Chapter 2 Overview
! all matter, whether liquid, solid, or gas,consists of atoms,
which form molecules
! the identity of an atom and its chemical behavior is
strictly defined by the number of electrons in the outer
electronic sphere, which equals to the number of protons in
the nucleus
! the atomic weight is the combined mass of protons and
neutrons in the nucleus. Each contributes one unit of mass.
The mass of the electrons is negligible
Chapter 2 Overview
6 electrons
6 electrons
6 no+
6p
12
7 no+
6p
13
C
C
ISOTOPES OF CARBON
number of
neutrons
N = Mass - Z
atomic number
number of protons
Periodic Table
Main Group Elements:Group Number corresponds to ionic charge
Chapter 2 Overview
Ionic compounds: Simple salts. Charges Balanced
NaCl
Na+
Cl–
LiI
Li+
I–
Cs2SO4
Cs+
SO42–
Fe2(CO3)3
Fe3+
CO32–
(NH4)3PO4
NH4+
PO43–
Chapter 2 Overview
Molecular Weight = Sum of Atomic Weights
MW(NaCl) = Atomic Mass of Na + At. Mass of Cl
Chapter 2 Sample Problem
2.132. Succinic acid is an important metabolite in biological
energy production. Give the molecular formula, empirical
formula, and molecular mass of succinic acid.
O H H
O
C C C C
H O H H O H
succinic acid
Chapter 2 Sample Problem
2.132. Succinic acid is an important metabolite in biological
energy production. Give the molecular formula, empirical
formula, and molecular mass of succinic acid.
O H H
• mol. formula:count all atoms: C4H6O4
O
C C C C
H O H H O H
succinic acid
Chapter 2 Sample Problem
2.132. Succinic acid is an important metabolite in biological
energy production. Give the molecular formula, empirical
formula, and molecular mass of succinic acid.
O H H
• mol. formula:count all atoms: C4H6O4
O
C C C C
H O H H O H
succinic acid
• empirical formula: C2H3O2 (whole numbers!)
Chapter 2 Sample Problem
2.132. Succinic acid is an important metabolite in biological
energy production. Give the molecular formula, empirical
formula, and molecular mass of succinic acid.
O H H
• mol. formula:count all atoms: C4H6O4
O
C C C C
H O H H O H
succinic acid
• empirical formula: C2H3O2 (whole numbers!)
• MW = 4!12.01 + 6!1.008 + 4!16.00 = 118.09
Chapter 2 Sample Problem
2.128. How can iodine (Z = 53) have a higher atomic number yet
lower atomic weight than tellurium (Z = 52)?
Stoichiometry: The Mole
describes quantitative aspects of chemical reactions
One mole (1 mol) is 6.022!1023 molecules (ions, atoms, etc.)
Avogadro’s number
M = MW
molar
weight (g/mol)
molecular
weight
Stoichiometry
MW(H2) = 2.02
M = 2.02 g/mol
MW(NaCl) = 58.44
M = 58.44 g/mol
MW(AgNO3) = 169.91
M = 169.91 g/mol
1 : 1 ratio of NaCl and AgNO3:
58.44 g of NaCl and 169.91 g of AgNO3
or
etc
0.584 g of NaCl and 0.169 g of AgNO3
Stoichiometry
M (g/mol) =
mass in grams
m (g)
n (mol)
molar
weight (g/mol)
number of mols
Stoichiometry
mass in grams
m (g)
M (g/mol) =
n (mol)
molar
weight (g/mol)
n (mol) =
number of mols
m (g)
M (g/mol)
Stoichiometry
mass in grams
m (g) = n (mol) ! M (g/mol)
number of moles (mol)
molar
weight (g/mol)
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
• divide given mass by mol weight
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
• divide given mass by mol weight
• what is ammonium carbonate?
n (mol) =
m (g)
M (g/mol)
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
n (mol) =
m (g)
M (g/mol)
• divide given mass by mol weight
• what is ammonium carbonate? NH4+ ,CO32– " (NH4)2CO3
• what is its mol. weight?
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
n (mol) =
m (g)
M (g/mol)
• divide given mass by mol weight
• what is ammonium carbonate? NH4+ ,CO32– " (NH4)2CO3
• what is its mol. weight?
