1. Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ
Chapter 3
Stoichiometry of Formulas and Equations
ปริมาณสัมพันธ์ของสูตรและสมการ

2. Mole - Mass Relationships in Chemical Systems
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating the Amounts of Reactant and Product
3.5 Fundamentals of Solution Stoichiometry

3. The Mole
mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.
This amount is 6.022x The number is called Avogadro’s number and is abbreviated as N.
One mole (1 mol) contains 6.022x1023 entities (to four significant figures)

4. Counting objects of fixed relative mass.
Figure 3.1
Counting objects of fixed relative mass.
12 red 7g each = 84g
12 yellow each=48g
55.85g Fe = x 1023 atoms Fe
32.07g S = x 1023 atoms S

5. One mole of common substances.
Figure 3.2
Oxygen
32.00 g
One mole of common substances.
Water
18.02 g
CaCO3 g
Copper
63.55 g

6. Table 3.1 Summary of Mass Terminology
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
amu
Atomic mass
Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance
(also called
atomic weight)
Molecular
(or formula) mass (also called molecular weight)
Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit)
amu
Molar mass (M)
Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units)
g/mol
(also called
gram-molecular weight)

7. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol)
Table 3.2
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles of atoms
12 moles of atoms
6 moles of atoms
Atoms/mole of
compound
6(6.022 x 1023) atoms
12(6.022 x 1023) atoms
6(6.022 x 1023) atoms
Mass/molecule of compound
6(12.01 amu) =72.06 amu
12(1.008 amu) =12.10 amu
6(16.00 amu) =96.00 amu
Mass/mole of
compound
72.06 g
12.10 g
96.00 g

8. Interconverting Moles, Mass, and Number of Chemical Entities
Mass (g) = no. of moles x
no. of grams
1 mol g
No. of moles = mass (g) x
no. of grams
1 mol M
No. of entities = no. of moles x
6.022x1023 entities
1 mol
No. of moles = no. of entities x
6.022x1023 entities
1 mol

9. Summary of the mass-mole-number relationships for elements.
Figure 3.3
MASS(g)
of element
M (g/mol)
AMOUNT(mol)
of element
Avogadro’s number
(atoms/mol)
ATOMS
of element

10. Sample Problem 3.1 = 3.69 g Ag = 1.04x1024 atoms Fe
Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element
Sample Problem 3.1
PROBLEM:
(a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe?
PLAN:
(a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M.
amount(mol) of Ag
mass(g) of Ag
multiply by M of Ag (107.9g/mol)
mol Ag
107.9 g Ag
= 3.69 g Ag
SOLUTION:
mol Ag x
PLAN:
(b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms.
mass(g) of Fe
amount(mol) of Fe
atoms of Fe
divide by M of Fe (55.85g/mol)
55.85g Fe
mol Fe
SOLUTION:
95.8g Fe x
= 1.72 mol Fe
6.022x1023atoms Fe
mol Fe
multiply by 6.022x1023 atoms/mol
1.72mol Fe x
= 1.04x1024 atoms Fe

11. Summary of the mass-mole-number relationships for compounds.
Figure 3.3
MASS(g)
of compound
M (g/mol)
chemical
formula
AMOUNT(mol)
of elements in
compound
AMOUNT(mol)
of compound
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound

12. number of (NH4)2CO3 formula units amount(mol) of (NH4)2CO3 divide by M
Calculating the Moles and Number of Formula Units in a Given Mass of a Compound
Sample Problem 3.2
Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6 g of ammonium carbonate?
PROBLEM:
PLAN:
After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6 g to mols using M and then the mols to formula units with Avogadro’s number.
mass(g) of (NH4)2CO3
number of (NH4)2CO3 formula units
amount(mol) of (NH4)2CO3
divide by M
multiply by 6.022x1023 formula units/mol
SOLUTION:
The formula is (NH4)2CO3.
M = (2 x g/mol N)+(8 x g/mol H)
+(12.01 g/mol C)+(3 x g/mol O)
= g/mol
mol (NH4)2CO3
96.09 g (NH4)2CO3
6.022x1023 formula units (NH4)2CO3
mol (NH4)2CO3 x
41.6 g (NH4)2CO3 x =
2.61x1023 formula units (NH4)2CO3

