UNIT 6

 Reactions


Write formula/word equations
Balance Equations
 Identify Types


Predict Products
Write Net Ionic Equations



Stoichiometry
 Conversions
Limiting & Excess Reagent
Percent Yield

 Process in which one or more pure substances are converted into one or more different pure substances
Reactants: Zn + I
2
Product: Zn I
2

 Temperature Change
 Color Change
 Production of gas
 Formation of a precipitate
 Production of light

4 Al(s) + 3 O
2
(g) ---> 2 Al
2
O
3
(s)
(Reactants) (Products)

Reactants react to produce products
 The letters (s), (g), (l), and (aq) are the physical states of compounds.
 “aq” represents aqueous meaning dissolved in water (solution)
 The numbers in the front are called coefficients
.

Subscripts represent the number of each atom in a compound

Symbol Meaning
+ used to separate one reactant or product from another
→ used to separate the reactants from the products - it is pronounced "yields" or "produces" when the equation is read
↔ used when the reaction can proceed in both directions - this is called an equilibrium arrow and will be used later in the course
↑ an alternative way of representing a substance in a gaseous state
↓ an alternative way of representing a substance in a solid state
∆ indicates that heat is applied to make the reaction proceed

 Elements that cannot exist by themselves (always occur in pairs)
 Bromine (Br
2
)
 Iodine (I
2
)
 Nitrogen (N
2
)
 Chlorine (Cl
2
)
 Hydrogen (H
2
)
 Oxygen (O
2
)
 Fluorine (F
2
)

1. When lithium hydroxide pellets are added to a solution of sulfuric acid, lithium sulfate and water are formed.
2 LiOH
(s)
+ H
2
SO
4(aq)
 Li
2
SO
4(aq)
+ 2 H
2
O
(l)
2. When crystalline C
6
H
12
O
6 is burned in oxygen, carbon dioxide and water vapor are formed.
C
6
H
12
O
6(s)
+ 6 O
2(g)
 6 CO
2(g)
+ 6 H
2
O
(g)


Law of Conservation of Mass
 Matter cannot be destroyed (atoms of reactants must equal products)
 Balance equations to get same number of each atom on the left and right in an equation
2HgO(s) ---> 2 Hg(l) + O
2
(g)
2 Hg atoms, 2 O atoms 2 Hg atoms, 2 O atoms

___ Al(s) + ___ Br
2
(l) ---> ___ Al
2
Br
6
(s)
__C
3
H
8
(g) + __ O
2
(g)  __ CO
2
(g) + __ H
2
O(g)
__ B
4
H
10
(g) + __ O
2
(g)  __ B
2
O
3
(g) + __ H
2
O(g)

 Synthesis (combination)
 Decomposition
 Single Replacement (displacement)
 Double Replacement (precipitation)
 Combustion
 Acid-Base Neutralization

Synthesis (Combination) Reactions
 Two or more substances combine to form a new compound.
A + X  AX
Synthesis of:
Binary compounds
Metal carbonates
Metal hydroxides
H
2
+ O
2
 H
2
CaO + CO
CaO + H
2
2
O
 CaCO
3
O  Ca(OH)
2
Metal chlorates
Oxyacids
KCl + O
CO
2
2
+ H
2
 KClO
3
O  H
2
CO
3

 A single compound breaks down into two or more simpler substances
AX  A + X
Decomposition of:
Binary compounds
Metal carbonates
Metal hydroxides
Metal chlorates
Oxyacids
H
2
O 
CaCO
3
KClO
H
2
CO
3
3
H

Ca(OH)
2
2
+ O
2
CaO + CO
 CaO + H
2
 KCl + O
2
 CO
2
+ H
2
O
2
O

Single Replacement
(displacement) Reactions
 One element replaces another in a reaction
 Metals replace metals
 Nonmetals replace nonmetals
A + BX  AX + B
BX + Y  BY + X

 Decide whether or not one element will replace another
 Metals can replace other metals provided that they are above the metal that they are trying to replace
 If the metal is not above what it is trying to replace, the result is “no reaction”

Double Replacement
(Precipitation) Reactions

Two elements or ions “switch partners”
AX + BY  AY + BX
 One of the compounds formed is usually a precipitate
, an insoluble gas that bubbles out of solution, or a molecular compound
, usually water.


