# PowerPoint

```CHAPTER 3
Chemical Equations &
Reaction Stoichiometry
Objectives
 Understand how to write
chemical equations
 Perform calculations based
on chemical equations
 Calculate percent yields
from chemical reactions
 Understand the concept of
sequential reactions
2
Equations

Consider a simple equation:
23  ?

The question is: what happens to 2
if we multiply it by 3?
23  6


The answer: it is converted to 6
In chemistry, we try to answer similar
problems using chemical equations
3
Chemical Equations
CH4 + O2  ?


We ask: what happens to methane
when it reacts with oxygen (burns)?
We know the answer from the
chemical experiment
CH4 + O2  CO2 + H2O

The only thing we have to check is
the Law of Conservation of Matter
4
Chemical Equations
The Law of Conservation of Matter:
in any physical or chemical change the
total mass of matter remains constant
which means
the number of atoms of each element
involved remains unchanged

CH4 + O2  CO2 + H2O
unbalanced equation
5
Chemical Equations
 We need to:
(1) balance the equation
CH4 + O2 
CO2 + H2O
(2) make sure that we have the same
number of atoms for each element
on the left and on the right side
CH4 + 2O2  CO2 + 2H2O
reactants
products
6
Chemical Equations
CH4 + 2O2  CO2 + 2H2O

Symbolic representation of a
chemical reaction that shows:
 reactants on the left side
 products on the right side
 relative amounts of each using
stoichiometric coefficients
7
Balancing Chemical Equations
 Fe + O2  Fe3O4
 Al + H2SO4  Al2(SO4)3 + H2
 C6H12O6 + O2  CO2 + H2O
8
Quantitative Aspects

Let’s go back to
23  6

If we multiply both sides of the equation
by some number, they still will be equal
8  (2  3)  8  6

We can treat chemical equations
in the same way
9
Quantitative Aspects

What happens in the reaction between
methane and oxygen numerically?
CH4
+ 2O2  CO2 + 2H2O
1 molecule 2 molecules 1 molecule

2 molecules
We can multiply both left and right sides by
the same number – they will remain equal:
CH4
+ 2O2  CO2 + 2H2O
1 molecule 2 molecules 1 molecule
8 molecules 16 molecules 8 molecules
2 molecules
x8
16 molecules
10
Example 1

How many O2 molecules are required
to react with 81 atoms of Fe?
Fe + O2  Fe3O4
11
Quantitative Aspects

Let’s multiply the equation by Avogadro’s number:
CH4
+ 2O2  CO2 + 2H2O
1 molecule 2 molecules 1 molecule
2 molecules
6.022x1023
molecules
2x(6.022x1023
molecules)
1 mole
=
2 moles
6.022x1023
molecules
=

=
=
1 mole
2x(6.022x1023
molecules)
2 moles
In the same way as we talk about chemical
equations in terms of molecules, we can
consider them in terms of moles
12
Example 2

How many moles of H2 is produced when 70 moles
of aluminum react with excess sulfuric acid?
Al + H2SO4  Al2(SO4)3 + H2
13
Quantitative Aspects

Since we know the molar mass of each substance
we can also establish the mass relationships:
CH4
+ 2O2  CO2 + 2H2O
1 mole
2 moles
1 mole
2 moles
x 16.0 g/mol
32.0 g/mol
44.0 g/mol
18.0 g/mol
16.0 g +

64.0 g
=
44.0 g
+ 36.0 g
The total mass of products should be
the same as the total mass of reactants
14
Example 3

What mass of CO is required to react
with 146 g of iron(III) oxide?
Fe2O3 + CO  Fe + CO2
15
Example 4

What mass of carbon dioxide can be produced
by the reaction of 0.540 mole of iron(III)
oxide with excess carbon monoxide?
Fe2O3 + 3CO  2Fe + 3CO2

We should be concerned only with iron(III) oxide
and carbon dioxide since carbon monoxide is in
excess. Therefore, we calculate the number of
moles of CO2 which can be produced from 0.540
mole of Fe2O3:
 1 mole of Fe2O3 produces 3 moles of CO2, hence
 0.540 mole of Fe2O3 will produce
(0.540 x 3) mole = 1.62 mole of CO2
16
Example 4 (continued)

Now we can calculate the mass of CO2 produced:

mass(CO2) = #moles(CO2) x molar mass (CO2)
mass(CO2) = 1.62 mole x 44.0 g/mol = 71.28 g
17
Example 5

What mass of iron(III) oxide reacted with excess
carbon monoxide if carbon dioxide produced by
the reaction had a mass of 8.65 grams?