13. Mass percent from the chemical formula
Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound (amu)
Mass % of element X =
moles of X in formula x molar mass of X (amu)
molecular (or formula) mass of compound (amu)
x 100

14. (a) What is the mass percent of each element in glucose?
Calculating the Mass Percents and Masses of Elements in a Sample of Compound
Sample Problem 3.3
PROBLEM:
Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy.
(a) What is the mass percent of each element in glucose?
(b) How many grams of carbon are in 16.55g of glucose?
amount(mol) of element X in 1mol compound
PLAN:
We have to find the total mass of glucose and the masses of the constituent elements in order to relate them.
multiply by M (g/mol) of X
mass(g) of X in 1mol of compound
SOLUTION:
divide by mass(g) of 1mol of compound
Per mole glucose there are
6 moles of C
12 moles H
6 moles O
(a)
mass fraction of X
multiply by 100
mass % X in compound

15. Sample Problem 3.3
Calculating the Mass Percents and Masses of Elements in a Sample of Compound
continued
1.008 g H
mol H
12.01 g C
mol C
12 mol H x
= g H
6 mol C x
= g C
16.00 g O
mol O
M = g/mol
6 mol O x
= g O
(b)
72.06 g C
g glucose
mass percent of C = =
x 100 = mass %C
g H
g glucose
mass percent of H = =
x 100 = mass %H
96.00 g O
g glucose
mass percent of O = =
x 100 = mass %O

16. Empirical and Molecular Formulas
Empirical Formula -
The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms.
Molecular Formula -
The formula of the compound as it exists, it may be a multiple of the empirical formula.

17. NaClO4 is sodium perchlorate.
Determining the Empirical Formula from Masses of Elements
Sample Problem 3.4
PROBLEM:
Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound?
PLAN:
Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula).
mol Na
22.99 g Na
SOLUTION:
2.82 g Na x
= mol Na
mass(g) of each element
mol Cl
35.45 g Cl
divide by M(g/mol)
4.35 g Cl x
= mol Cl
amount(mol) of each element
mol O
16.00 g O
use # of moles as subscripts
7.83 g O x
= mol O
preliminary formula
Na1 Cl1 O3.98
Na1 Cl1 O3.98
NaClO4
change to integer subscripts
NaClO4 is sodium perchlorate.
empirical formula

18. Determining a Molecular Formula from Elemental Analysis and Molar Mass
Sample Problem 3.5
Determining a Molecular Formula from Elemental Analysis and Molar Mass
PROBLEM:
During physical activity. lactic acid (M = g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
PLAN:
assume 100g lactic acid and find the mass of each element
divide each mass by mol mass(M)
amount(mol) of each element
molecular formula
use # mols as subscripts
preliminary formula
divide mol mass by mass of empirical formula to get a multiplier
convert to integer subscripts
empirical formula

19. C3H6O3 is the molecular formula
Determining a Molecular Formula from Elemental Analysis and Molar Mass
Sample Problem 3.5
continued
SOLUTION:
Assuming there are 100 g of lactic acid, the constituents are
40.0 g C x
mol C
12.01g C
6.71 g H x
mol H
1.008 g H
53.3 g O x
mol O
16.00 g O
= 3.33 mol C
= 6.66 mol H
= 3.33 mol O
C3.33
H6.66
O3.33
CH2O
empirical formula
3.33
3.33
3.33
mass of CH2O
molar mass of lactate
90.08 g
30.03 g 3
C3H6O3 is the molecular formula

20. CnHm + (n+ ) O2 = n CO(g) + H2O(g) m 2
Figure 3.4
Combustion train for the determination of the chemical composition of organic compounds.
CnHm + (n+ ) O2 = n CO(g) H2O(g) m 2

21. Determining a Molecular Formula from Combustion Analysis
Sample Problem 3.6
PROBLEM:
Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained:
mass of CO2 absorber after combustion = g
mass of CO2 absorber before combustion = g
mass of H2O absorber after combustion = g
mass of H2O absorber before combustion = g
What is the molecular formula of vitamin C?
difference (after-before) = mass of oxidized element
PLAN:
find the mass of each element in its combustion product
preliminary formula
empirical formula
molecular formula
find the mols