Solubility – ability to dissolve
 In a double replacement (precipitate) reaction, one of the products must be insoluble in water and form a precipitate

Precipitate solution
– insoluble solid formed by a reaction in
 If both products are soluble the result is “no reaction”
 Solubility rules help you determine whether or not a compound will form a precipitate or remain an aqueous solution

Soluble Ionic Compounds
Alkali metals, NH
4
+
NO
3
, C
2
H
3
O
2
, ClO
3
, ClO
4
-
Cl , Br , I -
SO
4
-2
Except with:
(no exceptions)
Ag + , Hg
2
+2 , Pb +2
Sr +2 , Ba +2 , Ca +2 , Ag + , Pb +2 , Hg
2
+2
Insoluble Ionic Compounds
CO
3
-2 , PO
4
-3 , SiO
3
-2 , O -2 , SO
3
-2 , CrO
4
-2
S -2
OH -
Except with:
NH
4
+ , alkali metals
NH
4
+ , alkali metals
Ca +2 , Sr +2 , Ba +2 , Mg +2 (group 2)
NH
4
+ , alkali metals,
(Ca +2 , Ba +2 , Sr +2 are slightly soluble)

 A substance combines with oxygen, releasing a large amount of energy in the form of light and heat.
 Produces a flame
 Fuel + oxygen produces carbon dioxide and water vapor
C x
H x
+ O
2
 CO
2
+ H
2
O

Acid-Base Neutralization Reactions
 When the solution of an acid and solution of a base are mixed


Products have no characteristics of either the acid or the base
Acid + Base (metal hydroxide)  salt + water
 Salt comes from cation of base and anion of acid
HY + XOH  XY + H
2
O

 Molecular Equation – shows complete chemical formulas of reactants and products
Pb(NO
3
)
2
(aq) + 2KI(aq)  PbI
2
(s) + 2KNO
3
(aq)
 Complete Ionic Equation – All soluble electrolytes shown as ions
Pb +2 (aq) + 2NO
3
(aq) + 2K + (aq) + 2I (aq)  PbI
2
(s) + 2K + (aq) + 2NO
3
(aq)
 Net Ionic Equation – shows only the ions and molecules directly involved in the equation
Pb +2 (aq) + 2I (aq)  PbI
2
(s)

Writing Complete Ionic Equations
1.
3.
Start with a balanced molecular equation.
2.



Break all soluble strong electrolytes
(compounds with (aq) beside them) into their ions. indicate the correct formula and charge of each ion indicate the correct number of each ion write (aq) after each ion
Bring down all compounds with (s), (l), or
(g) unchanged.

Writing Complete Ionic Equations
Example:
2Na
3
PO
4
(aq) + 3CaCl
2
(aq)  6NaCl(aq) + Ca
3
(PO
4
)
2
(s)
Becomes…
6Na + (aq) + 2PO
4
3(aq) + 3Ca 2+ (aq) + 6Cl (aq)  6Na + (aq) +
6Cl (aq) + Ca
3
(PO
4
)
2
(s)

 Appear in identical forms among both the reactants and products of a complete ionic equation
 When writing net ionic equations they cancel each other out
Pb +2 (aq) + 2NO
3
(aq) + 2K + (aq) + 2I (aq)  PbI
2
(s) + 2K + (aq) + 2NO
3
(aq)

 Cancel out spectator ions from complete ionic equation then write what’s left
6Na + (aq) + 2PO
4
6Cl -
3(aq) + 3Ca 2+ (aq) + 6Cl (aq)  6Na + (aq) +
(aq) + Ca
3
(PO
4
)
2
(s)
Becomes…
2PO
4
3(aq) + 3Ca 2+ (aq)  Ca
3
(PO
4
)
2
(s)

Write complete ionic and net ionic equations for the following:
1.
3(NH
4
)
2
CO
3
(aq) + 2Al(NO
3
)
3
(aq)  6NH
4
NO
3
(aq) +
Al
2
(CO
3
)
3
(s)
2.
2NaOH(aq) + H
2
SO
4
(aq)  Na
2
SO
4
(aq) + 2H
2
O(l)
3.
Zn(s) + CuSO
4
(aq) --> ZnSO
4
(aq) + Cu(s)

1.
Complete Ionic Equation:
6NH
4
+ (aq) + 3CO
3
2(aq) + 2Al 3+ (aq) + 6NO
3
(aq)

6NH
4
+ (aq) + 6NO
3
(aq) + Al
2
(CO
3
)
3
(s)
Net Ionic Equation: 2 Al 3+ (aq) + 3 CO
3
2(aq)  Al
2
(CO
3
)
3
(s)
2.
Complete Ionic Equation:
2Na + (aq) + 2OH (aq) + 2H + (aq) + SO
4
2(aq)  2Na + (aq) + SO
4
2(aq) + 2H
2
O(l)
Net Ionic Equation: OH (aq) + H + (aq)  H
2
O(l)
*Note: simplify net ionic equations if possible
3.
Complete Ionic Equation:
Zn(s) + Cu 2+ (aq) + SO
4
2(aq)  Zn 2+ (aq) + SO
4
2(aq) + Cu(s)
Net Ionic Equation: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

The study of the quantitative aspects of chemical reactions.

