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K8: Intensified Unit K Reading Notes
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UNIT K READING NOTES
K-1 Reading

Substances that exist before a chemical change are called reactants.

Substances that are formed during a chemical change are called products.

A chemical equation indicates the reactants and products of a reaction.

A word equation describes a chemical change using the names of the reactants and products:
Silver + sulfur  silver sulfide



reactants
product
would read: “Silver reacts with sulfur to yield (or
to produce) silver sulfide.”
A word equation is helpful for telling us what is reacting to form products, but it doesn’t tell us how much
stuff reacts, or how much stuff is produced. That is, it is qualitative. In order to get information about how
much stuff is being reacted and produced, we need a quantitative relationship. For that we need something
called formula equations.
A chemical change is usually indicated if one of the following is observed:

Formation of a gas,

Formation of a solid product (a precipitate),

A color change, or

Release or absorption of energy in the form of heat, light, and/or electricity.
Some of the above observations could also indicate the occurrence of a physical change. In order to prove
the occurrence of a chemical change, a chemical analysis showing evidence of the formation of a substance
that is different from the reactants would be required. You should go the Holt Chemistry website on
Blackboard, and check out the Visual Concept for “Signs of a Chemical Reaction.”
Showing Energy Changes in Equations
Endothermic reactions are ones in which energy must be added in order for the reaction to occur. We
show this by writing the following: 2H2O + energy  2 H2 + O2 or
Exothermic reactions are ones that release energy when they occur. We show this by writing the
following: 2 H2 + O2  2H2O + energy
The energy that is involved can be in many different forms—energy involved in endothermic reactions is
usually thermal energy (heat) or electricity. Exothermic reactions usually involve the release of thermal
energy or light.
We know that we can write a word equation for the synthesis of water from hydrogen and oxygen:
Hydrogen gas + Oxygen gas  water
This can also be written as
H2 + O2  H2O
There is a phrase that you have definitely heard of before: “Matter is neither created nor destroyed.” This is
actually the result of work done by a guy named Lavoisier back in the early 1800’s, and it’s called the Law
of Conservation of Matter (or Mass).
Anyhow, if we write this equation in the form of particle diagrams, we can see whether the Law of
Conservation has been obeyed:
H2
+
O2

H2O
K8: Intensified Unit K Reading Notes
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In order to conserve matter this is what must actually be the case in a reaction:
2H2
+
O2

2H2O
+
The numbers in front of the chemical formulas are called coefficients. They tell the relative number of
molecules (or formula units) taking part in a chemical reaction. Just like empirical formulas coefficients
are the smallest whole numbers that satisfy the Law of Conservation of Mass. Chemical reaction
equations are balanced when the number of atoms of each element as a reactant is equal to the
number of atoms of that element as a product. THAT MEANS THAT THE NUMBER OF ATOMS TO
THE LEFT OF THE ARROW MUST EQUAL THE NUMBER OF ATOMS TO THE RIGHT OF THE
ARROW.
Showing Phases in Chemical Equations
Substances may exist in the form of a solid, a liquid, or a gas. These forms of matter are called physical
phases or states.

In the solid phase, a sample of a substance is relatively rigid and has a definite volume and shape. We
denote a solid in a chemical reaction equation by writing (s) after the chemical formula for each solid
substance.

In the liquid phase, a sample of a substance has a definite volume, but it is able to change its shape by
flowing. Under the action of gravity, a liquid will take the shape of its container. We denote a liquid in
a chemical reaction equation by writing (l) after the chemical formula for each liquid substance.

In the gaseous phase, a sample of a substance has no definite volume, and it shows very little response
to gravity. If it is not in a container, it spreads out indefinitely. If confined within a closed container, it
fills the container, but will escape through any opening. We denote a gas in a chemical reaction
equation by writing (g) after the chemical formula for each gaseous substance.

