1. Chemical Equations & Reaction Stoichiometry3
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. Chapter Three Goals Chemical Equations 化學方程式
Calculations Based on Chemical Equations
The Limiting Reactant Concept 限量反應物概念
Percent Yields from Chemical Reactions
Sequential Reactions 連續性反應
Concentrations of Solutions 溶液濃度
Dilution of solutions 溶液的稀釋
Using Solutions in Chemical Reactions

3. Chemical Equations 化學方程式
The substances that react, called reactants
reactants on left side of reaction 反應物
The substances formed, called products
products on right side of equation 產物
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+ 2H2O Reactants Products
Coefficients indicate the amount of each
Compound or element present and CAN be changed to balance the equation
CH4 + 2O 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

5. Low of Conservation of Matter
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

6. Chemical Equations C3H8 + 5O2 3CO2+ 4H2O
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 + 5O CO2+ 4H2O

7. Law of Conservation of Matter
reactants
products
C2H6O + O CO H2O 2 3
3 X 2 =6
2C, 6H, 3O
2C, 6H, 7O
atom count

8. Law of Conservation of Matter
4 oxygen atoms = 2O2 +
C2H6O + O CO2+ 3H2O
C2H6O + O CO2+ 3H2O 3
2C, 6H, 7O
2C, 6H, 7O
Final atom count
C2H6O + 3O CO2+ 3H2O

9. Law of Conservation of Matter
Al + HCl AlCl3+ H2 3
atom count:
Al, 3H, 3Cl
Al, 2H, 3Cl
Al + HCl AlCl3+ H2 6 3
Al, 6H, 6Cl
Al, 6H, 3Cl
atom count:
Al +6HCl AlCl3+ 3H2 2
atom count:
Al, 6H, 6Cl
2Al, 6H, 6Cl 2
Al +6HCl AlCl3+ 3H2
2Al, 6H, 6Cl
2Al, 6H, 6Cl
Final atom count
2Al + 6HCl AlCl3+ 3H2

10. Law of Conservation of Matter
NH3 burns in oxygen to form NO & water
NH3 + O NO2+ H2O
4NH3 + 5O NO2+ 6H2O
C7H16 burns in oxygen to form carbon dioxide and water.
C7H16 + O CO2+ H2O
C7H O CO2+ 8H2O
Balancing equations is a skill acquired only with lots of practice
work many problems

11. Law of Conservation of Matter
Example 3-1 Balancing Chemical Equation
Balance the following chemical equations:
(a) P4 + Cl PCl3
(b) RbOH + SO Rb2SO3 + H2O
(b) P4O10 + Ca(OH) Ca3(PO4)2 + H2O
(a) P4 + 6Cl PCl3
(b) 2RbOH + SO Rb2SO3 + H2O
(b) P4O10+6Ca(OH) Ca3(PO4)2+6 H2O
Exercise 8, 10 11

12. 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 Fe2+ 3CO2
reactants yields products
1 formula unit 3 molecules atoms molecules
1 mole moles moles moles
159.7 g g g g

13. Calculations Based on Chemical Equations
Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?
Fe2O3 +3CO Fe+3CO2
3CO molecules
1Fe2O3 formula unit
?CO molecules=25 formula units Fe2O3 x
=75 molecule of CO

14. 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 Fe+3CO2
2 Fe atoms
1 Fe2O3 formula unit
?Fe atoms=2.5x105 formula units Fe2O3 x
=5.0x105 Fe atoms

15. 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 Fe+3CO2
1mol Fe2O3
159.7g Fe2O3
3 mol CO
1 mol Fe2O3
28.0g CO
1 mol CO
?g CO=146g Fe2O3 x x x
=76.8g CO

16. Calculations Based on Chemical Equations
Example 3-4: What mass of carbon dioxide can be produced by the reaction of mole of iron (III) oxide with excess carbon monoxide?
Fe2O3 +3CO Fe+3CO2
3 mol CO2
1 mol Fe2O3
44.0g CO2
1 mol CO2
?g CO2=0.54mol Fe2O3 x x
=71.3g CO2

17. 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 Fe+3CO2
1mol CO2
44.0g CO2
1 mol Fe2O3
3 mol CO2
1 mol Fe2O3
159.7g Fe2O3
?g Fe2O3=8.65g CO2 x x x
=10.5g Fe2O3

