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Unit 3
Chemical Equations and Stoichiometry
This is the point in chemistry where you feel like you are “doing” chemistry! You are beginning to
see how all the little pieces fit together to make the big picture. Again, much of this information will
look familiar to you from your chemistry class.
In this unit, you will:
Balance chemical equations
Predict the products of chemical reactions
Perform stoichiometric calculations
Determine the limiting reactant in a system
Determine the percent yield of product
Work sequential reactions
Calculate solution concentrations
Perform calculations on solution equations
3.01 Balancing Chemical Equations
Before you begin balancing equations, let’s review some concepts.
1. Chemical reactions don’t always occur. Evidence that a chemical reaction
has occurred include:
Evolution of light and heat
Production of gas
Formation of precipitate
Color change
2. The following exist (as elements) as diatomics. They form a “7” on the periodic table (hydrogen
is off by itself.)
H2, N2, O2, F2, Cl2, Br2, I2
3. Some common symbols:
 Yields
 Reversible reaction

Gas produced

Precipitate produced

Change
(s) Solid
(cr) Crystal solid
(g) Gas
(l) Liquid
(aq) aqueous
4. Steps to writing equations:
1.Determine type of reaction
2. Write word equation
3. Translate into formulas
4. Post diatomics
5. Balance (t chart)
6. Rewrite
Go to the website below. Take notes on balancing each type of equation.
http://www.chemtutor.com/react.htm#bal
PRACTICE:
Balance the following equations. Scroll down to check your answers.
1.
2.
3.
4.
5.
CaC2(s) + H2O(l)  C2H2(g) + Ca(OH)2(aq)
H2S(g) + O2(g)  SO2 + H2O(g)
C6H12O6(s) + O2(g)  CO2(g) + H2O(l)
Mg(s) + H3PO4(aq)  Mg3(PO4)2(aq) + H2(g)
KClO3(s)  KCl(s) + O2(g)
1.
2.
3.
4.
5.
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
2H2S(g) + 3O2(g)  2SO2 + 2H2O(g)
C6H12O6(s) + 6O2(g)  6CO2(g) + 6 H2O(l)
3Mg(s) + 2H3PO4(aq)  Mg3(PO4)2(aq) + 3H2(g)
2 KClO3(s)  2KCl(s) + 3O2(g)
Answers:
3.02 Predicting the Products of Chemical Reactions
It is particularly important that you can predict what the products will be. It may
seem extremely difficult at first, but there is an easy system.
First, you will classify the type of reaction. Is it a synthesis, decomposition, single
replacement, double replacement, or combustion?
After you have classified the reaction, you will classify the reactants. A convenient chart will help
you to choose the correct products.
You should use a 4 line format for your equations:
Line 1: skeleton line
Line 2: word equation
Line 3: practice line
Line 4: formal equation
Print out the 3 following charts. You will be using them frequently.
Activity Series of the Elements
An element that is above will replace an element below in a reaction.
Activity of
metals
Li
Rb
K
Ba
Sr
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Cd
Co
Ni
Sn
Pb
H2
Sb
Bi
Cu
Hg
Ag
Pt
Au
Lithium to sodium: React with cold water and acids replacing
hydrogen.
React with oxygen forming oxides
Activity of
halogen
nonmetals
F2
Cl2
Br2
I2
Magnesium to cadmium: React with steam (but not cold
water) and acids replacing hydrogen.
React with oxygen, forming oxides
Cobalt to lead: Do not react with water. React with acids,
replacing hydrogen. React with oxygen, forming oxides.
Hydrogen to mercury: React with oxygen, forming oxides.
Silver to gold: Fairly unreactive, forming oxides only indirectly
Solubility Rules:
1. Most nitrate (NO3-) salts are soluble.
2. Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion
(NH4+) are soluble.
3. Most chloride, bromide, and iodide salts are soluble. Notable exceptions are salts
containing the ions Ag+, Pb2+, and Hg22+.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, Hg2SO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH
and KOH. The compounds Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble.
6. Most sulfide (S2-), carbonate (CO32-), chromate (CrO42-), and phosphate (PO43-) salts are
only slightly soluble.
Process Steps:
Step one in the equation writing process is to determine the type of reaction. You do this by
looking at your reactants.
Ask yourself, “what am I starting with?”
If you start with 2 elements or a metal oxide (like PbO) and water, you have a synthesis reaction.
If you start with only one compound, you have a decomposition reaction.
If you start with an organic compound (like CH4 or C6H12O6), you will have a combustion reaction.
If you start with one compound and an element, you have a single replacement reaction.
If you start with two compounds, you will have a double replacement reaction.
Let’s try an example of each type of reaction:
Example 1:
Write the balanced equation for the reaction between magnesium and oxygen.
1.Determine type of reaction You begin with two elements, so you have a synthesis reaction.
Write the reactants on line 2. See pink text
Write the correct skeleton equation on line 1. To do this, look at your chart. You have two
elements (magnesium and oxygen), so the skeleton equation is element + O 2  oxide. See
orange text
2. Write word equation
The product will be an oxide – since you have magnesium, the oxide will be magnesium oxide.
Write this on the product side of line 1. See dark red text.
3. Translate into formulas
On line 3, you will “play” with your equation. Start by writing the symbols. See green text.
4. Post diatomics
Oxygen is diatomic, make sure that you have a subscript “2” on oxygen.
5. Balance (t chart)
Many equations are easy to balance. If necessary, you can use a “t chart” to make sure your
products equal reactants. See blue text
6. Rewrite
Once you are satisfied that the equation is balanced, rewrite it neatly on line 4. (Since this work is
done on the computer. See bold text
Line 1: element + O2  oxide
Line 2: magnesium + oxygen  magnesium oxide
Line 3: 2 Mg + O2  2 MgO
Line 4: 2 Mg + O2  2MgO
Example 2:
