CHEMICAL REACTIONS
CHAPTER 4
Reactants: Zn + I2
Product: ZnI2
2
Chapter 4 Outline
• Chemical Equations
• Stoichiometry
–Limiting Reactants
–Chemical Analysis
Chemical Equations
Depict the kind of reactants and
products and their relative amounts in a
reaction
4 Al(s) + 3 O2(g)
2Al2O3(s)
The numbers in the front are called
stoichiometric coefficients
The letters (s), (g), and (l) are the
physical states of compounds.
3
4
Chemical Equations
4 Al(s) + 3 O2(g)
2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules
2 molecules of Al2O3
4 moles of Al + 3 moles of O2
2 moles of Al2O3
5
Chemical Equations
• Because the same atoms
are present in a reaction
at the beginning and at
the end, the amount of
matter in a system does
not change.
• The Law of the
Conservation of Matter
6
Chemical Equations
Because of the principle of the
conservation of matter, an
equation must be
balanced.
It must have the same number of
atoms of the
same kind on both sides.
Lavoisier, 1788
7
Balancing Equations
8
Balancing Equations
9
Balancing Equations
___C3H8 (g) + ___ O2 (g) ----> ___CO2 (g) + ___ H2O (g)
C3H8 (g) +
5 O2 (g) ---->
3 CO2 (g)
+
4 H2O (g)
10
Balancing Equations
___B4H10 (g) + ___ O2 (g) ---> ___ B2O3 (g) + ___ H2O (g)
2 B4H10 (g) + 11 O2 (g) ---> 4 B2O3 (g) + 10 H2O (g)
11
STOICHIOMETRY
- the study of the
quantitative
aspects of
chemical
reactions.
12
PROBLEM: If 454 g of NH4NO3
decomposes, how much H2O and N2O are
formed? What is the theoretical yield of
products?
STEP 1
Write the balanced chemical equation
NH4NO3
N 2 O + 2 H 2O
13
454 g of NH4NO3 --> N2O + 2 H2O
STEP 2
Convert reactant mass to moles
(454 g) --> moles
1 mol
454 g •
= 5.68 mol NH4 NO3
80.04 g
14
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3
Convert moles reactant --> moles product.
Relate moles NH4NO3 to moles product.
1 mol NH4NO3 --> 2 mol H2O
Express this relation as the
STOICHIOMETRIC FACTOR
2 mol H 2 O produced
1 mol NH 4 NO 3 used
15
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3
Convert moles reactant (5.68 mol)
moles product
2 mol H 2O produced
5.68 mol NH 4NO 3 •
1 mol NH 4NO 3 used
= 11.4 mol H2O produced
16
454 g of NH4NO3 --> N2O + 2 H2O
STEP 4
Convert moles product (11.4 mol) to mass
product.
This is called the
THEORETICAL YIELD
This is the “Expected” # of moles.
17
454 g of NH4NO3 --> N2O + 2 H2O
STEP 4
Convert moles prod. (11.4 mol) to mass prod.
18.02 g
11.4 mol H 2 O •
= 204 g H2O
1 mol
This is the “Expected” Mass!
ALWAYS FOLLOW THESE STEPS IN SOLVING
STOICHIOMETRY PROBLEMS!
Repeat to find the grams of N2O formed.
18
454 g of NH4NO3 --> N2O + 2 H2O
STEP 5
How much N2O is formed?
Total mass of reactants=total mass of
products
454 g NH4NO3 = ___ g N2O + 204 g H2O
mass of N2O = 250. g
This is an alternate method.
19
454 g of NH4NO3 --> N2O + 2 H2O
STEP 6
Calculate the percent yield.
If you isolated only 131 g of N2O, what is the
percent yield?
This compares the theoretical (250. g) and
actual (131 g) yields.
20
454 g of NH4NO3 --> N2O + 2 H2O
STEP 6
Calculate the percent yield.
actual yield
% yield =
• 100%
theoretical yield
131 g
% yield =
• 100% = 52.4%
250. g
21
General Plan For
Stoichiometry Calculations
Mass
reactant
Mass
product
Moles
reactant
Moles
product
22
PROBLEM: Using 5.00 g of H2O2, what mass of
O2 and of H2O can be obtained?