Fe2O3 + 3CO  2Fe + 3CO2

We should be concerned only with iron(III) oxide and
carbon dioxide. First, we calculate their molar masses:

Mr(Fe2O3) = 159.69 g/mol

Mr(CO2) = 44.01 g/mol
18
Example 5


From the equation we see that 1 mole of Fe2O3 affords
3 moles of CO2. We can convert these quantities to
grams:

m (Fe2O3) = 1 mole x 159.69 g/mole = 159.69 g

m (CO2) = 3 mole x 44.01 g/mole = 132.03 g
Now we can write the following relationship:

132.03 g CO2 is obtained from 159.69 g Fe2O3
8.65 g CO2 is obtained from

?
g Fe2O3
Therefore,
132.03g 159.69g

8.65 g
?g
?g 
159.69g x 8.65 g
 10.5 g Fe2O3
132.03g
19
Limiting Reactant Concept

Example: A box contains 87 bolts, 110 washers,
and 99 nuts. How many sets, each consisting
of one bolt, two washers, and one nut, can you
construct from the contents of one box?

87 bolts 

110 washers 

99 nuts 
20
Example 6

What is the maximum mass of sulfur dioxide
that can be produced by the reaction of 95.6 g
of carbon disulfide with 110. g of oxygen?
CS2  O2  CO2  SO2
21
Example 7

Calculate the mass of carbon tetrachloride
which can be produced by the reaction of 10.0 g
of carbon with 100.0 g of chlorine. Determine
the mass of excess reagent left unreacted.
C + Cl2  CCl4
1) Balance the equation and calculate molar masses
of all substances (since all of them are included
in the formulation of the problem):
C
M r:
12.01 g/mol
+
2Cl2

70.91 g/mol
CCl4
153.82 g/mol
22
Example 7 (continued)
2) Calculate the amount of reactants in moles by
dividing the mass of each by its molar mass:
C
m:
Mr :
10.0 g
12.01 g/mol
0.833 mol
+
2Cl2

CCl4
100.0 g
70.91 g/mol
1.41 mol
It follows from the equation that 1 mole of C reacts
with 2 moles of Cl2, therefore 0.833 mole of C should
react with 0.833 x 2 = 1.67 mole of Cl2. The amount of
Cl2 participating in the reaction is smaller (1.41 mol)
which means that Cl2 is the limiting reactant
23
Example 7 (continued)
3) Now, knowing the amount of the limiting
reactant, we can easily calculate the amount
of product. According to the equation,
C
+
2Cl2

CCl4
from 2 moles of Cl2 we can obtain 1 mole of
CCl4. Hence, from 1.41 mol of Cl2 we can
obtain 1.41/2 = 0.705 mol of CCl4.
m (CCl4) = #moles(CCl4) x Mr(CCl4) =
= 0.705 mol x 153.82 g/mol = 108 g
24
Example 7 (continued)
4)
According to the equation, the number of moles
of carbon consumed in the reaction equals to
the number of moles of CCl4 formed:
C
+
2Cl2

CCl4
#moles (C) = 0.705 mol
m (C) = #moles(C) x Mr(C) = 0.705 mol x 12.01 g/mol = 8.47 g
This means that the amount of carbon that didn’t react is
m (C-unreacted) = 10.0 g – 8.47 g = 1.53 g
25
Percent Yields from Reactions


Theoretical yield is calculated by assuming
that the reaction goes to completion
It is the maximum yield possible for the
given reaction
BUT



in many reactions the reactants are not
completely converted to products
a particular set of reactants may undergo
two or more reactions simultaneously
sometimes it is difficult to separate the
desired product from other products in
the reaction mixture
26
Percent Yields from Reactions


Actual yield is the amount of a specified pure
product actually obtained from a given reaction
Percent yield:
actual yield
% yield=
 100%
theoretical yield


There are many reactions that do not give
the 100% yield
When calculating the percent yield, always
make sure that the actual and theoretical
yields are expressed in the same units
27
Example 8

A 10.0 g sample of ethanol, C2H5OH, was boiled with
excess acetic acid, CH3COOH, to produce 14.8 g of ethyl
acetate, CH3COOC2H5. What is the percent yield?