Solids will often not react with each other in the solid state, so they are often dissolved in water to
form an aqueous solution. We use the notation (aq) to indicate this.
You will find other relevant symbols in Table 11.1 on page 323.
Balancing Chemical Equations

A balanced chemical equation tells the number of formula units or molecules of each substance that
reacts and is produced during a chemical reaction. Here are the steps for balancing an equation:
Step 1: Write a word equation for the reaction
Step 2: Write the correct formula for all reactants and products. Remember that the elements nitrogen,
oxygen, fluorine, chlorine, bromine, iodine, and hydrogen only appear as diatomic molecules
when they are by themselves (N2, O2, F2, Cl2, Br2, I2, and H2).
Step 3: Determine coefficients that make the equation balance. These coefficients should be used to make
sure that the number of atoms of an element on the left-hand-side of the arrow is equal to the
number of atoms to the right of the arrow. BECAUSE THE FORMULA OF A SUBSTANCE
REPRESENTS ITS IDENTITY, YOU CANNOT MESS WITH THE SUBSCRIPTS—ONLY
WITH THE COEFFICIENTS!
K8: Intensified Unit K Reading Notes
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Example:
Write the balanced equation for the reaction between aluminum sulfate and calcium chloride to produce
aluminum chloride and a white precipitate of calcium sulfate.
Step 1: Write the word equation:
aluminum
sulfate
+
calcium
chloride

aluminum
chloride
+
calcium sulfate
+
CaSO4
Step 2: Replace the words in the word equation with the correct formulas:
Al2(SO4)3
+
CaCl2

AlCl3
Step 3: Adjust the coefficients to make the equation balance (we can use particle diagrams to illustrate):
If: Al =
S=
Ca =
O=
Cl =

+
Al2(SO4)3
+
3 CaCl2

+
2 AlCl3
+
3 CaSO4
Another procedure for balancing equations could be the following:
1.
Balance the atoms of elements that are combined and that appear only once on each side of the
equation.
2.
Balance polyatomic ions as single units when they appear on both sides of the equation.
3.
Balance H atoms and O atoms after atoms of all other elements have been balanced.
4.
Once a chemical formula is written you must change the number of atoms of any element only
with a coefficient, NEVER BY CHANGING THE SUBSCRIPTS!!!!!!!!!!!!!!!!!!
K-2 Reading
Classifying Chemical Reactions, Predicting Products, and Determining if they Occur
You can classify chemical reactions by the types of substances that react and are produced. If we look at
whether the reactants and products are compounds or elements, most, but not all, chemical processes can
be classified into one of five categories:
K8: Intensified Unit K Reading Notes
1.
2.
3.
4.
Page 4 of 12
Combination, or synthesis:
element or
compound
+
element or
compound

compound
A
+
B

AB
Decomposition, or analysis:
compound

AB

two or more elements or compounds
A
+
B
Single displacement (or replacement):
element
+
compound

element
+
compound
A
+
BC

B
+
AC
Double displacement (or replacement), or exchange of ions:
compound
+
compound

compound
+
compound
AB
+
CD

AD
+
CB
+
oxygen gas
(O2)
+
energy (and
possibly another
compound like
CO2)
5) Combustion:
H2 or a
hydrocarbon

water (usually)
Direct Combination or Synthesis Reactions
In this type of reaction, two or more substances combine to produce a single, more complex substance. The
reactants are molecular substances that may be elements, compounds, or both. Examples follow:
2H2 (g) + O2 (g)  2H2O (l)
2CO (g) + O2 (g)  2CO2 (g)
CaO (s) + H2O (l)  Ca(OH)2 (g)
Decomposition or Analysis Reactions
In this type of reaction, a single substance is broken down into two or more simpler substances. These
simpler substances may be either elements or compounds. Most decomposition reactions are
endothermic—that is they require energy, usually in the form of heat or electricity. Sometimes they can
be sped up by using a catalyst—a substance that speeds up a reaction without itself being permanently
changed. Examples follow:
electricity , dilute_ H SO

