1. Chapter 8: Chemical Reactions
A chemical reaction is the process whereby atoms from one or more substances are rearranged to form different substances.
A B → C
Reactants are always on the left of the arrow!
Products are always on the right of the arrow!
Indicators that a chemical reaction has taken place:
heat or light produces (flame)
gas bubbles are produced in a liquid
a change in color
a solid precipitates from a solution
a new odor is produced

2. Chemical reactions can be described in word equations, formula equations, and chemical equations.
Word equations use chemical names to describe the reactants and products of a reaction.
Formula equations use chemical formulas to describe the reactants and products of a reaction.
Chemical equations use “balanced” chemical formulas to describe the reactants and products of a reaction so that the law of conservation of mass is obeyed.

3. Word Equation:
Iron (s) Chlorine (g) → Iron(III)chloride (s)
This equation says that solid iron reacts with chlorine gas to produce solid iron(III)chloride.
Iron and Chlorine are the reactants. Iron(III)chloride is the product.
Formula Equation:
Fe (s) Cl2 (g) → FeCl3 (s)
Notice that the number of Cl atoms is not equal on both sides of the arrow.
Chemical Equation:
2 Fe (s) Cl2 (g) → FeCl3 (s)
Notice that the number of Fe and Cl atoms are equal on both sides of the arrow.

4. Writing a Chemical Equation requires that we know how to “balance” an equation.
Once the compounds in a reaction are known, we can not change the formulas of the compounds-we can only change how many of each compound we use. This means we can only change the numbers that appear before the formula of a compound (called the coefficients).
Fe (s) Cl2 (g) → FeCl3 (s)
At the start, there are one atom of Fe on both sides of the arrow, and there are two atoms of Cl on the left and three atoms of Cl on the right.
Balancing can usually be completed by inspection-trial and error using different coefficients and counting the number of each type of atom on both sides.
Trial and Error can be time consuming, so I will give some hints to help you start. If we start in a good place, balancing is usually easy.

5. Odd-Even Conflict: If one side must be even while the other is odd, start by multiplying the “odd” thing by 2.
Fe (s) Cl2 (g) → FeCl3 (s)
Since the Cl on the left must always be even and the Cl on the right is currently odd, start by multiplying the FeCl3 by 2.
Fe (s) Cl2 (g) → FeCl3 (s)
Notice that even though there were too many Cl on the right, we decided to make even more. Now there are an even number of Cl on the right (6 to be exact). There are now also 2 Fe on the right.
What can we multiply the Cl2 by to give us a total of 6 Cl on the left?
Fe (s) Cl2 (g) → FeCl3 (s)
There are now 6 Cl on both sides, but one Fe on the left and two on the right. What can we do to fix that?
Multiply the Fe by 2 to give us a total of 2 Fe on the left?
Balanced!
2 Fe (s) Cl2 (g) → FeCl3 (s)

6. Atoms appearing in three or more places: If element appears in an equation in three or more places, balance that element last.
C3H8 (g) O2 (g) → CO2 (g) H2O (g)
Even though oxygen, hydrogen and carbon are not balanced, since oxygen atoms appear in three places in the reaction, we will balance the other elements first and then take care of the oxygen.
There are three C on the left and only one on the right. What should we do?
C3H8 (g) O2 (g) → CO2 (g) H2O (g)
There are eight H on the left and only two on the right. What should we do?
C3H8 (g) O2 (g) → CO2 (g) H2O (g)
Now that C and H are balanced, we can check the O. What do we need to do?
C3H8 (g) O2 (g) → CO2 (g) H2O (g)
Can you see that each time we corrected an element, the O changed, so trying to take care of it first would make the problem more difficult.

7. Polyatomic ions that do not change: If the same polyatomic ion appears on both sides of an equation, balance the polyatomic ion as a group rather than taking it apart and balancing each atom in it.
AgNO3 (aq) Ba(OH)2 (aq) → Ba(NO3)2 (aq) AgOH (s)
Note that the nitrate ion (NO3-1) and the hydroxide ion (OH-1) appear on both sides of the equation. Therefore we will not focus on balancing N, O, and H, but instead will focus on balancing NO3-1 and OH-1.
There are two OH-1 on the left and only one on the right. What should we do?
AgNO3 (aq) Ba(OH)2 (aq) → Ba(NO3)2 (aq) AgOH (s)
There are two NO3-1 on the right and only one on the left. What should we do?
2 AgNO3 (aq) Ba(OH)2 (aq) → Ba(NO3)2 (aq) AgOH (s)
Notice that we never actually counted how many O atoms were present, but instead focused on the polyatomic ions in order to balance the equation.

