STOICHIOMETRY

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STOICHIOMETRY

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Title: STOICHIOMETRY

1
Stoichiometric Calculations Stoicheion
(element) metron (measurement) Relating mass or
moles of reactants in a chemical reaction to mass
or moles of products.
2
USING EQUATIONS

Nearly everything we use is manufactured from
chemicals.
Soaps, shampoos, conditioners, cds, cosmetics,
medications, and clothes.
For a manufacturer to make a profit the cost of
making any of these items cant be more than the
money paid for them.
Chemical processes carried out in industry must
be economical, this is where balanced equations
help.

3
USING EQUATIONS

Equations are a chemists recipe.
Equations tell chemists what amounts of reactants
to mix and what amounts of products to expect.
When you know the quantity of one substance in a
reaction, you can calculate the quantity of any
other substance consumed or created in the
reaction.
Quantity meaning the amount of a substance in
grams, liters, molecules, or moles.

4
USING EQUATIONS

When you bake cookies you probably use a recipe.
A cookie recipe tells you the amounts of
ingredients to mix together to make a certain
number of cookies.
If you need more cookies than the yield of the
recipe, the amounts of ingredients can be doubled
or tripled.

5
USING EQUATIONS

In a way, a cookie recipe provides the same kind
of information that a balanced chemical equation
provides
Ingredients are the reactants
Cookies are the products.

6
USING EQUATIONS

Imagine you are in charge of manu-facturing for a
Bicycle Company.
The business plan for Rugged Rider requires the
production of 128 custom-made bikes each day.
One of your responsibilities is to be sure that
there are enough parts available at the start of
each day.

7
USING EQUATIONS

Assume that the major components of the bike are
the frame (F), the seat (S), the wheels (W), the
handlebars (H), and the pedals (P).
The finished bike has a formula of FSW2HP2.
The balanced equation for the production of 1
bike is.

F S2WH2P ? FSW2HP2
8
(No Transcript)
9
USING EQUATIONS

Now in a 5 day workweek, the bike company is
scheduled to make 640 bikes. How many wheels
should be in the plant on Monday morning to make
these bikes?
What do we know?
Number of bikes 640 bikes
1 FSW2HP22W
What is unknown?
of wheels ? wheels

The connection between wheels and bikes is 2
wheels per bike. We can use this information as
a conversion factor to do the calculation.

2 W
640 FSW2HP2
1 FSW2HP2
11

A CONVERSION FACTOR IS SIMPLY A THIS PER THAT
EQUATION
FOR EVERY TWO WHEELS (THIS) YOU GET ONE BIKE
(THAT)
ANOTHER WAY TO THINK OF A CONVERSION WOULD BE
FOR EVERY 12 INCHES YOU GET 1 FOOT
1 FOOT
12 INCHES

Another way to think of a conversion would be a
chemical equation.
2 H2 (g) O2 (g) ? 2 H20 (l)
This is the only recipe that makes water.
You need 2 moles of H2 and 1 mole of O2 to make 2
moles of water. As usual the 1 is understood with
coefficients and of atoms.
2 mol H2 2 mol H20
As a conversion you would write
2mol H2 or 1 mol H2O
1 mol H2O 2 mol H2

You can also use the reactant oxygen in the
formula to set up a conversion.
Try that at your seat.
If you wrote
1 mol O2 2 mol H2O
1 mol O2 or 2 mol H2O
2mol H2O 1 mol O2
Good Job!
You just did a stoichiometric conversion!
Big word, but you did it!

Take the following sentence
Nitrogen gas and hydrogen gas react to form
ammonia gas.
It would read in a chemical equation like this
N2(g) H2(g) ? NH3(g)
Is it balanced?
no

In order for it to be balanced, the first step
would be change the coefficients. A 2 in front of
the product will give the correct number of
nitrogen atoms.
Placing the coefficient 3 in front of the number
of hydrogen atoms in the reaction will balance
the number of hydrogen atoms.

N2(g) 3H2(g) ? 2NH3(g)

What can we learn from this equation?