22. Molecular formular = C6H8O6
Sample Problem 3.6
Determining a Molecular Formula from Combustion Analysis
continued
SOLUTION:
CO2
85.35 g g = 1.50 g
H2O
37.96 g g = 0.41 g
12.01 g CO2
44.01 g CO2
There are g C per mol CO2
1.50 g CO 2 x
= g C
2.016 g H2O
18.02 g H2O
There are g H per mol H2O
0.41 g H2O x
= g H
O must be the difference: g - ( ) = 0.545
0.409 g C
12.01 g C
0.046 g H
1.008 g H
0.545 g O
16.00 g O
= mol C
= mol H
= mol O
g/mol
88.06 g
C1H1.3O1
C3H4O3
= 2.000
Molecular formular = C6H8O6

23. Table 3.3 Some Compounds with Empirical Formula CH2O
(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)
Molecular Formula
Whole-Number Multiple M
(g/mol)
Name
Use or Function
formaldehyde
CH2O 1
30.03
disinfectant; biological preservative
acetic acid
C2H4O2 2
60.05
acetate polymers; vinegar (5%soln)
lactic acid
C3H6O3 3
90.09
sour milk; forms in exercising muscle
erythrose
C4H8O4 4
120.10
part of sugar metabolism
ribose
C5H10O5 5
150.13
component of nucleic acids and B2
glucose
C6H12O6 6
180.16
major energy source of the cell
CH2O
C2H4O2
C3H6O3
C4H8O4
C5H10O5
C6H12O6

24. Table 3.4 Two Pairs of Constitutional Isomers
C4H10
C2H6O
Property
Butane
2-Methylpropane
Ethanol
Dimethyl Ether
M(g/mol)
58.12
58.12
46.07
46.07
Boiling Point
-0.50C C
78.50C
-250C
Density at 200C
0.579 g/mL
(gas)
0.549 g/mL
(gas)
0.789 g/mL
(liquid)
g/mL
(gas)
Structural
formulas
Space-filling
models

25. The formation of HF gas on the macroscopic and molecular levels.
Figure 3.6

26. A three-level view of the chemical reaction in a flashbulb.
Figure 3.7
A three-level view of the chemical reaction in a flashbulb.

27. Balancing Chemical Equations
translate the statement
balance the atoms
adjust the coefficients
check the atom balance
specify states of matter

28. Balancing Chemical Equations Sample Problem 3.7
Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction.
PLAN:
SOLUTION:
translate the statement
C8H O CO H2O
balance the atoms
C8H O CO H2O
25/2 8 9
adjust the coefficients
2C8H O CO H2O
check the atom balance
2C8H O CO H2O
specify states of matter
2C8H18(l) + 25O2 (g) CO2 (g) + 18H2O (g)

29. Summary of the mass-mole-number relationships in a chemical reaction.
Figure 3.8
MASS(g)
of compound A
MASS(g)
of compound B
M (g/mol) of compound A
M (g/mol) of compound B
AMOUNT(mol)
of compound A
molar ratio from
balanced equation
AMOUNT(mol)
of compound B
Avogadro’s number
(molecules/mol)
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound A
MOLECULES
(or formula units)
of compound B

30. write and balance equation
Sample Problem 3.8
Calculating Amounts of Reactants and Products
PROBLEM:
In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide.
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
PLAN:
write and balance equation
find mols O2
find mols SO2
find mols Cu2O
find g SO2
find mols O2
find kg O2

31. Sample Problem 3.8
Calculating Amounts of Reactants and Products
continued
SOLUTION:
2Cu2S(s) + 3O2(g) Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
3mol O2
2mol Cu2S
10.0mol Cu2S x
(a)
= 15.0mol O2
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
10.0mol Cu2S x
2mol SO2
2mol Cu2S
64.07g SO2
mol SO2
(b) x
= 641g SO2
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
2.86kg Cu2O x
103g Cu2O
kg Cu2O
mol Cu2O
143.10g Cu2O
(c) x
= 20.0mol Cu2O
= 0.959kg O2
kg O2
103g O2
20.0mol Cu2O x
3mol O2
2mol Cu2O
32.00g O2
mol O2 x x

32. Writing an Overall Equation for a Reaction Sequence
Sample Problem 3.9
Writing an Overall Equation for a Reaction Sequence
PROBLEM:
Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process.
PLAN:
SOLUTION:
write balanced equations for each step
2Cu2S(s) + 3O2(g) Cu2O(s) + 2SO2(g)
2Cu2O(s) + 2C(s) Cu(s) + 2CO(g)
cancel reactants and products common to both sides of the equations
sum the equations
2Cu2S(s)+3O2(g)+2C(s) Cu(s)+2SO2(g)+2CO(g)

