ASP Chemistry - Ch 4 Notes.pdf

Chapter 4 – Notes 1 ASP Chemistry
Chapter 4
Chemical Quantities and Aqueous Reactions
 Sec 4.2 – Reaction Stoichiometry
 Define Stoichiometry
It is the relationship between chemical amounts in a balanced chemical equation.
Stoichiometry allows us to predict the amounts of products or reactants that will form in a chemical
reaction based on the amounts of reactants given. It also allows us to determine the amount of
reactants necessary to form a given amount of product.
 Role of coefficients in Stoichiometry
The coefficients in a chemical equation specify the relative amounts in moles of each of the
substances involved in the reaction.
 Example:
Consider the photosynthesis reaction
6 CO2 (g) + 6 H2O (l)
𝑆𝑢𝑛𝑙𝑖𝑔ℎ𝑡
→ 6 O2 (g) + C6H12O6 (aq)
6 moles 6 moles 6 moles 1 mole
6×44.01 g 6×18.016 g 6×32 g 1×180.2 g
264.04 g 108.096 g 192 g 180.2 g
372g 372 g
 Example:
In photosynthesis, plants convert carbon dioxide and water into glucose (C6H12O6) according to the
reaction: 6 CO2 (g) + 6 H2O (l)
𝑆𝑢𝑛𝑙𝑖𝑔ℎ𝑡
→ 6 O2 (g) + C6H12O6 (aq)
Suppose you determine that a particular plant consumes 44.0 g of CO2 in one week. Assuming that
there is more than enough water present to react with all of the CO2, what mass of glucose (in grams)
can the plant synthesize from the CO2?
6 CO2 C6H12O6
6 moles 1 mole
6×44. g 1×180. g
44.0 g ??? m C6H12O6 = 30.0 g

 Questions
- Solve Example 4.5
- Solve For Practice 4.4, 4.5
- Solve Conceptual Connection 4.3, 4.4, 4.5
- Solve Problems by Topic Questions 25 – 34
 Application
1. Manganese (IV) oxide reacts with aluminum to form elemental manganese and aluminum oxide:
3 MnO2 + 4 Al → 3 Mn + 2 Al2O3
What mass of Al is required to completely react with 25.0 g MnO2?
a) 7.76 g Al
b) 5.82 g Al
c) 33.3 g Al
d) 10.3 g Al
2. Consider the unbalanced equation for the combustion of hexane:
C6H14 (g) + O2 (g) → CO2 (g) + H2O (g)
Balance the equation and determine how many moles of O2 are required to react completely with 7.2
moles of C6H14.
3. How many moles of CO2 are produced when 3 moles of pentane react with excess oxygen
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)
a) 5 moles
b) 3/5 moles
c) 15 moles
d) 3 moles
4. The overall equation involved in photosynthesis is 6 CO2 + 6 H2O → C6H12O6 + 6 O2.
How many grams of glucose (C6H12O6, 180.1 g/mol) form when 4.40 g of CO2 react?
a) 18.0 g
b) 3.00 g
c) 108 g
d) 0.0167 g
5. The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate
airbags: 2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
How many grams of N2 are produced from 6.00 g of NaN3?
a) 3.88 g
b) 1.72 g
c) 0.138 g
d) 2.59 g

 Sec 4.3 – Limiting Reagent, Theoretical Yield and Percent Yield
- The limiting reagent is the reactant that is completely consumed in a chemical reaction and limits
the amount of product.
- The reactant in excess is any reactant that occurs in a quantity greater than is required to completely
react with the limiting reactant.
- The theoretical yield is the amount of product that can be made in a chemical reaction based on the
amount of limiting reactant.
- The actual yield is the amount of product actually produced by a chemical reaction.
- The percent yield is calculated as
𝐚𝐜𝐭𝐮𝐚𝐥 𝐲𝐢𝐞𝐥𝐝
𝐭𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐲𝐢𝐞𝐥𝐝
× 𝟏𝟎𝟎
IMPORTANT NOTE
Consider the following reaction:
Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g)
Initial amounts 1 mole 3 moles 0 moles 0 moles
Change -1 mole - 3 moles + 2 moles + 3 moles
Final 0 moles 0 moles 2 moles 3 moles
The stoichiometric coefficients do NOT tell us how much we have but they tell us the ratios in which
reactants are lost and products are formed.

 Example :
1. Ammonia, NH3, can be synthesized by the reaction:
2 NO (g) + 5 H2 (g) → 2 NH3 (g) + 2 H2O (g)
Starting with 90.0 g NO and 30.0 g H2, find the theoretical yield of ammonia in grams.
Steps for solving the question
2 NO 5 H2
90.0 g 30.0 g
calculate n n =
𝑚
𝑀
=
90.0
30
= 3.00 moles n =
𝑚
𝑀
=
30.0
2×1
= 15.0 moles
𝑛
𝑟𝑒𝑎𝑐𝑡𝑖𝑛𝑔 𝑟𝑎𝑡𝑖𝑜
3.00
2
= 1.50
15.0
5
= 3.00
Compare 1.50 < 3.00
Limiting reagent Excess reagent
(smaller value) (larger value)
After identifying the limiting reagent, reword the question as if you only have the limiting reagent
and solve the problem.
Given
m NO = 90.0 g
R.T.F
Theoretical yield? = ????
2 NO 2 NH3
2 moles 2 moles
2 × 30.0 g 2 × 17.0 g
90.0 g ??? Mass of NH3 = 51.0 g
Theoretical yield of ammonia is 51.0 g
2. Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the
equation:
3 NO2 (g) + H2O (l) → 2 HNO3 (l) + NO (g)
Suppose that 5 moles NO2 and 1 mole H2O combine and react completely. How many moles of the
reactant in excess are present after the reaction has completed?
The limiting reactant is the 1 mole H2O, which is completely consumed. The l mole of H2O requires
3 moles of NO2 to completely react; therefore, 2 moles NO2 remain after the reaction is complete.
 Questions
- Solve Example 4.7
- Solve Conceptual Connection 4.6, 4.7
- Solve Conceptual Problem 76, 77