Write the balanced equation for the reaction between potassium and iodine.
Again, you have two elements, so this will be a synthesis.
When a metal and a halogen combine, the product is a salt.
Line 1: metal + halogen  salt
Line2: calcium + iodide  calcium iodide
Line 3: 2 K + I2  2 KI
Line 4: 2 K + I2  2 KI
Example 3:
Write the balanced equation for the reaction between calcium oxide and water.
You have a metal oxide and water. This will be a synthesis reaction. The products will be a metal
hydroxide.
Line 1: metal oxide + water  metal hydroxide
Line 2: calcium oxide + water  calcium hydroxide
Line 3: CaO + HOH  Ca(OH)2
Line 4: CaO + HOH  Ca(OH)2
Example 4:
Write the balanced equation for the decomposition of sodium chloride.
You have one reactant. This is a decomposition reaction.
Since sodium chloride is a binary compound, you will get 2 element.
Line 1: binary compound  element A + element B
Line 2: sodium chloride  sodium + chloride
Line 3: 2 NaCl  2 Na + Cl2
Line 4: 2 NaCl  2 Na + Cl2
Example 5:
Write the balanced equation to show what happens to calcium carbonate when it breaks down.
You have one reactant. This is a decomposition reaction.
Since you start with a metal carbonate, you will end with a metal oxide + carbon dioxide
Line 1: metal carbonate  metal oxide + carbon dioxide
Line 2: calcium carbonate  calcium oxide + carbon dioxide
Line 3: CaCO3  CaO + CO2
Line 4: CaCO3  CaO + CO2
Example 6:
Write the balanced equation to show the breakdown of potassium hydroxide.
You have one reactant. This is a decomposition reaction.
Since you start with a metal hydroxide, you get a metal oxide and water.
Line 1 : metal hydroxide  metal oxide + water
Line 2: potassium hydroxide  potassium oxide + water
Line 3: 2 KOH  K2O + HOH
Line 4: 2 KOH  K2O + HOH
Example 7:
Write the balanced equation to show the decomposition of sodium chlorate.
You have one reactant. This is a decomposition reaction.
Since you start with a metal chlorate, you will end up with a metal chloride and oxygen.
Line 1: metal chlorate  metal chloride + oxygen
Line 2: sodium chlorate  sodium chloride + oxygen
Line 3: 2 NaClO3  2 NaCl + 3 O2
Line 4: 2 NaClO3  2 NaCl + 3 O2
Example 8:
Write the balanced equation to show the decomposition of sulfuric acid.
You have one reactant. This is a decomposition reaction.
Since you begin with an acid, you will end with a nonmetal oxide and water.
Here’s the catch: You have to make sure that the cation in the nonmetal oxide retains the same
oxidation number that it had in the acid.
The oxidation number of S in H2SO4 is +6. So the nonmetal oxide will have to be SO3.
Line 1: acid  nonmetal oxide + water
Line 2: sulfuric acid  sulfur VI oxide + water
Line 3: H2SO4  SO3 + HOH
Line 4: H2SO4  SO3 + HOH
Example 9:
Write the balanced equation for the combustion of C2H6.
You won’t be given the name of an organic compound at this point. You will be given the formula
because you have not learned how to name them yet.
Remember, all combustion equations must have oxygen added! You will always end up with
carbon dioxide and water.
These are a little tricky to balance, you just have to have patience.
Start with the carbons first. Notice, some of the numbers on line 3 are crossed out. I had to try this
equation a couple of different ways.
Line 1: organic + oxygen  carbon dioxide + water
Line 2: C2H6 + oxygen  carbon dioxide + water
Line 3: 2 C2H6 + 7O2  42 CO2 + 63 HOH
Line 4: 2 C2H6 + 7 O2  4 CO2 + 6 HOH
Example 10
Write the balanced equation for the reaction between sodium chloride and calcium.
You have an element and a compound. This will be a single replacement.
The hardest thing about replacement reactions is keeping the cations and anions straight.
I suggest drawing a line between the cation and anion. I will have a few more scratch marks on
line 3 for this equation.
Notice under “metal replaces another metal” you are instructed to see the activity series. If the
metal that is in elemental form is more active than the metal in the compound, a reaction will
occur. In this case, calcium is above sodium, so the reaction will occur.
Line 1: replacement of metal by metal
Line 2: calcium + sodium chloride  sodium + calcium chloride
Line 3: Ca + 2 NaCl  2 Na + CaCl2
Line 4: Ca + 2 NaCl  2 Na + CaCl2
Example 11
Write the balanced equation for the reaction between lithium and water.
You have an element and a compound. This will be a single replacement.
Check the activity chart. Will lithium replace hydrogen in water? Yes.
Line 1: metal + water  metal hydroxide + H2
Line 2: lithium + water  hydrogen + lithium hydroxide
Line 3: 2 Li + 2 HOH  2 LiOH + H2
Line 4: 2 Li + 2 HOH  2 LiOH + H2
Example 12:
Write the balanced equation for the reaction between sodium and sulfuric acid.
You have an element and a compound. This will be a single replacement.
Line 1: metal + acid  salt + hydrogen gas
Line 2: sodium + sulfuric acid  sodium sulfate and hydrogen gas
Line 3: 2 Na + H2SO4  Na2SO4 + H2
Line 4: 2 Na + H2SO4  Na2SO4 + H2
Example 13:
Write the balanced equation for the reaction between beryllium iodide and fluorine.
You have a compound and an element. This is a single replacement.
This is a halogen replacing a halogen. You are instructed to see the activity series. Is fluorine
above iodine? Yes. This means fluorine will replace iodine in this reaction.
Line 1: halogen replacement
Line 2: beryllium iodide + fluorine  beryllium fluoride + iodine
Line 3: BeI2 + F2  BeF2 + I2
Line 4: BeI2 + F2  BeF2 + I2
Example 14:
Write the balanced equation between silver nitrate and a solution of sodium chloride.
You begin with two compounds. This is a double replacement reaction. Double replacement
reactions that occur in solution may also be called precipitation reactions. You will need to look at
the solubility chart to see if a precipitate (solid) will form. You may also have to write the ionic
equations. We will cover that in another section.
