Making Quantitative Predictions About Chemical Reactions

1
Making Quantitative PredictionsAbout Chemical
Reactions
Chemical Stoichiometry
2
Qualitative and QuantitativePredictions
The classification of chemical reactions helps
us to predict what will be produced by a
reaction. We also need to know how much will be
produced by a reaction. It is also necessary
to know how much reactant is required to produce
a desired amount of product.
3
The study of the quantitative
Relationships between the reactants and
products in a chemical reaction
is called
chemical stoichiometry.
4
Mass-Mass Stoichiometry
The first predictions we will attempt to make
will involve the masses of reactants and
products.
How many grams of oxygen would be released by
the decomposition of 10.00 grams of potassium
chlorate?
5
The Balanced Equation
Calculations for chemical reactions must be
based upon balanced chemical equations.
Write the equation for the decomposition of
potassium chlorate to produce oxygen gas and
potassium chloride.
KClO3 gt KCl O
Remember that oxygen is a diatomic gas.
6
The Balanced Equation
Calculations for chemical reactions must be
based upon balanced chemical equations.
Write the equation for the decomposition of
potassium chlorate to produce oxygen gas and
potassium chloride.
KClO3 gt KCl O2
Thats better.
7
The Balanced Equation
Calculations for chemical reactions must be
based upon balanced chemical equations.
Write the equation for the decomposition of
potassium chlorate to produce oxygen gas and
potassium chloride.
KClO3 gt KCl O2
Now balance with coefficients.
8
The Balanced Equation
Every calculations for chemical reactions must
be based upon a balanced chemical equation.
Write the equation for the decomposition of
potassium chlorate to produce oxygen gas and
potassium chloride.
2 KClO3 gt 2 KCl 3 O2
9
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
Beneath the balanced equation, construct a flow
chart to organize the elements of the problem.
After reading the problem a second time, put a
question mark under the reactant or product you
need to find. Include the unit of
measurement.
10
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
?
g
11
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
?
g
After the third reading, put the measurements
or quantities that you were given under the
reactant or product to which that value pertains.
12
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
10.00 g
?
g
After the third reading, put the measurements
or quantities that you were given under the
reactant or product to which that value pertains.
13
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
10.00 g
?
g
Because the balanced equation only gives us
molar relationships but we are interested in
masses, we must convert what the equation gives
us into what we want.
14
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
10.00 g
?
g
Grams of KClO3 must be converted to moles of
KClO3. Moles of KClO3 then convert to moles of
O2 . Moles of O2 then convert to grams of O2 .
Use lines beneath the equation to show these
three steps.
15
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
10.00 g
?
g


mol KClO3 ----------- mol O2
Use lines beneath the equation to show these
three steps.
16
The Flow Chart

2 KClO3 gt 2 KCl 3 O2
10.00 g
?
g
/\

3

1
2
\/
mol KClO3 -----------gt mol O2
Each line represents a conversion factor. Arrows
show the sequence of conversion factors.
17
The Calculation Process

2 KClO3 gt 2 KCl 3 O2
10.00 g
?
g
/\

3

1
2
\/
mol KClO3 -----------gt mol O2
Now we set up three conversion factors. Each
conversion factor will accomplish one leg of the
flow chart.
18
The Calculation Process

2 KClO3 gt 2 KCl 3 O2
g
?
10.00 g

/\
3
1
2
\/

mol KClO3 -----------gt mol O2
mol KClO3
1
____________
1)
10.00 g KClO3
mol KClO3

x
0.0816
122.6
g KClO3
19
The Calculation Process

2 KClO3 gt 2 KCl 3 O2
g
?
10.00 g

/\
3
1
2
\/

mol KClO3 -----------gt mol O2
mol KClO3
1
____________
1)
10.00 g KClO3
mol KClO3

x
0.0816
122.6
g KClO3
The second conversion factor will convert moles
of potassium chlorate to moles of oxygen gas.
20
The Calculation Process

2 KClO3 gt 2 KCl 3 O2
g
?
10.00 g

/\
3
1
2
\/

mol KClO3 -----------gt mol O2
mol KClO3
1
____________
1)
10.00
mol KClO3

x
0.0816
g KClO3
122.6
g KClO3
3
molO2
2)
0.0816
mol O2
__________

mol KClO3
0.122
x
mol KClO3
2
21
The Calculation Process

2 KClO3 gt 2 KCl 3 O2
g
?
10.00 g

/\
3
1
2
\/

mol KClO3 -----------gt mol O2
mol KClO3
1
____________
1)
10.00
mol KClO3

x
0.0816
g KClO3
122.6
g KClO3
3
molO2
2)
0.0816
mol O2
__________

mol KClO3
0.122
x
mol KClO3
2
The third and final conversion factor converts
moles of oxygen to grams of oxygen.
22
The Calculation Process

2 KClO3 gt 2 KCl 3 O2
g
?
10.00 g

/\
3
1
2
\/

mol KClO3 -----------gt mol O2
mol KClO3
1
____________
1)
10.00
mol KClO3

x
0.0816
g KClO3
122.6
g KClO3
3
molO2
2)
0.0816
mol O2
__________

mol KClO3
0.122
x
mol KClO3
2
g O2
32.0
3.915
3)
0.122 mol O2
g O2
x
__________

mol O2
1.0
23
This calculation allows us to predict that the
decomposition of 10.00 grams of potassium
chlorate will produce 1.96 grams of oxygen gas.