18. 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 Fe+3CO2
1lb Fe2O3
454g Fe2O3
3 mol CO
159.7 Fe2O3
?lb CO =125lb Fe2O3 x x
1 mol CO
28g CO
454g CO
1lb CO
=65.7lb CO x x
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
How many O2 molecule react with 47 CH4 molecules according to the preceding equation?
How many moles of water could be produced by the reaction of 3.5mol of methane with excess oxygen?
CH4 +2O CO2+2H2O
2 O2 molecule
1 CH4 molecule
? O2 molecules =47 CH4 molecules x
=94 O2 molecule
2 mol H2 O
1 mol CH4
? mol H2 O =3.5mol CH4 x
=7.0 mol H2O
Exercise 12, 14,18

20. CH4 +2O2 CO2+2H2O =76.8g O2 =96.0g O2 =24.0g CH4 =264.0g CO2 2 mol O2
Example 3-4, 3-5, 3-6, 3-7 Mass of a Reactant Required, Mass of a Product Formed
What mass of O2 is required to react completely with 1.2mol of CH4?
What mass of O2 is required to react completely with 24.0g of CH4?
What mass of CH4, is required to react with 96.0g of O2?
Calculated the mass of CO2, in grams, that can be produced by burning 6.0mol of CH4 in excess O2.
CH4 +2O CO2+2H2O
2 mol O2
1 mol CH4
32.0g O2
1 mol O2
? O2 g =1.2 mol CH4 x
=76.8g O2 x
1mol CH4
16.0g CH4
2 mol O2
1 mol CH4
32.0g O2
1 mol O2
? O2 mol =24g CH4 x x x
=96.0g O2
1mol O2
32g O2
1 mol CH4
2 mol O2
16g CH4
1 mol CH4
? CH4 mol =96g O2 x x x
=24.0g CH4
1 mol CO2
1 mol CH4
44.0g CO2
1 mol CO2
? CO2 g =6.0 mol CH4 x x
=264.0g CO2
Exercise 22, 24, 26,28

21. 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 + O P4O10 5
1 mol P4
5 mol O2
124.0g P4
1 mol P4
? P4 g =1.5 mol O2 x
=37.2g P4 x
Exercise 28 21

22. 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 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

23. Limiting Reactant ConceptH
CO + 2H CH3OH
H2 molecule: excess
That is:
Not enough CO molecules to react with all H2 molecule
Limiting reactant
(limiting reagent)

24. =33.0g CO2 CH4 +2O2 CO2+2H2O CH4 excess =1.0mol CH4 =1.5mol 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?
CH4 +2O CO2+2H2O
CH4 excess
1mol
2mol
1mol
2mol
1mol CH4
16.0g CH4
? mol CH4 =16.0g CH4 x
=1.0mol CH4
1mol O2
32.0g O2
? mol O2 =48.0g O2 x
=1.5mol O2
1mol CH4
2mol O2
? mol CH4 =1.5mol O2 x
=0.75mol CH4
1.0mol CO2
2.0mol O2
44.0g CO2
1.0mol CO2
? g CO2 =1.5mol O2 x x
=33.0g CO2
Exercise 28 24

25. 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
1mol
2mol
1mol
2mol
NiCl2 excess
1mol NiCl2
129.6g NiCl2
? mol NiCl2 =25.9g NiCl2 x
=0.2mol NiCl2
1mol NaOH
40.0gNaOH
? mol NaOH =10.0g NaOH x
=0.25mol NaOH
1.0mol Ni(OH)2
2.0mol NaOH
92.7g Ni(OH)2
1.0mol Ni(OH)2
? g Ni(OH)2 =0.25mol NaOH x x
=11.6g Ni(OH)2
Exercise 34,36 25

26. 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 +3O CO2+2SO2
1mol
3mol
1mol
2mol
CS2 excess
1mol CS2
76.2g CS2
? mol CS2 =95.6g CS2 x
=1.25mol CS2
1mol O2
32.0g O2
? mol O2 =110.0g O2 x
=3.44mol O2
2.0mol SO2
3.0mol O2
64.0g SO2
1.0mol SO2
? g SO2 =3.44mol O2 x x
=146.6g SO2

27. 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.
Actual yield
Theoretical yield
% yield =
x100%

28. 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 +HNO C6H5NO2+H2O
1mol
1mol
1mol
1mol
Theoretical yield
123.0g 1C6H5NO2
78.0g C6H6
? g C6H5NO2=15.6g C6H6 x
=24.6g C6H5NO2
18.0g
24.6g
% yield =
x100%
=73.2%
Exercise 44 28

29. 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?
CH3COOH+C2H5OH CH3COOC2H5+H2O
Theoretical yield
88.0g CH3COOC2H5
46.0g C2H5OH
? g CH3COOC2H5 =10.0g C2H5OH x
=19.1g CH3COOC2H5
Actual yield
Theoretical yield
% yield =
x100%
14.8g
19.1g
% yield =
x100%
=77.5%