2!14.01 + 8!1.008 + 12.01 + 3!16.00 = 96.09 g/mol
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
n (mol) =
m (g)
M (g/mol)
• divide given mass by mol weight
• what is ammonium carbonate? NH4+ ,CO32– " (NH4)2CO3
• what is its mol. weight?
2!14.01 + 8!1.008 + 12.01 + 3!16.00 = 96.09 g/mol
• answer: 41.6 g : 96.06 g/mol = 0.433 mol
Sample Problem 3.2
Ammonium carbonate, is a white solid that decomposes with
warming. Among its many uses, it is a component of baking powder,
fire extinguishers, and smelling salts. How many moles are in 41.6 g
of ammonium carbonate?
n (mol) =
m (g)
M (g/mol)
• divide given mass by mol weight
• what is ammonium carbonate? NH4+ ,CO32– " (NH4)2CO3
• what is its mol. weight?
2!14.01 + 8!1.008 + 12.01 + 3!16.00 = 96.09 g/mol
• answer: 41.6 g : 96.06 g/mol = 0.433 mol
(number of formula units = 0.433 mol ! 6.022!1023 mol-1 = 2.61!1023)
Mass Percent from Chemical
Formula
Glucose (C6H12O6) is the most important nutrient in the living cell
for generating energy. What is the mass percent of each element in
glucose?
• let’s focus on carbon first
Mass Percent from Chemical
Formula
Glucose (C6H12O6) is the most important nutrient in the living cell
for generating energy. What is the mass percent of each element in
glucose?
• let’s focus on carbon first
mass of C in one mole(cule) .
mass of C6H12O6 in one mole(cule)
Mass Percent from Chemical
Formula
Glucose (C6H12O6) is the most important nutrient in the living cell
for generating energy. What is the mass percent of each element in
glucose?
• let’s focus on carbon first
mass of C in one mole(cule) .
mass of C6H12O6 in one mole(cule)
6!12.01
.= 72.06 .
= 0.400 = 40.0 mass%
6!12.01 + 12!1.008 + 6!16.00 180.156
Elemental Analysis and
Molecular Formula
• complete burning, than analysis of oxidation products
• used to determine empirical formula of an unknown
compound (gives an empirical formula only!)
• used to confirm the structure
• (mass spectroscopy is a superior modern technique
that gives a molecular formula directly)
Elemental Analysis and
Molecular Formula
Sample Problem 3.5: During physical activity, lactic acid (M =
90.08 g/mol) forms in muscle and is responsible for muscle
soreness. Elemental analysis shows that it 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 of lactic acid
Elemental Analysis and
Molecular Formula
Sample Problem 3.5: During physical activity, lactic acid (M =
90.08 g/mol) forms in muscle and is responsible for muscle
soreness. Elemental analysis shows that it 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 of lactic acid
• mass% # arbitrary weight (let’s take 100 g)
• convert grams to moles
Elemental Analysis and
Molecular Formula
Sample Problem 3.5: During physical activity, lactic acid (M =
90.08 g/mol) forms in muscle and is responsible for muscle
soreness. Elemental analysis shows that it 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 of lactic acid
• mass% # arbitrary weight (let’s take 100 g)
• convert grams to moles
C=
40.0 g
= 3.33 mol
12.01 g/mol
O=
6.71 g
H=
= 6.66 mol
1.008 g/mol
53.3 g
= 3.33 mol
16.00 g/mol
Elemental Analysis and
Molecular Formula
Sample Problem 3.5: During physical activity, lactic acid (M =
90.08 g/mol) forms in muscle and is responsible for muscle
soreness. Elemental analysis shows that it 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 of lactic acid
• mass% # arbitrary weight (let’s take 100 g)
• convert grams to moles