8. Five Types of Chemical Reactions:
Synthesis:
Two or more reactants combine to make a more complicated product.
AB X → ABX
Combustion:
Any rapid reaction with oxygen (O2 gas) to produce heat and/or light.
AB O2 → AO BO
Decomposition:
One complicated reactant breaking down to make two or more products.
ABX → AB X
Double Displacement:
Positive and negative ions switch partners (A and B are positive ions while X and Y are negative ions)
AX BY → AY BX
Single Displacement:
One element becomes part of a compound while another that is part of a compound becomes a free element (Y is a free element and becomes part of a compound …….)
AX Y → AY X

9. Practice: balance the equations and identify which type of reaction each one is.
Click to see Solutions
CH4 (g) O2 (g) → CO2 (g) H2O (g)
Balanced: CH4 (g) O2 (g) → CO2 (g) H2O (g)
Combustion:
SnBr4 (s) → Sn (s) Br2 (g)
Balanced: SnBr4 (s) → Sn (s) Br2 (g)
Decomposition:
AgNO3 (aq) NaCl (aq) → AgCl (s) NaNO3 (aq)
Already balanced!
Double Displacement:
Li (s) BaCl2 (g) → LiCl (s) Ba (s)
Balanced: Li (s) BaCl2 (g) → LiCl (s) Ba (s)
Single Displacement:

10. Practice: balance the equations and identify which type of reaction each one is.
Click to see Solutions
SO2 (g) O2 (g) → SO3 (g)
Balanced: SO2 (g) O2 (g) → SO3 (g)
Combustion and Synthesis
LiBr (s) F2 (g) → LiF (s) Br2 (l)
Balanced: 2 LiBr (s) F2 (g) → LiF (s) Br2 (l)
Single Displacement:
CaO (s) H2O (l) → Ca(OH)2 (s)
Already balanced!
Synthesis:
NaOH (aq) HCl (aq) → NaCl (aq) H2O (l)
Already balanced!
Double Displacement:

11. Synthesis:
Two or more reactants combine to make a more complicated product.
AB X → ABX
CaO (s) H2O (l) → Ca(OH)2 (s)
The reactants may be compounds (ionic or covalent) or may be elements.
The product must always be a compound (ionic or covalent).
If more than one kind of product is made, the reaction is not generally called synthesis.

12. Decomposition:
One complicated reactant breaking down to make two or more products.
ABX → AB X
The reactant must always be compounds (ionic or covalent).
The products may be compounds (ionic or covalent) or elements.
Example where the products are compounds .
Fe2(CO3) 3 (s) → Fe2O3 (s) CO2 (g)
Example where the products are elements.
2 Fe Cl3 (s) → Fe (s) Cl2 (g)

13. Single Displacement:
A BX → AX B
An ionic compound is reacting with a metallic element.
The metallic element (charge of zero) will become a cation and replace the cation in the ionic compound.
The cation that was replaced will become an element (charge of zero)
SnBr2 (s) Fe (s) → FeBr3 (s) Sn (s)
The element iron (Fe) is replacing the Sn+2 cation. When it does this, iron becomes a cation (Fe+3) while the Sn+2 cation becomes an element (Sn).

14. Single Displacement:
AX Y → AY X
An ionic compound is reacting with a nonmetal element.
The nonmetal element (charge of zero) will become an anion and replace the anion in the ionic compound.
The anion that was replaced will become an element (charge of zero)
SnBr2 (s) F2 (g) → SnF2 (s) Br2 (g)
The element fluorine (F2) is replacing the Br-1 anion. When it does this, fluorine becomes a fluoride ion (F-1) while the Br-1 anion becomes an element (Br2).

