Transcript Types of Chemical Reactions

A Study of Chemical Reactions
Equations, Mole Conversions, &
Stoichiometry
Types of Reactions

Many chemical
reactions have
defining
characteristics
which allow
them to be
classified as to
type.
Types of Chemical Reactions

The five types of chemical reactions in this
unit are:
 Combination/Synthesis
 Decomposition/Analysis
 Single Replacement/Displacement
 Double Replacement/Metathesis
 Combustion
Combination Reactions


Two or more substances combine to form
one substance.
 The general form is A + X AX
Example:
 Magnesium + oxygen  magnesium
oxide
 2Mg + O2  2MgO
Magnesium + Oxygen
Combination Reactions


Combination reactions may also be called
composition or synthesis reactions.
Some types of combination reactions:

Combination of elements
 K + Cl2 
 One product will be formed
Combination Reactions

K + Cl2 

Write the ions: K+ Cl-

Balance the charges: KCl

Balance the equation: 2K + Cl2  2KCl
Combination Reactions

Some types of combination reactions:
 Oxide + water 
 Nonmetal
oxide + water  acid
 SO2 + H2O  H2SO3
 Metal oxide + water  base
 BaO + H2O  Ba(OH)2
Combination Reactions

Some types of combination reactions:
 Metal oxides + nonmetal oxides
Na2O + CO2  Na2CO3
CaO + SO2  CaSO3
Decomposition Reactions


One substance reacts to form two or more
substances.
 The general form is AX  A + X
Example:
 Water can be decomposed by
electrolysis.
 2H2O  2H2 + O2
Electrolysis of Water
Decomposition Reactions

Types of Decomposition Reactions:

Decomposition of carbonates
 When heated, some carbonates
break down to form an oxide and
carbon dioxide.
 CaCO3  CaO + CO2
 H2CO3  H2O + CO2
Decomposition Reactions

Types of decomposition reactions:

Some metal hydroxides decompose
into oxides and water when heated.
 Ca(OH)2  CaO + H2O
Note that this is the reverse of a similar
combination reaction.
Decomposition Reactions

Types of decomposition reactions:

Metal chlorates decompose into
chlorides and oxygen when heated.
 2KClO3
 2KCl + 3O2
 Zn(ClO3)2  ZnCl2 + 3O2

Some of these reactions are used in
explosives.
Decomposition Reactions

Some substances can easily
decompose:
Ammonium hydroxide is actually
ammonia gas dissolved in water.
 NH4OH  NH3 + H2O
 Some acids decompose into water and
an oxide.
 H2SO3  H2O + SO2

Decomposition Reactions


Some decomposition reactions are
difficult to predict.
The decomposition of nitrogen
triiodide, NI3, is an example of an
interesting decomposition reaction.
Nitrogen triiodide
Single Replacement Reactions

Cationic: A metal will replace a metal
ion in a compound.


The general form is A + BX  AX + B
Anionic: A nonmetal will replace a
nonmetal ion in a compound.

The general form is Y + BX  BY + X
Single Replacement Reactions

Examples:
 Ni + AgNO3 
 Nickel replaces the metallic ion Ag+.
 The silver becomes free silver and the
nickel becomes the nickel(II) ion.
 Ni + AgNO3  Ag + Ni(NO3)2
 Balance the equation:
 Ni + 2AgNO3  2Ag + Ni(NO3)
Activity Series
Single Replacement Reactions



Not all single replacement reactions that
can be written actually happen.
The metal must be more active than the
metal ion.
Aluminum is more active than iron in Al +
Fe2O3 in the following reaction:
Thermite Reaction
Thermite Reaction
Al + Fe2O3 
 Aluminum will replace iron(III)
 Iron(III) becomes Fe and aluminum
metal becomes Al3+.
 2Al + Fe2O3  2Fe + Al2O3

Single Replacement Reactions

An active nonmetal can replace a less
active nonmetal.
 The halogen (F2, Cl2, Br2, I2)
reactions are good examples.
 F2 is the most active and I2 is the
least.
Cl2 +2 NaI  2 NaCl + I2
Double Replacement Reactions


Ions of two compounds exchange
places with each other.
 The general form is AX + BY  AY
+ BX
Metathesis is an alternate name for
double replacement reactions.
NaOH + CuSO4
Metathesis (sink or float?)