 Conversion factor that relates amount in moles of any two substances involved in a chemical reaction
2 Al
2
O
3
(l)  4 Al(s) + 3 O
2
(g)
Mole ratio Al
2
O
3 to O
2
= 2:3
Mole ratio Al to Al
2
O
3
Mole ratio Al to O
2
= 4:2 or 2:1
= 4:3

 Solved just like conversions!
 You must start with a balanced chemical equation
 Types:
 Mole  Mole
 Mass

Mass
 Mass  Mole or Mole  Mass


2 Al
2
O
3
(l)  4 Al(s) + 3 O
2
(g)
 How many moles of O
2
3.5 moles of Al
2
O
3
?
are produced from
3.5 mol Al
2
O
3
×
3 mol O
2
2 mol Al
2
O
3
= 5.25 mol O
2
* Use mole ratio to convert between moles!


2 Al
2
O
3
(l)  4 Al(s) + 3 O
2
(g)
 How many grams of Al are produced from
4.56 grams of Al
2
O
3
?
 Molar Mass Al
2
O
3
= 101.96 g/mol Molar Mass Al = 26.98 g/mol
4.56 g Al
2
O
3
×
1 mol Al
2
O
3
101.96 g Al
2
O
3
×
4 mol Al
2 mol Al
2
O
3
×
26.98 g Al
1 mol Al
= 2.41 g Al

Recipe makes 10 pancakes
 3 eggs
 2 cups bisquik
 1 cup milk
 1 cup chocolate chips
What “limits” how many pancakes I can make and what will be left over?
 What is the most amount of pancakes that I can make with 6 eggs and 5 cups of milk?
 What is the most amount of pancakes that I can make with 3 cups of chocolate chips and 8 cups of milk?

 The limiting reactant is the reactant that is consumed first , limiting the amounts of products formed.
 The excess reactant is the reactant that is leftover after the reaction has gone to completion.

Reactants Products
2 NO(g) + O
2
(g) 2 NO
2
(g)
Limiting reactant = ___________
Excess reactant = ____________

Calculating Limiting/Excess Reagent
2 NO(g) + O
2
(g) 2 NO
2
(g)
 Given 12.4 grams of NO and 9.40 grams of O
2
, which is the limiting and which is the excess reagent?
12.4 g NO ×
1 mol NO
30.01 g NO
×
2 mol NO
2
2 mol NO
×
46.01 g NO
2
1 mol NO
2
= 19.01 g NO
2
9.40 g O
2
×
1 mol O
2
16.00 g O
2
× 2 mol NO
2
1 mol O
2
×
46.01 g NO
2
1 mol NO
2
= 54.06 g NO
2

Calculating Limiting/Excess Reagent
2 NO(g) + O
2
(g) 2 NO
2
(g)
12.4 g NO ×
1 mol NO
30.01 g NO
×
2 mol NO
2
2 mol NO
×
46.01 g NO
2
1 mol NO
2
= 19.01 g NO
2
9.40 g O
2
×
1 mol O
2
32.00 g O
2
× 2 mol NO
2
1 mol O
2
×
46.01 g NO
2
1 mol NO
2
= 27.03 g NO
2
 NO limits the amount of NO
2
 Limiting reagent = NO that is made
 O
2 will be leftover once the reaction is complete
 Excess reactant = O
2

Calculating Limiting/Excess Reagent
2 NO(g) + O
2
(g) 2 NO
2
(g)
 How much O
2 will be in excess once the reaction is complete?
12.4 g NO ×
1 mol NO
30.01 g NO
×
1 mol O
2
2 mol NO
×
32.00 g O
2
1 mol O
2
= 6.61 g O
2
6.61 grams of O
2 will be used in the reaction.
You have 9.40 grams to start with.
9.40 – 6.61 = 2.79 grams O
2 in excess (leftover)

 If the equation has 2 or more products, when determining the limiting/excess reactants, simply pick one of the products and convert both reactants to that product.
 You MUST use the same product for both.

Actual Yield
Theoretical Yield
× 100
 Percentage comparing how much product was actually produced compared to what should have been produced.
 Calculate theoretical yield using stoichiometry.
 If you know how much of each reactant you start out with, use stoichiometry to calculate how much of the given product you should produce.

AgNO
3
(aq) + KCl (aq) → AgCl(s) + KNO
3
(aq)
An experiment was performed combining using 3.4 g of
AgNO
3 and an unlimited supply of KCl. If the experiment yielded 2.7 g of AgCl, what is the percent yield of the experiment?
3.4 g AgNO
3
× 1 mol AgNO
3
169.88 g AgNO
3
× 1 mol AgCl
1 mol AgNO
3
× 143.32 g AgCl
1 mol AgCl
= 2.9 g AgCl
2.7
2.9
× 100 = 93%