Stoichiometry
Agenda
• Day 57 - Stoichiometry Mass to Moles
• Lesson: PPT- Stoichiometry Mass to Moles
• Handouts: 1. Stoichiometry Part 1Handout; 2.
Stoichiometry Worksheet
• Text: 1. P. 322-324- Stoichiometry Mass to
Moles
• HW: 1. Finish all the worksheets; 2. P. 325 #
3-10
•
Stoichiometry
Consider: 2 Mg (s) + O2 (g)  2 MgO (s)
The balanced equation gives us the following information:
•
2 Mg (s)
+ O2 (g)
 2 MgO (s)
2atoms
1 molecule
2 formula units
2 moles
1 mole
2 moles
(2 mol)(MM of Mg)
(1 mol)(MM of O2)
(2 mol)(MM of MgO)
(2 mol)( 24g/mol)
(1 mol)( 32g/mol)
(2 mol)( 40g/mol)
48 g
32 g
80 g
Note that the Law of Conservation of Mass is always
obeyed.
Stoichiometry
Consider: 4NH3 + 5O2  6H2O + 4NO
• Recall that many conversion factors exist:
4 mol NH3/5 mol O2, 6 mol H2O/4 mol NH3, etc
• In words, this tells us that for every 4 moles of NH3, 5
moles of O2 are required, etc.
•
•
“Stoichiometry” refers to the relative quantities of moles.
It also refers to calculations that make use of mole ratios.
•
•
•
Recall also that molar masses provide factors:
1 mol NH3 / 17 g NH3, 32 g O2 / 1 mol O2
Is 4 g NH3 / 5 g O2 a conversion factor?
•
•
No. The equation tells us moles not grams.
Notice that stoichiometry requires precision
Stoichiometry questions (1) -Factor-Label
Method
Consider : 4NH3 + 5O2  6H2O + 4NO
• How many moles of H2O are produced if 0.176 mol of O2
are used?
# mol H2O= 0.176 mol O2 x 6 mol H2O = 0.211
5 mol O2 mol
H2Oif
• How many moles of NO are produced in the reaction
17.00 mol of H2O are also produced?
# mol NO=17.00 mol H2O x 4 mol NO = 11.33
6 mol H2O mol NO
Notice that a correctly balanced equation is essential to get
the right answer
Stoichiometry questions (2)
Consider : 4NH3 + 5O2  6H2O + 4NO
• How many grams of H2O are produced if 1.9 mol of NH3
are combined with excess oxygen?
# g H2O=
1.9 mol NH3 x 6 mol H2O x18.02 g H2O= 51 g
H2O
4 mol NH3 1 mol H2O
•
How many grams of O2 are required to produce 0.3 mol of
H2O?
# g O2=
0.3 mol H2O x 5 mol O2 x 32 g O2 = 8 g O2
6 mol H2O
1 mol O2
Stoichiometry questions (3)
Consider : 4NH3 + 5O2  6H2O + 4NO
• How many grams of NO is produced if 12 g of O2 is
combined with excess ammonia?
# g NO=
12 g O2 x 1 mol O2 x 4 mol NO x 30.01 g NO
32 g O2
5 mol O2
1 mol NO
= 9.0 g NO
Converting grams to grams
Notice that we cannot directly convert from grams of
one compound to grams of another. Instead we have to
go through moles.
• Many stoichiometry problems follow a pattern:
grams(x)  moles(x)  moles(y)  grams(y)
• We can start anywhere along this path depending on the
question we want to answer
Q- for the reaction 2H2 + O2  2H2O what is the path we
would take for the following
• Given 2 moles H2O, calculate grams H2O?
• Moles O2 required for 36 g H2?
• Grams of H2O produced from 6 grams O2?
•
Moving along the stoichiometry path
•
We always use the same type of information to make the
jumps between steps:
Molar mass of x
Molar mass of y
grams (x)  moles (x)  moles (y)  grams (y)
Mole ratio from
balanced equation
Given: 4NH3 + 5O2  6H2O + 4NO
a) How many moles of H2O can be made using 0.50 mol NH3?
b) What mass of NH3 is needed to make 1.50 mol NO?
c) How many grams of NO can be made from 120.0 g of NH3?
