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Ch 4. Chemical Quantities and
Aqueous Reactions
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
1 mol
2 mol
1 mol
2 mol
Stoichiometry of the reaction
FIXED ratio for each reaction
Can calculate how much other chemicals are required
or produced if the amount of one chemical is known.
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
1 mol
2 mol
1 mol
2 mol
2 mol
4 mol
2 mol
4 mol
3 mol
6 mol
3 mol
6 mol
5.22 mol
10.44 mol
5.22 mol
10.44 mol
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
5.02 mol of NH3 is used in the above reaction.
How many moles of O2 is required to react with all the NH3?
How many moles of H2O will be produced?
2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g)
2 mol
5.02 mol
3 mol
1 mol
3 mol
3 mol
x
y
z
u
2 mol 5.02 mol 2


3 mol
x
3
 x  7.53 mol
2 5.02 mol

1
y
 y  2.51 mol
2 5.02 mol

3
z
 z  7.53 mol
2 5.02 mol

3
u
 u  7.53 mol
2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g)
2 mol
3 mol
1 mol
6.04 g
mass?
mass?
3 mol
3 mol
6.04 g ÷ (14.01 g/mol x 1 + 1.008 g/mol x 3) = 0.355 mol
0.355 mol
2 0.355 mol

3
x
x
y
 x  0.533 mol
Mass of CuO = 0.533 mol x 79.55 g/mol = 42.4 g
2 0.355 mol

1
y
 y  0.178 mol
Mass of N2 = 0.178 mol x 28.02 g/mol = 4.99 g
Other Examples: page 131 ― 132
CH4 + 2O2  CO2 + 2H2O
initial:
1 mol
2 mol
0 mol
0 mol
final:
0 mol
0 mol
1 mol
2 mol
The actual amount of reactants consumed and actual amount
of products generated agree with the stoichiometry.
initial:
1 mol
1 mol
0 mol
0 mol
final:
? mol
? mol
? mol
? mol
CH4 + 2O2  CO2 + 2H2O
initial:
consumed:
final:
1 mol
1 mol
x = 0.5 mol
1 mol
(1 − 0.5) mol
0 mol
1
x