 Application
1. Ammonia is produced using the Haber process: 3 H2 + N2 → 2 NH3
Calculate the mass of ammonia produced when 35.0 g of nitrogen react with 12.5 g of hydrogen.
a) 47.5 g
b) 42.6 g
c) 35.0 g
d) 63.8 g
e) 70.5 g
2. Ammonia is produced using the Haber process: 3 H2 + N2 → 2 NH3
What percent yield of ammonia is produced from 15.0 kg of each of H2 and N2, if 13.7 kg of product
are recovered? Assume the reaction goes to completion.
a) 7.53 × 10–2
%
b) 1.50 × 10–1
%
c) 75.3%
d) 15.0%
e) 16.2%
3. What mass of TiCl4 is needed to produce 25.0 g of Ti if the reaction proceeds with an 82% yield?
TiCl4 + 2Mg  Ti + 2 MgCl2
a) 30.5 g
b) 121 g
c) 99.1 g
d) 81.2 g
4. Zinc (II) sulfide reacts with oxygen according to the reaction: 2 ZnS(s) + 3O2(g)→2 ZnO(s) + 2
SO2(g)
A reaction mixture initially contains 4.2 moles ZnS and 6.8 moles O2. Once the reaction has occurred
as completely as possible, what amount (in moles) of the excess reactant is left?

More Applications
1. How many moles of KBrO3 are required to
prepare 0.0700 moles of Br2 according to the
reaction:
KBrO3+5KBr+6HNO3→6KNO3+3Br2+3H2O
a) 0.0732
b) 0.0704
c) 0.220
d) 0.0233
2. Which of the following statements is
FALSE for the chemical equation given
below in which nitrogen gas reacts with
hydrogen gas to form ammonia gas assuming
the reaction goes to completion?
N2 + 3H2→ 2NH3
a) One mole of N2 will produce two moles of
NH3.
b) One molecule of nitrogen requires three
molecules of hydrogen for complete
reaction.
c) The reaction of 14 g of nitrogen produces
17 g of ammonia.
d) The reaction of three moles of hydrogen
gas will produce 17 g of ammonia.
3. Calcium carbide, CaC2, is an important
preliminary chemical for industries
producing synthetic fabrics and plastics.
CaC2 may be produced by heating calcium
oxide with coke:
CaO + 3C → CaC2 + CO
What is the amount of CaC2 which can be
produced from the reaction of excess
calcium oxide and 10.2g
of carbon? (Assume 100% efficiency of
reaction for purposes of this problem.)
a) 18.1 g
b) 28.4 g
c) 20.8 g
d) 19.8 g
4. Calculate the mass of hydrogen formed when
25 grams of aluminum reacts with excess
hydrochloric acid.
2Al + 6HCl → Al2Cl6 + 3H2
a) 0.41 g
b) 1.2 g
c) 1.8 g
d) 2.8 g
5. When 12 g of methanol (CH3OH) was
treated with excess oxidizing agent (MnO4
−
),
14 g of formic acid (HCOOH) was obtained.
Using the following chemical equation,
calculate the percent yield. (The reaction is
much more complex than this; please ignore
the fact that the charges do not balance.)
3CH3OH+4MnO4
−
→3HCOOH+4MnO2
a) 100%
b) 92%
c) 82%
d) 70%
6. A commercially valuable paint and adhesive
material, dimethyl sulfoxide (DMSO),
(CH3)2SO, can be prepared by the reaction of
oxygen with dimethyl sulfide, (CH3)2S,
using a ratio of one mole oxygen to two
moles of the sulfide:
O2 + 2(CH3)2S → 2(CH3)2SO
If this process is 83% efficient, how many
grams of DMSO could be produced from 65
g of dimethyl sulfide and excess O2?
a) 68 g
b) 75 g
c) 83 g
d) 51 g

7. The formation of ethyl alcohol (C2H5OH) by
the fermentation of glucose (C6H12O6) may
be represented by:
C6H12O6 → 2C2H5OH + 2CO2
If a particular glucose fermentation process
is 87.0% efficient, how many grams of
glucose would be required for the production
of 51.0 g of ethyl alcohol (C2H5OH)?
a) 68.3 g
b) 75.1 g
c) 115 g
d) 229 g
8. The limiting reagent in a chemical reaction is
one that:
a) has the largest molar mass (formula
weight).
b) has the smallest molar mass (formula
weight).
c) has the smallest coefficient.
d) is consumed completely.
9. If 5.0 g of each reactant were used for the
following process, the limiting reactant
would be:
2KMnO4+5Hg2Cl2+16HCl
→10HgCl2+2MnCl2+2KCl+8H2O
a) KMnO4
b) HCl
c) H2O
d) Hg2Cl2
10. What mass of ZnCl2 can be prepared from
the reaction of 3.27 grams of zinc with 3.30
grams of HCl?
Zn +2HCl → ZnCl2 + H2
a) 6.89 g
b) 6.82 g
c) 6.46 g
d) 6.17 g
11. How many grams of NH3 can be prepared
from 77.3 grams of N2 and 14.2 grams of
H2? (Hint: Write and balance the equation
first.)
a) 93.9 g
b) 79.7 g
c) 47.0 g
d) 120.0 g
12. Silicon carbide, an abrasive, is made by the
reaction of silicon dioxide with graphite.
SiO2 +3C →SiC + 2CO
If 100 g of SiO2 and 100 g of C are reacted
as far as possible, which one of the following
statements will be correct?
a) 111 g of SiO2 will be left over.
b) 44 g of SiO2 will be left over.
c) 82 g of C will be left over.
d) 40 g of C will be left over.

 Sec 4.4 – Solution Concentration and Solution Stoichiometry
 Introduction
- Homogeneous mixtures are called solutions.
- The component of the solution that changes state is called the solute.
- Solute is the substance being dissolved.
- The component that keeps its state is called the solvent.
- Solvent is the substance doing the dissolving, and in water solutions, the solvent is the water.
- An aqueous solution is one in which water acts as the solvent.
 Solution Concentration
- A dilute solution is one containing only a few solute particles in a large amount of solvent.
- A concentrated solution has much more solute than a dilute solution.
- The solubility limit depends on several factors and is different for different substances; there is no
standard rule.
- Molarity (M) is used to indicate the concentration of a solution in moles solute per liter of solution.
(It is the amount of solute (in moles) divided by the volume of solution (in liters))
Molarity (M) =
Amount of solute (in mol)
Volume of solution (in L)
A concentrated solution contains a
relatively large amount of solute
relative to solvent.
A dilute solution contains a relatively
small amount of solute relative to
solvent.