30. Sequential Reactions =6.25mol CO =267.2g Ni(CO)4 C + H2O CO+H2
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 CO+H2
Ni + 4CO Ni(CO)4
1mol
1mol
1mol
1mol
4mol
1mol
1mol C
12.0g C
1mol CO
1mol C
? g CO=75.0g C x
=6.25mol CO X
1mol Ni(CO)4
4mol CO
171g Ni(CO)4
1mol Ni(CO)4
? g Ni(CO)4=6.25mol CO x X
=267.2g Ni(CO)4
Exercise 50

31. =558g P4O10 =746g H3PO4 P4 +5O2 P4O10 P4O10+6H2O 4H3PO4 1mol P4
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:
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?
P4 +5O P4O10
P4O10+6H2O H3PO4
1mol
5mol
1mol
1mol
6mol
4mol
1mol P4
124g P4
1mol P4O10
1mol P4
284g P4O10
1mol P4O10
? g P4O10=272g P4x x x X
89.5%
=558g P4O10
1mol P4O10
284g P4O10
4mol H3PO4
1mol P4O10
98.0g H3PO4
1mol H3PO4
? g H3PO4= 558g P4O10x x x x
96.8%
=746g H3PO4
Exercise 52,54

32. 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
Mass of solution
% by mass of solute =
x100%
Mass of solution= mass of solute + mass of solvent
% by mass of solute has the symbol % w/w

33. Concentration of Solutions
Example 3-11: What mass of NaOH is required to prepare g of solution that is 8.00% w/w NaOH?
x100%
% by mass of solute =
Mass of solute
Mass of solution
8.0g NaOH
100.0g solution
250.0g solution x
=20.0g NaOH
Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.
100.0g solution
8.0g NaOH
? g solution=
32.0g NaOH x
=400.0g solution

34. Concentration of Solutions
Example 3-13: Calculate the mass of NaOH in mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL.
1.09g sol’n
1ml sol’n
8.0g NaOH
100.0g sol’n
? g solution= 300.0ml sol’n x X
=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.
100.0g sol’n
12.0g KOH
1mL sol’n
1.1g sol’n
? ml solution= 40.0g KOH x X
=300ml solution
Sol’n weight

35. 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.
100.0g solution
6.0g NiSO4
=12.0g NiSO4
? g NiSO4=
200.0g sol’n x
Example 3-15 Mass of Solution
A 6% NiSO4 solution contained 40.0g NiSO4. Calculate the mass of the solution.
6.0g NiSO4
100.0g soln
=667g soln
? g sol’n=
40.0g NiSO4 x

36. Concentration of Solutions
Example 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.
1.0ml sol
1.06g sol
100.0g solution
6.0g NiSO4
? g NiSO4=
200.0ml sol’n x X
=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.
15.0g Fe(NO3)3
100.0g sol’n
1.0ml sol
1.16g sol
? ml sol’n=
30.0g Fe(NO3) 3x x
=172ml soln
Exercise 60

37. Concentrations of Solutions
Second common unit of concentration:
Molarity is defined as the number of moles of solute per literof solution.
Number of moles of solute
Number of liters of solution
Molarity =
M =
moles L
M =
mmoles mL

38. 0.010M KMnO4 sol.
0.395g KMnO4 (MW158) in 250ml distilled H2O

39. 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.
98.1g H2SO4
12.5g H2SO4
L sol’n
? mol H2SO4
= 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 .
1.0L sol
0.8 mol Ca(NO4) 2
1mol Ca(NO4) 2
164.0g Ca(NO4) 2
? g Ca(NO4) 2=
3.50L x x
=459.0g Ca(NO4) 2

40. Concentrations of Solutions
Example 3-17: The specific gravity of concentrated HCl is and it is 36.31% w/w HCl. What is its molarity?
Specific gravity =1.185
 density=1.185g/mL of 1185g/L
1.0L sol
1185g sol
100g sol
36.31g HCl
36.46g HCl
1mol HCl
? M HCl= x x
=11.80M HCl

41. Concentration of Solutions
Example Molarity
Calculated the Molarity (M) a solution that contains 3.65g of HCl in 2.00L of solution.
36.5g HCl
3.65g HCl
L sol’n
? mol HCl =
= 0.05 M HCl
2.00L sol’n
Exercise 62
Example 3-19 Mass of solute
Calculate the mass of Ba(OH)2 required to prepare 2.50L of M solution of barium hydroxide.
1.0L sol
0.06 mol Ba(OH) 2
1mol Ba(OH) 2
171.3g Ba(OH) 2
? g Ba(OH) 2=
2.50L x X
=25.7g Ba(OH) 2
Exercise 64

42. Concentration of Solutions
Example 3-20 Molarity
A sample of commercial sulfuric acid is 96.4% H2SO4 by mass, and its specific gravity is Calculate the molarity of this sulfuric acid solution.
Specific gravity =1.84
 density=1.84g/mL of 1840g/L
1.0L sol
1840g sol
100g sol
96.4g H2SO4
98.1g H2SO4
1mol H2SO4
? M H2SO4= x x
=18.1 M H2SO4
Exercise 70

43. 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

44. 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.