INTERPRETING CHEMICAL EQUATIONS
Open up your text book to page 239 and look at
figure 9.4.
You can interpret this balanced equation in two
ways
(1.) the number of particles (1 molecule of
nitrogen will react with 3 molecules of hydrogen
to produce 2 molecules of ammonia.)
(2.) or by the number of moles

The coefficients of the balanced chemical
equation indicate the numbers of moles of
reactants and products in a chemical reaction.
1 mole of N2 reacts with 3 moles of H2 to produce
2 moles of NH3.
N2 and H2 will always react to form ammonia in
this 132 ratio of moles.
N2(g) 3H2(g) ? 2NH3(g)

Using the coefficients, to make connections
between reactants and products, is the most
important information that a chemical equation
provides.

Using the formula for water provided you with 2
possible conversions if you wanted to work with
the molar ratio (how many moles of H2 and moles
of O2 produce how many moles of H2O)
2 H2 (g) O2 (g) ? 2 H20 (l)
2 H2 2 H2O and 1 O2 2 H2O
2H2 1 O2
2H2O 2 H2O
What two conversions can you use for this
formula?
N2(g) 3H2(g) ? 2NH3(g)

N2(g) 3H2(g) ? 2NH3(g)
1 N2 2 NH3
1 N2
2 NH3
and
3 H2 2 NH3
3 H2
2 NH3

The mass of 1 mol of N2 molecules is 28 g the
mass of 3 mols of H2 molecules is 6 g for a
total mass of reactants of 34 g.
The mass of 2 moles of NH3 molecules is 2 X 17g
or 34 g.
Reactants mass must products mass

Review
Where did I get those values?
1 mol N 14.0 g there are 2 moles28.0 g
There is 1 mole of N2 so 1 X 28.0 g 28.0 g
1mol H 1.0g there are 2 moles 2 g
The coefficient tells you there are 3 X 2g
3 X 2 6g in all
For NH3
14.0g3g 17.0g there are 2 mol 34.0 g

N2(g) 3H2(g) ? 2NH3(g)
23

N2(g) 3H2(g) ? 2NH3(g)
28.0g 6.0 g 34.0g 34.0 g
mass of reactants mass of products
The law of conservation of mass is obeyed.
Note the law of conservation of mass is always
obeyed.
A different mass would mean that the equation was
not properly balanced.

Using molar information, you can calculate
amounts of reactants and products in a number of
ways.
For example
You can make a similar connections with volume of
a gas at STP .
What is that relationship?

1 mole of a gas at STP is equal to 22.4 L
This is referred to as molar volume
That means, that a mole of any gas will occupy
that amount of space as long as the temperature
and pressure in the environment is at STP.

It also means that if you counted how many
molecules are present when you have a mole of
that gas it would add up to
6.02 X 1023 molecules

The following slide is similar to the picture on
page 239 of your text book.
You can open to that page while we talk about the
different relationships.

28
22.4 L
29

Remember that 1 mole of any gas at STP occupies
22.4 L of space.
It follows that 22.4 L of N2 reacts with 67.2 L
of H2 to form 44.8 L of Ammonia gas. (notice
volumes dont balance)

Unlike mass and atoms, - molecules, formula
units, moles, and volumes of gases will not
necessarily be conserved- although in some
situations they may.
For example in the formation of hydrogen iodide
all are conserved
H2(g) I2 (g) ? 2HI (g)

NOTE VERY IMPORTANT CONCEPT
Only mass and atoms are conserved in every
chemical reaction.

Using the coefficients of balanced reaction
equations and our knowledge of mole conversions
we can perform powerful calculations. A.K.A.
stoichiometry.
A balanced equation is essential for all
calculations involving amounts of reactants and
products.

If you know the number of moles of 1 substance,
the balanced equation allows you to calculate the
number of moles of all other substances in the
equation.

34
9.2 MOLE MOLE CALCULATIONS

The following reaction shows the synthesis of
aluminum oxide.

3O2(g) 4Al(s) ? 2Al2O3(s)
3O2(g) 4Al(s) ? 2Al2O3(s)

How many moles of aluminum are needed to form 3.7
mol Al2O3?