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Reaction is catalyzed by MnO2
Step 1: moles of H2O2
Step 2: use STOICHIOMETRIC FACTOR to
calculate moles of O2
Step 3: mass of O2
Repeat for H2O.
Reactions Involving a
LIMITING REACTANT
• In a given reaction, there is not
enough of one reagent to use up the
other reagent completely.
• The reagent in short supply LIMITS
the quantity of product that can be
formed.
24
LIMITING REACTANTS
Reactants
2 NO(g) + O2 (g)
Products
2 NO2(g)
NO
Limiting reactant = ___________
O2
Excess reactant = ____________
25
LIMITING
REACTANTS
LIMITING REACTANTS
26
React solid Zn with 0.100
mol HCl (aq)
Zn + 2 HCl ---> ZnCl2 + H2
mass Zn (g)
mol Zn
mol HCl
mol HCl/mol Zn
Rxn 1
7.00
0.107
0.100
0.93
Rxn 2
3.27
0.050
0.100
2.00
Rxn 3
1.31
0.020
0.100
5.00
27
PROBLEM: Mix 5.40 g of Al with 8.10 g
of Cl2. How many grams of Al2Cl6 can
form?
28
Step 1 of LR problem:
compare actual mole
ratio of reactants to
theoretical mole ratio.
29
Step 1 of LR problem: compare
actual mole ratio of reactants to
theoretical mole ratio.
2 Al + 3 Cl2 ---> Al2Cl6
Reactants must be in the mole ratio
mol Cl2
3
=
mol Al
2
30
Deciding on the Limiting Reactant
2 Al + 3 Cl2
If
Al2Cl6
mol Cl2
3
>
mol Al
2
then there is not enough Al
to use up all the Cl2, and the
limiting reagent is
Al.
31
Deciding on the Limiting Reactant
2 Al + 3 Cl2 ---> Al2Cl6
If
mol Cl2
3
<
mol Al
2
then there is not enough Cl2
to use up all the Al, and the
limiting reagent is
Cl2
32
Step 2 of LR problem: Calculate
moles of each reactant
We have 5.40 g of Al and 8.10 g of Cl2
1 mol
5.40 g Al •
= 0.200 mol Al
27.0 g
1 mol
8.10 g Cl 2 •
= 0.114 mol Cl 2
70.9 g
33
Find mole ratio of reactants
mol Cl2
0.114 mol
=
mol Al
0.200 mol
= 0.57
This
should be 3/2 or 1.5/1 if
reactants are present in the
exact stoichiometric ratio.
Limiting reagent is
Cl2
34
Mix 5.40 g of Al with 8.10 g of Cl2. What
mass of Al2Cl6 can form?
Limiting reactant = Cl2
Base all calculations on Cl2
grams
Cl2
moles
Cl2
grams
Al2Cl6
1 mol Al 2 Cl 6
3 mol Cl 2
moles
Al2Cl6
35
CALCULATIONS: calculate mass of
Al2Cl6 expected.
Step 1: Calculate moles of Al2Cl6
expected based on LR.
0.114 mol Cl2 •
1 mol Al 2Cl 6
= 0.0380 mol Al2Cl 6
3 mol Cl2
Step 2: Calculate mass of Al2Cl6
expected based on LR.
266.4 g Al 2Cl6
0.0380 mol Al 2Cl6 •
= 10.1 g Al 2Cl6
mol
36
How much of which reactant will
remain when reaction is complete?
• Cl2 was the limiting reactant.
Therefore, Al was present in excess.
But how much?
• First find how much Al was required.
• Then find how much Al is in excess.