C=
40.0 g
= 3.33 mol
12.01 g/mol
O=
6.71 g
H=
= 6.66 mol
1.008 g/mol
53.3 g
= 3.33 mol
16.00 g/mol
CH2O
Elemental Analysis and
Molecular Formula
Sample Problem 3.5: During physical activity, lactic acid (M =
90.08 g/mol) forms in muscle and is responsible for muscle
soreness. Elemental analysis shows that it 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 of lactic acid
Empirical formula:
"
CH2O
“M = 30.03 g/mol”
Actual M is 3 (three) times greater, therefore:
Molecular formula:
C3H6O3
Chemical Equations
Typical Structure:
Starting Material(s) # Products(s)
H2 + Cl2 # 2HCl
Cl
H
H
+
Cl
Cl
H
=
Cl
H
Chemical Equations
H2 + Cl2 # 2HCl
n
m (g)
(mol) = M (g/mol)
0.124 mol H2
0.124 mol Cl2
2!0.124 mol HCl
23.12 mol H2
23.12 mol Cl2
2!23.12 mol HCl
0.905 mol H2
0.745 mol Cl2
Chemical Equations
H2 + Cl2 # 2HCl
n (mol) =
m (g)
M (g/mol)
0.124 mol H2
0.124 mol Cl2
2!0.124 mol HCl
23.12 mol H2
23.12 mol Cl2
2!23.12 mol HCl
0.905 mol H2
0.745 mol Cl2
2!0.745 mol HCl
Chemical Equations
H2 + Cl2 # 2HCl
n
m (g)
(mol) = M (g/mol)
Global Strategy:
Input (g, ml, kg, etc) " moles
moles (starting)
(starting) "
"
moles (final) " Output (g, ml, kg, etc)
Balancing Chemical Equations
Starting Material(s) #
Products(s)
space for coefficients
! the same number of atoms of each element
on both sides
Balancing Chemical Equations
Unbalanced equation:
PF3 + HCl # PCl3 + HF
Balancing Chemical Equations
Unbalanced equation:
PF3 + HCl # PCl3 + HF
PF3 + HCl # PCl3 + 3HF
Balancing Chemical Equations
Unbalanced equation:
PF3 + HCl # PCl3 + HF
PF3 + HCl # PCl3 + 3HF
PF3 + 3HCl # PCl3 + 3HF
Balancing Chemical Equations
Unbalanced equation:
PF3 + HCl # PCl3 + HF
PF3 + HCl # PCl3 + 3HF
Balanced equation!:
PF3 + 3HCl # PCl3 + 3HF
Check
Balancing Chemical Equations
Unbalanced equation:
C8H18 + O2 # CO2 + H2O
Balancing Chemical Equations
Unbalanced equation:
C8H18 + O2 # CO2 + H2O
C8H18 + O2 # 8CO2 + H2O
C8H18 + O2 # 8CO2 + 9H2O
C8H18 + 12.5O2 # 8CO2 + 9H2O 16+9=25
Balanced equation:
2C8H18 + 25O2 # 16CO2 + 18H2O
Check
Stoichiometric Equivalents
C3H8 + 5O2 # 3CO2 + 4H2O
(propane)
1 mol of C3H8 is equivalent to 4 mol of H2O
3 mol of CO2 is equivalent to 5 mol of O2
3 mol of CO2 is equivalent to 4 mol of H2O
Stoichiometric Equivalents
C3H8 + 5O2 # 3CO2 + 4H2O
In the combustion of propane, how many moles of carbon dioxide
are produced along with 10.0 moles of water?
10 = 2.5
4
2.5!3 = 7.5 " 7.5 moles of CO2
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
Fundamental Strategy:
Input (g, ml, kg, etc) " moles
moles(starting)
(initial) "
"
moles
" Output (g, ml, kg, etc)
moles (final)
(final) "
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
1 m3 = 1000 L, 1kg = 1000 g, d = 1.34 kg/m3
d=m
V
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
1 m3 = 1000 L, 1kg = 1000 g, d = 1.34 kg/m3
d=m
V
n=
m
M
m(O2) = 25 m3!1.34 kg/m3 = 33.5 kg = 33500 g
n(O2) = 33500 g ÷ 32.00 g/mol = 1047 mol
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
n(O2) = 1047 mol
n(C6H12O6) = (1/6)!n(O2) = (1/6)!1047 mol = 174.5 mol
n=
m
M
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
n(O2) = 1047 mol
n(C6H12O6) = (1/6)!n(O2) = (1/6)!1047 mol = 174.5 mol
n=
m
M
m(C6H12O6) = 174.5 mol ! 180.16 g/mol = 31438 g
Stoichiometric Equivalents
6CO2 + 6H2O # C6H12O6 + 6O2
glucose
(MW = 180.16)
How many grams of glucose are produced during
photosynthesis that generates 25 m3 of oxygen (at normal
conditions, d = 1.34 g/L).