Answers
4NH3 + 5O2  6H2O + 4NO
a)
# mol H2O= 0.50 mol NH3x 6 mol H2O = 0.75
mol
4 mol NH3
b)
H
O
2
# g NH3=
1.50 mol NO x 4 mol NH3 x 17.04 g NH3= 25.6 g
NH3
c)
4 mol NO 1 mol NH3
# g NO=
120.0 g NH3 x1 mol NH3 x 4 mol NO x 30.01 g NO
17.04g
4 mol NH3 1 mol NO
NH3
=
211.3 g NO
The steps to solving stoichiometric
problems are as follows- CHART METHOD
1.
2.
3.
4.
5.
6.
7.
8.
Write the chemical equation
Balance the chemical equation
Write the molar ratio for the equation
Write all given masses of substances in the
equation
Write the molar masses for all substances
Find the number of moles of each substance
Find the new molar ratio
Solve for the unknown
Given: 4NH3 + 5O2  6H2O + 4NO
How many grams of NO can be made from 120.0 g of
NH3?
Molar Ratio
(MR)
Mass (m)
Molar Mass
(MM)
Moles (n)
4 NH3
5 O2
6 H2O
4 NO
4
5
6
4
120.0 g
? m = n x MM
= 7.042 x30.01
= 211.3 g
30.01 g/mol
17.04 g/mol
n = m/ MM
7.042 mol
4:4
= 4/4 x7.042
= 7.042 mol
More Stoichiometry Questions
Follow the rules for significant digits. Show all calculations.
1. 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
a) what mass of O2 will react with 400.0 g C4H10?
b) how many moles of water are formed in a)?
2. 3 HCl + Al(OH)3 -> 3 H2O + AlCl3
How many grams of aluminum hydroxide will react with
5.30 moles of HCl?
3. Ca(ClO3)2 -> CaCl2 + 3 O2
What mass of O2 results from the decomposition of 1.00
kg of calcium chlorate?
4. The reaction of Ca with water can be predicted using the
activity series. What mass of water is needed to
completely react with 2.35 g of Ca?
5. Fe2O3 + 3CO -> 2Fe + 3CO2.
a) How many moles of carbon monoxide are required to
react with 163.0 g of iron(III) oxide?
b) How many grams of CO2 are produced from a reaction
that also produces 23.9 grams of Fe?
6.
3Cu + 8HNO3  3Cu(NO3)2 + 4H2O + 2NO
a) how many moles of copper(II) nitrate can be prepared
from 17.0 moles of Cu?
b) how many grams of copper(II) nitrate can be prepared
using 3.8 moles of HNO3?
c) what mass of water results from the reaction of 8.50
kg of copper metal?
Practice Problems:
8. According to the following equation how many moles of lithium hydroxide are required
to react with 20.0 moles of CO2?
CO2 + 2LiOH  Li2CO3 + H2O
9. How many moles of ammonia, NH3 are produced when 6.00 moles of hydrogen gas react
with excess nitrogen gas?
3H2 + N2  2NH3
10. What mass, in grams, of glucose is produced when 3.00 moles of water react with
carbon dioxide?
6CO2 + 6H2O  C6H12O6 + 6O2
11. What mass in grams of magnesium oxide is produced when 2.00 moles of magnesium
react with oxygen?
12. The reaction below is run using 824 grams of NH3 and excess oxygen, how many moles
of NO are formed? How many moles of H2O are formed?
4NH3 + 5O2  4NO + 6H2O
13. If mercury (II) oxide decomposes, how many grams of mercury (II) oxide are needed to
produce 125 grams of oxygen?
14. How many grams of SnF2 are produced from the reaction of 30.0 grams of HF with Sn?
Sn + 2HF  SnF2 + H2
15. What mass of aluminum is produced by the decomposition of 5000. grams of Al2O3?
16. How many molecules of hydrogen chloride gas will be produced when 33.5 grams of
chlorine combine with excess hydrogen?
Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.
Agenda
• Day 58 - Limiting Factor
• Lesson: PPT- Stoichiometry Mass to Moles
• Handouts: 1. Stoichiometry Part 1(Limiting
Factor)Handout; 2. Limiting Factor
Stoichiometry Worksheet
• Text: 1. P. 326-334- Stoichiometry - Limiting
Factor
• HW: 1. Finish all the worksheets; 2. P. 330 #
2-10; P.334 # 1-3; P. 335 # 2-11
Let’s say you need the following
to make one hamburger:
2 buns
1 piece of meat
1 piece of lettuce
If you have 16 buns, 16 pieces of meat, and 20 pieces of
lettuce, how many hamburgers can we make? What limits
how many we can make? What is in excess?