 Preparing a Solution of Specified Concentration (IMPORTANT LAB SKILL)
- To make an aqueous solution of a specified molarity, we usually put the solute into a flask and then
add water to reach the desired volume of solution.
Example: Preparing a 1 Molar NaCl solution
Calculations:
Mass of NaCl required = n × M = C × V × M = 1 × 1 × (22.99 + 35.45) = 58.44 g
Steps:
1. Measure the solute in grams. Dissolve the solute in a small amount of water (solvent) in a beaker.
2. Pour the solute in the appropriate volumetric flask.
3. Add water (solvent) to the mark in the flask.
 Example:
1. If you dissolve 11.9 g KBr in enough water to make 2.00 L of solution, what is the molarity of the
solution?
𝐶 =
𝑛
𝑉
=
𝑚
𝑀 ×𝑉
=
11.9
119 × 2.00
= 0.0500 𝑀
Check
The units of the answer (M) are correct. The magnitude is reasonable since common solutions range
in concentration from 0 to about 18 M. Concentrations significantly above 18 M are suspect and
should be double-checked.

2. How many liters of a 0.125 M NaOH solution contain 0.250 mol of NaOH?
V =
n
C
=
0.250
0.125
= 2.00 𝐿
 Questions
- Solve For More Practice 5.1, 5.2
- Solve Conceptual Connection 5.1, 5.2
 Application :
What is the molarity of a solution containing 55.8 g of MgC12 dissolved in 1.00 L of solution?
a) 55.8 M
b) 1.71 M
c) 0.586 M
d) 0.558 M
 Solution Dilution
- The concentrated solution is called a stock solution.
- The equation for dilution is
Number of moles of concentrated solution = Number of moles of dilute solution
M1V1 = M2V2
 Solution dilution steps :
1. Measure out the specific volume of stock solution.
2. Pour into the appropriate volumetric flask which already contains ½ or 2/3 of the needed solvent
to be added.
3. Add water to the mark.
IMPORTANT NOTES
If you are diluting an acid solution, pour ½ or 2/3 of the water needed in the appropriate flask. Then
SLOWLY add the acid while swirling it. Finish by adding the last amount of water to the mark. This
is considered a safety issue.
When diluting acids, always add the concentrated acid to the water. Never add water to concentrated
acid solutions, as the heat generated may cause the concentrated acid to splatter and burn your skin.

Example: Preparing 3.00 L of 0.500 M CaCl2 from a 10.0 M Stock Solution
Calculations:
M1V1 = M2V2
10.0 × V1 = 0.500 × 3.00 L
V2 = 0.150 L
We make the solution by diluting 0.150 L of the stock solution to a total volume of 3.00 L (V2).
The resulting solution will be 0.500 M in CaC12.
 Example:
To what volume should you dilute 0.200 L of a 15.0 M NaOH solution to obtain a 3.00 M NaOH
solution? What is the volume of water needed?
M1V1 = M2V2
15.0 × 0.200 L = 3.00 × V2
V2 = 1.00 L
Volume of water = V2 – V1 = 1.00 – 0.200 = 0.80 L = 800 mL
Check
The final units (L) are correct. The magnitude of the answer is reasonable because the solution is diluted
from 15.0 M to 3.00 M, a factor of five. Therefore the volume should increase by a factor of five.
 Questions
- Solve For Practice 5.3
- Solve For More Practice 5.3
- Solve Conceptual Connection 5.3

 Application:
What volume of a 1.50 M HCl solution should you use to prepare 2.00 L of a 0.100 M HCl solution?
a) 0.300 L
b) 0.133 L
c) 30.0 L
d) 2.00 L
 Solution Stoichiometry
Because molarity relates the moles of solute to the liters of solution, it can be used to convert between
amount of reactants and/or products in a chemical reaction.
The general conceptual plan for these kinds of calculations begins with the volume of a reactant or
product.
 Example:
What volume (in L) of 0.150 M KCl solution will completely react with 0.150 L of a 0.150 M
Pb(NO3)2 solution according to the following balanced chemical equation?
2 KCl (aq) + Pb(NO3)2 (aq) → PbCl2 (s) + 2 KNO3 (aq)
Steps:
1. Find number of moles Pb(NO3)2 = C × V = 0.150 × 0.150 = 0.0225 moles
2. Compare number of moles of each
2 KCl Pb(NO3)2
2 moles 1 mole
???? 0.0225 moles
Number of moles of KCl = 0.0450 moles
3. Find V of KCl 𝑉 =
𝑛
𝐶
=
0.0450
0.150
= 0.300 𝐿
 Questions
 Application
Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction:
2 Kl(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + Pbl2(s)
What minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all
of the lead in 155.0 mL of a 0.112 M lead (ll) nitrate solution?
a) 348 mL
b) 86.8 mL
c) 174 mL
d) 43.4 mL

 Sec 4.5 – Types of Aqueous Solutions and Solubility
 Introduction
- Consider two familiar aqueous solutions: salt water and sugar water.
Salt water is a homogeneous mixture of NaCl and H2O.
Sugar water is a homogeneous mixture of C12H22O11 and H2O.
- As you stir either of these two substances into the water, it seems to disappear.
How do solids such as salt and sugar dissolve in water?
 What Happens When a Solute Dissolves?
- There are attractive forces between the solute particles holding them together.
- There are also attractive forces between the solvent molecules.
- When we mix the solute with the solvent, there are attractive forces between the solute particles and the
solvent molecules.
- If the attractions between solute and solvent are strong enough, the solute will dissolve.
 Charge Distribution in a Water Molecule
There is an uneven distribution of electrons within the water molecule.
This causes the oxygen side of the molecule to have a partial negative charge (𝛿–
) and the hydrogen side
to have a partial positive charge (𝛿+
).
When a solid is put into a solvent, the
interactions between solvent and solute
particles compete with the interactions
between the solute particles themselves.
An uneven distribution of electrons
within the water molecule causes the
oxygen side of the molecule to have a
partial negative charge and the
hydrogen side to have a partial positive
charge.