4

 2 H2 (g)+ O2 (g): The decomposition of water using electricity and a
2H2O (l)     2
dilute sulfuric acid catalyst.

 2 Hg (l) + O2 (g): The decomposition of Mercury (II) oxide by heat. The Greek letter, ,
2HgO (s) 
over the arrow shows that the reactant is heated.
K8: Intensified Unit K Reading Notes
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Single Displacement (Replaeement) Reactions
In this type of reaction, a free element becomes an ion, and an ion in solution becomes a neutral atom.
Examples:
Zn (s)
+
H2SO4 (aq)


Zinc replaces hydrogen

ZnSO4 (aq)
+
H2 (g)
Cu (s)
+
2AgNO3 (aq)


Copper replaces silver

Cu(NO3)2 (aq)
+
2 Ag (s)
Cl2 (g)
+
2 NaBr (aq)


Chlorine replaces bromine

2 NaCl (aq)
+
Br2 (l)
It is important to realize that all of these reactions involve the simultaneous occurrence of the following
type of ionization reactions:
Cu Cu2+ + 2 electrons
and
2 Ag+ +2 electrons  2 Ag
This is because ionic compounds tend to separate into charged particles when they are in water. The
different particles formed from AgNO3, for example would be:
AgNO3 = Ag+ + (NO3)1One could now ask if the reverse reaction can occur. We find our answer in a property called the activity of
an element. The process of losing electrons is called oxidation, while the process of gaining electrons is
called reduction. A good way to remember this is the memory device LEO goes GER (Lose Electrons
Oxidized, Gain Electrons Reduced).
The tendency of an element to give up its electrons to reduce the cation of another element is called its
activity. An element that is more active will give up its electrons (be oxidized) to a less active species so
that it can be reduced to its elemental form. Table 11.2 on page 333 shows the relative activities of some
common metal elements, and a more detailed chart of the relative activities of halogens appears in the
activity series found in you Unit K Outline Packet. The relative activities of copper and silver explain why
the following single replacement reaction cannot occur.
2 Ag (s)
+
Cu(NO3)2 (aq)



Silver replaces copper?
Cu(s)  2 AgNO3 (aq)

Does _ not _ occur
Double Displacement (Replacement) Reactions
This type of reaction occurs between two ionic compounds that are dissolved in water. The cation of one
compound replaces the cation in the other compound to produce two new compounds:
AgNO3 (aq)

+
NaCl (aq)


AgCl (s)
+
NaNO3 (aq)

NaCl (aq)
+
H2O (l)
Ag+ replaces Na+
NaOH (aq)
+
HCl (aq)
K8: Intensified Unit K Reading Notes

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
Na+ replaces H+
Like a single replacement reaction, a double replacement reactions must satisfy certain conditions in
order for it to occur. One of the following must be true:
1.
A cation of one reactant is insoluble in the presence of the anion of the other reactant. Insoluble
means that you can’t dissolve something in water. If you look at the solubility table that is in your
Unit K Outline Packet, you will see on the chart that Ag+ is not soluble in water in the presence of
chloride ion (Cl1-). Your text provides solubility rules in the Table 11.3 on page 344. This is why
silver chloride, a white solid, is formed in the first of the above reactions. When a solid is formed from
a physical or chemical change in a liquid or gas phase, the solid is called a precipitate. This means
that if a solid is predicted to form in a double replacement reaction then the reaction will occur.
2.
A double replacement reaction will occur if hydroxide anion (OH 1-) from one reactant combines with
hydrogen cation (H+) from the other reactant to form water. This can be seen in the second of the
above reactions.
3.
Finally, if a gas is predicted to form during a double replacement reaction, then the reaction will occur.
An example of this is the following:
FeS (s)
+
2 HCl (aq)
FeCl2 (aq)
+
H2S (g)