Notice, that in these equations, you will have to keep track of the phase of each reactant and
product. In the solubility chart, you will see that silver chloride is a solid.
Line 1: cation A replaces cation B
Line 2: silver nitrate (aq) + sodium chloride(aq)  silver chloride(s) + sodium nitrate(aq)
Line 3: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Line 4: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Example 15:
Write the balanced equation between sodium hydroxide and hydrochloric acid.
You begin with two compounds. This is a double replacement reaction. When a double
replacement reaction occurs between an acid and a base, it is called a neutralization reaction.
You will end up with a salt and water. You may also have to write net ionic equations.
Line 1: acid + base  salt + water
Line 2: sodium hydroxide + hydrochloric acid  salt + water
Line 3: NaOH  HCl  NaCl + HOH
Line 4: NaOH  HCl  NaCl + HOH
Practice:
Write the balanced equations.
Save your work as an .rtf file.
Place in the drop box for your instructor to grade.
Synthesis:
1. zinc plus oxygen
2. magnesium plus sulfur
3. aluminum plus phosphorous
Decomposition
4.
5.
6.
7.
8.
9.
potassium oxide yields
copper II sulfide yields
magnesium carbonate yields
magnesium hydroxide yields
aluminum chlorate yields
sulfuric acid yields
Combustion
10. C2H6 plus oxygen
11. C6H12O6 plus oxygen
Single Replacement
12. bromine + lead IV chloride 
13. chlorine + lead IV bromide 
Double Replacement
14. hydrochloric acid + magnesium hydroxide 
15. barium sulfate + potassium phosphate 
16. Iron II chloride + silver phosphate 
After your instructor has graded this assignment, take the quiz titled “Writing Equations.”
3.03 Performing Stoichiometric Calculations
You will be performing a lot of stoichiometry in this course. The word
stoicheion means element. Metron means measure. When you
perform stoichiometry, you are measuring how many elements are
present under given circulstances.
There are 4 Types of problems
1. mole to mole calculations
]2. mole to mass calculations
3. mass to mole calculations
4. mass to mass calculations
The mole ratio is a conversion factor that will help you make these determinations.
To find the mole ratio, you must correctly write the balanced equation.
2 Al2O3(l)  4 Al(s) +3 O2(g)
If you have 2 moles of aluminum oxide, you will get 4 moles of aluminum and 3 moles of oxygen.
2:4:3
Or
2 mol Al2O3
4 mol Al
or
2mol Al2O3
3 mol O2
or
4 mol Al
3 mol O2
or
4 mol Al
2 mol Al2O3
3 mol O2
2 mol Al2O3
3 mol O2
4 mol Al
Ideal Stoichiometric Calculations
How to set up your stoichiometry problems
First, you must correctly balance the equation.
You will find a chart below. This chart is an excellent tool for organizing your information. Each line
contains information that can be used to answer questions.
Lines 1 through 3 are answer lines. You will find answers to your specific problems on these line.
Lines 5 through 7 are standard lines. These lines contain the standard conditions.
Again, copy this chart so that you can use it over and over again.
Line 1: Volume (L)
Line 2: Moles (mol)
Line 3: Mass (g)
Line 4: Equation
Line 5: GAW/GMW (g/mol)
Line 6: Moles (mol)
Line 7: Mass (g)
Check:
Standards
mole ratio:
Let’s try working some problems using this tool.
Example1:
What mass of hydrochloric acid is consumed by its reaction with 2.50 mol of magnesium? What is
the mass of each product produced?
1. Write the correctly balanced equation on line 4. See bold text in chart.
Notice the equation is spaced out. This helps to keep track of the information about each
element/compound.
2. On line 5, write the formula mass (or gram molecular weight, GMW) of each participant in the
equation. Write it directly under each term. See violet text in chart.
3. On line 6, write the number of moles of each substance in the above equation. (This will be the
coefficient in front of the term.) See red text in chart.
4. On line 7, write the standard mass. This is the mass of each participant in the standard
equation. You get these numbers by multiplying line 5 x line 6. See blue text in chart.
5. Check to see if your equation is balanced properly. Add the mass of the reactants on line 7.
Add the mass of the products on line 7. Write the answer next to Check. These numbers should
equal. See orange text in chart.
6. Write the given information on the appropriate line above the equation. Lines 1 through 3 are
your givens. Later, we will call these theoretical lines.
In this example, we were given 2.50 mol of magnesium. Moles go on line 2. Write 2.50 above
magnesium on line 2. See violet text in chart.
7. Calculate the mole ratio. In order to get a ratio, you have to divide two numbers that have the
same unit. We were given moles, so we can divide line 2 by line 6. If we were given grams, we
would divide the given number on line 3 by line 7. Write this number next to mole ratio. 2.5 mol/ 1
mol = 2.50. See grey text in chart.
8. Multiply the mole ratio by line 6 if you want the answer in moles. Multiply the mole ratio by the
number on line 7 if you want the answer in grams. Write the answers above the appropriate
participant. If your answers are in L, write the answer on line 1. If your answers are in moles, write
them on line 2. If your answers are in g, write the answers on line 3.
We were asked to find the mass of hydrochloric acid. Multiply 2.50 x 72 g. Also, we were asked to
find the mass of each product. Multiply the mole ratio by 94.3g to find the mass of magnesium
chloride produced. Multiply the mole ratio by 2 g to find the mass of hydrogen gas produced. Write
the answer on line 3. See green text in chart.
You can write only the answer that you are looking for.
Line 1: Volume (L)
Line 2: Moles (mol)
2.50
Line 3: Mass (g)
Line 4: Equation
Mg(s) +
Line 5: GAW/GMW (g/mol) 24.3
Line 6: Moles (mol) :
1
Line 7: Mass (g)
24.3
Check: 96.3 g = 96.3 g
181.53
2 HCl(aq)
36
2
72