 Interactions in a Sodium Chloride Solution
- For example, when sodium chloride is put into water, there is a competition between the attraction of
Na+
cations and Cl -
anions to each other (due to their opposite charges) and the attraction of Na+
and Cl-
to water molecules.
- The attraction of Na+
and Cl -
to water is based on the polar nature of the water molecule.
- The oxygen atom in water is electron-rich, giving it a partial negative charge (𝛿-). The hydrogen atoms,
in contrast, are electron-poor, giving them a partial positive charge (𝛿+).
- As a result, the positively charged sodium ions are strongly attracted to the oxygen side of the water
molecule (which has a partial negative charge), and the negatively charged chloride ions are attracted to
the hydrogen side of the water molecule (which has a partial positive charge).
- In the case of NaCl, the attraction between the separated ions and the water molecules overcomes the
attraction of sodium and chloride ions to each other, and the sodium chloride dissolves in the water
The attraction between water molecules
and the ions of sodium chloride causes
NaCl to dissolve in the water.
When sodium chloride is put into
water, the attraction of Na+
and Cl-
ions
to water molecules competes with the
attraction between the oppositely
charged ions themselves.

 Electrolyte and Nonelectrolyte Solutions
- Electrolytes are materials that dissolve in water to form a solution that will conduct electricity
- Nonelectrolytes are materials that dissolve in water to form a solution that will NOT conduct
electricity.
- Ionic substances such as sodium chloride that completely dissociate into ions
when they dissolve in water are strong electrolytes.
- In contrast to sodium chloride, sugar is a molecular compound.
- Most molecular compounds (except for acids), dissolve in water as intact
molecules.
- Compounds such as sugar that do NOT dissociate into ions when dissolved in water are called
nonelectrolytes, and the resulting solutions-called nonelectrolyte solutions-do not conduct electricity.
 Interactions between Sugar and Water Molecules
- In contrast to sodium chloride, sugar is a molecular compound.
- Most molecular compounds-with the important exception of acids, which we discuss shortly-dissolve in
water as intact molecules.
- Sugar dissolves because the attraction between sugar molecules and water molecules overcomes the
attraction of sugar molecules to each other.
- So unlike a sodium chloride solution (which is composed of dissociated ions), a sugar solution is
composed of intact C12H22O11 molecules homogeneously mixed with the water molecules.
A solution of salt (an electrolyte)
conducts electrical current.
A solution of sugar (a nonelectrolyte)
does not

 Binary Acids
- Acids are molecular compounds that ionize when they dissolve in water.
The molecules are pulled apart by their attraction for the water.
When acids ionize, they form H+
cations and also anions.
- The percentage of molecules that ionize varies from one acid to another.
- Acids that ionize virtually 100% are called strong acids.
HCl(aq)  H+
(aq) + Cl−
(aq)
- Acids that only ionize a small percentage are called weak acids.
HF(aq) ⇌ H+
(aq) + F−
(aq)
Partial charges on sugar molecules and
water molecules result in attractions
between the sugar molecules and water
molecules.
Do not have enough ions, doesn’t
conduct electricity well

 Strong and Weak Electrolytes
- Strong electrolytes are materials that dissolve completely as ions.
Examples: Ionic compounds and strong acids where their solutions conduct electricity well
- Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions.
Examples: Weak acids where their solutions conduct electricity, but not well
- When compounds containing a polyatomic ion dissolve, the polyatomic ion stays together.
HC2H3O2(aq) ⇌ H+
(aq) + C2H3O2
−
(aq)
 Classes of Dissolved Materials
Unlike soluble ionic compounds, which
contain ions and therefore dissociate in
water, acids are molecular compounds
that ionize in water
Conduct electricity

 Dissociation and Ionization
- When ionic compounds dissolve in water, the anions and cations are separated from each other. This is
called dissociation.
Na2S (aq)  2 Na+
(aq) + S2–
(aq)
- When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion.
Na2SO4 (aq)  2 Na+
(aq) + SO4
2−
(aq)
- When strong acids dissolve in water, the molecule ionizes into H+
and anions.
H2SO4 (aq)  2 H+
(aq) + SO4
2−
(aq)
 Questions
- Solve Problems by Topic Questions 39 and 40
 The Solubility of Ionic Compounds
- When an ionic compound dissolves in water, the resulting solution contains not the intact ionic
compound itself, but its component ions dissolved in water.
- However, not all ionic compounds dissolve in water.
If we add AgCl to water, for example, it remains solid and appears as a white powder at the bottom of
the water.
- In general, a compound is termed soluble if it dissolves in water and insoluble if it does not.
 Solubility of Salts
- If we mix solid AgNO3 with water, it dissolves and forms a strong electrolyte solution.
- Silver chloride, on the other hand, is almost completely insoluble.
- If we mix solid AgCl with water, virtually all of it remains as a solid within the liquid water.
AgCl does not dissolve in water; it
remains as a white powder at the
bottom of the beaker
Big ion breaks into smaller ion

 When Will a Salt Dissolve?
Whether a particular compound is soluble or insoluble depends on several factors. We can follow a set
of empirical rules that chemists have inferred from observations on many ionic compounds. These
solubility rules are summarized in the table below.
 The solubility rules state that:
1. Compounds containing the sodium ion are soluble. That means that compounds such as NaBr, NaNO3,
Na2SO4, NaOH, and Na2CO3 all dissolve in water to form strong electrolyte solutions.
2. Compounds containing the N03
-
ion are soluble. That means that compounds such as AgNO3, Pb(NO3)2,
NaNO3, Ca(NO3)2, and Sr(NO3)2 all dissolve in water to form strong electrolyte solutions.
When compounds containing polyatomic ions such as NO3
-
dissolve, the polyatomic ions dissolve as
intact units.
3. Compounds containing the CO3
2-
ion are insoluble. Therefore, compounds such as CuCO3, CaCO3,
SrCO3, and FeCO3 do not dissolve in water.
Exceptions: compounds containing CO3
2-
are soluble when paired with Li +
, Na+
, K+
, or NH4
+
.
Thus Li2CO3, Na2CO3, K2CO3, and (NH4)2CO3 are all soluble.
 Example
Predict whether each compound is soluble or insoluble.
Compound Solubility Explanation
PbCl2 Insoluble Compounds containing Cl−
are normally soluble, but Pb2+
is an exception.
CuCl2 Soluble Compounds containing Cl−
are normally soluble and Cu2+
is not an exception.
Ca(NO3)2 Soluble Compounds containing NO3
−
are always soluble
BaSO4 Insoluble Compounds containing SO4
2−
are normally soluble, but Ba2+
is an exception.