37
Calculating Excess Al
2 Al + 3 Cl2
products
0.200 mol 0.114 mol = LR
38
Calculating Excess Al
2 Al + 3 Cl2
0.200 mol
products
0.114 mol = LR
0.114 mol Cl2 •
2 mol Al
= 0.0760 mol Al req' d
3 mol Cl2
Excess Al = Al available - Al required
= 0.200 mol - 0.0760 mol
= 0.124 mol Al in excess
39
Nitrogen and iodine react to form
nitrogentriiodide. If 50.0 g of nitrogen is
mixed with 350.0 g iodine, calculate the
number of grams of product formed and
the grams of reactant remaining.
N2
+
3 I2
28.0 g/mol
1.79 mole
2 NI3
253.8 g/mol
1.38 mole
394.7 g/mol
L.R. (0.460 S.U.)
50.0g -12.9g
37.1g left
0g left
363 g
350.0g + 50.0g = 400.0g = 363g + 37.1g
Using Stoichiometry to
Determine a Formula
Burn 0.115 g of a hydrocarbon, CxHy, and produce
0.379 g of CO2 and 0.1035 g of H2O.
What is the empirical formula of CxHy?
CxHy + oxygen
0.379 g CO2 + 0.1035 g H2O
40
41
Chemical Molecular Analysis
in the lab
42
Using Stoichiometry to
Determine a Formula
CxHy + some oxygen
0.379 g CO2 + 0.1035 g H2O
First, recognize that all C in CO2 and all H in H2O
is from CxHy.
1.
2.
Calculate moles of C in CO2
8.61 x 10-3 mol C
Calculate moles of H in H2O
1.149 x 10 -2 mol H
Using Stoichiometry to
Determine a Formula
CxHy + some oxygen
0.379 g CO2 + 0.1035 g H2O
Now find ratio of mol H/mol C to find values
of x and y in CxHy.
1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C
= 1.33 mol H / 1.00 mol C
= 4 mol H / 3 mol C
Empirical formula = C3H4
43
44
Chemical Analysis
-
Combustion Determine a Formula
CxHy + some oxygen
0.379 g CO2 + 0.1035 g H2O
Recognize that all C in CO2 and all H in H2O is from CxHy.
1. Calculate moles of C in CO2
0.379 g CO2
mole CO2
mole C
44.0 g CO2
mole CO2
= 0.00861 mole C
45
Chemical Analysis
-
Combustion Determine a Formula
CxHy+some oxygen
0.379gCO2+0.1035 g H2O
2. Calculate moles of H in H2O
0.1035 g H2O mole H2O 2 mole H
18.0 g H2O mole H2O
= 0.0115 mole H
Chemical Analysis
46
-
Combustion Determine a Formula
CxHy + some oxygen
0.379 g CO2 + 0.1035 g H2O
Now find ratio of mol H/mol C to find values
of x and y in CxHy.
0.0115 mol H/ 0.00861 mol C
= 1.34 mol H / 1.00 mol C
= 4.02 mol H / 3.00 mol C
Empirical formula = C3H4
Sample Problems
47
1) A 0.537 g sample of an unknown compound
containing only carbon, hydrogen, and oxygen was
burned to produced 1.030 g of CO2 and 0.632 g of H2O.
Determine the empirical formula. Given that the
molecular weight is approximately 90 g/mole,
determine the molecular formula.
1.030 g CO2 mole CO2
44.0g CO2
mole C
mole CO2
0.632 g H2O mole H2O 2 mole H
18.0 g H2O mole H2O
12.0 g C
mole C
= 0.281 g C
1.0 g H
mole H
= 0.070 g H
Sample Problems
48
1) A 0.537 g sample of an unknown compound
containing only carbon, hydrogen, and oxygen was
burned to produced 1.030 g of CO2 and 0.632 g of H2O.
Determine the empirical formula. Given that the
molecular weight is approximately 90 g/mole,
determine the molecular formula.
0.537 g C, H, O
- 0.281 g C
- 0.070 g H
0.186 g O
49
Sample Problems
C
H
O
.281 g 1 mole .070 g 1 mole
12.0 g
.186 g 1 mole
1.0 g
16.0 g
0.0234
0.070
0.0116
0.0116
0.0116
0.0116
2.02
6.0
Empirical formula
1.00
C2H6O
Sample Problems
50
Empirical formula C2H6O
90
46
=2
Molecular formula C4H12O2
Alternate method. Use grams of cmpd. and
its molar mass to determine moles of cmpd.
