ml solution

? ml solution= 40.0g KOH x 100.0g sol’n12.0g KOH

1mL sol’n1.1g sol’nX

Sol’n weight

Sol’n weight

35

Example 3-14 Percent of SoluteCalculated the mass of nickel(II) sulfide, NiSO4, contained in 200.0g of a 6% solution of NiSO4.

Example 3-14 Percent of SoluteCalculated the mass of nickel(II) sulfide, NiSO4, contained in 200.0g of a 6% solution of NiSO4.

Concentration of Solutions

? g NiSO4= 100.0g solution6.0g NiSO4 200.0g sol’n x =12.0g NiSO4

Example 3-15 Mass of SolutionA 6% NiSO4 solution contained 40.0g NiSO4. Calculate the mass of the solution.

Example 3-15 Mass of SolutionA 6% NiSO4 solution contained 40.0g NiSO4. Calculate the mass of the solution.

? g sol’n= 6.0g NiSO4

100.0g soln 40.0g NiSO4 x =667g soln

Example 3-16 Mass of SoluteCalculated 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.

Example 3-16 Mass of SoluteCalculated 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= 100.0g solution6.0g NiSO4 200.0ml sol’n x

=12.70g NiSO4

Example 3-17 Percent solute and DensityWhat 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.

Example 3-17 Percent solute and DensityWhat 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.

? ml sol’n= 15.0g Fe(NO3)3

100.0g sol’n 30.0g Fe(NO3) 3x

=172ml soln

1.0ml sol1.06g sol X

36

Exercise 60

1.0ml sol1.16g sol x

Exercise 58

37

Concentrations of SolutionsConcentrations of Solutions•Second common unit of concentration:Molarity is defined as the number of moles of

solute per literof solution.

Molarity = Number of moles of soluteNumber of liters of solution

M = molesL M = mmoles

mL

38

0.010M KMnO4 sol. 0.395g KMnO4 (MW158) in 250ml distilled H2O

39

Concentrations of SolutionsConcentrations of SolutionsExample 3-15: Calculate the molarity of a solution that contains

12.5 g of sulfuric acid in 1.75 L of solution.

1.75L sol’n98.1g H2SO4

12.5g H2SO4

= 0.0728 M H2SO4L sol’n? mol H2SO4=

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= 1mol Ca(NO4) 2

164.0g Ca(NO4) 2 3.50L x

=459.0g Ca(NO4) 2

1.0L sol0.8 mol Ca(NO4) 2 x

40

Concentrations of SolutionsConcentrations of SolutionsExample 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=100g sol

36.31g HCl

=11.80M HCl

1.0L sol1185g sol x 36.46g HCl

1mol HCl x

41

Example 3-18 MolarityCalculated the Molarity (M) a solution that contains 3.65g of HCl in 2.00L of solution.

Example 3-18 MolarityCalculated the Molarity (M) a solution that contains 3.65g of HCl in 2.00L of solution.


Example 3-19 Mass of soluteCalculate the mass of Ba(OH)2 required to prepare 2.50L of 0.0600M solution of barium hydroxide.

Example 3-19 Mass of soluteCalculate the mass of Ba(OH)2 required to prepare 2.50L of 0.0600M solution of barium hydroxide.

Exercise 64

Exercise 622.00L sol’n36.5g HCl3.65g HCl

= 0.05 M HClL sol’n? mol HCl =

? g Ba(OH) 2= 1mol Ba(OH) 2

171.3g Ba(OH) 2 2.50L x

=25.7g Ba(OH) 2

1.0L sol0.06 mol Ba(OH) 2 X

42

Concentration of SolutionsExample 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.

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.

Exercise 70

Specific gravity =1.84 density=1.84g/mL of 1840g/L

? M H2SO4= 100g sol96.4g H2SO4

1.0L sol1840g sol x

98.1g H2SO4

1mol H2SO4 x

=18.1 M H2SO4

43

Concentrations of SolutionsConcentrations of Solutions• One of the reasons that molarity is commonly

used is because:M x L = moles solute

and M x mL = mmol solute

44

Dilution of Solutions Dilution of Solutions 溶液的稀釋溶液的稀釋• 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.