2 1 mol
0 mol
0 mol
y = 0.5 mol z = 1 mol
 x = 0.5 mol
1 mol CH4 requires 2 mol O2, available O2 is 1 mol: limiting reagent.
Result: 1 mol O2 will be consumed completely and CH4 will
have leftover: excess reagent.
The reactant of which there are fewer moles than the
stoichiometry requires is the limiting reagent.
The reactant of which there are more moles than the
stoichiometry requires is the excess reagent.
Chemical reactions always occur according to the
stoichiometry, therefore the limiting reagent is consumed
and the excess reagent has leftover. The amount of products
is determined by the amounts of reagents that are actually
consumed.
excess reagent
limiting reagent
CH4 + 2O2  CO2 + 2H2O
initial:
1 mol
1 mol
consumed: 0.5 mol
final:
(1 − 0.5) mol
0.5
= 1
0 mol
0 mol
0.5 mol
1 mol
1 mol
0 mol
:
1
:
0.5
:
1
:
2
:
1
:
2
+
+
+
+
+
+
+
+
2 slices of bread + 1 slice if ham  1 sandwich
4 slices of bread + 1 slice if ham  1 sandwich + 2 slices of bread
excess reagent
limiting reagent
amount of product
excess reagent
leftover
For the following reaction, if a sample containing 18.1 g of NH3
is reacted with 90.4 g of CuO, which is the limiting reagent? How
many grams of N2 will be formed?
2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g)
How many grams of excess reagent will be leftover?
If 6.63 g of N2 is actually produced, what is the percent yield?
actual yield
percent yi eld 
 100 %
theoretica l yield
Procedure for limiting/excess reagent calculations
aA + bB  cC + dD
1) Make sure the equation is balanced.
2) Find the moles of each reactant:
moles = mass in gram / molar mass
3) Pick up any reactant, say A, and use the stoichiometry to
calculate the required amount of the other reactant B.
4) Compare the required amount of B with the available
amount of B.
a) If required > available, then B is the limiting reagent and A
is the excess reagent.
b) If required < available, then B is the excess reagent and A
is the limiting reagent.
5) Use the amount of the limiting reagent and the stoichiometry
to calculate the amount of any product and the amount of the
excess reagent that has been consumed.
6) Leftover excess reagent = available − consumed
7) If actual yield is given
percent yield = (actually yield / theoretical yield) x 100%
68.5 g CO reacts with 8.60 g H2 in the following reaction.
What is the limiting reagent? How many grams of excess
reagent is leftover? What is the theoretical yield of CH3OH?
If 35.7 g CH3OH is actually produced, what is the percent
yield of CH3OH?
H2(g) + CO(g)  CH3OH(g)
Procedure for limiting/excess reagent calculations
aA + bB  cC + dD
1) Make sure the equation is balanced.
2) Find the moles of each reactant:
moles = mass in gram / molar mass
3) Pick up any reactant, say A, and use the stoichiometry to
calculate the required amount of the other reactant B.
4) Compare the required amount of B with the available
amount of B.
a) If required > available, then B is the limiting reagent and A
is the excess reagent.
b) If required < available, then B is the excess reagent and A
is the limiting reagent.
5) Use the amount of the limiting reagent and the stoichiometry
to calculate the amount of any product and the amount of the
excess reagent that has been consumed.
6) Leftover excess reagent = available − consumed
7) If actual yield is given
percent yield = (actually yield / theoretical yield) x 100%
1.50 g of ammonia reacts with 2.75 g of oxygen gas to produce
nitrogen monoxide and water.
NH3(g) + O2(g) NO(g) + H2O(g)
a)
b)
c)
d)
e)
f)
g)
Balance the equation.
What is the mass of O2 in grams required by NH3?
Which reactant is the limiting reagent?
How many grams of NO will be produced in theory?
How many grams of H2O will be produced in theory?
How many grams of the excess reagent remain unreacted?
If only 1.80 g of NO are produced, what is the percent yield?
Classification of Matter
Homogeneous (Solutions)
(visibly indistinguishable)
Mixtures
(multiple components)
Heterogeneous
(visibly distinguishable)
Matter
Elements
Pure Substances
(one component)
Compounds
Solute + Solvent  Solution
Solvent = water, aqueous solution
Water can dissolve many substances
H2O
O
H
H
C12H22O11
Based on the electrical conductivity in aqueous solution
salts
strong electrolytes
strong acids
strong bases
electrolytes
weak acids
solutes
weak electrolytes
weak bases
nonelectrolytes
strong electrolytes: dissociate 100 % into ions
weak electrolytes: only a small fraction dissociate
into ions
nonelectrolytes: no dissociation
salts: NaCl, Na2SO4, Fe(ClO4)3 ……
strong acids: HCl, HNO3, H2SO4, HClO4
remember
strong bases: NaOH, KOH
Base: compounds that give OH− when dissolved in water.
weak acids: acetic acid: HC2H3O2
remember
weak bases: ammonia: NH3