Divide moles of C and H by these moles to find
the subscripts for C and H. The subscript for O
can be determinded by difference.
51
Sample Problems
2) A 0.1247 g sample of ascorbic acid, vitamin C, was
burned to produce 0.1869 g of CO2 and 0.0510 g H2O.
Ascorbic acid contains only carbon, hydrogen, and
oxygen. Determine the empirical formula.
Given that the molecular weight is approximately 180
g/mole, determine the molecular formula.
.1869 g CO2 mole CO2
44.0 g CO2
mole C
mole CO2
12.0 g C
mole C
= 0.0510 g C
.0510 g H2O mole H2O
18.0 g H2O
2 mole H
mole H2O
1.0 g H
mole H
= 0.0057 g H
Sample Problems
52
2) A 0.1247 g sample of ascorbic acid, vitamin C, was
burned to produce 0.1869 g of CO2 and 0.0510 g H2O.
Ascorbic acid contains only carbon, hydrogen, and
oxygen. Determine the empirical formula. Given that
the molecular weight is approximately 180 g/mole,
determine the molecular formula.
0.1247 g C, H, O
- 0.0510 g C
- 0.0057 g H
0.0680 g O
53
Sample Problems
C
H
.0510g 1 mole
12.0 g
O
.0057g 1 mole .0680g 1 mole
1.0 g
16.0 g
.00425
.0057
.00425
.00425
.00425
.00425
1.00
1.3
1.00
3.00
3.9
3.00
Empirical formula
C3H4O3
Sample Problems
Empirical formula C3H4O3
180
88
=2
Molecular formula C6H8O6
54
Chemical Analysis
Mixtures-Determine a Percent
55
1. The amount of calcium present in milk can be
determined by adding oxalate ion, C2O42-(in the
form of its water-soluble sodium salt, Na2C2O4); the
insoluble compound calcium oxalate is
precipitated. Suppose you take a 75.0 g sample of
milk and isolate 0.288 g of calcium oxalate from it.
What is the weight percentage of calcium in the
milk?
MILK (Ca2+)
Na2C2O4
CaC2O4
Chemical Analysis
Mixtures-Determine a Percent
MILK (Ca2+)
Na2C2O4
75.0 g
CaC2O4
0.288 g
? % Ca
% Ca =
g Ca
75.0 g milk
X 100
56
57
Chemical Analysis
Mixtures-Determine a Percent
MILK (Ca2+)
Na2C2O4
75.0 g
CaC2O4
0.288 g
? % Ca
.288 g CaC2O4
mole CaC2O4
mole Ca
128.1 g CaC2O4 mole CaC2O4
mole Ca
40.1 g Ca
= 0.0902 g Ca
% Ca =
0.0902 g Ca
75.0 g milk
X 100 = 0.120%
Chemical Analysis
Mixtures-Determine a Percent
58
2. A 4.22 g sample of calcium chloride and
sodium chloride was dissolved in water,
and the solution was treated with sodium
carbonate to precipitate the calcium as
calcium carbonate. After isolating the solid
calcium carbonate, it was heated to drive
off the carbon dioxide and form 0.959 g of
calcium oxide. What is the weight percent
of calcium chloride in the original 4.22 g
sample?