33. An ice cream sundae analogy for limiting reactions.
Figure 3.10
An ice cream sundae analogy for limiting reactions.

34. Table 3.5 Information Contained in a Balanced Equation
Viewed in Terms of
Reactants
C3H8(g) + 5O2(g)
Products
3CO2(g) + 4H2O(g)
molecules
3 molecules CO2 + 4 molecules H2O
1 molecule C3H8 + 5 molecules O2
amount (mol)
1 mol C3H8 + 5 mol O2
3 mol CO2 + 4 mol H2O
44.09 amu C3H amu O2
mass (amu)
amu CO amu H2O
mass (g)
44.09 g C3H g O2
g CO g H2O
total mass (g) g

35. Using Molecular Depictions to Solve a Limiting-Reactant Problem
Sample Problem 3.10
PROBLEM:
Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine.
(a) Suppose the box shown at left represents a container of the reactant mixture before the reaction occurs (with chlorine colored green). Name the limiting reactant, and draw the container contents after the reaction is complete.
(b) When the reaction is run again with mol of Cl2 and 3.00 mol of F2, what mass of chlorine trifluoride will be prepared?
PLAN:
Write a balanced chemical equation. Compare the number of molecules you have to the number needed for the products. Determine the reactant that is in excess. The other reactant is the limiting reactant.

36. Sample Problem 3.10 continued 3.00 mol F2 0.750 mol Cl2 139 g ClF3
Using Molecular Depictions to Solve a Limiting-Reactant Problem
continued
SOLUTION:
Cl2(g) + 3F2(g) ClF3(g)
(a) You need a ratio of 2 Cl and 6 F for the reaction.
You have 6 Cl and 12 F.
6 Cl would require 18 F.
12 F need only 4 Cl (2 Cl2 molecules). F2
Cl2
There isn’t enough F, therefore it must be the limiting reactant.
You will make 4 ClF2 molecules (4 Cl, 12 F) and have 2 Cl2 molecules left over.
(b) We know the molar ratio of F2/Cl2 should be 3/1.
Since we find that the ratio is 4/1, that means F2 is in excess and Cl2 is the limiting reactant.
3.00 mol F2
0.750 mol Cl2 = 4 1
2 mol ClF3
1 mol Cl
92.5 g ClF3
1 mol ClF3
0.750 mol Cl2 x x =
139 g ClF3

37. Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant
Sample Problem 3.11
PROBLEM:
A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed?
PLAN:
We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given.
In this case one of the reactants is in molar excess and the other will limit the extent of the reaction.
mass of N2H4
mass of N2O4
limiting mol N2
divide by M
multiply by M
mol of N2H4
mol of N2O4
g N2
molar ratio
mol of N2
mol of N2

38. Sample Problem 3.11
Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant
continued
SOLUTION: 2
N2H4(l) + N2O4(l) N2(g) H2O(l) 3 4
mol N2H4
32.05g N2H4
1.00x102g N2H4 x
N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.
= 3.12 mol N2H4
3 mol N2
2mol N2H4
3.12mol N2H4 x
= 4.68mol N2
mol N2
28.02g N2
4.68mol N2 x
= 131g N2
mol N2O4
92.02g N2O4
2.00x102g N2O4 x
= 2.17mol N2O4
3 mol N2
mol N2O4
2.17mol N2O4 x
= 6.51mol N2

39. A + B (reactants) C (main product) D (side products)
Figure 3.11
The effect of side reactions on yield.
A + B
(reactants) C
(main product) D
(side products)

40. Sample Problem 3.12 40.10 g SiC mol SiC 103g
Calculating Percent Yield
PROBLEM:
Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process?
PLAN:
SOLUTION:
write balanced equation
SiO2(s) + 3C(s) SiC(s) + 2CO(g)
103 g SiO2
kg SiO2
mol SiO2
60.09 g SiO2
find mol reactant & product
100.0 kg SiO2 x x
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
40.10 g SiC
mol SiC kg
103g
1664 mol SiC x x
= kg
actual yield/theoretical yield x 100
51.4 kg
66.73 kg
percent yield x
100
=77.0%