Given 3.7 moles of Al2O3
Uknown ____ moles of Al
35

Solve for the unknown

3O2(g) 4Al(s) ? 2Al2O3(s)
4 mol Al
3.7 mol Al2O3
7.4 mol Al
2 mol Al2O3
36

Mathmatically, the 7.4 mol of Al was solved by
the following
3.7 mol X 4 mol 14.8 mol
That number is then divided by 2
14.8 2 7.4 mol.
The given, (3.7) is always first multiplied by
the numerator (4) and then divided by the divisor
(2)
You always put the given first when solving
stoiciometric problems)
The units you need to end up with should be in
the numerator

3O2(g) 4Al(s) ? 2Al2O3(s)
Suppose you were asked how many moles of Al2O3
would be produced with 4 mol of O2?
What is the given?
The given is 4 mol O2
What do you want to find out?
mol of Al2O3

3O2(g) 4Al(s) ? 2Al2O3(s)
4 mol O2 X 2 mol Al2O3 2.6 mol Al2O3
3 mol O2

IMPORTANT NOTE
MOLE TO MOLE CALCULATIONS DO NOT REQUIRE YOU TO
REFER TO THE PERIODIC TABLE.
YOU SIMPLY SET UP THE PROBLEM BY LOOKING AT THE
MOLAR RATIO OF THE BALANCED CHEMICAL EQUATION

3O2(g) 4Al(s) ? 2Al2O3(s)
If asked how many moles of O2 are required to
produce 5 moles of Al2O3?
Mol of given X mol O2
mol of Al2O3
No need to determine molar mass
How would you set this up?

3O2(g) 4Al(s) ? 2Al2O3(s)
5 mol Al2O3 X 3 O2
2 Al2O3
7.5 mol O2

42
MASS MASS CALCULATIONS

No lab balance measures moles directly, instead
the mass of a substance is usually measured.
From the mass of 1 reactant or
product, the mass of any other reactant or
product in a given chemical equation can be
calculated
As in mole-mole calculations, the unknown can be
either a reactant or a product.

In mass to mass conversions you use the molar
mass ratio of what 1 mol of a compound is
(periodic table)
For example the ratio conversion for H2O
from mol to H2O to g H2O would be
1 mol H2O 18.0 g
1 mol H2O
18.0 g H2O
Where did I get 18.0 g ?
2 X 1.0g 16.0 g 18.0 g

3O2(g) 4Al(s) ? 2Al2O3(s)
NOTE The molar mass is calculated by
(1.) determine what 1 mole is equal to first
(2.) then multiply that number by the coefficient
Example using the above
3O2
(1.) O2 16.0 g X 2 since there are 2 atoms of
oxygen in the molecule
(2.) 3 X 32.0g 96.0 g

45
MASS MASS CALCULATIONS
Acetylene gas (C2H2) is produced by adding water
to calcium carbide (CaC2).
CaC2 2H2O ? C2H2 Ca(OH)2
How many grams of C2H2 are produced by adding
water to 5.00 g CaC2?
46
CaC2 2H2O ? C2H2 Ca(OH)2

What do we know?
Given mass 5.0 g CaC2
Mole ratio 1 mol CaC2 1 mol C2H2
Molar mass of CaC2 64.0 g CaC2
How do we know that?
By calculating molar mass
Molar mass of C2H2 26.0g C2H2
What do we want to know?
grams of C2H2 produced

47
5.0CaC2
26.0 g C2H2
2.03 g C2H2
64.0 g CaC2
48

It can also be set up like this

5.0 g CaC2
1 mol CaC2
1 mol C2H2
64.0 g CaC2
1mol CaC2
26.0 g C2H2
1mol C2H2
2.03 g C2H2
49

Answer to 19
a. 15.3 mol O2
b. 10.2 mol CO2, 13.6 mol H2O
Solve for a.
3.40 mol C3H7OH X 9 mol O2
2 mol C3H7OH
15.3 mol O2
Solve for b
3.4 mol C3H7OH X6 molCO2 10.2 mol CO2
2 mol C3H7OH