238
5.05
MgCl2(aq) + H2(g)
94.3
2
1
1
94.3
2
mole ratio: 2.50
182 g of hydrochloric acid is consumed when reacted with 2.50 mol of magnesium. The
amount of magnesium chloride produced is 238 g. 5.05 g of hydrogen gas is produced.
PRACTICE
Check your answers below.
1. Acetylene gas (C2H2) is produced as a result of the following reaction:
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
a. If 32.0 g of CaC2 are consumed in this reaction, how many moles of water are
needed?
b. How many moles of each product would be formed?
2. When sodium chloride reacts with silver nitrate, silver chloride precipitates. What mass of
silver chloride is produced from 75.0 g of silver nitrate?
3. Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry. It is
produced when ammonium nitrate is decomposed according to the following reaction.
NH4NO3(s)  N2O(g) + 2H2O(l)
a. How many grams of ammonium nitrate are required to produce 33.0 g of nitrous oxide?
b. How many grams of water are produced in this reaction?
ANSWERS
1.
Line 1: Volume (L)
Line 2: Moles (mol)
Line 3: Mass (g)
32.0
Line 4: Equation
CaC2(s) +
Line 5: GAW/GMW (g/mol) 64.1
Line 6: Moles (mol)
1
Line 7: Mass (g)
64.1
Check:
100 = 100
0.998
.499
.499
2H2O(l) 
18
2
36
C2H2(g) +
Ca(OH)2(aq)
26
74.1
1
1
26
74.1
mole ratio: 32.0 / 64.1 = 0.499
0.998 moles of water are needed. 0.499 moles of acetylene are produced. 0.499 moles of
calcium hydroxide are produced.
2. 63.3 g AgCl
3. a. 60.0 g NH4NO3
.b. 27.0 g H2O
ASSIGNMENT:
Work the following problems. SHOW YOUR WORK!
Save your work as an .rtf file.
Submit it to your drop box for grading.
1. Sodium chloride is produced from its elements through a synthesis reaction. What mass of
each reactant would be required to produce 25.0 mol of sodium chloride?
2. If 4.50 mol of ethane, C2H6, undergo combustion, how many moles of oxygen are required?
How many moles of each product are formed?
3. Iron is generally produced from iron ore through the following reaction in a blast furnace:
Fe2O3(s) + CO(g)  Fe(s) + CO2(g)
a. If 4.00 kg of Fe2O3 are available to react, how many moles of CO are needed?
b. How many moles of each product are formed?
3.04 Limiting Reactants
We all inherently understand the concept of the limiting reactant. The
limiting reactant is the reactant that will run out first. For example, you
want to make a cake that requires three eggs. You only have 2 eggs.
You cannot make one cake. Eggs are your limiting reactant.
Excess reactant – the reactant that you have plenty of. You don’t
have to worry about this. Using the cake example, if the recipe calls
for ¼ cup oil and you have a gallon of oil in your kitchen cabinet, oil is
in excess.
Determining the limiting reactant in chemical reactions is just as easy. Again, you will use the
chart for stoichiometry.
You will set up the chart in the same way. The only difference is, you will have two given amounts.
You will find the mole ratio of each. The reactant with the lowest mole ratio is the limiting reactant.
You will use that mole ratio to do your calculations.
How to determine the limiting reactant
Example 1:
Silicon dioxide (quartz) is usually quire unreactive but reacts readily with hydrogen fluoride
according to the following equation.
SiO2(s) + 4HF (g)  SiF4(g) + 2H2O(l)
If 2.0 mol of HF are exposed to 4.5 mol of SIO2, which is the limiting reactant? How much of each
product in grams will be formed?
In this example, you will find the mole ratio of both reactants. See violet text in chart.
Notice the mole ratio for HF is lower. HF is your limiting reactant. SiO2 is your excess reactant. It
doesn’t matter how much SiO2 you have, HF limits the reaction. You will use the mole ratio for HF
to do the rest of the calculations.
1.
2.
2.0 mol
3.
4.
SiO2(s) +
4HF(g) 
5.
60.1 g/mol 20 g/mol
6.
1 mol
4 mol
7.
60.1 g
80 g
Check: 140.1 = 140.1
52 g
SiF4(g) +
104.1 g/mol
1 mol
104.1 g
18 g
2H2O(l)
18 g/mol
2 mol
36 g
Index:
HF: 2.0mol/4mol = .5
SiO2: 4.5 mol/1 mol = 4.5
This reaction will result in the production of 52 g of SiF4 and 18 g of water.
Practice:
1. Zinc and sulfur react to form zinc sulfide according to the following equation.
8 Zn(s) + S8(s)  8 ZnS
a. If 2.00 mol of Zn are heated with 1.00 mol of S8, identify the limiting reactant.
b. How many moles of excess reactant remain?
c. How many moles of the product are formed?
Answer:
a. Zn
b. 0.75 mol of S8 remain
c. 2.00 mol ZnS
PRACTICE
Perform the following calculations. Show all your work.
Save as an .rtf file.
Place in your drop box for grading.
1. a. If 2.50 mole of copper and 5.50 mol of silver nitrate are available to react by single
replacement, identify the limiting reactant.
b. Determine the amount of moles of excess reactant remaining.
c. Determine the amount in moles of each product formed.
d. Determine the mass of each product formed.
2. Sulfuric acid reacts with aluminum hydroxide by double replacement.
a. If 30.0 g of sulfuric acid react with 25.0 g of aluminum hydroxide, identify the limiting reactant.
b. Determine the mass of excess reactant remaining.
c. Determine the mass of each product formed. Assume a 100% yield.
3. The energy used to power one of the Apollo lunar missions was supplied by the following
overall reaction:
2N2H4 + (CH3)2N2H2 + 3N2O4  6N2 + 2CO2 + 8H2O
For the phase of the mission when the lunar module ascended from the surface of the moon, a
total of 1200. kg of N2H4 were available to react with 1000. kg of (CH3)2N2H2 and 4500. kg of
N2O4.
a. For this portion of the flight, which of the allocated components was used up first?
b. How much water, in kilograms, was put into the lunar atmosphere through this reaction?
3.05. Percent Yield
Using stoichiometry, we can predict how much we should
get of something in a reaction. As you have probably seen in