K-3 Reading
Let us begin with defining stoichiometry. Most texts break it down into 2 categories. The first one we’ve
already seen—it’s composition stoichiometry, which is the study of mass relationships of elements in
compounds. Examples of this include calculating percentage composition, and determination of empirical
and molecular formulas. The second one is reaction stoichiometry, which deals with the mass, mole,
particle, and volume relationships between reactants and products in a chemical reaction. This means that
we will always need to have a balanced chemical reaction equation, such as those we worked with earlier in
the unit, before we can proceed with establishing mass relationships between reactants and products.
The text goes into the different types of reaction stoichiometry problems you can expect to see in terms of
the information that is known, and the information that is unknown. What I want you take away from this
unit is that a balanced chemical reaction equation is an extremely powerful tool that will permit you to
calculate mass, particle, and volume information.
Before we discuss the mole ratios, we need to discuss STP and Avogadro’s Hypothesis (Law). As we
discussed in Unit J, chemists define standard temperature as T = 0C (Celcius) = 273 K (Kelvins), and
standard pressure as any of the following equivalent values: P = 101.3 kPa = 14.7 psi ( lb2 ) = 1 atmosphere
in
(atm) = 760 mm Hg. Standard temperature and pressure together are designated as STP. You don’t have to
convert between these values at this time, but you need to be able to recognize that they are standard
temperature and pressure. Avogadro’s Hypothesis (or Law) states that equal volumes of any two gas
samples held at the same temperature and pressure will have the same number of gas particles (atoms if
they are ideal gases, molecules if they are any other gases) in those samples. If a mole of any gas is held at
STP conditions, its volume can be estimated with reasonable accuracy to be 22.4 dm3 (L). We can use this
knowledge to solve problems involving moles of gases and volume:
EXAMPLE: What is the volume of 3.30 mol of a gas sample held at STP?
KNOWN: 1 mol gas @ STP = 22.4 L
SET-UP & SOLUTION:

3.30_ mol
1

22.4 _ L _ gas
1_ mol
UNKNOWN: 3.30 mol gas ? L gas
 73.9 L gas
The text discusses mole ratios as conversion factors that relate the amounts in moles of any two substances
involved in a chemical reaction. As we discuss in class, there is so much more information that you can
take away from a balanced chemical equation, so let me go through the whole thought process with you. If
you are given
a balanced chemical reaction equation, there are three pieces of information that can be
related directly to the coefficients in that equation:
K8: Intensified Unit K Reading Notes
Page 7 of 12
3 H2
+
N2

2 NH3
Information that can come directly from the
coeficients
3 molecules H2
=
1 molecule N2
=
2 molecules NH3
relative # of particles of reactants and
products
3 mol H2
=
1 mol N2
=
2 mol NH3
relative # of moles of reactants and products
3 dm3 (L) H2
=
1 dm3 (L) N2
=
2 dm3 (L) NH3
relative volumes of gas phase reactants and
products at the same T and P
You will notice that the values in each row are considered to be equal to each other, because 3 molecules of
H2 will always require 1 molecule of N2 to produce 3 molecules of NH3. The relative volume information
in the third equivalency comes straight from Avogadro’s Law, and we can use it to answer questions like
the following:
What volume of hydrogen gas is required to produce 13.7 L NH 3 if all reactants and products are held at
the same temperature and pressure?
13.7 L NH3  ? L H2

13.7 _ L _ NH 3
1


3 _ L _ H2
2 _ L _ NH 3
 20.6 L H required to produce 13.7 L NH .
2
3
The same type of calculation could be done for relative numbers of moles or relative numbers of particles.
IT’S IMPORTANT TO NOTE THAT THE ONLY PIECE OF INFORMATION THAT YOU CANNOT
TAKE directly FROM THE COEFFICIENTS IN A PERIODIC TABLE IS THE RELATIVE MASSES
OF REACTANTS AND PRODUCTS. We can, however, use the relative mole values to determine the
relationships between masses, moles, volumes @ STP, and # of particles. We will call these relationships
between masses, moles, volumes @ STP, and the number of particles the recipe, and we will use it as the
universal basis for solving all of the problems in this unit in a manner that is much simpler than the
methods outlined in the text.
Let us first do a mass calculation:
EXAMPLE: Determine the mass of NaCl that will decompose to yield 355 g Cl 2 (Na is the other product of
the reaction).
1) Set up a formula equation for the reaction and balance it:

2 NaCl
2 Na
+
Cl2
2) We know that a balanced chemical reaction equation yields information about the relative number of
moles that react ad are produced, so let’s start to write a recipe immediately beneath the equation:
2 NaCl

2 Na
+
Cl2
2 mol NaCl
2 mol Na
1 mol Cl2
relative #
moles
117.0 g NaCl
46.0 g Na
71.0 g Cl2
relative
masses
3) Leave room under the mole recipe, so that terms can be added as we calculate them. We can now find
the molar masses of each of the reactants and products that I have already put under the mole recipe:
NaCl:
Cl2
Na: 23.0 g
Cl: 35.5 g  2 = 71.0 g Cl2 = 1 mol Cl2
Cl: 35.5 g
58.5 g NaCl = 1 mol Na Cl
and
1 mol Na = 23.0 g Na
117.0 g NaCl = 2 mol NaCl
and
2 mol Na = 46.0 g Na
If we look in the recipe, we see that we don’t actually have one mole of either sodium chloride or sodium,
so we must multiply both sides of the molar mass equivalencies by two (2).
K8: Intensified Unit K Reading Notes
Page 8 of 12
Once the “relative masses” row is complete in the recipe, we can do a factor-label set-up and solution. You
should note that EACH ITEM IN THE RECIPE IS EQUIVALENT TO ALL OTHER ITEMS, SO THEY
CAN ALL APPEAR IN A CONVERSION FACTOR!
355 g Cl2  ? g NaCl
355gCl2  117.0gNaCl 

 
 585 g NaCl
 1
  71.0gCl2 

Now we can expand this concept to add particles and volumes at STP (remember that 1 mol gas = 22.4 L@
STP, and 1 mol = 6.02  1023 particles [atoms, F. U., or molecules]).
EXAMPLE:
Calcium carbonate (CaCO3) solid reacts with dilute (read aqueous) hydrochloric acid (HCl) to form carbon
dioxide, calcium chloride, and water.
A) Write the balanced chemical reaction equation, leaving room underneath for a large recipe:
CaCO3(s)
+
2 HCl(aq)

CO2(g)
+
CaCl2(aq)
+
H2O(l)
B) Do molar mass calculations, and build a recipe with rows for relative masses, volumes of gases @
STP, and number of particles. I will show the molar mass calcs first, and then re-write the equation
with the recipe beneath it:
CaCO3
Ca: 40.1 g  1 = 40.1 g
C: 12.0 g  1 = 12.0 g
O: 16.0 g  3 = 48.0 g
1 mol CaCO3 = 100.1 g CaCO3 = 6.02  1023 F.U CaCO3
HCl
H: 1.0 g  1 =
1.0 g
Cl: 35.5 g  1 = 35.5 g
1 mol HCl = 36.5 g HCl = 6.02  1023 molecules HCl
2 mol HCl = 73.0 g HCl = 1.204  1024 molecules HCl
CO2
C: 12.0 g  1 = 12.0 g
O: 16.0 g  2 = 32.0 g
1 mol CO2 = 44.0 g CO2 = 6.02  1023 molecules CO2 = 22.4 L CO2 @ STP
CaCl2
Ca: 40.1 g  1 = 40.1 g
Cl: 35.5 g  2 = 71.0 g
1 mol CaCl2 = 111.1 g CaCl2 = 6.02  1023 F.U CaCl2
H2O
H:
1.0 g  2 = 2.0 g
O: 16.0 g  1 = 16.0 g
1 mol HCl = 18.0 g H2O = 6.02  1023 molecules H2O
K8: Intensified Unit K Reading Notes
Page 9 of 12
Now, fill in the recipe:
CaCO3(s)
+
2 HCl(aq)