NaOH + CuSO4 
The Na+ and Cu2+ switch places.
Na+ combines with SO42- to form Na2SO4.
Cu2+ combines with OH- to form Cu(OH)2
NaOH + CuSO4  Na2SO4 + Cu(OH)2
2NaOH + CuSO4  Na2SO4 + Cu(OH)2
CuSO4 + Na2CO3
Double Replacement
CuSO4 + Na2CO3 
 Cu2+ combines with CO32- to form
CuCO3.
 Na+ combines with SO42- to form
Na2SO4.
 CuSO4 + Na2CO3  CuCO3 + Na2SO4

Na2CO3 + HCl
Double Replacement
Na2CO3 + HCl 
 Notice that gas bubbles were
produced rather than a precipitate.
 What was the gas?
 Write the double replacement
reaction first.

Double Replacement
Na2CO3 + HCl 
 Na+ combines with Cl- to form NaCl.
 H+ combines with CO32- to form
H2CO3.
 Na2CO3 + 2HCl  2NaCl + H2CO3
 H2CO3 breaks up into H2O and CO2.

Double Replacement
The gas formed was carbon dioxide.
 The final balanced reaction is:
Na2CO3 + HCl  NaCl + H2O + CO2.
 Balance the equation.
 Na2CO3 + 2HCl  2NaCl + H2O +
CO2

Combustion Reaction
When a substance combines with
oxygen, a combustion reaction
results.
 The combustion reaction may also be
an example of an earlier type such as
2Mg + O2  2MgO.
 The combustion reaction may be
burning of a fuel.

Combustion Reaction
Methane, CH4, is natural gas.
 When hydrocarbon compounds are
burned in oxygen, the products are
water and carbon dioxide.
 CH4 + O2  CO2 + H2O
 CH4 + 2O2  CO2 + 2H2O

Combustion Reactions
Combustion reactions involve light
and heat energy released.
 Natural gas, propane, gasoline, etc.
are burned to produce heat energy.
 Most of these organic reactions
produce water and carbon dioxide.

Practice


Classify each of the following as to
type:
H2 + Cl2  2HCl


Combination
Ca + 2H2O  Ca(OH)2 + H2

Single replacement
Practice

2CO + O2  2CO2


Combination and combustion
2KClO3  2KCl + 3O2

Decomposition
Practice

FeS + 2HCl  FeCl2 + H2S


Double replacement
Zn + HCl  ?
Single replacement
 Zn + 2HCl  ZnCl2 + H2

How molecules are symbolized
Cl2
2Cl
2Cl2
• Molecules may also have brackets to
indicate numbers of atoms. E.g. Ca(OH)2
• Notice that the OH is a group
O Ca O
H
• The 2 refers to both H and O H
• How many of each atom are in the following?
a) NaOH
Na = 1, O = 1, H = 1
b) Ca(OH)2 Ca = 1, O = 2, H = 2
c) 3Ca(OH)2 Ca = 3, O = 6, H = 6
Balancing equations: MgO
The law of conservation of mass states that
matter can neither be created or destroyed
 Thus, atoms are neither created or destroyed,
only rearranged in a chemical reaction
 Thus, the number of a particular atom is the
same on both sides of a chemical equation
 Example: Magnesium + Oxygen (from lab)
 Mg + O2  MgO
Mg + O O  Mg O

• However, this is not balanced
• Left:
Mg = 1, O = 2
• Right: Mg = 1, O = 1
Balance equations by “inspection”
Mg + O2  MgO
2Mg + O2  2MgO is correct
Mg + ½O2