lab, rarely do we do everything correctly and get exactly
what we are supposed to get. The way of expressing how far
off you are from what you are supposed to get is by
calculating percent yield.
For example, you make a home made sugar cookies from a
recipe that should make 3 dozen cookies. You, however,
only end up with 25 cookies. Several things could have happened. You could have made the
cookies too big. You may have put slightly less flour or sugar in that what was listed. At any rate,
your percent yield is:
25 cookies x 100% = 69%
36 cookies
percent yield =
what you actually get in lab x 100%
What you are supposed to get
You will need to determine what you are supposed to get by performing stoichiometric
calculations.
Example 1:
The percentage yield for the production of phosphorus pentachloride from the reaction between
phosphorus trichloride and chlorine gas is 83.2%. What mass of PCl5 is expected from the
reaction of 73.7 g of PCl3 with excess chlorine?
The numbers on lines 1 through 3 can be called theoretical numbers. They are the numbers you
would get if your reaction went perfectly. So, theoretically, you will obtain 111.6 g PCl 5.
Actually, you only get 83.2% yield. So you multiply 111.6 g x (0.832).
1.
2.
3.
73.7 g
4.
PCl3 +
5. 136 g/mol
6.
1 mol
7. 136 g
check
51 g
Cl2 
70 g/mol
1 mol
70 g
111.6 g
PCl5
206 g/mol
1 mol
206 g
Index: .5419
If everything went perfectly, you would get 111.6 g PCl5. The percent yield, however is 83.2% of
that. Thus, you obtain 93 g PCl5.
PRACTICE
Perform the following calculations. Show all your work.
Save as an .rtf file.
Place in your drop box for grading.
1. Coal gasification is a process that converts coal into methane gas, CH4. If this reaction has a
percent yield of 85.0%, how much methane can be obtained from 1250. g of carbon?
2C(s) + 2H2O(l)  CH4(g) + CO2(g)
2. In the reaction between CO and Fe3O4, the theoretical yield in an experiment is calculated to be
47.2 g Fe. When a careless chemistry student carries out the experiment, the actual yield is 42.9 g
Fe. Calculate the percent yield.
3. When ammonia, NH3, is prepared from 28 g N2 and excess H2, only 30. g is produced. What is
the percent yield?
3.06 Sequential Reactions
Sequential reactions are just series of reactions. The key to
performing calculations on them is just to put one foot in front of the
other!
You aren’t performing anything new, you are just adding on.
Example 1:
The Ostwald Process for producing nitric acid from ammonia consists of the following steps:
4 NH3(g) + 5O2(g)  4 NO(g) + 6H2O(g)
2NO(g) + O2(g)  2NO2(g)
3NO2(g) + H2O(g)  2HNO3(aq) + NO(g)
If the yield in each step is 94%, how many grams of nitric acid can be produced from 5.00 kg of
ammonia?
What we finally want to know is the amount in kg of nitric acid we will end up with. We have to
know how much NO2 we start out with. We get that number from the above reaction.
We have to perform the first reaction to get the second.
Do this one step at a time.
1. Perform stoichiometry on the 1st equation to find out how much NO goes into the second
equation.
1
2.
3.
5 x 103 g
4.
4 NH3(g) +
5O2(g) 
5.
17 g/mol
32 g/mol
6.
4 mol
5 mol
7.
68 g
160 g
check
228 = 228
8.82 x 103 g
4 NO(g) +
30 g/mol
4 mol
120 g
6H2O(g)
18 g/mol
6 mol
108 g
Index: 5000 g/ 68g = 73.53
2. Remember, you only get 94% yield. 8.82 x 103 g (0.94) = 8.29 x 103 g NO.
3. This becomes the given for your second equation.
1
2.
3.
4.
5.
6.
7.
check
8.29 x 103 g
2NO(g) +
30 g/mol
2 mol
60 g
92 = 92
O2(g) 
32 g/mol
1 mol
32 g
1.27 x 104 g
2NO2(g)
46 g/mol
2 mol
92 g
Index: 8.29 x 103 g/ 60g = 1.38 x 103 g
4. The percent yield from reaction 2 is also 94%. 1.27 x 10 4 g ( 0.94) = 1.2 x 104 g NO2(g)
5. This becomes the given for the 3rd equation.
1
2.
3.
4.
5.
6.
7.
check
1.2 x 104 g
3NO2(g) + H2O(g) 
46 g/mol
18 g/mol
3 mol
1 mol
138 g
18 g
156 = 156
1.09 x 104 g
2HNO3(aq) +
NO(g)
63 g/mol
30 g/mol
2 mol
1 mol
126 g
30 g
Index: 1.27 x 104 g/138 g = 86.6
6. Finally, calculate the percent yield for nitric acid.
1.09 x 104 g (0.94) = 1.02 x 104 g
7. Remember, you want to give the answer in kg
10.2 kg of HNO3
ASSIGNMENT
Perform the following calculations. Show all your work.
Save as an .rtf file.
Place in your drop box for grading.
1. The aspirin substitute, acetaminophen (C8H9O2N), is produced by the following 3 step process.
I.
II.
III.
C6H5O3N(s) + 3H2(g) + HCl(aq)  C6H8ONCl(s) + 2H2O(l)
C6H8ONCl(s) + NaOH(aq)  C6H7ON(s) + H2O(l) + NaCl(aq)
C6H7ON(s) + C4H6O3(l)  C8H9O2N(s) + HC2H3O2(l)
The first two reactions have percent yields of 87% and 98% by mass, respectively. The overall
reaction yields 3 mol of acetaminophen product for every 4 mol of C6H5O3N reacted.
a. What is the percent yield by mass for the overall process?
b. What is the percent yield by mass of step III?
3.07 Solution Concentrations
Many reactions occur in solution. The amount of a substance present in a reaction
depends on its concentration. If the concentration is high, there will be a lot of solute present to
participate in the reaction.
Solute is the substance being dissolved.
Solvent is the substance doing the dissolving.
For this reason, you need to be able to calculate solution concentrations and determine how much
of a substance is present in solution based on its concentration.
There are many ways to calculate the concentration of solutions. Two that you should remember
are molarity and molality.
moles _ solute
L _ of _ solution
moles _ solute
Molality, m
kg _ of _ solvent
Molarity , M =
Molarity is the expression most frequently used. You will be doing more extensive calculations
with solutions in later units.
You must know how to find the concentration of a solution.
Example 1:
What is the molarity of a solution in which 58 g of NaCl are dissolved in 1.0 L of solution?
Convert grams to moles:
58 g 1mol 1mol
x