 Questions
Solve Conceptual Connection 5.6
 Sec 4.6 - Precipitation Reactions
- Precipitation reactions are reactions in which a solid forms when we mix two solutions.
- Reactions between aqueous solutions of ionic compounds produce an ionic compound that is insoluble
in water.
The insoluble product is called a precipitate.
 Example: Precipitation of Lead(II) Iodide

 No Precipitation Means No Reaction
- Precipitation reactions do not always occur when two aqueous solutions are mixed.
- Combine solutions of KI and NaCl and nothing happens.
KI(aq) + NaCl(aq)  No Reaction

 Predicting Precipitation Reactions
1. Determine what ions each aqueous reactant has.
2. Determine formulas of possible products.
Exchange ions: (+) ion from one reactant with (–) ion from other
Balance charges of combined ions to get the formula of each product.
3. Determine solubility of each product in water.
Use the solubility rules.
If product is insoluble or slightly soluble, it will precipitate.
4. If neither product will precipitate, write no reaction after the arrow
5. If any of the possible products are insoluble, write their formulas as the products of the reaction using (s)
after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate
aqueous.
6. Balance the equation. Remember to only change coefficients, not subscripts.
 Questions
- Solve Example 5.6, 5.7
- Solve For More Practice

 Hard Water
- Have you ever taken a bath in hard water?
- Hard water contains dissolved ions such as Ca2+
and Mg2+
that diminish the effectiveness of soap. These
ions react with soap to form a gray curd that may appear as "bathtub ring" after you drain the tub.
- Hard water is particularly troublesome when washing clothes. Consider how your white shirt would look
covered with the gray curd from the bathtub and you can understand the problem.
- Consequently, most laundry detergents include substances designed to remove Ca2+
and Mg2+
from the
laundry mixture.
- The most common substance used for this purpose is sodium carbonate, which dissolves in water to
form sodium cations (Na+
) and carbonate (C03
2-
) anions:
Na2CO3(aq)→ 2 Na+
(aq) + CO3
2-
(aq)
- Sodium carbonate is soluble, but calcium carbonate and magnesium carbonate are not.
- Consequently, the carbonate anions react with dissolved Mg2+
and Ca2+
ions in hard water to form solids
that precipitate from (or come out of) solution:
Mg2+
(aq) + CO3
2-
(aq) → MgCO3(s)
Ca2+
(aq) + CO3
2-
(aq) → CaCO3(s)
- The precipitation of these ions prevents their reaction with the soap, eliminating curd and preventing
white shirts from turning gray.

 Sec 4.7 - Representing Aqueous Reactions: Molecular, Ionic, and Complete Ionic
Equations
1. Molecular Equation
An equation showing the complete neutral formulas for each compound in the aqueous reaction as if
they existed as molecules is called a molecular equation.
2 KOH(aq) + Mg(NO3)2(aq) → 2 KNO3(aq) + Mg(OH)2(s)
2. Complete Ionic Equation
In actual solutions of soluble ionic compounds, dissolved substances are present as ions. Equations that
describe the material’s structure when dissolved are called complete ionic equations.
 Rules of writing the complete ionic equation:
- Aqueous strong electrolytes are written as ions (Soluble salts, strong acids, strong bases)
- Insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form. (Solids,
liquids, and gases are not dissolved, hence molecule form)
2 K+
(aq) + 2 OH−
(aq) + Mg2+
(aq) + 2 NO3
−
(aq) → 2 K+
(aq) + 2 NO3
−
(aq) + Mg(OH)2(s)
Spectator Ions
- Notice that in the complete ionic equation, some of the ions in solution appear unchanged on both sides
of the equation.
- These ions are called spectator ions because they do not participate in the reaction.
3. Net Ionic Equation
2 K+
(aq) + 2 OH−
(aq) + Mg2+
(aq) + 2 NO3
−
(aq) → 2 K+
(aq) + 2 NO3
−
(aq) + Mg(OH)2(s)
An ionic equation in which the spectator ions are removed is called a net ionic equation.
Mg2+
(aq) + 2 OH−
(aq) → Mg(OH)2(s)
 Summarizing Aqueous Equations
- A molecular equation is a chemical equation showing the complete, neutral formulas for every
compound in a reaction.
- A complete ionic equation is a chemical equation showing all of the species as they are actually present
in solution.
- A net ionic equation is an equation showing only the species that actually change during the reaction.

 Questions
- Solve Example 5.8
 Sec 4.8 – Acid-Base and Gas Evolution Reactions
• Two other important classes of reactions that occur in aqueous solution are
1. Gas-evolution reactions.
In a gas-evolution reaction, a gas forms, resulting in bubbling
2. Acid–base Reaction.
It is also called a neutralization reaction.
An acid reacts with a base and the two neutralize each other, producing water (or in some cases a
weak electrolyte).
• In both acid–base and gas-evolution reactions, as in precipitation reactions, the reactions occur when the
anion from one reactant combines with the cation of the other.
• Many gas-evolution reactions are also acid–base reactions.
 Acid-Base Reactions
 Arrhenius Definitions:
1. Acid: Substance that produces H+
Example: HCl(aq) → H+
(aq) + Cl–
(aq)
- Some acids—called polyprotic acids
• These acids contain more than one ionizable proton and release them sequentially.
• For example, sulfuric acid, H2SO4 is a diprotic acid.
• It is strong in its first ionizable proton, but weak in its second.
H2SO4 (aq) → H+
(aq) + HSO4
-
(aq)
HSO4
-
(aq) ⇌ H+
(aq) + SO4
2-
(aq)
2. Base: Substance that produces OH ions in aqueous solution
Examples:
NaOH(aq) → Na+
(aq) + OH–
(aq)
Sr(OH)2 (aq) → Sr2+
(aq) + 2 OH-
(aq)

 Relation between Hydrogen ion and Hydronium ion
- An H+
ion is a bare proton.
- In solution, bare protons normally associate with water molecules to form hydronium ions
H+
(aq) + H2O(l) → H3O +
(aq)
- Chemists use H+
(aq) and H3O +
(aq) interchangeably to mean the same thing-a hydronium ion.
- The chemical equation for the ionization of HCl and other acids is often written to show the association
of the proton with a water molecule to form the hydronium ion:
HCl(aq) + H2O (l) → H3O +
(aq) + Cl –
(aq)
 Examples of Acids and Bases
The Hydronium ion
Protons normally associate with water
molecules in solution to form H3O+
ions, which in turn interact with other
water molecules.