59
Chemical Analysis
Mixtures-Determine a Percent
CaCl2/NaCl
Na2CO3
CaCO3
4.22 g
% CaCl2 =
g CaCl2
H
e
a
t
X 100
4.22 g sample
CaO
0.959 g
60
Chemical Analysis
Mixtures-Determine a Percent
0.959 g CaO
mole CaO
mole CaCl2
56.1 g CaO
mole CaO 4
mole CaCl2
111.1 g CaCl2
= 1.90 g CaCl2
% CaCl2 =
1.90 g CaCl2
4.22 g sample
X 100 = 45.0%
61
Practice Problems
1. Balance the following equations:
CS2 + Cl2 --> CCl4 + S2Cl2
N2 + O2 --> NO
C8H18 + O2 --> CO2 + H2O
2. Write the formula equation for each of the following:
sodium + water --> sodium hydroxide + hydrogen
magnesium + oxygen --> magnesium oxide
aluminum + hydrochloric acid -->
aluminum chloride + hydrogen
62
Practice Problems
2. (continue)
aluminum + hydrochloric acid -->
aluminum chloride + hydrogen
sodium chlorate --> sodium chloride + oxygen
mercury(II) sulfate + ammonium sulfide -->
mercury(II) sulfide + ammonium sulfate
iron + cupric sulfate --> iron(III) sulfate + copper
63
Practice Problems
For problems 3-6
3 H2 + N2 g 2 NH3
3. How many moles of H2 are required to
react 4.2 moles of N2?
4. How many moles of H2 are required to react
74 grams of N2?
5. How many grams of NH3 would be
produced from 45 g of H2?
6. How many moles of NH3 would be
produced from the reaction of 18.5 g H2 and
95 g of N2?
64
Practice Problems
7. Phosphine, PH3, is formed when calcium phosphide is added to
water. How many grams of phosphine can be obtained from 205 g
of calcium phosphide?
8. How many grams of iron will be required to release all of the
antimony from 10.0 g antimony trisulfide? (Ferrous sulfide is
formed)
9. If calcium oxide were prepared by heating calcium carbonate, how
many grams of the carbonate would be required to produce the
15.0 g of the oxide?
65
Practice Problems
10. How many grams of cupric sulfate are needed to
completely react with 145 g of sodium chloride? How
many grams of sodium sulfate could be produced by
this reaction?
11. How many grams of sulfuric acid will react with 40.0 g
of aluminum metal?
12. 75 grams of zinc are added to 120 grams of nitrous
acid. How many grams of hydrogen gas are evolved?
66
Practice Problems
13. 10.0 grams of hydrogen and 75 grams of oxygen are
exploded together in a reaction tube. How many grams
of water are produced? What other gas is found in the
tube(besides water vapor) after the reaction, and how
many grams of this gas are there?
67
Practice Problems
14. A white powder was a mixture of NaBr and
NaNO3. A sample of the powder weighing
1.341 g was dissolved in water and a solution of
AgNO3 was added until the precipitation of AgBr
was complete. The reaction mixture was filtered
and dried and the precipitate of AgBr weighed
1.896 g. What was the percentage by weight of
NaBr in the original sample?
68
Practice Problems
15. A 4.81 g sample of an unknown compound
containing only carbon, hydrogen, and nitrogen was
burned to produce 13.74 g of CO2 and 1.68 g of H2O.
Determine the empirical formula.