2 KClO3 gt 2 KCl 3 O2
g
?
10.00 g

/\
3
1
2
\/

mol KClO3 -----------gt mol O2
mol KClO3
1
____________
1)
10.00
mol KClO3

x
0.0816
g KClO3
122.6
g KClO3
3
molO2
2)
0.0816
mol O2
__________

mol KClO3
0.122
x
mol KClO3
2
g O2
32.0
3.915
3)
0.122 mol O2
g O2
x
__________

mol O2
1.0
24
To avoid rounding off three times, it is often
more convenient for calculator use to arrange
our three conversion factors in a linear
sequence. So, Instead of three separate
calculations, like this, it would look more like

mol KClO3
1
____________
1)
10.00
mol KClO3

x
0.0816
g KClO3
122.6
g KClO3
3
molO2
2)
0.0816
mol O2
__________

mol KClO3
0.122
x
mol KClO3
2
g O2
32.0
3.915
3)
0.122 mol O2
g O2
x
__________

mol O2
1.0
25
This.

1
molO2
3
g O2
32.0
mol KClO3
____________
__________
__________
10.00
x
g KClO3
x
x
g KClO3
122.6
mol KClO3
2
mol O2
1.0

g O2
3.915
Once set up in this format, the series of
calculator keystrokes used is quite simple.
26

1
molO2
3
g O2
32.0
mol KClO3
____________
__________
__________
10.00
g KClO3
x
x
x
g KClO3
122.6
mol KClO3
2
mol O2
1.0

g O2
3.915
Remember to multiply by numerators (the numbers
on the top) and divide by denominators (the
numbers on the bottom).
27

1
molO2
3
g O2
32.0
mol KClO3
____________
__________
__________
10.00
g KClO3
x
x
x
g KClO3
122.6
mol KClO3
2
mol O2
1.0

g O2
3.915
Remember to multiply by numerators (the numbers
on the top) and divide by denominators (the
numbers on the bottom).
28

1
molO2
3
g O2
32.0
mol KClO3
____________
__________
__________
10.00
g KClO3
x
x
x
g KClO3
122.6
mol KClO3
2
mol O2
1.0

g O2
3.915
Remember to multiply by numerators (the numbers
on the top) and divide by denominators (the
numbers on the bottom).
29

1
molO2
3
g O2
32.0
mol KClO3
____________
__________
__________
10.00
g KClO3
x
x
x
g KClO3
122.6
mol KClO3
2
mol O2
1.0

g O2
3.915
Remember to multiply by numerators (the numbers
on the top) and divide by denominators (the
numbers on the bottom).
30

1
molO2
3
g O2
32.0
mol KClO3
____________
__________
__________
10.00
x
g KClO3
x
x
g KClO3
122.6
mol KClO3
2
mol O2
1.0

g O2
3.915
10 / 122.6 x 3 / 2 x 32
3.91517289
Rounding to four significant figures (because the
mass data was given that way) gives us
3.915
g O2
31
Determine the mass of iron which
could be produced by the reaction of 100.0
grams of iron(III) oxide with enough carbon to
use it all.
Fe2O3 C gt Fe CO2
32
Determine the mass of iron which
could be produced by the reaction of 100.0
grams of iron(III) oxide with enough carbon to
use it all.
2 Fe2O3 3 C gt 4 Fe 3 CO2
33
Determine the mass of iron which
could be produced by the reaction of 100.0
grams of iron(III) oxide with enough carbon to
use it all.
2 Fe2O3 3 C gt 4 Fe 3 CO2
? g
34
Determine the mass of iron which
could be produced by the reaction of 100.0
grams of iron(III) oxide with enough carbon to
use it all.
2 Fe2O3 3 C gt 4 Fe 3 CO2
? g
100.0 g
35
Determine the mass of iron which
could be produced by the reaction of 100.0
grams of iron(III) oxide with enough carbon to
use it all.
2 Fe2O3 3 C gt 4 Fe 3 CO2
? g
100.0 g

/\

\/
mol Fe2O 3 ------------------------gt mol
Fe
36
Determine the mass of iron which
could be produced by the reaction of 100.0
grams of iron(III) oxide with enough carbon to
use it all.
2 Fe2O3 3 C gt 4 Fe 3 CO2
? g
100.0 g

/\

1
3
\/
2
mol Fe2O 3 ------------------------gt mol
Fe
37
g Fe
1 mol Fe2O 3
mol Fe
4
55.8
100.0 g Fe2O 3 x _________ x
________ x ________
1 mol Fe
mol Fe2O 3
2
g Fe2O 3
159.6
100 / 159.6 x 2 x 55.8
69.9
g Fe
38
g Fe
1 mol Fe2O 3
mol Fe
4
55.8
100.0 g Fe2O 3 x _________ x
________ x ________
1 mol Fe
mol Fe2O 3
2
g Fe2O 3
159.6
100 / 159.6 x 2 x 55.8
69.9
g Fe
39
g Fe
1 mol Fe2O 3
mol Fe
4
55.8
100.0 g Fe2O 3 x _________ x
________ x ________
1 mol Fe
mol Fe2O 3
2
g Fe2O 3
159.6
100 / 159.6 x 2 x 55.8
69.9
g Fe
40
g Fe
1 mol Fe2O 3
mol Fe
4
55.8
100.0 g Fe2O 3 x _________ x
________ x ________
mol Fe2O 3
1 mol Fe
2
g Fe2O 3
159.6
100 / 159.6 x 2 x 55.8
69.9
g Fe