 Acids and Bases in Solution
• Acids ionize in water to form H+
ions.
More precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion,
H3O+.
Most chemists use H+
and H3O+
interchangeably.
• Bases dissociate in water to form OH ions.
Bases, such as NH3, that do not contain OH
ions, produce OH
by pulling H off water molecules.
NH3 (aq) + H2O (l) → NH4
+
(aq) + OH-
(aq)
• In the reaction of an acid with a base, the H+
from the acid combines with the OH from the base to
make water.
The cation from the base combines with the anion from the acid to make the salt.

 Acid–Base Reactions
- Also called neutralization reactions because the acid and base neutralize each other’s properties
- Acid- base reactions generally form water and an ionic compound-called a salt that usually remains
dissolved in the solution. The net ionic equation for many acid-base is H+
(aq) + OH
(aq) → H2O(l)
(as long as the salt that forms is soluble in water).
- Examples:
2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)
H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)
 Example
Write a molecular and net ionic equation for the reaction between aqueous HI and aqueous Ba(OH)2
2 HI(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaI2 (aq)
2 H+
(aq) + 2 OH-
→ 2 H2O (l) or simply H+
(aq) + OH-
→ H2O (l)
 Questions
- Solve Example 5.10
- Solve For More Practice
 Acid–Base Titrations
– In a titration, a substance in a solution of known concentration (titrant) is reacted with another
substance in a solution of unknown concentration (analyte).
At this point, called the endpoint, the reactants are in their stoichiometric ratio.
The unknown solution (titrant) is added slowly from an instrument called a burette.
– In acid–base titrations, because both the reactant and product solutions are colorless, a chemical is
added that changes color when the solution undergoes large changes in acidity/alkalinity.
This chemical is called an indicator.
– At the endpoint of an acid–base titration, the number of moles of H+
equals the number of moles of
OH. This is also known as the equivalence point.
– In most laboratory titrations, the concentration of one of the reactant solutions is unknown, and the
concentration of the other is precisely known. By carefully measuring the volume of each solution
required to reach the equivalence point, we can determine the concentration of the unknown solution.

 Acid-Base Indicators and their Colors
Indicator Color in Acidic medium
Color in Neutral
Medium
Color in Basic
Medium
Phenolphthalein Colorless Colorless Pink (fuschia)
Methyl orange Red orange Yellow
Red litmus paper stays the same stays the same turns blue
Blue litmus paper turns red stays blue stays blue
pH paper red green blue or purple
In this titration, NaOH is added to a dilute
HCl solution.
When the NaOH and HCl reach
stoichiometric proportions (the equivalence
point), the phenolphthalein indicator changes
color to pink.

IMPORTANT NOTE
Different Acid-Base Reaction
 Case 1
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
1 : 1
na : nb
na = n b
CaVa = Cb Vb
 Case 4
2 HCl(aq) + Ca(OH)2(aq) → CaCl2(aq)+2H2O(l)
2 : 1
na : nb
na = 2 n b
CaVa = 2 Cb Vb
 Case 2
H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O(l)
1 : 2
na : nb
2 na = n b
2CaVa = CbVb
 Case 5
H2SO4(aq) +Ca(OH)2(aq) → CaSO4(s) + 2H2O (l)
1 : 1
na : nb
na = n b
CaVa = Cb Vb
 Case 3
H3PO4 (aq) + 3 NaOH (aq) → Na3PO4 (aq) + 3 H2O(l)
1 : 3
na : nb
3 na = n b
3 CaVa = CbVb
 Steps to solve these type of questions
1. Write balanced equation.
2. Write relation between acid and base in
terms of moles.
3. Write given and required to find.
4. Find the unknown.
 Example

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?
Given: 12.54 mL of NaOH solution, 0.100 M NaOH solution, 10.00 mL of HCl solution
RTF: concentration of HCl solution
HCl(aq) + NaOH (aq) → NaCl (aq) + H2O (l)
Number of moles of acid = number of moles of base
C1V1 = C2 V2
C1 ×
10.00
1000
= 0.100 ×
12.54
1000
C1 = 0.125 M
Concentration of HCl = 0.125 M
 Questions
- Solve Problems by Topic Questions 57, 58
 Gas-Evolution Reactions
- In a gas-evolution reaction, two aqueous solutions mix to form a gaseous product that bubbles out of
solution.
- Some gas-evolution reactions form a gaseous product directly when the cation of one reactant combines
with the anion of the other.
- Examples:
1. When sulfuric acid reacts with lithium sulfide, dihydrogen sulfide gas forms:
H2SO4(aq) + Li2S(aq) → H2S(g) + Li2SO4(aq)
2. When sulfuric acid reacts with potassium sulfide, dihydrogen sulfide gas forms:
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
- Other gas-evolution reactions often form an intermediate product that then decomposes (breaks down
into simpler substances) to form a gas.
For example, when aqueous hydrochloric acid is mixed with aqueous sodium bicarbonate, gaseous CO2
bubbles out of the reaction mixture.
NaHCO3(aq) + HCl(aq)  NaCl(aq) + H2CO3(aq)
The intermediate product, H2CO3, is not stable and decomposes into H2O and gaseous CO2.
H2CO3(aq)  H2O(l) + CO2(g)
The equation becomes: NaHCO3(aq) + HCl(aq)  NaCl (aq) + H2O (l) + CO2 (g)

Other important gas-evolution reactions form either H2SO3 or NH4OH as intermediate products:
HCl(aq) + NaHSO3(aq) → H2SO3(aq) + NaCl(aq)→H2O (l) + SO2(g) + NaCl(aq)
Intermediate product gas
NH4Cl(aq) + NaOH(aq) → NH4OH(aq) + NaCl(aq) → H2O (l) + NH3(g) + NaCl(aq)
Intermediate product gas
 Types of Compounds that Undergo Gas-Evolution Reactions
 Example
Write a molecular equation for the gas-evolution reaction that occurs when you mix aqueous nitric acid
and aqueous sodium carbonate
2 HNO3 (aq) + Na2CO3 (aq) → 2 NaNO3 (aq) + CO2 (g) + H2O (l)

 Questions
- Solve Example 5.12
- Solve Problems by Topic Questions 59, 60
 Sec 4.9 - Oxidation-Reduction Reactions
 Introduction
- To convert a free element into an ion, the atoms must gain or lose electrons. Of course, if one atom loses
electrons, another must accept them.
- Reactions where electrons are transferred from one atom to another are redox reactions.
- Atoms that lose electrons are being oxidized, while atoms that gain electrons are being reduced.
2 Na(s) + Cl2(g) → 2 NaCl(s) (Redox reaction)
0 0 +1 -1
Na → Na+
+ 1 e–
(Oxidation)
Cl2 + 2 e–
→ 2 Cl–
(Reduction)
- Oxidation is loss of electrons, while reduction is gain of electrons (OIL RIG)
- Oxidation-reduction reactions or redox reactions are reactions in which electrons transfer from one
reactant to the other
- Many redox reactions involve the reaction of a substance with oxygen.
a) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) (Rusting)
b) 2 C8H18(l) + 25 O2(g) →16 CO2(g) + 18 H2O(g) (Combustion of octane)
c) 2 H2(g) + O2(g) → 2 H2O(g) (Combustion of hydrogen)
 Combustion as Redox
The hydrogen in the balloon reacts
with oxygen upon ignition to form
gaseous water (which is dispersed
in the flame).