STOICHIOMETRY- MOLE AND LIMITING
FACTOR
• A balanced chemical equation indicates the number
of moles of each REACTANT that will react and the
number of moles of each PRODUCT that will be
produced in the reaction
• Even if one reactant is present in excess only the
amount need to react, dictated by the molar ratio,
will actually react
• The amount reacting will be determined by the
reactant that is in the lesser amount (the limiting
factor). Problems of this type may be recognized by
the fact that information will be given about at least
two reactants
Limiting reagent defined
Given: 4NH3 + 5O2  6H2O + 4NO
Q - How many moles of NO are produced if
__ mol NH3
are burned in __ mol O2?
4 mol NO, works out exactly
4 mol NH3, 5 mol O2
4 mol NH3, 20 mol O2
4 mol NO, with leftover O2
8 mol NH3, 20 mol O2
8 mol NO, with leftover O2
Here, NH3 limits the production of NO; if there was more
NH3, more NO would be produced
• Thus, NH3 is called the “limiting reagent”
2 mol NO, leftover NH3
4 mol NH3, 2.5 mol O2
•
•
In limiting reagent questions we use the limiting reagent
as the “given quantity” and ignore the reagent that is in
excess …
Limiting reagents in stoichiometry
4NH3 + 5O2  6H2O + 4NO
E.g. How many grams of NO are produced if 4 moles NH3
are burned in 20 mol O2?
Since NH3 is the limiting reagent we will use this as our
“given quantity” in the calculation
# g NO=
4 mol NH3 x 4 mol NO x 30.0 g NO= 120 g NO
4 mol NH3 1 mol NO
• Sometimes the question
is more complicated. For
example, if grams of the two reactants are given instead of
moles we must first determine moles, then decide which
is limiting …
Solving Limiting Reagents: g to mol
4NH3 + 5O2  6H2O + 4NO
Q - How many g NO are produced if 20 g NH3 is burned in
30 g O2?
A - First we need to calculate the number of moles of each
reactant
# mol NH3= 20 g NH3 x 1 mol NH3 = 1.176
mol NH3
17.0 g NH3
# mol O2=
30 g O2
1 mol O2
0.9375
x
=
mol O2
32.0 g O2
A – Once the number of moles of each is calculated we can
determine the limiting reagent via a chart …
Comparison chart- A quick method to determine the
LIMITING REAGENT Use this amount in
the calculations
NH3
O2
4
5
What we Have
1.176 mol
0.937 mol
Limiting Factor Test
1.176/4 =
0.294 mol
0.937/5 =
0.1874mol
Molar Ratio
* The smaller value indicates the Limiting Reactant
A - There is more NH3 (what we have) than needed (what
we need). Thus NH3 is in excess, and O2 is the limiting
reagent.
Determine the reactant in excess and the excess
amount if 20 g NH3 is burned in 30 g O2? Use this amount in
the calculations
Reactants
Molar Ratio
NH3
O2
4
5
What we have
1.176 mol
0.937 mol
Limiting Factor
Test
4/5 x 0.937 mol
5/4 x1.176 mol
What we need
Excess
0.749 mol
1.47 mol
1.176 – 0749
0.937 – 1.47 =
= + 0.427 mol
=
-
Limiting
Factor
Stoichiometry (given = limiting)
So far we have followed two steps …
1) Expressed all chemical quantities as moles
2) Determined the limiting reagent via a chart
Finally we need to …
3) Perform the stoichiometry using the limiting reagent as
the “given” quantity
Q - How many g NO are produced if 20 g NH3 is burned in
30 g O2?
4NH3 + 5O2  6H2O + 4NO
# g NO=
30 g O2 x 1 mol O2 x 4 mol NO x 30.0 g NO
32.0 g O2
5 mol O2 1 mol NO
=
22.5 g NO
Given: 4NH3 + 5O2  6H2O + 4NO
How many g NO are produced if 20 g NH3 is burned in
30 g O2?
Molar Ratio
(MR)
Mass (m)
Molar Mass
(MM)
Moles (n)
4 NH3
5 O2
6
H2O
4 NO
4
5
6
4
20.0 g
30 g
17.04 g/mol
1.176 mol
32.0
g/mol
0.937mol
L. F.
? m = n x MM
= 0.745 x30.01
= 22.5 g
30.01 g/mol
5:4
= 4/5 x 0.937
= 0.745mol
Practice questions
1.
2.
3.
4.
5.
2Al + 6HCl  2AlCl3 + 3H2
If 25 g of aluminum was added to 90 g of HCl, what mass
of H2 will be produced (try this two ways – with a chart &
using the shortcut)?