n(O2) = 1047 mol
n(C6H12O6) = (1/6)!n(O2) = (1/6)!1047 mol = 174.5 mol
n=
m
M
m(C6H12O6) = 174.5 mol ! 180.16 g/mol = 31438 g
Example
2H2 + O2 # 2H2O
16.0 g
n (mol) =
16.0 g
?g
m (g)
M (g/mol)
Example
2H2 + O2 # 2H2O
16.0 g
16.0 g
7.94 mol 0.5 mol
n (mol) =
m (g)
M (g/mol)
?g
Example
2H2 + O2 # 2H2O
16.0 g
16.0 g
7.94 mol 0.5 mol
n (mol) =
?g
1.0 mol
m (g)
M (g/mol)
Example
2H2 + O2 # 2H2O
16.0 g
16.0 g
7.94 mol 0.5 mol
n (mol) =
m (g)
M (g/mol)
?g
1.0 mol " 18.02 g
Yields: Theoretical, Percent
2A + B # A2B + c + d + ....
0.343 g should give 0.772 g
actually, 0.645 g was obtained
Yield = Y = 0.645 g !100% = 84%
0.772 g
n (mol) =
m (g)
M (g/mol)
Yields: Theoretical, Percent
2A + B # A2B + c + d + ....
0.200 mol should give 0.100 mol
actually, 0.084 mol was obtained
Yield = Y = 0.084 mol !100% = 84%
0.100 mol
n (mol) =
m (g)
M (g/mol)
Yields: Overall
90%
75%
85%
A # B # C # D
Overall Y = 0.90 ! 0.75 ! 0.85 ! 100% = 57%
57%
A # D
Solutions
Solute
Solvent
concentration: a quantitative description
of a solution
Solutions
Molarity =
moles of solute
liters of solution
M = n solute (mol)
Vsolvent (L)
3 M means 3 mol/L
Sample Problem 3.14
“Isotonic saline” is 0.15 M aqueous solution of NaCl that simulates
the total concentration of ions found in many cellular fluids. Its uses
range from a cleansing rinse for contact lenses to a washing medium
for red blood cells. How do you prepare 0.80 L of isotonic saline from
6.0 M stock solution of NaCl?
M=
n
V
Sample Problem 3.14
“Isotonic saline” is 0.15 M aqueous solution of NaCl that simulates
the total concentration of ions found in many cellular fluids. Its uses
range from a cleansing rinse for contact lenses to a washing medium
for red blood cells. How do you prepare 0.80 L of isotonic saline from
6.0 M stock solution of NaCl?
We’ll need:
M=
n(NaCl) = 0.15 mol/L ! 0.80 L = 0.12 mol
n
V
Sample Problem 3.14
“Isotonic saline” is 0.15 M aqueous solution of NaCl that simulates
the total concentration of ions found in many cellular fluids. Its uses
range from a cleansing rinse for contact lenses to a washing medium
for red blood cells. How do you prepare 0.80 L of isotonic saline from
6.0 M stock solution of NaCl?
We’ll need:
n(NaCl) = 0.15 mol/L ! 0.80 L = 0.12 mol
V(6M_soln) = 0.12 mol ÷ 6.0 mol/L = 0.02 L
M=
n
V
0.78 L
Example
*3.141. 50 mL of a solution of NaOH with an unknown concentration
reacts with an excess of an acid solution of oxalic acid prepared by
dissolving 0.588 g of oxalic acid dihydrate (H2C2O4•2H2O) in water.
The excess of oxalic acid then reacts completely with 9.65 ml of 0.115
M of stock NaOH solution. What is the concentration of the unknown
NaOH solution.