CO2(g)
+
CaCl2(aq)
+
H2O(l)
1 mol CaCO3
2 mol HCl
1 mol CO2
1 mol CaCl2
1 mol H2O
100.1 g CaCO3
73.0 g HCl
44.0 g CO2
111.1 g CaCl2
18.0 g H2O
1.204  1024
molecules
HCl
6.02  1023
molecules
CO2
6.02  1023
F.U CaCl2
6.02  1023
molecules
H2O
6.02  1023
F.U CaCO3
22.4 L CO2 @ STP
C) You can now answer the following questions, using the numbers in the recipe for conversion factors:
1.
What volume of CO2 will be produced @ STP when 80.0 g CaCO3 reacts?
80.0 g CaCO3  ? L CO2 @ STP
80.0 _ g _ CaCO3  22.4 _ L _ CO2 @STP 

 
 = 17.9 L CO2 @ STP will be produced

  100.1_ g _ CaCO3 
1
2.

How many molecules of HCl are needed to produce 50.0 L CO 2 @ STP?
50.0 L @ STP  ? molecules HCl
50.0 _ L _ CO2 @STP  1.204 10 24 molecules _ HCl 

 
 = 2.69  1024 molecules HCl needed

  22.4 _ L _ CO2 @STP
1

3.

80.0 _ g _ CaCO3  6.02 10 23 molecules _ CO2 

 
 = 4.80  1023 molecules are produced

 
1
100.1_ g _ CaCO3

4.

How many molecules of CO2 are produced from 80.0 g CaCO3?
What mass of CaCl2 is prodced when 50.0 L CO2 @ STP is also produced?
50.0 _ L _ CO2 @STP   111.1_ g _ CaCl2 

 
= 248 g CaCl2 (SF = 3)

 22.4 _ L _ CO2 @STP 
1
K-4 Reading: Limiting and Excess Reactants and Percent Yield

Sometimes you are given starting quantities of more than one reactant (none of our previous problems have
done that). This gives rise to a different type of stoichiometry problem in which one of the reactants is
completely consumed before the other one. When this occurs, the reaction must stop, and no more product
is made. We say that the reactant that is used up first “limits” the reaction, and we call that reactant the
limiting reactant. The reactant that is not used up is called the excess reactant. BECAUSE NO MORE
PRODUCT CAN BE MADE IF ONE OF THE REACTANTS RUNS OUT, ALL PRODUCT
CALCULATIONS MUST BE BASED ON THE LIMITING REACTANT ONLY.
K8: Intensified Unit K Reading Notes
Page 10 of 12
The hardest part to doing these types of problems is figuring out which reactant limits (runs out first). Once
you do that successfully, the calculations are the same as we did in the K-3 R discussion. Let’s look at how
we approach these problems:
A 5.0-gram strip of magnesium is ignited in a jar containing 0.50 L of oxygen gas @ STP to produce solid
magnesium oxide.
A) Which reactant limits the reaction?
B) What mass of solid magnesium oxide is formed by the reaction?
C) Determine how much of the excess reactant is left over at the end.
Step 1: Write the balanced reaction equation and the recipe. A particle row won’t be needed:
2 Mg(s)
+
O2(g)

2 MgO(s)
2 mol Mg
1 mol O2
2 mol MgO
48.6 g Mg
32.0 g O2
80.6 g MgO
22.4 L O2@ STP
The molar mass charts for determining the above recipe rows are as follows:
1 mol Mg = 24.3 g Mg
O2
2 mol Mg = 48.6 g Mg
MgO:
O: 16.0 g  2 = 32.0 g O2 = 1 mol O2
Mg:
24.3 g  1 = 24.3 g
O:
16.0 g  1 = 16.0 g
1 mol MgO = 40.3 g MgO
2 mol MgO = 80.6 g MgOS
Step 2: Part A: Which reactant limits?
Plan of Attack: Choose one of the starting reactant values from the problem statement, and use the recipe to
calculate how much of the second reactant is needed to completely consume the first one. There are two
possible calculations that can be done here, but they should both lead you to the same answer:
Choice 1: 5.0 g Mg  ? L O2 @ STP
5.0g _ Mg  22.4L _ O2 @STP 