N2 + 3H2  2NH3: If you have 20 g of N2 and 5.0 g of H2,
which is the limiting reagent?
What mass of aluminum oxide is formed when 10.0 g of
Al is burned in 20.0 g of O2?
When C3H8 burns in oxygen, CO2 and H2O are produced.
If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2
is produced?
How can you tell if a question is a limiting reagent
question vs. typical stoichiometry?
1
# mol Al = 25 g Al x 1 mol Al = 0.926 mol
27.0 g Al
# mol HCl = 90 g HCl x 1 mol HCl = 2.466 mol
36.5 g HCl
What we
have
What we
need
Al
0.926
0.926/0.926
= 1 mol
2
2/2 = 1 mol
HCl
2.466
2.466/0.926 HCl is
limiting.
= 2.7 mol
6
6/2 = 3 mol
# g H2 =
1 mol HCl 3 mol H2 2.0 g H2
90 g HCl x
x
x
= 2.47 g H2
36.5 g HCl 6 mol HCl 1 mol H2
Question 1: shortcut
2Al + 6HCl  2AlCl3 + 3H2
If 25 g aluminum was added to 90 g HCl, what
mass of H2 will be produced?
# g H2= 25 g Al x 1 mol Al x 3 mol H2 x 2.0 g H2 = 2.78 g H2
27.0 g Al 2 mol Al 1 mol H2
# g H2 =90 g HCl x1 mol HClx 3 mol H2 x 2.0 g H2 = 2.47 g H2
36.5 g HCl 6 mol HCl 1 mol H2
Question 2: shortcut
N2 + 3H2  2NH3
If you have 20 g of N2 and 5.0 g of H2, which is
the limiting reagent?
# g NH3=
20 g N2 x1 mol N2x 2 mol NH3 x17.0 g NH3= 24.3 g H2
28.0 g N2 1 mol N2 1 mol NH3
# g NH3 =
5.0 g H2 x 1 mol H2 x 2 mol NH3 x17.0 g NH=3 28.3 g H2
2.0 g H2 3 mol H2 1 mol NH3
N2 is the limiting reagent
Question 3: shortcut
4Al + 3O2  2 Al2O3
What mass of aluminum oxide is formed when
10.0 g of Al is burned in 20.0 g of O2?
# g Al2O3=
10.0 g Al x 1 mol Al x 2 mol Al2O3 x102.0 g Al2O3 = 18.9 g Al2O3
1 mol H2
27.0 g Al 4 mol Al
# g Al2O3=
20.0 g O2x 1 mol O2 x2 mol Al2O3x102.0 g Al2O3 = 42.5 g Al2O3
32.0 g O2 3 mol O2 1 mol H2
Question 4: shortcut
C3H8 + 5O2  3CO2 + 4H2O
When C3H8 burns in oxygen, CO2 and H2O are
produced. If 15.0 g of C3H8 reacts with 60.0 g
of O2, how much CO2 is produced?
# g CO2=
15.0 g C3H8x1 mol C3H8 x 3 mol CO2x44.0 g CO2 = 45.0 g CO2
44.0 g C3H8 1 mol C3H8 1 mol CO2
# g CO2=
60.0 g O2x 1 mol O2 x 3 mol CO2 x 44.0 g CO2 = 49.5 g CO2
32.0 g O2 5 mol O2 1 mol CO2
5. Limiting reagent questions give values for
two or more reagents (not just one)
Question 2
# mol N2= 20 g N2 x 1 mol N2 = 0.714 mol N2
28 g N2
# mol H2= 5.0 g H2 x 1 mol H2 = 2.5 mol H2
2 g H2
N2
H2
0.714 mol
2.5 mol
What we have
0.714/0.714
2.5/0.714
= 1 mol
= 3.5 mol
What we need
1 mol
3 mol
We have more H2 than what we need, thus H2
is in excess and N2 is the limiting factor.