 Redox without Combustion
- Consider the following reactions:
2 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
- The reactions involve a metal reacting with a nonmetal.
The metal (which has a tendency to lose electrons) reacts with a nonmetal (which has a tendency to gain
electrons) where metal atoms lose electrons to non-metal atoms
 Redox Reaction
- The transfer of electrons does not need to be a complete transfer (as occurs in the formation of an ionic
compound) for the reaction to qualify as oxidation–reduction.
- For example, consider the reaction between hydrogen gas and chlorine gas:
H2(g) + Cl2(g) → 2 HCl(g)
- When hydrogen bonds to chlorine, the electrons are unevenly shared, resulting in
a) an increase of electron density (reduction) for chlorine
b) a decrease in electron density (oxidation) for hydrogen.

 Oxidation States
- For reactions that are not metal and nonmetal, or do not involve O2, we need a method for determining
how the electrons are transferred.
- Chemists assign a number to each element in a reaction called an oxidation state that allows them to
determine the electron flow in the reaction.
- Even though they look like them, oxidation states are not ion charges!
• Oxidation states are imaginary charges assigned based on a set of rules.
• Ion charges are real, measurable charges.
 Rules for Assigning Oxidation States
These rules are hierarchical. If any two rules conflict, follow the rule that is higher on the list.
– The following rules are in order of priority:
Rule Example
1. The oxidation state of an atom in a free
element is 0
Cu 0
Na 0
Cl2 0
2. The oxidation state of a monoatomic ion is
equal to its charge
Ca2+
+ 2
Cl-1
- 1
Al3+
+ 3
3.
a) The sum of the oxidation states of all atoms in
a neutral molecule is 0.
b) The sum of the oxidation states of all the
atoms in a polyatomic ion equals the charge
on the ion.
H2O
2(+1) - 2 2(+1) + (-2) = 0
NO3
-
1 (x) + 3 (-2) = -1 x = + 5
4. In their compounds, metals have positive
oxidation states.
a) Group 1 A metals always have an
oxidation state of +1.
b) Group 1 A metals always have an
oxidation state of +2.
NaCl Na +1
CaF2 Ca +2

5. In their compounds, nonmetals are assigned
oxidation states according to the table shown.
Entries at the top of the table take precedence
over entries at the bottom of the table.
 Exceptions:
1) Oxidation state of O in peroxides is – 1
2) Oxidation state of H in metal hydrides is - 1
Na2O2
2 (+1) + 2x = 0 x = - 1
NaH
+ 1 + x = 0 x = -1
 Example
Assign an oxidation state to each atom in each element, ion, or compound.
a) Cl2 0
b) Na+1
+1
c) KF K +1 F -1
d) CO2

e) SO4
2-
f) K2O2
 Questions
- Solve Conceptual connection 5.10
 Identifying Redox Reactions
– Oxidation: An increase in oxidation state
– Reduction: A decrease in oxidation state
 Carbon changes from an oxidation state of 0 to an oxidation state of +4.
Carbon loses electrons and is oxidized.
 Sulfur changes from an oxidation state of 0 to an oxidation state of –2.
Sulfur gains electrons and is reduced.

 Redox Reactions
 Oxidation and reduction must occur simultaneously. If an atom loses electrons another atom must take
them.
 The reactant that reduces an element in another reactant is called the reducing agent. The reducing
agent contains the element that is oxidized.
 The reactant that oxidizes an element in another reactant is called the oxidizing agent. The oxidizing
agent contains the element that is reduced.
 Example
Use oxidation states to identify the element that is oxidized and the element that is reduced in the
following redox reaction.
Since Mg increased in oxidation state, it was oxidized. Since H decreased in oxidation state, it was
reduced.
 Questions
 Example
Determine whether each reaction is an oxidation–reduction reaction. For each oxidation–reduction
reaction, identify the oxidizing agent and the reducing agent
a)
This is a redox reaction because magnesium increases in oxidation number (oxidation) and oxygen
decreases in oxidation number (reduction).
2 Na(s) + Cl2(g) → 2 Na+
Cl–
(s)
Na is oxidized, while Cl is reduced.
Na is the reducing agent, and Cl2 is the oxidizing agent.

b)
This is not a redox reaction because none of the atoms undergo a change in oxidation number
c)
This is a redox reaction because zinc increases in oxidation number (oxidation) and iron (II) ion
decreases in oxidation number (reduction).
IMPORTANT NOTE
Acid- base Reactions are NOT REDOX REACTIONS
 Questions
-
 Combustion Reactions
 Combustion reactions are characterized by the reaction of a substance with oxygen to form one or more
oxygen-containing compounds, often including water.
Combustion reactions are Exothermic (they emit heat).
 Examples:
1. Write the balanced equation for the combustion of natural gas (CH4). It reacts with oxygen to form
carbon dioxide and water. Assign the oxidation state to each element.