15. Activity Series for Single Displacement Reactions
Metals
Most Reactive
Halides (halogens) Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H2 Sb Bi Cu Hg Ag Pt Au
Most Reactive F Cl Br I
Least Reactive
Only a more reactive element can replace another element in a compound!
The element produced must be less reactive than the starting element!
Least Reactive

16. Will the reaction below will work?
Click for Solution
2 LiBr (s) F2 (g) → LiF (s) Br2 (l)
Yes because F is more reactive than Br
Will the reaction below will work?
2 LiBr (s) I2 (g) → LiI (s) Br2 (l)
No because I is less reactive than Br
Will the reaction below will work?
AgBr (s) Li (s) → LiBr (s) Ag (s)
Yes because Li is more reactive than Ag
Will the reaction below will work?
CaBr2 (s) Zn (s) → ZnBr2 (s) Ca (s)
No because Zn is less reactive than Ca

17. Notice that Halogen atoms (group 17) always replace halogen ions!
2 LiBr (s) F2 (g) → LiF (s) Br2 (l)
Here the element fluorine (F2) is replacing the ion Br-1
Notice that Metal atoms almost always replace metal ions!
AgBr (s) Li (s) → LiBr (s) Ag (s)
Here the element lithium (Li) is replacing the ion Ag+1
Notice that some metal ions can replace hydrogen!
HBr (s) Li (s) → LiBr (s) H2 (g)
Here the element lithium (Li) is replacing the ion H+1
Notice that the ion that was replaced always becomes an element!

18. Double Displacement:
AX (aq) BY (aq) → AY BX
The reactants are almost always aqueous ionic compounds.
One of the products must be a solid, a liquid (usually water), or a gas while the other product will be a new aqueous ionic compound.
A reaction that makes a solid: (these are also called precipitation reactions)
Na2SO4 (aq) BaCl2 (aq) → NaCl (aq) BaSO4 (s)
BaSO4 (s) is called a precipitate
A reaction that makes water: (these are also called acid-base reactions)
H2SO4 (aq) Sr(OH)2 (aq) → H2O (l) SrSO4 (aq)
A H+1 ion (acid) will react with an OH-1 ion (base) to make HOH (or H2O).
A reaction that makes a gas:
Na2S (aq) HCl (aq) → H2S (g) NaCl (aq)

19. For double replacement reactions that produce precipitates, switch ions and write correct ionic formulas for the products-then balance the equation.
For Example Mix:
Al2(SO4)3 (aq) BaCl2 (aq)
Take the ionic compounds apart:
Al+3 ions
SO4-2 ions
Ba+2 ions
Cl-1 ions
Switch the ions around-remember positive always pairs up with negative:
Al+3 ions
Cl-1 ions
Ba+2 ions
SO4-2 ions
Write correct formulas for ionic compounds-remember the net charge is 0:
AlCl
BaSO4
These are the possible products
Write a skeleton equation:
Al2(SO4)3 (aq) BaCl2 (aq) → AlCl3 (aq) BaSO4 (s)
Balance the skeleton equation to get the chemical equation:
Al2(SO4)3 (aq) BaCl2 (aq) → AlCl3 (aq) BaSO4 (s)

20. Two pictures of the polar water molecule.
Dissolving a salt in water:
Two pictures of the polar water molecule.
Polar water molecules surround ions in ionic compounds to allow ionic compounds to dissolve in water.
The negative side of the water molecule is attracted to the cation while the positive side is attracted to the anion.

21. An aqueous solution of ions will conduct electricity.
a) Solution of a strong electrolyte
like salt
b) Solution of a weak electrolyte
like a weak acid
c) Solution of a non-electrolyte
like sugar

22. NaCl (a salt) breaks apart completely into ions when it dissolves in water. Therefore it is a strong electrolyte.
HCl (a strong acid) breaks apart completely into ions when it dissolves in water. Therefore it is a strong electrolyte.
This is also the definition of a strong acid.

23. All ionic compounds that will dissolve completely in water break apart to form ions in solution-they are all strong electrolytes.
An ionic compound that does not dissolve in water can be a precipitate for a double replacement reaction. Precipitates are non-electrolytes because they do not produce ions when they mix with water.
Dissolving NaOH in water:
H2O
NaOH (s) Na+1 (aq) OH-1 (aq)
Since NaOH (s) dissolves in water to make ions, NaOH is a strong electrolyte

24. Illustration of a precipitation reaction between Ba(NO3)2 (aq) and K2CrO4 (aq)
a) Ions present when the two solutions are first mixed.
b) Ions remaining after formation of precipitate (shown on the bottom of the beaker).
K+1 and NO3-1 are spectator ions
The reaction in chemical equation form:
Ba(NO3)2 (aq) K2CrO4 (aq) → BaCrO4 (s) KNO3 (aq)