4Al + 3O2  2 Al2O3
# mol Al = 10 g Al x 1 mol Al = 0.37 mol Al
27 g Al
# mol O2 = 20 g O2 x 1 mol O2 = 0.625 mol O2
32 g O2
There is
Al
O2
more
0.37 mol
0.625 mol
What we
than
0.37/.37
0.625/0.37
have
enough
= 1 mol
= 1.68 mol
O
;
Al
is
2
What we
4 mol
3 mol
limiting
need
4/4 = 1 mol 3/4 = 0.75 mol
3
# g Al2O3 = 0.37 mol Al x 2 mol Al2O3 x 102 g Al2O3
4 mol Al
1 mol Al2O3
=
18.87 g Al O
C3H8 + 5O2  3CO2 + 4H2O
# mol C3H8 =15 g C3H8 x 1 mol C3H8 = 0.34 mol
C 3H 8
44 g C3H8
# mol O2 = 60 g O2 x 1 mol O2 = 1.875 mol O2
32 g O2
We have
C 3H 8
O2
more
than
0.34
mol
1.875
mol
What we
enough O2,
0.34/.34
1.875/0.34
have
C3H8 is
= 1 mol
= 5.5 mol
limiting
Need
1 mol
5 mol
# g CO2 =
44
g
CO
3
mol
CO
2
2 x
0.34 mol C3H8 x
1 mol C3H8 1 mol CO2
=
45 g CO2
4
Limiting Reagents: shortcut
MgCl2 + 2AgNO3  Mg(NO3)2 + 2AgCl
If 25 g magnesium chloride was added to 68 g
silver nitrate, what mass of AgCl will be produced?
# g AgCl=
25 g MgCl2 x 1 mol MgCl2 x 2 mol AgCl x 143.3 g AgCl
95.21 g MgCl2 1 mol MgCl2 1 mol AgCl
=
75.25 g AgCl
# g AgCl=
68 g AgNO3 x 1 mol AgNO3 x 2 mol AgCl x 143.3 g AgCl
169.88 g AgNO3 2 mol AgNO3 1 mol AgCl
=
57.36 g AgCl
Agenda
• Day 59 - % Yield
• Lesson: PPT- Stoichiometry Mass to Moles
• Handouts: 1. Stoichiometry Part 1( Percent
Yield) Handout; 2. Percent Yield Worksheet
• Text: 1. P. 336-338 - % Yield
• HW: 1. Finish all the worksheets; 2. P. 339
#4-14
•
PERCENTAGE YIELD
Yield: the amount of product
Theoretical yield: the amount of product we expect,
based on stoichiometric calculations
Actual yield: amount of product from a procedure or
experiment (this is given in the question)
actual
yield
Percent yield =
theoretical yield
x 100%
Give 4 possible reasons why the actual yield in a
chemical reaction often falls short of the theoretical
yield.
• Not all product is recovered (e.g. spattering)
• Reactant impurities (e.g. weigh out 100 g of
chemical which has 20 g of junk)
• A side reaction occurs (e.g. MgO vs. Mg3N2)
• The reaction does not go to completion
Sample problem
Q - What is the % yield of H2O if 138 g H2O is produced from
16 g H2 and excess O2?
Step 1: Write the balanced chemical equation
2H2 + O2  2H2O
Step 2: Determine actual and theoretical yield. Actual is
given, theoretical is calculated:
# g H2O= 16 g H2 x1 mol H2 x2 mol H2Ox 18.02 g H2O= 143 g
2.02 g H2 2 mol H2 1 mol H2O
Step 3: Calculate % yield
% yield =
actual
138 g H2O
x 100% =
x 100% = 96.7%
theoretical
143 g H2O
Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is produced
from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield. Actual is
given, theoretical is calculated:
# g NH3= 20.0 mol H2 x 2 mol NH3 x 17.04 g NH3 = 227 g
3 mol H2
1 mol NH3
Step 3: Calculate % yield
% yield = actual x 100% = 40.5 g NH3 x 100% = 17.8%
theoretical
227 g NH3
Practice problem
Q- When 5.00 g of KClO3 is heated it decomposes
according to the equation: 2KClO3  2KCl + 3O2
a) Calculate the theoretical yield of oxygen.
b) Give the % yield if 1.78 g of O2 is produced.
c) How much O2 would be produced if the
percentage yield was 78.5%?
Answers
a)
# g O2= (also works if you use mol O2)
5.00 g KClO3 x 1 mol KClO3 x 3 mol O2 x 32 g O2
122.55 g KClO3 2 mol KClO3 1 mol O2
= 1.958 g
b)
% yield = actual x 100% = 1.78 g O2 x 100% = 90.9%
theoretical
1.958 g O2
c)
actual
x g O2
x 100% =
x 100% = 78.5%
theoretical
1.958 g O2
x g O2 = 78.5% x 1.958 g O2 = 1.537 g O2
100%
% yield =
Challenging question
2H2 + O2  2H2O
What is the % yield of H2O if 58.0 g H2O are produced by
combining 60.0 g O2 and 7.00 g H2?