MgO
is
incorrect
Mg2 + O2  2MgO is incorrect
4Mgwith
+ 2elements
O2  4MgO
is incorrect
Hints: start
that occur
in one
compound on each side. Treat polyatomic ions
that repeat as if they were a single entity.
a) P4 + 5 O2  P4O10
b) 2 Li + 2 H2O  H2 + 2 LiOH
c) 2 Bi(NO3)3 + 3 K2S  Bi2S3 + 6 KNO3
d) C2H6 +3.5 O2  2 CO2 + 3 H2O
2 C2H6 + 7 O2  4 CO2 + 6 H2O
From
Balance these skeleton equations:
a)
b)
c)
d)
e)
f)
g)
h)
i)
Mg + 2HCl  MgCl2 + H2
3Ca + N2  Ca3N2
NH4NO3  N2O + 2H2O
2BiCl3 + 3H2S  Bi2S3 + 6HCl
2C4H10 + 13O2  8CO2 + 10H2O
6O2 + C6H12O6  6CO2 + 6H2O
3NO2 + H2O  2HNO3 + NO
Cr2(SO4)3+ 6NaOH  2Cr(OH)3+ 3Na2SO4
Al4C3 + 12H2O  3CH4 + 4Al(OH)3
The Mole
Q: how long would it take to spend a mole of $1 coins if
they were being spent at a rate of 1 billion per second?
Background: atomic masses
Look at the “atomic masses” on the periodic
table. What do these represent?
 E.g. the atomic mass of C is 12 (atomic # is 6)
 We know there are 6 protons and 6 neutrons
 Protons and neutrons have roughly the same
mass. So, C weighs 12 u (atomic mass units).
 What is the actual mass of a C atom?
 Answer: approx. 2 x 10-23 grams (protons and
neutrons each weigh about 1.7 x10-24 grams)
Two problems
1. Atomic masses do not convert easily to grams
2. They can’t be weighed (they are too small)

The Mole
With these problems, why use atomic mass at all?
1. Masses give information about # of p+, n0, e–
2. It is useful to know relative mass
E.g. Q - What ratio is needed to make H2O?
A - 2:1 by atoms, but 2:16 by mass
 It is useful to associate atomic mass with a
mass in grams. It has been found that
1g H, 12g C, or 23g Na have 6.02x1023 atoms
 6.02 x 1023 is a “mole” or “Avogadro’s number”
 “mol” is used in equations, “mole” is used in
writing; one gram = 1 g, one mole = 1 mol.
Mollionaire
Q: how long would it take to spend a mole of
$1 coins if they were being spent at a rate of
1 billion per second?
A: $ 6.02 x 1023 / $1 000 000 000
= 6.02 x 1014 payments = 6.02 x 1014 seconds
6.02 x 1014 seconds / 60 = 1.003 x 1013 minutes
1.003 x 1013 minutes / 60 = 1.672 x 1011 hours
1.672 x 1011 hours / 24 = 6.968 x 109 days
6.968 x 109 days / 365.25 = 1.908 x 107 years
A: It would take 19 million years
Comparing sugar (C12H22O11) & H2O
Same
1 gram each
1 mol each
No, they have dif. No, molecules
volume?
densities.
have dif. sizes.
Yes, that’s what
No, molecules
mass?
grams are.
have dif. masses
No, they have dif.
Yes.
# of moles? molar masses
No, they have dif. Yes (6.02x1023
# of
in each)
molecules? molar masses
No, sugar has
# of atoms?
No
more (45:3 ratio)
Molar mass
The mass of one mole is called “molar mass”
 E.g. 1 mol Li = 6.94 g Li
 This is expressed as 6.94 g/mol
 What are the following molar masses?
S 32.06 g/mol SO2 64.06 g/mol
Cu3(BO3)2 308.27 g/mol
Calculate molar masses (to 2 decimal places)
CaCl2
Cu x 3 = 63.55 x 3 = 190.65
(NH4)2CO3 B x 2 = 10.81 x 2 = 21.62
O2
O x 6 = 16.00 x 6 = 96.00
308.27
Pb3(PO4)2
C6H12O6