1
58 g
1
Find molarity:
1mol _ NaCl
1mol

 1M
1.0 L _ solution
L
Go to the website below to learn about solution stoichiometry.
http://antoine.frostburg.edu/chem/senese/101/moles/slides/sld021.htm
Advance through this presentation. Work the problems as you go.
When you finish working the problems, check your answers below:
Slide 21 (First slide for you to view)

What mass of AgNO3 (MW 169.87) is required to make 25.0 mL of 0.100 M AgNO3?

What is the molarity of a NaCl solution prepared by dissolving 1.00 g NaCl in enough
water to produce 500.0 mL of solution?

What volume of 0.100 M NaCl solution can be prepared from 1.00 g of NaCl?
Answers:
1.
0.025L 0.100mol 2.5 x10 3 mol
x

1
L
1
You need 2.5 x 10-3 mol of silver nitrate to make 25 mL of this solution.
2.5 x10 3 mol 169.87 g 0.425 g
x

1
mol
1
You will place 0.425 g silver nitrate into a 25 ml volumetric flask. You will fill to the mark with
distilled water.
2.
1.00 gNaCl
1mol
0.017mol
x

1
58.5 gNaCl
1
You have 0.017 moles of NaCl dissolved in 0.500 L of solution.
0.17mol 0.034mol