HCl is
Completely
Ionized
Aqueous
Solution of
NaOH
Reaction
of NH3 in
Water
concentrations
mass of component
mass percent 
 100 %
mass of whole sample
mass of solute
mass percent 
 100 %
mass of solution
no unit
10. g of sugar is dissolved in 40. g of water.
What is the mass percent of sugar in this solution?
moles of solute
Molarity (M) 
liters of solution
Unit: mol/L or M
0.50 mol of KBr is dissolved in water and forms a
solution of 12 L. What is the molarity of the solution?
Molarity (M) 
Example 4.5, page 141
moles of solute
liters of solution
25.5 g of KBr is dissolved in water and forms a
solution of 1.75 L. What is the molarity of the solution?
Example 4.6, page 142
How many liters of a 0.125 mol/L NaOH solution
contains 0.255 mol of NaOH?
How to prepare 1.00 L of NaCl aqueous
solution with a molarity of 1.00 mol/L?
1.00 mol NaCl + 1.00 L of H2O = 1.00 mol/L NaCl (aq)
Solution Dilution
Concentrated solutions for storage, called stock solutions
stock solution + water  desired solution
moles of solute before dilution = moles of solute after dilution
moles of solute
Molarity (M) 
liters of solution
M1V1 = M2V2
M1: molarity of concentrated solution
V1: volume of concentrated solution
M2: molarity of diluted solution
V2: volume of diluted solution
Example on page 143
A lab procedure calls for 3.00 L of a 0.500 mol/L CaCl2
solution. How should we prepare it from a 10.0 mol/L stock
solution?
Example 4.7, page 144
To what volume should you dilute 0.200 L of a 15.0 mol/L
NaOH solution to obtain a 3.00 mol/L NaOH solution?
Types of reactions
Precipitation reactions
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
formula equation
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq)
complete ionic equation
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq)
spectator ions
Cl−(aq) + Ag+(aq)  AgCl(s)
net ionic equation
EXAMPLE 4.9 Predicting whether an Ionic Compound Is Soluble
Predict whether each compound is soluble or insoluble.
(a) PbCl2
(b) CuCl2
(c)Ca(NO3)2 (d) BaSO4
BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2KCl(aq)
BaCl2(aq)  Ba2+(aq) + 2Cl−(aq)
K2SO4(aq)  2K+ (aq) + SO42− (aq)
Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + SO42− (aq)  BaSO4(s) + 2Cl−(aq) + 2K+(aq)
Ba2+(aq) + SO42− (aq)  BaSO4(s)
Fe(NO3)3(aq) + KOH(aq) 
Fe(NO3)3(aq) + 3KOH(aq)  Fe(OH)3(s) + 3KNO3(aq)
Fe(NO3)3(aq)  Fe3+(aq) + 3NO3−(aq)
KOH(aq)  K+ (aq) + OH− (aq)
3KOH(aq)  3K+ (aq) + 3OH− (aq)
Fe3+(aq) + 3NO3−(aq) +3K+ (aq) +3OH− (aq)  Fe(OH)3(s) + 3NO3−(aq) + 3K+(aq)
Fe3+(aq) + 3OH− (aq)  Fe(OH)3(s)
BaCl2(aq) + KNO3(aq) 
BaCl2(aq) + 2KNO3(aq)  Ba(NO3)2(aq) + 2KCl(aq)
BaCl2(aq)  Ba2+(aq) + 2Cl−(aq)
KNO3(aq)  K+ (aq) + NO3− (aq)
2KNO3(aq)  2K+ (aq) + 2NO3− (aq)
Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + 2NO3− (aq) 
Ba2+(aq) + 2NO3− (aq) + 2Cl−(aq) + 2K+(aq)
Types of reactions
Precipitation reactions
Acid-base reactions
Acid: Substance that produces H+ ions in aqueous solution
Base: Substance that produces OH− ions in aqueous solution
HCl(aq)  H+(aq) + Cl−(aq)
NaOH(aq)  Na+(aq) + OH−(aq)
H+(aq) + OH−(aq)  H2O(l)
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
H+(aq) + Cl−(aq) +Na+(aq) +OH−(aq)  H2O(l) + Na+(aq) + Cl−(aq)
H+(aq) + OH−(aq)  H2O(l)
acidic
basic
neutral
neutralization
HCl(aq)  H+(aq) + Cl−(aq)
What is the molarity of
HCl(aq) or H+(aq)?
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
When reaction completes
nNaOH = nHCl
MNaOHVNaOH = MHClVHCl
prepared,
known
measured
by buret, known
unknown
measured by
pippet, known
MNaOHVNaOH = MHClVHCl
Read acid-base titration starting on page 158 and the online instruction
for next week’s titration
Example 4.14
The titration of a 10.00-mL sample of an HCl
solution of unknown concentration requires 12.54
mL of a 0.100 M NaOH solution to reach the
equivalence point. What is the concentration of the
unknown HCl solution in M?
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
MNaOHVNaOH = MHClVHCl
Types of reactions
Precipitation reactions
Acid-base reactions
Oxidation-Reduction reactions
2Mg(s) + O2(g)  2MgO(s)
Reactions that involve electron transfer are called
oxidation-reduction reactions, or redox reactions.
Oxidation numbers (states)
A way to keep track of the electrons gained or lost
1) For atoms in its elemental form, oxidation number = 0
Na, Ag, Ar, O2, N2, P4
2) For monatomic ion, oxidation number = charge of the ion
Na+, Ca2+, Co2+, Co3+, Cl−, O2−
NaCl, Na2O, CaCl2, CaO, CoCl2, CoCl3, Co2O3, CoO
H2O
O
H
H
3) In covalent compounds
O: −2
H: +1
F: −1
In a neutral compound, the sum of the oxidation number = 0
CO, CO2, SF6, SF4, H2S, NH3, P2O5, N2O3
In a polyatomic ion, the sum of the oxidation number = ion charge
NO3−, SO42−, NH4+, Cr2O72−, MnO4−