Molar mass
The mass of one mole is called “molar mass”
 E.g. 1 mol Li = 6.94 g Li
 This is expressed as 6.94 g/mol
 What are the following molar masses?
S 32.06 g/mol SO2 64.06 g/mol
Cu3(BO3)2 308.27 g/mol
Calculate molar masses (to 2 decimal places)
CaCl2
110.98 g/mol (Cax1, Clx2)
(NH4)2CO3 96.11 g/mol (Nx2, Hx8, Cx1, Ox3)
O2
32.00 g/mol (Ox2)
Pb3(PO4)2 811.54 g/mol (Pbx3, Px2, Ox8)
C6H12O6 180.18 g/mol (Cx6, Hx12, Ox6)

Converting between grams and moles
If we are given the # of grams of a compound
we can determine the # of moles, & vise-versa
 In order to convert from one to the other you
must first calculate molar mass

Formula
HCl
H2SO4
NaCl
Cu
g/mol
36.46
98.08
58.44
63.55
g
mol (n)
9.1
0.25
53.15 0.5419
207
3.55
1.27 0.0200
The Mole
Convert 36.0 grams of carbon into atoms.
Convert 30 molecules of methane into Liters
of gas.
Simplest and molecular formulae
Consider NaCl (ionic) vs. H2O2 (covalent)
Na Cl Na Cl
Cl Na Cl Na
• Chemical formulas are either “simplest” (a.k.a.
“empirical”) or “molecular”. Ionic compounds
are always expressed as simplest formulas.
• Covalent compounds can either be molecular
formulas (I.e. H2O2) or simplest (e.g. HO)
Q - Write simplest formulas for propene (C3H6),
C2H2, glucose (C6H12O6), octane (C8H14)
Q - Identify these as simplest formula, molecular
formula, or both H2O, C4H10, CH, NaCl
Answers
Q - Write simplest formulas for propene (C3H6),
C2H2, glucose (C6H12O6), octane (C8H14)
Q - Identify these as simplest formula, molecular
formula, or both H2O, C4H10, CH, NaCl
A - CH2 CH CH2O
C 4H 7
A - H2O is both simplest and molecular
C4H10 is molecular (C2H5 would be simplest)
CH is simplest (not molecular since CH can’t
form a molecule - recall Lewis diagrams)
NaCl is simplest (it’s ionic, thus it doesn’t
form molecules; it has no molecular formula)
Calculating percentage mass
If you can work out Mr then this bit is easy…
Percentage mass (%) =
Mass of element Ar
Relative formula mass Mr
x100%
Calculate the percentage mass of magnesium in magnesium oxide, MgO:
Ar for magnesium = 24
Ar for oxygen = 16
Mr for magnesium oxide = 24 + 16 = 40
Therefore percentage mass = 24/40 x 100% = 60%
Calculate the percentage mass of the following:
1) Hydrogen in hydrochloric acid, HCl
2) Potassium in potassium chloride, KCl
3) Calcium in calcium chloride, CaCl2
4) Oxygen in water, H2O
Empirical formulae
Empirical formulae is simply a way of showing how many atoms are in a
molecule (like a chemical formula). For example, CaO, CaCO3, H20 and KMnO4
are all empirical formulae. Here’s how to work them out:
A classic exam question:
Find the simplest formula of 2.24g of iron
reacting with 0.96g of oxygen.
Step 1: Divide both masses by the relative atomic mass:
For iron 2.24/56 = 0.04
For oxygen 0.96/16 = 0.06
Step 2: Write this as a ratio and simplify:
0.04:0.06 is equivalent to 2:3
Step 3: Write the formula:
2 iron atoms for 3 oxygen atoms means the formula is Fe2O3
21/07/2015
Example questions
1)
Find the empirical formula of magnesium oxide which contains 48g of
magnesium and 32g of oxygen.
2)
Find the empirical formula of a compound that contains 42g of nitrogen and 9g
of hydrogen.
3)
Find the empirical formula of a compound containing 20g of calcium, 6g of
carbon and 24g of oxygen.
21/07/2015
Stoichiometry
Molar Mass of Compounds