0.500 L
L
The solution is 0.034 M.
3.
1g _ NaCl
1mol
0.017mol
x

1
58.5 g _ NaCl
1
You have 0.017 moles NaCl. You want the solution to be 0.100 M
0.17mol
L
x
 0.17 L
1
0.100mol
You will dissolve the 1 gram NaCl in enough water to make 170 mL of solution.
Slide 23:
You will often find instances where solutions need to be diluted to reach a certain concentration.
This is the example in slide 23. Acids are stored in large containers (stock) at a very high
concentration. In order to get the necessary concentration, they must be diluted.
To determine the amount of stock solution (1) to make the diluted solution (2), use this equation:
M1V1=M2V2
In this example, 500 mL of a 0.500 M solution of HCl is prepared from a 12.1 M stock solution.
How much of the stock solution is needed?
Rearrange the equation to solve for V1, the volume of stock solution:
V1 
M 2V2 M 2 V2 1

x x
M1
1
1 M1
Plug in the known information:
V1 
0.500mol 0.5L
L
x
x
 0.02 L
L
1
12.1mol
You will place 0.02 L of the 12.1M stock solution into a volumetric flask. You will fill the flask with
water until it reaches 500 mL.
Slide 24:
How many mL of 0.100 M NaOH solution are needed to react with 10.0 mL of 0.41 M HC2H3O2?
Here, you are doing stoichiometry for a neutralization reaction. The only difference between this
and the problems you were doing earlier, is that the amount is given indirectly through the
concentrations.
First, write the correctly balanced equation:
NaOH +
1mol
HC2H3O2  NaC2H3O2 + HOH
1 mol
1mol
1mol
You have a 1:1 ratio between the acid and the base.
Next, determine the number of moles of acetic acid are present.
0.010 L 0.41mol
x
 0.0041mol
1
L
0.0041 mol
NaOH +
1mol
0.0041 mol
HC2H3O2  NaC2H3O2 + HOH
1 mol
1mol
1mol
Since you have 0.0041 mol of acetic acid, you need 0.0041 mol of sodium hydroxide to completely
neutralize it.
Finally, determine how many mL of sodium hydroxide are needed to provide that number of
moles.
0.0041mol
L
x
 0.041L
1
0.100mol
You will need 41 mL of 0.1M NaOH.
Slide 25:
1.500 M NaOH
(the titrant)
25.0 mL of
battery acid
What is the concentration of H2SO4 in battery acid, if it consumes 40.0 mL of the titrant?
This slide is showing you a titration. A titration is a method for determining unknown
concentrations through a neutralization reaction.
You will work with titrations in later units.
Even though this problem uses words that aren’t the same as we’ve been using, it is asking
something very familiar.
In other words,
What is the concentration of the sulfuric acid if 25.0 mL of it reacts with 40 mL of 1.500 M sodium
hydroxide?
Again, write the balanced neutralization equation.
H2SO4 +
1 mole
2NaOH 
2 mole
Na2SO4 +
2HOH
Notice there is a 1:2 ratio. You need twice as many moles of sodium hydroxide.
Next, determine how many moles of sodium hydroxide were used in the reaction.
0.040 L 1.500mol
x
 0.06mol
1
L
0.06 moles of NaOH were consumed in this reaction.
Next, determine how many moles of sulfuric acid were used.
0.03 mol
H2SO4 +
1 mole
0.06 mol
2NaOH 
2 mole
Na2SO4 +
2HOH
There were 0.03 moles of sulfuric acid present in the 25 mL of sulfuric acid solution.
Determine the molarity of the sulfuric acid solution:
0.03mol
 1.2M
0.025 L
The concentration of sulfuric acid is 1.2 M.
ASSIGNMENT
Take the quiz titled Molarity
3.08 Writing Net Ionic Equations
As mentioned previously, sometimes double replacement reactions are
called precipitation reactions. This means that two solutions are added
together and a solid (precipitate) forms.
When writing these types of equations, it is necessary to know the species
(or ions) present in the solution.
For this reason, there are three types of equations:
The molecular equation – gives the overall reaction stoichiometry but not necessarily the actual
forms of the reactants and products in solution.
The complete ionic equation – represents as ions all reactants and products that are strong
electrolytes.
The net ionic equation – includes only those solution components undergoing a change. Spectator
ions not included.
Writing net ionic equations is an integral component of the AP Chemistry exam. You will need to
have a thorough understanding of the solubility rules. If you have not already printed out the
solubility chart from section 3.02, do so.
We will practice writing net ionic equations first. Then, we will learn how to perform stoichiometric
calculations on precipitation reactions.
Example 1:
Write the net ionic equation for the reaction between aqueous potassium chloride and aqueous
silver nitrate.
1. Write the balanced molecular equation. You have two compounds, so this will be a double
replacement reaction.
KCl + AgNO3  KNO3 + AgCl
2. Add the subscripts to the molecular equation. Refer to the solubility chart to determine
subscripts.
You are given the information that the reactants are aqueous. From the solubility chart, you can
see that KNO3 will be soluble (Rule 1). AgCl is not (Rule 3).
KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
3. Write the complete ionic equation.
You will show each aqueous term as an ion.
K+(aq) +Cl-(aq) + Ag+(aq) +NO3-(aq)  K+(aq) +NO3-(aq) + AgCl(s)
Notice that AgCl is not written as ions because it is solid.
4. Find the spectator ions. These are the ions that do not directly participate in the reaction.
In this case, the participating ions are Ag+ and Cl- because the form a precipitate.
K+ and NO3- are note directly participating; they are the spectator ions.
.5. Cross out the spectator ions
K+(aq) +Cl-(aq) + Ag+(aq) +NO3-(aq)  K+(aq) +NO3-(aq) + AgCl(s)
6. Write the net ionic equation. This includes only the participating species.
Cl-(aq) + Ag+(aq)  AgCl(s)
PRACTICE:
Write the net ionic equations for the following.
Check your answers below when you are finished.
Write the molecular equation, write the complete ionic equation , identify the spectator ions and
possible precipitates, and the net ionic equation for the following reactions.
1. Aqueous mercury II chloride and aqueous potassium sulfide
2. Aqueous sodium carbonate and aqueous calcium chloride
3. Aqueous copper II chloride and aqueous ammonium phosphate
ANSWERS
1.
a. molecular equation: HgCl2(aq) + K2S(aq)  HgS(s) + 2KCl(aq)
b. complete ionic equation: Hg+2(aq) + 2Cl-(aq) + 2K+(aq) + S-2(aq)  HgS(s) + 2K+(aq) + 2Cl-(aq)
c. spectator ions: K+, Cld. precipitate: HgS
e. net ionic equation: Hg+2(aq) + S-2(aq)  HgS(s)
2.
a. molecular equation: Na2CO3(aq) + CaCl2(aq)  2NaCl(aq) + CaCO3(s)
b. complete ionic equation:
2Na+(aq) + CO3-2(aq) + Ca+2(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + CaCO3(s)
c. spectator ions: Na+, Cld. precipitate: CaCO3
e. net ionic equation: CO3-2(aq) + Ca+2(aq)  CaCO3(s)
3. a. molecular equation: 3CuCl2(aq) + 2(NH4)3PO4(aq)  Cu3(PO4)2(s) + 6NH4Cl
b. complete ionic equation:
3Cu+2(aq) + 6Cl-(aq) + 6NH4+(aq) + 2PO4-3 (aq)  6NH4+(aq) + 6Cl-(aq) + Cu3(PO4)2(s)
c. spectator ions: Cl-, NH4+
d. precipitate: Cu3(PO4)2
e. net ionic equation:
3Cu+2(aq) + 2PO4-3 (aq)  Cu3(PO4)2(s)
3.09 Perform Calculations on Solution Equations
Now, we will add everything we have done in this unit to perform
stoichiometric calculations on reactions that occur in solution.
Example 1:
Calculate the mass of solid sodium chloride that must be added to 1.50 L of a 0.100 M AgNO 3
solution to precipitate all the Ag+ ions in the form of AgCl.
We will use the stoichiometry chart that we used before. The only difference, however, is that we
will have to calculate the number of moles of Ag+ present from the concentration.
Write the concentration and volume on the volume line above silver nitrate. See violet text
Calculate the number of moles of silver nitrate are present by multiplying the volume by the
concentration. Write the answer on line 2. See blue text.
The mole ratio is 0.15. See dark red text.
Multiply the mole ratio to determine the amount of NaCl needed in grams. See red text.
Line 1: Volume (L)
Line 2: Moles (mol)
Line 3: Mass (g)
Line 4: Equation
Line 5: GAW/GMW (g/mol)
Line 6: Moles (mol) :
Line 7: Mass (g)
Check: 228.5 g = 228.5 g
1.50 L (0.100 M)
.15 mol
8.78 g
NaCl(s) +
58.5 g/mol
1
58.5 g
AgNO3-(aq)
170g/mol
1
170g