69
Practice Problems Answers
1. 1,3,1,1
1,1,2
2,25,16,18
2. 2 Na + 2 HOH --> 2 NaOH + H2
2 Mg + O2 --> 2 MgO
2 Al + 6 HCl --> 2 AlCl3 + 3 H2
2 NaClO3 --> 2 NaCl + 3 O2
HgSO4 + (NH4)2S --> Hg + (NH4)2SO4
2 Fe + 3 CuSO4 --> Fe2(SO4)3 + 3 Cu
3. 13 mole H2
4. 7.9 mole H2
5. 260 g NH3
6. 6.2 mole NH3
70
Practice Problems Answers
7. 76.5 g PH3
8. 7.68 g Fe
9. 26.8 g CaCO3
10. 198 g CuSO4, 176 g Na2SO4
11. 218 g H2SO4
12. 2.3 g H2
13. 84 g H2O, 0.6 g H2
14. 77.48%
15. C5H3N
71
Writing Equations
zinc + chlorine
--->
zinc chloride
Zn (s) + Cl2 (g) --> ZnCl2 (s)
Combination, Synthesis
Writing Equations
72
potassium nitrate --> potassium nitrite + oxygen
KNO3 (s) --> KNO2 (s) + O2 (g)
2 KNO3 (s) --> 2 KNO2 (s)
Decomposition
+
O2 (g)
Writing Equations
73
magnesium bromide + chlorine --> magnesium chloride + bromine
MgBr2 (s) + Cl2 (g) --> MgCl2 (s) + Br2 (g)
Single Displacement
Writing Equations
calcium hydroxide + hydrochloric acid -->
calcium chloride + water
Ca(OH)2 (aq) + HCl (aq) --> CaCl2 (aq) + H2O (l)
Ca(OH)2 (aq) + 2HCl (aq) --> CaCl2 (aq) + 2 H2O (l)
Double Displacement
74
Writing Equations
75
zinc chloride + ammonium sulfide -->
zinc sulfide + ammonium chloride
(aq) + (NH4)2S (aq) --> ZnS (s) + NH4Cl (aq)
(aq) + (NH4)2S (aq) --> ZnS (s) + 2 NH4Cl (aq)
ZnCl2
ZnCl2
Writing Equations
aluminum + cupric chloride -->
copper + aluminum chloride
Al (s) + CuCl2 (aq) --> Cu (s) + AlCl3 (aq)
2 Al (s) + 3 CuCl2 (aq) --> 3 Cu (s) + 2 AlCl3 (aq)
76
77
How many moles of H2
are required to produce 9 moles of
NH3?
PROBLEM:
STEP 1
Write the balanced chemical equation
3 H2 + N2 --> 2 NH3
78
STEP 2
Write the given and requested information
below the equation.
3 H2
? mole
+
N2
-->
2 NH3
9 mole
79
STEP 3
Calculate using the information.
3 H2 +
? mole
9 mole NH3
N2
-->
2 NH3
9 mole
3 mole H2
2 mole NH3 = 10 mole H2
80
How many moles of
NH3 can be produced from 10.4
moles of N2?
PROBLEM:
3 H2 + N2
10.4 mole ? mole
10.4 mole N2
-->
2 NH3
2 mole NH3
= 20.8 mole NH3
mole N2
81
How many grams
H2 are required to produce
moles of NH3?
PROBLEM:
?g
3 H2
+
N2
-->
of
8
2 NH3
8 mole
8 mole NH3 3 mole H2
2 mole NH3
2.0 g H2
mole H2
= 20 g H2
82
How many moles of
NH3 can be produced from 55
grams of N2?
PROBLEM:
55 g
3 H2
55 g N2
+ N2
? mole
-->
2 NH3
1 mole N2
2 mole NH3
28.0 g N2
mole N2
= 3.9 mole
NH3
83
How many grams
H2 are required to react
grams of N2?
PROBLEM:
?g
3 H2
+ N2
24 g
24 g N2 mole N2
-->
2 NH3
3 mole H2 2.0 g H2
28.0 g N2 mole N2
of
24
mole H2
= 5.1 g H2
84
How many grams of
N2 are required to produce 155
grams of NH3?
PROBLEM:
?g
3 H2
+ N2
155 g
-->
2 NH3
155 g NH3 mole NH3 mole N2
28.0 g N2
17.0 g NH3 2 mole NH3 mole N2
= 128 g N2
85
Sample Problems
1) Sulfur dioxide may be oxidized to sulfur trioxide.
How many grams of sulfur dioxide could be converted
by this process if 100.0 g of oxygen are available for
the oxidation?
2 SO2
?g
+
100.0 g O2 mole O2
32.0 g O2
O2 -->
100.0 g
2 SO3
2 mole SO2
64.1 g SO2
mole O2
mole SO2
= 401 g SO2
86
Sample Problems
2) Lightning discharges in the atmosphere catalyze the
conversion of nitrogen to nitrogen dioxide. How many
grams of nitrogen would be required to make 25.0 g of
nitrogen dioxide in this way?