45. 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

46. Dilution of Solutions V1 = 0.333L M1V1 = M2V2
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 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
18.0M
2.40M x 2.5L
V1 =
V1 = 0.333L

47. Dilution of Solutions M1V1 = M2V2 18.0M x V1 ml = 0.9 M x 1000.0 ml
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 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.
M =
moles V
? mol H2SO4 = 0.5 L x 0.324M
=0.162mole
? g H2SO4 =0.162 mole x 98.1
=15.9g

48. 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.

49. Using Solutions in Chemical Reactions
Example 3-20: What volume of 0.500M BaCl2 is required to completely react with 4.32 g of Na2SO4?
Na2SO4+BaCl BaSO4+ 2 NaCl
1mol Na2SO4
142g Na2SO4
1mol BaCl2
1mol Na2SO4
? mol BaCl2=4.32g Na2SO4 x x
=0.03 mol BaCl2
M =
moles V
moles M =
0.03mol BaCl2
0.5M BaCl2
V =
=0.06 L BaCl2

50. Al(NO3)3+3NaOH Al(OH)3+ 3 NaNO3 =0.01 mol Al(NO3)3
Example 3-21: (a)What volume of M NaOH will react with 50.0 mL 0f 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
3mol NaOH
1mol Al(NO3)2
? mol NaOH =0.01 mole Al(NO3) 3 x
=0.03 mol NaOH
moles M
0.03 mol NaOH
0.2 M NaOH
L = =
=0.15 L NaOH
1mol Al(OH)3
1mol Al(NO3)2
78g Al(OH)3
1mol Al(OH)2
? g Al(OH)3 =0.01 mole Al(NO3)3 x x
=0.78g Al(OH)3

51. 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

52. H2SO4+Na2CO3 Na2SO4+ CO2+H2O =0.026mole =0.026mole H2SO4 =0.08L
Example 3-23 Solution Stoichiometry
Calculate the volume in liters and in milliliters of a M solution of surfuric acid required to react completely with 2.792g of Na2CO3 according to the equation.
H2SO4+Na2CO Na2SO4+ CO2+H2O
1.0 mol Na2CO3
106.0g Na2CO3
? mol Na2CO3 =2.792g x
=0.026mole
1mol H2SO4
1mol Na2CO3
? mol H2SO4 =0.026 mol Na2CO3 x
=0.026mole H2SO4
M =
moles V
0.026mole
0.324M
? L H2SO4 =
=0.08L
=80.0ml

53. H2SO4+2NaOH Na2SO4+ 2H2O =20.2x10-3mole =40.2x10-3mole NaOH =0.08L
Example 3-24 Volume of Solution Required
Find the volume in liters and in milliliters of a M NaOH solution required to react 40.0ml of 0.505M H2SO4 solution according to the reaction.
H2SO4+2NaOH Na2SO4+ 2H2O
=20.2x10-3mole
? mol H2SO4 =0.505g x 40.0x10-3 L
2mol NaOH
1mol H2SO4
? mol NaOH =20.2x10-3 mol H2SO4 x
=40.2x10-3mole NaOH
M =
moles V
40.2x10-3mole
0.505 M
? L NaOH =
=0.08L
=80.0ml

54. KOH+ HCl KCl + H2O =9.63x10-3mol HCl =9.63x10-3 mol KOH =0.249 M KOH
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 M HCl?
KOH+ HCl KCl + H2O
? mol HCl = (43.2ml/1000ml) x 0.223M
=9.63x10-3mol HCl
1mol KOH
1mol HCl
? mol KOH = 9.63x10-3 mol HCl x
=9.63x10-3 mol KOH
moles V
9.63x10-3 mol KOH
38.7 x10-3L KOH
M = =
=0.249 M KOH

55. Ba(OH)2 + 2HCl BaCl2+ 2 H2O = 4.54x10-3mol HCl = 2.27x10-3 mol Ba(OH)2
Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 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
2mol HCl
? mol HCl =4.54x10-3 mol HCl x
= 2.27x10-3 mol Ba(OH)2
moles V
2.27x10-3 mol Ba(OH)2
38.3 x10-3L HCl
M = =
= 5.93x10-2 M KOH