The molar mass (MM) of a compound is
determined the same way, except now you add
up all the atomic masses for the molecule (or
compound)




Ex. Molar mass of CaCl2
Avg. Atomic mass of Calcium = 40.08g
Avg. Atomic mass of Chlorine = 35.45g
Molar Mass of calcium chloride =
40.08 g/mol Ca + (2 X 35.45) g/mol Cl
 110.98 g/mol CaCl2
20
Ca
40.08
17
Cl
35.45
Liters of
Gas
Mole
Roadmap
22.4
Atoms or
Molecules
6.02 X 1023
Moles
molar mass from
periodic table
Mass
(grams)
Practice

Calculate the Molar Mass of
calcium phosphate
 Formula =
Ca3(PO4)2

Masses elements:

Molar Mass =
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through
Moles!!!
Chocolate Chip Cookies!!
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
How many eggs are needed to make 3 dozen cookies?
How much butter is needed for the amount of chocolate chips
used?
How many eggs would we need to make 9 dozen cookies?
How much brown sugar would I need if I had 1 ½ cups white
Cookies and Chemistry…Huh!?!?




Just like chocolate chip
cookies have recipes,
chemists have recipes as well
Instead of calling them
recipes, we call them reaction
equations
Furthermore, instead of using
cups and teaspoons, we use
moles
Lastly, instead of eggs, butter,
sugar, etc. we use chemical
compounds as ingredients
Chemistry Recipes


Looking at a reaction tells us how much of
something you need to react with
something else to get a product (like the
cookie recipe)
Be sure you have a balanced reaction
before you start!
Example: 2 Na + Cl2  2 NaCl
 This reaction tells us that by mixing 2 moles of
sodium with 1 mole of chlorine we will get 2 moles
of sodium chloride
 What if we wanted 4 moles of NaCl? 10 moles?
50 moles?

Practice

Write the balanced reaction for hydrogen gas
reacting with oxygen gas.




2 H2 + O 2  2 H2 O
How many moles of reactants are needed?
What if we wanted 4 moles of water?
What if we had 3 moles of oxygen, how much hydrogen
would we need to react and how much water would we
get?
What if we had 50 moles of hydrogen, how much oxygen
would we need and how much water produced?
Mole Ratios


These mole ratios can be used to calculate
the moles of one chemical from the given
amount of a different chemical
Example: How many moles of chlorine is
needed to react with 5 moles of sodium
(without any sodium left over)?
2 Na + Cl2  2 NaCl
5 moles Na 1 mol Cl2
2 mol Na
= 2.5 moles Cl2
Mole-Mole Conversions

How many moles of sodium chloride will
be produced if you react 2.6 moles of
chlorine gas with an excess (more than
you need) of sodium metal?
Mole-Mass Conversions


Most of the time in chemistry, the amounts are
given in grams instead of moles
We still go through moles and use the mole
ratio, but now we also use molar mass to get to
grams

Example: How many grams of chlorine are required
to react completely with 5.00 moles of sodium to
produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na 1 mol Cl2
70.90g Cl2
2 mol Na 1 mol Cl2
= 177g Cl2
Practice

Calculate the mass in grams of Iodine
required to react completely with 0.50
moles of aluminum.
Mass-Mole


We can also start with mass and convert to
moles of product or another reactant
We use molar mass and the mole ratio to get
to moles of the compound of interest