N2
?g
+ 2 O2
25.0 g NO2 mole NO2
46.0 g NO2
-->
2 NO2
25.0 g
1 mole N2
28.0 g N2
2 mole NO2 mole N2
= 7.61 g N2
87
Sample Problems
3) Ferric oxide may be reduced to pure iron
with coke (pure carbon). Suppose that 150.0 g
of ferric oxide is available. How many grams of
carbon would be needed?
2 Fe2O3
+ 3C
4 Fe + 3 CO2 150.0 g
?g
150.0 g Fe2O3 mole Fe2O3 3
mole C
159.6 g Fe2O3 2 mole Fe2O3
12.0 g C
= 16.9 g C
mole C
88
Sample Problems
4) Zinc metal will react with hydrochloric acid to produce
hydrogen gas. If 50.0 g of zinc is to be used in the reaction,
how many grams of acid would be needed to completely
react with all of the zinc? How many grams of hydrogen gas
would be produced?
Zn + 2 HCl
50.0 g
?g
50.0 g Zn
mole Zn
-->
ZnCl2
+
H2
?g
2 mole HCl 36.5 g HCl
65.4 g Zn mole Zn
mole HCl
= 55.8gHCl
89
Sample Problems
4) Zinc metal will react with hydrochloric acid to produce
hydrogen gas. If 50.0 g of zinc is to be used in the reaction,
how many grams of acid would be needed to completely
react with all of the zinc? How many grams of hydrogen gas
would be produced?
Zn + 2 HCl
50.0 g
?g
50.0 g Zn mole Zn
-->
mole H2
65.4 g Zn mole Zn
ZnCl2
+
H2
?g
2.0 g H2
mole H2
= 1.5 g H2
90
Sample Problems
5) Phosphoric acid, H3PO4, is produced in the reaction
between calcium phosphate and sulfuric acid. How many
grams of phosphoric acid would be produced from 55 grams
of calcium phosphate? What other product is formed and in
what quantity?
Ca3(PO4)2 + 3 H2SO4 --> 3 CaSO4 + 2 H3PO4
?g
?g
55 g Ca3(PO4)2 mole Ca3(PO4)2
3 mole CaSO4
55 g
136.2 g CaSO4
310.3 gCa3(PO4)2 mole Ca3(PO4)2 mole CaSO4
= 72 g CaSO4
91
Sample Problems
5) Phosphoric acid, H3PO4, is produced in the reaction
between calcium phosphate and sulfuric acid. How many
grams of phosphoric acid would be produced from 55 grams
of calcium phosphate? What other product is formed and in
what quantity?
Ca3(PO4)2 + 3 H2SO4 --> 3 CaSO4 + 2 H3PO4
?g
?g
55 g Ca3(PO4)2
mole Ca3(PO4)2
2 mole H3PO4
310.3 gCa3(PO4)2 mole Ca3(PO4)2
55 g
98.0 g H3PO4
mole H3PO4
= 35 g H3PO4
92
Percent Yield
actual yield
% yield =
• 100%
theoretical yield
93
If 19.3 g H2 produces
78.5 g NH3, what is the percent
yield?
PROBLEM:
3 H2 +
19.3 g
N2
--> 2 NH3
?g
19.3 g H2 mole H2 2 mole NH3 17.0 g NH3
2.0 g H2 3 mole H2
% yield =
78.5g
110 g
mole NH3
x 100 = 71 %
= 110 g
NH3
94
If the yield obtained is
75%, how many grams of NH3
would be obtained from 10.4 g of
N2?
PROBLEM:
3 H2 +
10.4 g
10.4 g N2
mole N2
28.0 g N2
N2
-->
2 NH3
?g
2 mole NH3 17.0 g NH3
mole N2
mole NH3
12.6 g NH3 x .75 = 9.4 g NH3
= 12.6 g
NH3
95
PROBLEM: 75.0 grams of potassium hydroxide
are permitted to react with 50.0 grams of
hydrochloric acid. How many grams of potassium
chloride are formed?
STEP 1
Write the balanced chemical equation
KOH + HCl --> KCl + HOH
96
STEP 2
Write the given and requested information
below the equation.