2. Write the balanced equation for the combustion of ethanol, the alcohol in alcoholic beverages. It reacts
with oxygen to form carbon dioxide and water.
Oxidation state: + 4 + 1 - 2 + 1 0 + 4 - 2 + 1 - 2
 Example
Assign an oxidation state to each atom in each element, ion, or compound.
Steps:
1. Balance the equation
2. Assign the oxidation number to each element
Oxidation state + 4 + 1 - 2 + 1 0 + 4 - 2 + 1 - 2
 Application
1. Write a balanced equation for the complete combustion of liquid C2H5SH
2. Complete and balance each combustion reaction equation.
a) S(s) + O2(g)→
b) C3H6(g) + O2(g)→
c) Ca(s) + O2(g)→
d) C5H12S(l) + O2(g)→

Types of Salts
Neutral Salt Acidic Salt Basic Salt
Made up of strong acid and strong
base.
Made up of strong acid and
weak base.
Made up of strong base and
weak acid.
Example
NaCl
NaOH HCl
strong base strong acid
Example
CuSO4
Cu(OH)2 H2SO4
weak base strong acid
Example
CH3COONa
NaOH CH3COOH
strong base weak acid
The resulting solution is neutral. The resulting solution is acidic. The resulting solution is basic.
Does not change the color of red
or blue litmus paper
Turns blue litmus paper into
red.
Turns red litmus paper into
blue.
The resulting solution contains
equal moles of H+
(aq) than OH-
(aq)
more H+
(aq) than OH-
(aq)
less H+
(aq) than OH-
(aq)

 Example:
1. Classify the salts below as acidic, basic or neutral:
NaCl Neutral salt
NaCl
NaOH HCl
KNO3 Neutral salt
KNO3
KOH HNO3
NH4Cl Acidic salt
NH4Cl
NH3 HCl
CH3COONa Basic salt
CH3COONa
NaOH CH3COOH
AlCl3 Acidic salt
AlCl3
Al(OH)3 HCl
2. Classify the salts below as acidic, basic or neutral:
KCl Neutral salt
KCl
KOH HCl
CH3COOK Basic salt
CH3COOK
KOH CH3COOH
Mg(NO3)2 Acidic salt
Mg(NO3)2
Mg(OH)2 HNO3
Na2CO3 Basic salt
Na2CO3
NaOH H2CO3
CuCl2 Acidic salt
CuCl2
Cu(OH)2 HCl
KNO3 Neutral salt
KNO3
KOH HNO3

Types of Chemical Reactions
1. Synthesis Reaction (Combination reaction)
Generally: A + B → AB
Examples:
2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s)
2 Na (s) + Cl2(g) → 2 NaCl (s)
2 Zn (s) + O2 (g) → 2 ZnO (s)
2 K (s) + 2 H2O (l) → 2 KOH (aq) + H2 (g)
2. Single Replacement Reaction (Displacement Reaction)
Generally: AB + C → AC + B
Examples:
2 Al (s) + Fe2O3 (s) → Al2O3(s) +2 Fe (l)
CuO (s) + C (s) → Cu (s) +CO2 (g)
Cu (s) + 2 AgNO3 (aq) → Cu(NO3)2 (aq) + 2 Ag (s)
Zn (s) + SnCl2 (aq) → ZnCl2(aq) + Sn (s)
3. Double Replacement Reaction
Generally: AB + CD → AC + BD
Examples:
CaCO3 (s) + H2SO3 (aq) → CaSO3 (s) + CO2 (g) + H2O (l)
2 KI (aq) + Pb(NO3)2 (aq) → PbI2(s) + 2 KNO3(aq)
NaCl (aq) + AgNO3(aq) → NaNO3 (aq) + AgCl (s)
4. Decomposition Reaction
Generally: AB → A + B
Examples:
2 H2O (l) → 2 H2(g) + O2 (g)
2 NI3 (s) → N2(g) + 3 I2 (g)
5. Combustion Reaction
In a combustion reaction, a substance combines with oxygen, releasing a large amount of energy in the
form of light and heat.
For organic compounds, such as hydrocarbons, the products of the combustion reaction are carbon
dioxide and water.

Examples:
CH4 (g) + O2 (g) → CO2 (g) + H2O (l)
2 Mg (s) + O2 (g) → 2 MgO (s)
S (s) + O2 (g) → SO2 (g)
P4 (s) + 5 O2 (g) → P4O10 (s)
 Applications:
Classify each of the following reactions as:
(a) Synthesis reaction
(b) Decomposition reaction
(c) Single replacement reaction
(d) Double replacement reaction
(e) Combustion reaction
1. 2 Na + 2 H2O → 2 NaOH + H2
2. C3H8 + 5 O2 → 3 CO2 + 4 H2O
3. 3 AgNO3 + K3PO4 → Ag3PO4 + 3 KNO3
4. P4 + 5 O2 → 2 P2O5
5. 2 C6H14 + 19 O2 → 12 CO2 + 14 H2O
6. 4 Cr + 3 O2 → 2 Cr2O3
7. Ca + 2 HCl → CaCl2 + H2
8. Ca(C2H3O2)2 + Na2CO3 → CaCO3 + 2 Na2C2H3O2
9. Cu(OH)2 + 2HC2H3O2 → Cu(C2H3O2)2 + 2H2O
10. BaCO3 →BaO + CO2
11. 2 NaClO3 → 2 NaCl + 3O2
12. 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
13. 2 AgNO3 + Ni → Ni(NO3)2 + 2 Ag
14. H2CO3 → H2O + CO2
15. 8 Cu + S8 → 8 CuS

Different Types of Reactions that will Produce Gas
Reaction Example
Acid + Metal → Salt + hydrogen
 Metal is not an alkali metal as Na, K or Li or Ca because the
reaction is highly explosive
 Metal is not Cu or Ag because they are highly Unreactive
2 HCl (aq) + Mg (s) → MgCl2 (aq) + H2(g)
Mg (s)+2H+
(aq) →Mg2+
(aq)+H2(g)
Acid+ Metal carbonate→ salt+ water+ carbon dioxide
(or bicarbonate)
2HCl(aq)+MgCO3(aq)→MgCl2(aq)+H2O(l)+CO2(g)
2 H+
(aq) + CO3
2-
(aq) → H2O(l)+CO2(g)
Acid+ Metal bicarbonate →salt+ water+ carbon dioxide
HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g)
H+
(aq) + HCO3
-
(aq) → H2O(l)+CO2(g)
Acid + Metal sulfite → Salt + Water + Sulfur dioxide
2HCl(aq)+Na2SO3(aq)→2NaCl(aq)+H2O(l)+SO2(g)
2 H+
(aq) + SO3
2-
(aq) → H2O(l)+SO2(g)
Acid + Metal sulfide → Salt + hydrogen sulfide
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
S2-
(aq) + 2 H+
(aq) →H2S (g)
Base + ammonium salt → Salt + water + ammonia
NH4Cl(aq)+NaOH(aq)→ NaCl(aq)+H2O(l)+NH3(g)
NH4
+
(aq)+OH-
(aq)→ H2O(l)+NH3(g)

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