Calculate the number of moles of ethane (C2H6)
needed to produce 10.0 g of water
2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O 1 mol H2O
2 mol C2H6 = 0.185
18.0 g H2O 6 mol H20 mol C2H6
Practice

Calculate how many moles of oxygen are
required to make 10.0 g of aluminum
oxide
Mass-Mass Conversions


Most often we are given a starting mass
and want to find out the mass of a
product we will get (called theoretical
yield) or how much of another reactant we
need to completely react with it (no
leftover ingredients!)
Now we must go from grams to moles,
mole ratio, and back to grams of
compound we are interested in
Mass-Mass Conversion
Ex. Calculate how many grams of
ammonia are produced when you react
2.00g of nitrogen with excess hydrogen.
 N2 + 3 H2  2 NH3
2.00g N2 1 mol N2 2 mol NH3 17.06g NH3

28.02g N2 1 mol N2
= 2.4 g NH3
1 mol NH3
Practice

How many grams of calcium nitride are
produced when 2.00 g of calcium reacts
with an excess of nitrogen?
Limiting Reactant: Cookies
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4
dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make
9 dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?
Limiting Reactant



Most of the time in chemistry we have more of
one reactant than we need to completely use
up other reactant.
That reactant is said to be in excess (there is
too much).
The other reactant limits how much product we
get. Once it runs out, the reaction
s.
This is called the limiting reactant.




Limiting
Reactant
To find the correct answer, we have to try all of
the reactants. We have to calculate how much
of a product we can get from each of the
reactants to determine which reactant is the
limiting one.
The lower amount of a product is the correct
answer.
The reactant that makes the least amount of
product is the limiting reactant. Once you
determine the limiting reactant, you should
ALWAYS start with it!
Be sure to pick a product! You can’t compare to
see which is greater and which is lower unless
the product is the same!
Limiting
Reactant
Limiting


10.0g of aluminum reacts with 35.0 grams of
chlorine gas to produce aluminum chloride. Which
reactant is limiting, which is in excess, and how
much product is produced?
2 Al + 3 Cl2  2 AlCl3
Start with Al:
10.0 g Al

Reactant: Example
1 mol Al
2 mol AlCl3
27.0 g Al
2 mol Al
133.5 g AlCl3
1 mol AlCl3
= 49.4g AlCl3
Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
LR Example Continued

We get 49.4g of aluminum chloride from the
given amount of aluminum, but only 43.9g of
aluminum chloride from the given amount of
chlorine. Therefore, chlorine is the limiting
reactant. Once the 35.0g of chlorine is used up,
the reaction comes to a complete
.
Limiting Reactant Practice

15.0 g of potassium reacts with 15.0 g of
iodine. Calculate which reactant is limiting
and how much product is made.
Finding the Amount of Excess


By calculating the amount of the excess
reactant needed to completely react with
the limiting reactant, we can subtract that
amount from the given amount to find the
amount of excess.
Can we find the amount of excess
potassium in the previous problem?
Finding Excess Practice

15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2  2 KI

We found that Iodine is the limiting reactant,
and 19.6 g of potassium iodide are produced.
15.0 g I2
K
1 mol I2
2 mol K
39.1 g
= 4.62 g K
USED!
254 g I2
1 mol I2
1
mol K 15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given amount
of excess
reactant
Amount of
excess
reactant
actually
used
Note that we started with
the limiting reactant! Once
you determine the LR, you
should only start with it!
Limiting Reactant: Recap
1.
2.
3.
4.
5.
6.
7.
You can recognize a limiting reactant problem because
there is MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product (pick
any product you choose.)
The lowest answer is the correct answer.
The reactant that gave you the lowest answer is the
LIMITING REACTANT.
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the amount used
from the given amount.
If you have to find more than one product, be sure to
start with the limiting reactant. You don’t have to
determine which is the LR over and over again!