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
M=
n
V
Example
*3.141. 50 mL of a solution of NaOH with an unknown concentration
reacts with an excess of an acid solution of oxalic acid prepared by
dissolving 0.588 g of oxalic acid dihydrate (H2C2O4•2H2O) in water.
The excess of oxalic acid then reacts completely with 9.65 ml of 0.115
M of stock NaOH solution. What is the concentration of the unknown
NaOH solution.
MWoad = 126.08
MWNaOH = 40.00
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
M=
n
V
Example
*3.141. 50 mL of a solution of NaOH with an unknown concentration
reacts with an excess of an acid solution of oxalic acid prepared by
dissolving 0.588 g of oxalic acid dihydrate (H2C2O4•2H2O) in water.
The excess of oxalic acid then reacts completely with 9.65 ml of 0.115
M of stock NaOH solution. What is the concentration of the unknown
NaOH solution.
MWoad = 126.08
MWNaOH = 40.00
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
n(H2C2O4) = 0.588g ÷ 126.08g/mol = 4.66 mmol
M=
n
V
Example
*3.141. 50 mL of a solution of NaOH with an unknown concentration
reacts with an excess of an acid solution of oxalic acid prepared by
dissolving 0.588 g of oxalic acid dihydrate (H2C2O4•2H2O) in water.
The excess of oxalic acid then reacts completely with 9.65 ml of 0.115
M of stock NaOH solution. What is the concentration of the unknown
NaOH solution.
MWoad = 126.08
MWNaOH = 40.00
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
n(H2C2O4) = 0.588g ÷ 126.08g/mol = 4.66 mmol
n(NaOH-known) = 9.65 mL ! 0.115mol/L = 1.11 mmol
M=
n
V
Example
*3.141. 50 mL of a solution of NaOH with an unknown concentration
reacts with an excess of an acid solution of oxalic acid prepared by
dissolving 0.588 g of oxalic acid dihydrate (H2C2O4•2H2O) in water.
The excess of oxalic acid then reacts completely with 9.65 ml of 0.115
M of stock NaOH solution. What is the concentration of the unknown
NaOH solution.
MWoad = 126.08
MWNaOH = 40.00
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
n(H2C2O4) = 0.588g ÷ 126.08g/mol = 4.66 mmol
n(NaOH-known) = 9.65 mL ! 0.115mol/L = 1.11 mmol
M=
n
V
n(oxalic-unknown) = 4.66 - 1.11/2 = 4.11 mmol
Example
*3.141. 50 mL of a solution of NaOH with an unknown concentration
reacts with an excess of an acid solution of oxalic acid prepared by
dissolving 0.588 g of oxalic acid dihydrate (H2C2O4•2H2O) in water.
The excess of oxalic acid then reacts completely with 9.65 ml of 0.115
M of stock NaOH solution. What is the concentration of the unknown
NaOH solution.
MWoad = 126.08
MWNaOH = 40.00
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
n(NaOH-unknown) = 4.11mmol!2 = 8.22 mmol
M=
n
V
M = 8.22 mmol = 0.164 M
50 mL
Look at equation
Chapter 3 Overview
Stoichiometry
•
The mole: proportional to the number of molecules
a convenient measure of amount
n(mol) = m(g) / M(g/mol)
M = MW
•
Chemical equations:
balanced: same number of atoms on both sides
show stoichiometric relationships between
reactants and products
•
Solutions:
concentration
M (mol/L) = n(mol) / V(L)
Chapter 3 Overview
Practice Problems
(3.44) Cortisol (M = 362.47 g/mol), one of the major steroid
hormones, is a key factor in the biosynthesis of protein.
Cortisol is 69.6% carbon, 8.34% H, 22.1% O by mass.
What is its molecular formula?
Chapter 3 Overview
Practice Problems
(3.44) Cortisol (M = 362.47 g/mol), one of the major steroid
hormones, is a key factor in the biosynthesis of protein.
Cortisol is 69.6% carbon, 8.34% H, 22.1% O by mass.
What is its molecular formula?