41. Calculating the Molarity of a Solution
Sample Problem 3.13
Calculating the Molarity of a Solution
PROBLEM:
Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains mol of glycine in 495 mL?
PLAN:
Molarity is the number of moles of solute per liter of solution.
mol of glycine
SOLUTION:
divide by volume
0.715 mol glycine
495 mL soln
1000mL
1 L x
concentration(mol/mL) glycine
103mL = 1L
= 1.44 M glycine
molarity(mol/L) glycine

42. mass-mole-number-volume relationships in solution.
Figure 3.12
MASS (g)
of compound
in solution
Summary of
mass-mole-number-volume relationships in solution.
M (g/mol)
AMOUNT (mol)
of compound
in solution
Avogadro’s number
(molecules/mol)
M (g/mol)
MOLECULES
(or formula units)
of compound
in solution
VOLUME (L)
of solution

43. Sample Problem 3.14
Calculating Mass of Solute in a Given Volume of Solution
PROBLEM:
A “buffered” solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. How many grams of solute are in 1.75 L of M sodium monohydrogen phosphate?
PLAN:
Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na2HPO4.
volume of soln
SOLUTION:
multiply by M
0.460 moles
1 L
moles of solute
1.75 L x
= mol Na2HPO4
multiply by M
g Na2HPO4
mol Na2HPO4
0.805 mol Na2HPO4 x
grams of solute
= 114 g Na2HPO4

44. Laboratory preparation of molar solutions.
Figure 3.12 A
Weigh the solid needed.
Transfer the solid to a volumetric flask that contains about half the final volume of solvent.
C Add solvent until the solution reaches its final volume.
B Dissolve the solid thoroughly by swirling.

45. Converting a concentrated solution to a dilute solution.
Figure 3.13
Converting a concentrated solution to a dilute solution.

46. MdilxVdil = #mol solute = MconcxVconc
Preparing a Dilute Solution from a Concentrated Solution
Sample Problem 3.14
PROBLEM:
“Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotomic saline from a 6.0 M stock solution?
PLAN:
It is important to realize the number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume.
MdilxVdil = #mol solute = MconcxVconc
volume of dilute soln
multiply by M of dilute solution
SOLUTION:
moles of NaCl in dilute soln = mol NaCl in concentrated soln
0.15 mol NaCl
L soln
0.80 L soln x
= 0.12 mol NaCl
divide by M of concentrated soln
L solnconc
6 mol
0.12 mol NaCl x
= L soln
L of concentrated soln

47. Sample Problem 3.15
Calculating Amounts of Reactants and Products for a Reaction in Solution
PROBLEM:
Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide?
PLAN:
Write a balanced equation for the reaction; find the grams of Mg(OH)2; determine the mol ratio of reactants and products; use mols to convert to molarity.
mass Mg(OH)2
divide by M
mol HCl
L HCl
divide by M
mol Mg(OH)2
mol ratio

48. Sample Problem 3.15
Calculating Amounts of Reactants and Products for a Reaction in Solution
continued
SOLUTION:
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
mol Mg(OH)2
58.33g Mg(OH)2
0.10g Mg(OH)2 x
= 1.7x10-3 mol Mg(OH)2
2 mol HCl
1 mol Mg(OH)2
= 3.4x10-3 mol HCl
1.7x10-3 mol Mg(OH)2 x 1L
0.10mol HCl
3.4x10-3 mol HCl x
= 3.4x10-2 L HCl

49. Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions in Solution
PROBLEM:
Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(II) sulfide form?
PLAN:
As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as we would for a limiting reactant problem. Find the amount of product which would be made from each reactant. Then choose the reactant that gives the lesser amount of product.

50. Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions in Solution
continued
L of Hg(NO3)2
mol Hg(NO3)2
mol HgS
multiply by M
mol ratio
Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq)
SOLUTION:
0.050 L Hg(NO3)2
0.020 L Hg(NO3)2
L of Na2S
mol Na2S
mol HgS
multiply by M
mol ratio
x mol/L
x 0.10 mol/L
1 mol HgS
1 mol Hg(NO3)2
1 mol HgS
1 mol Na2S x x
= 5.0x10-4 mol HgS
= 2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
232.7g HgS
1 mol HgS
5.0x10-4 mol HgS x
= 0.12 g HgS

51. An overview of the key mass-mole-number stoichiometry relationships.
Figure 3.15