For mol H2O
3.40 C3H7OH X 8 mol H2O
2 C3H7OH
13.6 mol H2O
Now try 20 0n page 250

Open to page 250
At your seat,
work on problem 20 a
and 21. a, b, and c
For 21 convert every reactant and product to
molar mass

52
PRACTICE TIME Problem 20 a on and 21 a, b, c on
page 250
53
16.0 g CH4 1 mol CH4
54

21.
a. 176 g CO2, 36.0 g H2O
2C2H2(g) 5O2(g) ? 4CO2(g) 2H20(g)
52 g 160g ? 176g
36 g
b. 160g 0f O2
c. 212 g total of reactants 212 g total of
products

A balanced reaction equation indicates the
relative numbers of moles of reactants and
products.
We can expand our stoichiometric calculations to
include any unit of measure that is related to
the mole.
The given quantity can be expressed in numbers of
particles, units of mass, or volumes of gases at
STP.
The problems can include mass-volume,
volume-volume, and particle-mass calculations.

Sn(s) 2HF(g) ? SnF2(s) H2(g)
What does this balanced equation tell me in
words?
Tin reacts with hydrogen fluoride gas to yield
Tin(II)fluoride and hydrogen gas.

Sn(s) 2HF(g) ? SnF2(s) H2(g)
What does it tell me in terms of moles
1 mol Sn 2 mol HF ? 1 mol SnF2 1 mol H2
1 mol Sn 1 mol SnF2
1 mol Sn 1 mol H2
2 mol HF 1 mol SnF2
2 mol HF 1 mol H2
1 mol Sn 2 mol HF (for reactants ratio)
1 mol SnF2 1 mol H2 (for product ratio)
There are 3 mol of reactants and 2 mol of products

Sn(s) 2HF(g) ? SnF2(s) H2(g)
What does it tell me in terms of volume at STP?
22.4 L Sn 1 mol SnF2
22.4 L Sn 22.4 L H2
44.8 L HF 1 mol SnF2
44.8 L HF 22.4 L H2
22.4 L Sn 44.8 L HF for reactant ratio
1 mol SnF2 22.4 L H2 for product ratio
67.2 L(22.4 44.8) on reactant side 22.4 L on
product side

Sn(s) 2HF(g) ? SnF2(s) H2(g)
In terms of mass
118.7 g Sn 40.0 g HF ? 156.7 SnF2 2.0g H2
118.7 g Sn 156.7 g SnF2
118.7 g Sn 2.0 g H2
40.0 g HF 156.7 g SnF2
40.0 g HF 2.0 g H2
158.7 g reactants(118.7 40.0) 158.7 g
products(156.7 2.0)

Sn(s) 2HF(g) ? SnF2(s) H2(g)
In terms of of molecules
. 6.02 X 1023 molecules Sn 6.02 X 1023
molecules SnF2
6.02 X 1023 molecules Sn 6.02 X 1023 molecules
H2
1.2 X 1024 molecules HF 6.02 X 1023 molecules
SnF2
1.2 X 1024 molecules HF 6.02 X 1023 molecules H2

Now try 20 b and c on page 250

62
PRACTICE TIME Problem 20 b and c on page 250
63

b. 1 mol CH4
22.4 L CH4
c. 1 mol CH4
6.02 X 1023 molecules CH4

Sn(s) 2HF(g) ? SnF2(s) H2(g)
How many liters of H2 will be produced with 12.0
L HF?
Given 12.0 L HF
Unknown ? L of H2
Conversion 22.4 L Sn / 22.4 L H2
12.0 L Sn X 22.4 L H2 6.0 L H2
44.8 L HF

Now try 22 a on page 250
at your seat

66
22.a Sn(s) 2HF(g) ? SnF2(s) H2(g) 9.40
L H2 X 44.8 L HF 18.8 L HF
22.4 L H2
67

22 b would require you to combine two different
equalities
44.8 L HF 6.02 X 1023 molecules H2
Given 20.L HF
? How many molecules of H2
20.0 L HF X 6.02 X 1023 molecules of H2
44.8 L HF
2.69 X 1023 molecules H2

c.
7.42 X 1024 molecules of HF
X 156.7 X 100 g SnF2
1.2 X 1024 molecules
of HF
969 g SnF2