C2H5OH + CH3COOH  CH3COOC2H5 + H2O
1) We start by calculating the molar masses
of C2H5OH and CH3COOC2H5:
Mr(C2H5OH) = 46.08 g/mol
Mr(CH3COOC2H5) = 88.12 g/mol
2) The amount of C2H5OH in moles is
10.0 g
# moles(C2H5OH) =
 0.217 mol
46.08g/mol
28
Example 8 (continued)
C2H5OH + CH3COOH  CH3COOC2H5 + H2O
3)
According to the equation, from 1 mole of C2H5OH
we can obtain 1 mole of CH3COOC2H5, therefore
from 0.217 mol of C2H5OH we can obtain 0.217 mol
of CH3COOC2H5, or
theoretical yield = 0.217 mol x 88.12 g/mol = 19.1 g
The actual yield, however, is 14.8 g and
14.8 g
percent yield=
 100%  77.5%
19.1 g
29
Example 9

10.6 g of Fe reacts with 25 g of Br2 to form
18 g of FeBr3. What is the percent yield?
2Fe + 3Br2  2FeBr3
1) Calculate the number of moles of each reactant
to determine which reactant is the limiting one:
10.6 g
m(Fe)
# moles(Fe) =

 0.190 mol
Mr (Fe) 55.85g/mol
25 g
m(Br2 )
# moles(Br2 ) =

 0.156 mol
Mr (Br2 ) 159.81g/mol
According to the equation, we need 3 moles of
Br2 per each 2 mole of Fe. The calculation shows
that we have more moles of Fe than of Br2.
Obviously, Br2 is the limiting reagent.
30
Example 9 (continued)
2) In this case we don’t even need to perform any
further calculations in order to see that Br2 is the
limiting reactant. Here’s why:
According to the equation, we need 3 moles of Br2
per each 2 mole of Fe. The calculation performed
in step (1) shows that we have more moles of Fe
than of Br2. Obviously, Br2 is the limiting reactant.
2Fe + 3Br2  2FeBr3
3) From the equation we see that each 3 moles of Br2
result in 2 moles of FeBr3. We can convert these
quantities to grams (multiplying by the molar mass):

m (Br2) = 3 mole x 159.81 g/mole = 479.43 g

m (FeBr3) = 2 mole x 295.56 g/mole = 591.12 g
31
Example 9 (continued)
4)
Now we can write the following relationship:
479.43 g Br2 gives 591.12 g FeBr3
25 g Br2 gives
? g FeBr3,
Therefore,
479.43g 591.12g

25 g
?g
?g 
591.12g x 25 g
 30.8 g FeBr3
479.43g
this is theoretical yield
18 g
percent yield=
 100%  58%
30.8 g
32
Sequential Reactions


A set of reactions required to convert
starting materials into the desired product
The amount of the desired product from
each reaction is taken as the starting
material for the next reaction.
magnesium
perchlorate
sodium
hydroxide
33
Example 10

13 g of decane, C10H22, is burned in a C-H combustion
train. The CO2 gas formed reacts with sodium
hydroxide, NaOH, and is converted into sodium
carbonate, Na2CO3. What mass of Na2CO3 will be
formed if all reactions proceed to completion?
1) The balanced sequential equations are:
2C10H22 + 31O2  20CO2 + 22H2O
2NaOH + CO2  Na2CO3 + H2O
2) We calculate the amount of C10H22 in moles:
# moles(C10H22 ) =
m(C10H22 )
13 g

 0.091 mol
Mr (C10H22 ) 142.28g/mol
34
Example 10 (continued)
2C10H22 + 31O2  20CO2 + 22H2O
2NaOH + CO2  Na2CO3 + H2O
3) Now, looking at the first equation, we can see that
when 2 moles of C10H22 are burned, 20 moles of CO2
are formed, that is to find the number of moles of
CO2 formed we need to multiply the given number of
moles of C10H22 by 10:
#moles(CO2) = 0.091 mol x 10 = 0.91 mol
4)
From the second equation it follows that every
mole of CO2 produces 1 mole of Na2CO3, that is
the amount of Na2CO3 obtained from 0.91 mol of
CO2 will also be equal to 0.91 mol.
35
Example 10 (continued)
5) Finally, knowing the number of moles of Na2CO3
we can calculate its mass:
m(Na2CO3) = #moles (Na2CO3) x Mr(Na2CO3)
m(Na2CO3) = 0.91 mol x 105.99 g/mol = 96 g
36
Example 11

What mass of (NH4)3PO4 can be produced in the
result of the following reactions if we start with
10 moles of N2 and excess hydrogen?
N2 + H2  NH3 (yield = 44%)
NH3 + H3PO4  (NH4)3PO4 (yield = 95%)
37