AgCl(s)+
NaNO3(aq)
143.5 g/mol 85g/mol
1
1
166.5 g
85 g
mole ratio: .15
You need 8.78 g of solid sodium chloride to precipitate the silver.
Example 2:
Calculate the mass of precipitate formed when 1.25 L of 0.0500 M lead II nitrate and 2.00 L of
0.0250 M sodium sulfate are mixed.
In this example, you are given two solutions. You will need to calculate the number of moles
present in each. See violet and blue text.
Next, find the limiting reactant. See dark red text.
Sodium sulfate is the limiting reactant. Perform calculations using that index.
To determine the mass of precipitate produced, multiply the index by the mass
Line 1: Volume (L)
Line 2: Moles (mol)
Line 3: Mass (g)
Line 4: Equation
Line 5: GAW/GMW (g/mol)
Line 6: Moles (mol) :
Line 7: Mass (g)
Check: 172 g = 172 g
1.25L(0.05M)
2.00L(0.025)M
0.0625 mol
0.05 mol
Pb(NO3)2(aq) +
332 g/mol
1
332 g
.
15.2g
Na2SO4(aq) 
2NaNO3(aq) +
PbSO4(s)
142g/mol
86 g/mol
304g/mol
1
2
1
142g
172 g
304g
mole ratio: Pb(NO3)2 = 0.0625/1 = 0.0625
Na2SO4 = 0.05/1 = .05
15.2 g of lead II sulfate will precipitate.
Unit 3 Free Response Question
Answer this question completely. Show all your work!
Save as an .rtf file.
Submit it to the drop box for grading.
You have a 3.00 L of a 3.00 M solution of sodium chloride solution. You also have 2.00 L of a 2.00
M solution of silver nitrate.
A.
B.
C.
D.
E.
F.
G.
H.
How many grams of sodium chloride are present in 3.00 L of a 3.000 M solution?
How many grams of silver nitrate are present in 2.00 L of a 2.00 M solution?
Write the molecular equation. (include the (s), (l), (g), (aq) symbols)
Write the complete ionic equation (include the (s), (l), (g), (aq) symbols)
What are the spectator ions?
Is there a precipitate?
Write the net ionic equation. (include the (s), (l), (g), (aq) symbols)
Calculate the concentrations of the following ions when the two solutions are added
together.
 Sodium ion
 Chlorine ion
 Silver ion
 Nitrate ion
I. What is the amount in grams of precipitate formed (assume a 100% yield)?
J. What is the amount in grams of precipitate formed if you have an 85% yield?