NOTE
Mole to mole and volume to volume problems do not
require you to refer to the periodic table.
For mol to mol look at coefficients
For volume to volume(STP) know that 1 mol is
equal to 22.4 L
The coefficient tells you what to multiply 22.4 L
by
No coefficient always means 1 so you multiply
22.4 L by 1

OPEN UP TEXT TO PAGE 262 PROBLEM 37
2.7 mol C X 1mol CS2
5 mol C
.54 mol CS2
5.44 mol SO2 X 5 mol C
2 mol SO2
13.6 mol C

c. .246 mol CS2 X 4 mol CO
1 mol CS2
.984 mol
End of chapter 9 for period 2

In any of these problems, the given quantity is
first converted to moles.
Then the mole ratio from the balanced equation is
used to convert from the moles of given to the
number of moles of the unknown
Then the moles of the unknown are converted to
the units that the problem requests.
The next slide summarizes these steps for all
typical stoichiometric problems

20.
a. 16.0 g CH4
1 mol CH4
b. 1 mol CH4
22.4 L CH4
c. 1 mol CH4
6.02 X 1023 molecules CH4
Try 21 on page 250 hint convert reactants and
products to molar mass(g)

74
Limiting Reactants Suppose you do not
have enough of one reactant to carry the
reaction to completion? To determine the
moles or masses of products, we
need to know which reactant will run
out first and which is in excess. We
cannot use coefficients or masses to make the
determination. As usual, everything is
measured in moles.
75
Limiting Reactants 1. Write the balanced
equation. 2. Convert masses to moles. 3. The
reactant producing the lowest moles of product
will be consumed (limiting reactant).
76
Take the following balanced chemical
equation Zn(s) 2HCl(aq) ? ZnCl2(aq)
H2(g) If 0.30 mol Zn is added to 0.52 mol HCl,
how many moles of H2 will be produced?
77
Zn(s) 2HCl(aq) ? ZnCl2(aq) H2(g)
0.30 mol Zn x 1 mol H2 0.30 mol H2
1 mol Zn 0.52 mol HCl x 1
mol H2 0.26 mol H2 2 mol
HCl The mole ratio conversions come directly
from coefficients in the balanced equation.
0.26 mol is the maximum amount of H2
produced when 0.30 mol of Zn and 0.52 mol
of HCl are mixed.
78
The HCL is the limiting reagent and the Zn will
be in excess. The limiting reagent determines the
amount of product possible for a reaction and is
used up entirely during a reaction. Excess
reagent will be what was not used up entirely
during the reaction
79

What is the limiting reagent when 4.3 mol of SO2
react with 2.9 mol O2 according to the equation
2 SO2(g) O2(g) ? 2SO3(g)
Solution
4.3 mol SO2 X 2 mol SO3 4.3 mol SO3
2 mol SO2
2.9 mol O2 X 2 mol SO3 5.8 mol SO3
1 mol O2
SO2 is the limiting reagent O2 will be in excess

Step two
Solve for percent yield by using the following
formula
Percent yield actual yield X 100
theoretical yield
4.65 g of Cu X 100
6.61 g of Cu
70.5

OPEN UP YOUR TEXT AND WORK ON PROBLEM 32 ON PAGE
259
32 SOLUTION
step one compare weight ratio of reactants in
the balanced equation
the weight ratio is 480 g to 480 g 11
The available O2 is the limiting reagent and
FeS2 will be in excess, only 16 g will react.

82
How much of the reactant O2 will be in
excess? Solution First solve for how much O2
will react with 4.3 mol SO2 4.3 mol SO2 X 1 mol
O2 2.2 mol O2 2 mol
SO2 5.8mol - 2.2mol 3.6 mol O2 in excess
83
Period 5 only, period 7 and 8 will get to this
after the test
84

1. 2Na(s) Cl2 ?2NaCl(s),
If 7.4 mol Na and 3.2 mol HCl are allowed to
react,
a. what is the limiting reagent?
b. How many moles of NaCl are produced?
c. How many moles of excess reagent remain?
2. C0(g) 2H2(g) ? CH3OH(g)
CH3OH(g) is called mehtanol
Interpret the above equation using the following
1. name, including physical properties. 2. volume
at STP 3. moles ratio 4. molar mass 5. any other
combination