Hint: determine limiting reagent first
# g H2O= 60 g O2 x1 mol O2 x2 mol H2Ox 18.02 g H2O= 68 g
32 g O2 1 mol O2 1 mol H2O
# g H2O= 7.0 g H2 x1 mol H2 x2 mol H2Ox 18.02 g H2O= 62.4 g
2.02 g H2 2 mol H2 1 mol H2O
% yield =
actual
58 g H2O
x 100% =
x 100% = 92.9%
theoretical
62.4 g H2O
More Percent Yield Questions
1. The electrolysis of water forms H2 and O2.
2H2O  2H2 + O2
What is the % yield of O2 if 12.3 g of O2 is produced from the
decomposition of 14.0 g H2O?
2. 107 g of oxygen is produced by heating 300 grams of
potassium chlorate. Calculate % yield.
2KClO3  2KCI + 3O2
3What is the % yield of ferrous sulphide if 3.00 moles of Fe
reacts with excess sulfur to produce 220 grams of ferrous
sulphide? Fe + S  FeS
More Percent Yield Questions
4.
Iron pyrites (FeS2) reacts with oxygen according to the
following equation:
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of iron pyrites is burned in 200 g of O2, 143
grams of ferric oxide is produced. What is the percent
yield of ferric oxide?
5.
70 grams of manganese dioxide is mixed with 3.5 moles
of hydrochloric acid. How many grams of Cl2 will be
produced from this reaction if the % yield for the
process is 42%?
MnO2 + 4HCI  MnCl2 + 2H2O + Cl2
1. The electrolysis of water forms H2 & O2.
2H2O  2H2 + O2
Give the percent yield of O2 if 12.3 g O2 is produced
from the decomposition of 14.00 g H2O?
• Actual yield is given: 12.3 g O2
• Next, calculate theoretical yield
# g O2=
14.00 g H2Ox 1 mol H2O x 1 mol O2 x 32.00 g O2= 12.43 g
18.02 g H2O 2 mol H2O 1 mol O2
Finally, calculate % yield
% yield =
actual
12.3 g O2
x 100% =
x 100% = 98.9%
theoretical
12.43 g O2
2. 107 g of oxygen is produced by heating 300 grams of
potassium chlorate.
2KClO3  2KCI + 3O2
• Actual yield is given: 107.0 g O2
• Next, calculate theoretical yield
# g O2=
300 g KClO3 x 1 mol KClO3 x 3 mol O2 x 32 g O2
122.55 g KClO3 2 mol KClO3 1 mol O2
= 117.5 g
Finally, calculate % yield
% yield = actual x 100% = 107 g O2 x 100% = 91.1%
theoretical
117.5 g O2
3. What is % yield of ferrous sulfide if 3 mol Fe produce
220.0 grams of ferrous sulfide?
Fe + S  FeS
• Actual yield is given: 220.0 g FeS
• Next, calculate theoretical yield
# g FeS=3.00 mol Fe x 1 mol FeS x 87.91 g FeS= 263.7 g
1 mol Fe 1 mol FeS
Finally, calculate % yield
% yield = actual x 100% = 220 g O2 x 100% = 83.4%
theoretical
263.7 g O2
4. 4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300.0 g of FeS2 is burned in 200.0 g of O2, 143 g Fe2O3
results. % yield Fe2O3?
First, determine limiting reagent
# g Fe2O3=
300.0 g FeS2x 1 mol FeS2 x 2 mol Fe2O3 x159.7 g Fe2O3
119.97 g FeS2 4 mol FeS2 1 mol Fe2O3
=
199.7 g Fe2O3
200.0 g O2 x 1 mol O2 x 2 mol Fe2O3 x159.7 g Fe2O3
32 g O2
11 mol O2
1 mol Fe2O3
= 181.48 g Fe2O3
% yield = actual x 100% = 143 g Fe2O3 x 100% = 78.8%
theoretical
181.48 g Fe2O3
5. 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How
many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 +
2H2O + Cl2
# g Cl2=
70 g MnO2 x 1 mol MnO2 x 1 mol Cl2 x 70.9 g Cl2
86.94 g MnO2 1 mol MnO2
1 mol Cl2
=
57.08 g Cl2
3.5 mol HCl x 1 mol Cl2 x 71 g Cl2 = 62.13 g Cl2
4 mol HCl
1 mol Cl2
x g Cl2
% yield = actual x 100%=
x 100% = 42%
theoretical
57.08 g Cl2
x g Cl2 = 42% x 57.08 g Cl2 = 24 g Cl2
100%