45

Dilution of SolutionsDilution 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

46

Dilution of SolutionsDilution of SolutionsExample 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 = M2V212.0M 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 LV1 = 18.0M

2.40M x 2.5L V1 = 0.333L

Example 3-21 Dilution How many milliliters of 18.0M H2SO4 are required to prepare 1.00L of a 0.9M solution of H2SO4.

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 mlV1 = 50.0ml

Dilution of Solutions

Example 3-22 Amount of soluteCalculate (a) the number of moles of H2SO4 and (b) the number of grams of H2SO4 in 500ml of 0.324M H2SO4 solution.

Example 3-22 Amount of soluteCalculate (a) the number of moles of H2SO4 and (b) the number of grams of H2SO4 in 500ml of 0.324M H2SO4 solution.

=0.162moleM = molesV ? mol H2SO4 = 0.5 L x 0.324M

? g H2SO4 =0.162 mole x 98.1 =15.9g

48

Using Solutions in Chemical Using Solutions in Chemical ReactionsReactions

• Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.

49


Example 3-20: What volume of 0.500M BaCl2 is required to completely react with 4.32 g of Na2SO4?

Na2SO4+BaCl2 BaSO4+ 2 NaCl? mol BaCl2=4.32g Na2SO4 x

=0.03 mol BaCl2

1mol BaCl21mol Na2SO4

x1mol Na2SO4142g Na2SO4

M = molesV

V = molesM

= 0.03mol BaCl20.5M BaCl2

=0.06 L BaCl2

50

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) 3

L = molesM

= 0.03 mol NaOH0.2 M NaOH

3mol NaOH1mol Al(NO3)2

x =0.03 mol NaOH

=0.15 L NaOH

? g Al(OH)3 =0.01 mole Al(NO3)3x

=0.78g Al(OH)3

1mol Al(OH)31mol Al(NO3)2

78g Al(OH)31mol Al(OH)2

x

51


• 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

Example 3-23 Solution StoichiometryCalculate 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.

Example 3-23 Solution StoichiometryCalculate 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.

=0.026mole? mol Na2CO3 =2.792g x

? L H2SO4 = =0.08L=80.0ml

H2SO4+Na2CO3 Na2SO4+ CO2+H2O

1.0 mol Na2CO3106.0g Na2CO3

1mol H2SO41mol Na2CO3

? mol H2SO4 =0.026 mol Na2CO3 x

=0.026mole H2SO4

M = molesV

0.026mole 0.324M

Example 3-24 Volume of Solution RequiredFind 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.

Example 3-24 Volume of Solution RequiredFind 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.

=20.2x10-3mole? mol H2SO4 =0.505g x 40.0x10-3 L

? L NaOH = =0.08L=80.0ml

H2SO4+2NaOH Na2SO4+ 2H2O

2mol NaOH1mol H2SO4

? mol NaOH =20.2x10-3 mol H2SO4 x

=40.2x10-3mole NaOH

M = molesV

40.2x10-3mole 0.505 M

54

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 KOH1mol HCl

=9.63x10-3 mol KOH

M = molesV

= 9.63x10-3 mol KOH38.7 x10-3L KOH

=0.249 M KOH

55

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? mol HCl =4.54x10-3 mol HCl x 1mol Ba(OH)2

2mol HCl= 2.27x10-3 mol Ba(OH)2

M = molesV

= 2.27x10-3 mol Ba(OH)2

38.3 x10-3L HCl= 5.93x10-2 M KOH

3 chemical equations & reaction stoichiometry 化學方程式 化學平衡及化學計量 ch 4...

Documents

3 chemical equations & reaction stoichiometry 化學方程式化學平衡及化學計量 ch 4...