(1) get molar ratio of C ÷ H ÷ O " get empirical formula
(2) look at molecular weight, empirical formula " get
molecular formula
Chapter 3 Overview
Practice Problems
(3.51) Balance the following equations:
(a)
Cu + S8 # Cu2S
Chapter 3 Overview
Practice Problems
(3.51) Balance the following equations:
(a)
Cu + S8 # Cu2S
16Cu + S8 # 8Cu2S
Chapter 3 Overview
Practice Problems
(3.51) Balance the following equations:
(a)
Cu + S8 # Cu2S
16Cu + S8 # 8Cu2S
(b)
P4O10 + H2O # H3PO4
Chapter 3 Overview
Practice Problems
(3.51) Balance the following equations:
(a)
Cu + S8 # Cu2S
16Cu + S8 # 8Cu2S
(b)
P4O10 + H2O # H3PO4
(3.52) Balance the following equations:
(c)
CaSiO3 + HF # SiF4 + CaF2 + H2O
Chapter 3 Overview
Practice Problems
(3.56) Convert the following into balanced equations:
(a) When lead (II) nitrate solution is added to potassium iodide
solution, solid lead (II) iodide forms and potassium nitrate
solution remains.
Chapter 3 Overview
Practice Problems
(3.56) Convert the following into balanced equations:
(a) When lead (II) nitrate solution is added to potassium iodide
solution, solid lead (II) iodide forms and potassium nitrate
solution remains.
Pb(NO3)2 + 2KI # PbI2 + 2KNO3
Chapter 3 Overview
Practice Problems
( 3.61) Chlorine gas can be made in the laboratory by the
reaction of hydrochloric acid and manganese (IV) dioxide:
4HCl + MnO2 # MnCl2 + Cl2 + 2H2O
When 1.82 mol of HCl reacts with excess MnO2, (a) how
many mols of Cl2 form? (b) How many grams of Cl2 form?
Chapter 3 Overview
Practice Problems
( 3.61) Chlorine gas can be made in the laboratory by the
reaction of hydrochloric acid and manganese (IV) dioxide:
4HCl + MnO2 # MnCl2 + Cl2 + 2H2O
When 1.82 mol of HCl reacts with excess MnO2, (a) how
many mols of Cl2 form? (b) How many grams of Cl2 form?
n (mol) =
m (g)
M (g/mol)
Chapter 3 Overview
Practice Problems
( 3.74) Calculate the maximum numbers of moles and
grams of H2S that can form when 158 g of aluminum sulfide
reacts with 131 g of water:
Al2S3 + H2O # Al(OH)3 + H2S
What mass of the excess reactant remains?
[unbalanced]
Chapter 3 Overview
Practice Problems:
Solutions
( 3.93) Calculate each of the following quantities:
(a) Volume in liters of 2.26 M potassium hydroxide
that contains 8.42 g of solute.
(c) molarity of 275 mL solution containing 135 mmol
of glucose.
Chapter 3 Overview
Practice Problems:
Solutions
( 3.103) How many moles of which reactant are in excess
when 350.0 mL of 0.210 M sulfuric acid reacts with 0.500 L
of 0.196 M sodium hydroxide to form water and aqueous
sodium sulfate.
Chapter 3 Overview
Practice Problems:Comprehensive
( 3.133) To 1.20 L of 0.325 M HCl, you add 3.37 L of a
second HCl solution of unknown concentration. The
resulting solution is 0.893 M HCl. Assuming the volumes
are additive, calculate the molarity of the second HCl
solution.
Chapter 3 Overview
Practice Problems:Comprehensive
( *3.141) In the chemical analysis of unknown, chemsits
often add an excess of a reactant, determine the amount of
that reactant remaining after the reaction with the unknown,
and use those amounts to calculate the amount of the
unknown. For analysis of an unknown NaOH solution, you
add 50.0 mL of the solution to 0.150 L of an acid solution
prepared by dissolving 0.588 g of solid oxalic acid dihydrate
(H2C2O4•2H2O, MW = 126.08) in water:
2NaOH + H2C2O4 # Na2C2O4 + 2H2O
The unreacted NaOH then reacts completely with 9.65 mL
of 0.116 M HCl:
NaOH + HCl # NaCl + H2O
What is the molarity of the original NaOH solution?