85
Percent Yield In most reactions not all reactants
go forward to form products. Some are lost to
byproducts and some do not react. The actual
yield is usually less than the anticipated,
theoretical yield. Percent yield actual
yield x 100 theoretical yield
86
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
87
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
88
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
89

Calcium carbonate is decomposed by heating, as
shown in the following equation
CaCO3(s)? CaO(s) CO2
What is the theoretical yield of 24.8 g CaCo3 is
heated?
B. What is the percent yield if 13.0 g CaO is
produced?

90
knowns Mass of CaCO3 24.8g CaCO3 1mol CaCO3
1mol CaO 1 mol CaCO3 100.1 g CaCO3 (molar
mass) 1 mol CaO 56.1 g (molar mass) Unknowns
theoretical yield CaO with using 24.8 g
CaCO3 mass A ? mol A ? mol B ? mass B 24.8 g
CaCO3 X 1 mol CaCO3 X 1mol CaO X
56.1 g CaO 100.1 g
CaCO3 1 mol CaCO3 1 mol CaO
91
13 .9 g CaO unknowns for b is percent
yield KnownActual yield 13.0 g
CaO KnownTheoretical yield 13.9 g CaO Percent
yield actual X 100
theoretical 13.1 g CaO X 100
94.2 13.9 g CaO
92

OPEN UP YOUR TEXT AND WORK ON PROBLEM 30 ON PAGE
259
SOLUTION
Step 1 determine theoretical yield
from the balanced chemical equation
Solution 1.87g of Al X 190.62 g of CuSO4
53.96 g of Al
6.61 g CuSO4

20.0 g FeS2 X 640.5 g of SO3
480g of FeS2
26.7 g SO3
16 g O2 X 640.5 g of SO3
480g of O2
21.4 g SO3

6.6 g of Cu produced (theoretical yield)
Where did I get my gram amounts?
What do I know from the balanced equation?
That 2 moles of Aluminum will equal 3 Moles of
Copper.
2 moles of Aluminum53.96 g (2 X 23.96)
3 moles of Copper 190.62 g (3 X 63.54)
Use that ratio to determine the expected yield
for only 1.87 g of aluminum.

29.
A limiting reagent determines the amount of
product that can be produced from a reaction.
Excess reagent is not used up by a reaction.
30. Step 1 determine theoretical yield
Solution 1.87g of Al X 190.62 g of Cu
53.96 g of Al

6.6 g of Cu produced (theoretical yield)
Where did I get my gram amounts?
What do I know from the balanced equation?
That 2 moles of Aluminum will equal 3 Moles of
Copper.
2 moles of Aluminum53.96 g (2 X 23.96)
3 moles of Copper 190.62 g (3 X 63.54)
Use that ratio to determine the expected yield
for only 1.87 g of aluminum.

97
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
98

Step two
Solve for percent yield by using the following
formula
Percent yield actual yield X 100
theoretical yield
4.65 g of Cu X 100
6.6 g of Cu
70.5

32.
step one compare weight ratio of reactants in
the balanced equation
the weight ratio is 480 g to 480 g 11
The available O2 is the limiting reagent and
FeS2 will be in excess, only 16 g will react.
640.5 g of SO3 are produced with 480 g of each
reactant

100
16.0 g O2 X 640.5 g of SO3
480 g of O2 21.4 g SO3 20.0 g FeS2 X
640.5 g of SO3 480 g of
FeS2 26.7 g SO3
101
Grams of A
Use molar mass as a conversion factor
Moles of A
Use coefficients from balanced equation as a
conversion factor
Moles of B
Use molar mass as a conversion factor
Grams of B
102

SOLUTIONS TO WORKSHEET

103

If you start with 10 grams of lithium hydroxide,
how many grams of lithium bromide will be
produced?
LiOH HBr ? LiBr H2O
What do you know?
1mol LiOH 23.9g ( gfm)
1 mol LiOH 1 mol LiBr
1 mol LiBr 86.8 g (gfm)

104