25. Chemical Equation (balanced)
Ba(NO3)2 (aq) K2CrO4 (aq) → BaCrO4 (s) KNO3 (aq)
Complete Ionic Equation (must show all aqueous ions separately)
(reactant side)
Ba+2 (aq) NO3-1 (aq) + 2 K+1(aq) + CrO4-2 (aq) →
→ BaCrO4 (s) K+1 (aq) NO3-1 (aq)
(product side)
Net Ionic Equation (spectator ions are removed)
Ba+2 (aq) + CrO4-2 (aq) →
BaCrO4 (s)
Spectator ions:
K+1 (aq) and NO3-1 (aq)

26. KCl (aq) + AgNO3 (aq) → AgCl (s) + KNO3 (aq)
1st solution
2nd solution
mixing
Final solution
K+1 (aq) + Cl-1 (aq) + Ag+1 (aq) + NO3-1(aq) → AgCl (s) + K+1 (aq) + NO3-1 (aq)
Can you identify the spectator ions?

27. There are three types of double replacement reactions:
Those that form a solid ionic precipitate.
Those that form water.
Those that produce a gas.
These are also called precipitation reactions.
These are also called acid-base reactions.
These have no other specific name.
If switching the ions in a double replacement reaction does not do one or more of these three things, then no reaction is taking place.

28. Double replacement reactions that produce water (acid-base)
Hydrobromic acid plus sodium hydroxide
HBr (aq) NaOH (aq) → H2O (l) NaBr (aq)
H+1 (aq) + Br-1 (aq) + Na+1 (aq) + OH-1 (aq) → H2O (l) + Na+1 (aq) + Br-1 (aq)
H+1 (aq) + OH-1 (aq) → H2O (l)
The key to identifying these reactions is to recognize acids and bases.
Adding H+1 ions to most of the negative ions we have used so far will produce an acid.
The only bases we have at this point are OH-1 and CO3-2
Therefore, hydroxide compounds and carbonate compounds are bases.

29. Double replacement reactions that produce a gas.
2 HI (aq) Li2S (aq) → H2S (g) LiI (aq)
2 H+1 (aq) I-1 (aq) + 2 Li+1 (aq) + S-2 (aq) → H2S (g) + 2 Li+1 (aq) I-1 (aq)
2 H+1 (aq) + S-2 (aq) → H2S (g)
2 HCl (aq) Na2CO3 (aq) → H2CO3 (aq) NaCl (aq)
But carbonic acid is not stable and falls apart!
H2CO3 (aq) → H2O (l) CO2 (g)
Therefore:
2 HCl (aq) Na2CO3 (aq) → H2O (l) CO2 (g) NaCl (aq)
The are not very many of these reactions. Common ones produce H2S, HCN or CO2.

30. For each of the following: Identify the type of reaction, identify the spectator ions, and write the net ionic equation. You may want to write the complete ionic equation to help you identify the spectator ions.
Click to see Solution
AgNO3 (aq) CaBr2 (aq) → AgBr (s) Ca(NO3) 2 (aq)
Spectators: Ca+2 and NO3-1, double replacement: precipitation
Ag+1 (aq) Br-1 (aq) → AgBr (s)
HNO3 (aq) Ca(OH)2 (aq) → H2O (l) Ca(NO3) 2 (aq)
Spectators: Ca+2 and NO3-1, double replacement: acid-base
H+1 (aq) OH-1 (aq) → H2O (l)
HClO4 (aq) Na2S (aq) → H2S (g) NaClO4 (aq)
Spectators: Na+1 and ClO4-1, double replacement: produces a gas
2 H+1 (aq) S-2 (aq) → H2S (g)

31. Rules for predicting precipitates:
Slightly soluble means that they will be a precipitate.
If a compound contains ions other than those mentioned, assume they are insoluble (form a precipitate) unless told otherwise.

32. Practice:
Use the rules on the previous page to predict whether the following ions will be soluble in water. If they are not soluble, then they would be precipitates in double replacement reactions.
Click to see Solution
NaCl Na2CO K2CO AgCl
soluble
soluble
soluble
insoluble
AgNO CaBr AgBr BaSO4
soluble
soluble
insoluble
insoluble
Fe2(CO3) Li2S Al(OH) (NH4)3PO4
insoluble
soluble
insoluble
soluble
Mg3(PO4) PbBr AgI MgSO4
insoluble
insoluble
insoluble
soluble