 
 = 2.3 L O2 @STP

  48.6g _ Mg 
1
You will notice that 2.3 L O2 @ STP is required to completely consume 5.0 g Mg, but you are only given
0.50 L in the jar. This means you don’t have enough O 2, so O2 limits and Mg is the excess reactant

Choice 2: 0.50 L O2 @ STP  ? g Mg
0.50L _ O2 @STP  48.6g _ Mg 

 
 = 1.1 g Mg

 22.4L _ O2 
1

K8: Intensified Unit K Reading Notes
Page 11 of 12
You will notice that the 1.1 g Mg needed to completely consume the given 0.50 L O 2 @ STP is much less
than the 5.0 g Mg provided in the jar. Because all of the O2 is used up, and not all of the Mg is used up, we
can again conclude that O2 limits, and Mg is the excess reactant.
Step 3: Part B: How much MgO is produced?
Because O2 runs out first (limits), the amount of MgO produced will be determined by the starting amount
of O2:
0.50 L O2 @ STP  ? g MgO
0.50L _ O2 @STP   80.6g _ MgO 

 
 = 1.8 g MgO produced

 22.4L _ O2 @STP 
1
Step 4: Part C: How much of the excess reactant is left over?

We already calculated in (A) that 1.1 g Mg is consumed in the reaction with 0.50 L of O 2 @ STP. The mass
of Mg in excess can then be found by a simple subtraction:
5.0 g Mg @ start

1.1 g Mg consumed during reaction
3.9 g Mg left over (in excess)
You might ask yourself the question: “Why are we even talking about limiting and excess reactants?,”
which actually does NOT make you a SLACKER. The truth is that many chemical processes do not
achieve what we call the calculated, or theoretical yield, because they are not occurring under perfect
conditions. When we are using very expensive reactants, this is a problem, so we make the cheaper
reactants available in excess so that the reaction is forced to use up the more expensive reactant, and the
manufacturer will not be throwing out expensive unused reactants.
Percent Yield
The amount of products we have been calculating so far have been THEORETICAL YIELDS, or what
should be produced given certain quantities of reactants. The reality is that many reactions actually fall far
short of the theoretical yield due to less-than-ideal reaction conditions. We can calculate percent yield with
the following equation:
 actual _ yield 
100
theoretical _ yield 
% YIELD = 
EXAMPLE: Methanol, CH3OH, cam be produced through the reaction of carbon monoxide and hydrogen
gas:

CO(g) + 2 H2(g)  CH3OH(l)
If 75.0 g CO reacts to produce 68.4 g CH3OH what is the percent yield?
First, we need to recognize that the mass of methanol that is provided as the product amount is the actual
yield. This means that we must therefore build a recipe and calculate the theoretical yield before we can
determine the percent yield. In order to do this, we must write the balanced chemical equation, and build a
recipe beneath it:
K8: Intensified Unit K Reading Notes
CO(g)
+
Page 12 of 12
2 H2(g)
1 mol CO

2 mol H2
28.0 g CO
CH3OH(l)
1 mol CH3OH
32.0 g CH3OH
Necessary molar mass calculations (NOTE THAT H2 ISN’T NECESSARY):
CO
CH3OH
C: 12.0 g
C: 12.0 g  1 = 12.0 g
O: 16.0 g
O: 16.0 g  1 = 16.0 g
H: 1.0 g  3 = 3.0 g
1 mol CH3OH = 32.0 g CH3OH
28.0 g CO = 1 mol CO
USE THE RECIPE TO CALCULATE THE THEORETICAL YIELD:
75.0 g CO  ? g CH3OH
75.0g _ CO  32.0g _ CH3OH 

 
= 85.7 g CH3OH

  28.0g _ CO 
1
CALCULATE PERCENT YIELD:

68.4g _ CH3OH 
 actual _ yield 
100 = 79.8 % YIELD
100 = 
theoretical _ yield 
85.7g _ CH3OH 
PERCENT YIELD = 