KOH + HCl
75.0 g
50.0 g
--> KCl + HOH
?g
STEP 3
97
Calculate the product assuming
that each reactant is the limiting
reagent.
KOH + HCl
75.0 g
--> KCl + HOH
50.0 g
?g
75.0 g KOH mole KOH mole KCl 74.6 g KCl
56.1 g KOH mole KOH mole KCl
50.0 g HCl
mole HCl
36.5 g HCl
mole KCl
mole HCl
= 99.7g KCl
74.6 g KCl
mole KCl
= 102 g KCl
STEP 4
Determine the limiting
reactant and the actual amount of
product.
KOH + HCl
75.0 g
50.0 g
--> KCl + HOH
?g
75.0 g KOH can produce 99.7 g KCl
KOH is the limiting reactant, HCl is
the excess reactant.
50.0 g HCl can produce 102 g KCl
99.7 g KCl
98
STEP 5
99
Determine the amount of excess
reactant by calculating the amount
used and subtracting from the starting
amount.
KOH + HCl
75.0 g
50.0 g
--> KCl + HOH
?g
75.0 g KOH mole KOH mole HCl 36.5 g HCl
56.1 g KOH mole KOH mole HCl
= 48.8g HCl
used
50.0 g - 48.8 = 1.2 g HCl left
100
Sample Problems
1) How many grams of carbon dioxide can be
obtained from the action of 100.0 grams of
hydrobromic acid on 100.0 grams of calcium
carbonate?
2 HBr + CaCO3 --> HOH + CO2 + CaBr2
100.0 g 100.0 g
?g
100.0 g HBr mole HBr
mole CO2
44.0 g CO2
80.9 g HBr 2 mole HBr mole CO2
= 27.2 g
COg2 CO2
100.0 g CaCO3 mole CaCO3
mole CO2
44.0
100.1 g CaCO3 mole CaCO3 mole CO2
= 44.0 g CO2
Sample Problems
2) How many grams of ammonia are evolved
when 34 grams of ammonium chloride are
added to 140 grams of barium hydroxide?
2 NH4Cl + Ba(OH)2 --> BaCl2 + 2 NH3 +2 HOH
34 g
140 g
?g
34 g NH4Cl mole NH4Cl 2
mole NH3
17.0 g NH3
53.5 g NH4Cl 2 mole NH4Cl mole NH3
= 11 g NH3
140 g Ba(OH)2 mole Ba(OH)2 2 mole NH3
17.0 g NH3
171.3 g Ba(OH)2 mole Ba(OH)2 mole NH3
= 28.0 gNH3
101
102
Sample Problems
3) 100.0 grams of lithium metal is dropped into
1.000 liter of water. How many grams of
hydrogen are produced?
2 Li + 2 HOH --> 2 LiOH + H2
100.0 g 1000. g
?g
100.0 g Li
6.9 g Li
mole Li
2 mole Li
mole H2
mole H2
2.0 g H2
= 14 g H2
1000. g HOH
18.0 g HOH
mole HOH mole H2
2 mole HOH mole H2
2.0 g H2
= 56 g H2
Sample Problems
103
4) .50.0 grams of oxygen are available for the
combustion of 25.0 grams of carbon. How many grams
in excess is the oxygen or carbon ?
O2
+ C --> CO2
? g excess
50.0 g
25.0 g
50.0 g O2
mole O2
32.0 g O2
mole C
mole O2
12.0 g C
mole C
= 18.8g C used
25.0 g - 18.8 =
6.2 g C left
Sample Problems
104
5) 140.0 grams of sulfuric acid is added to 230.0 grams
of barium peroxide. Which reactant is in excess and by
how many grams?
g excess
H2SO4 + BaO2 -->
140. g
230. g
BaSO4 + H2O2
140.gH2SO4 mole H2SO4 mole BaO2 169.3 g BaO2
98.1 g H2SO4 mole H2SO4 mole BaO2
= 242 g BaO2
Since only 230. g available,
BaO2 is the limiting reactant
Sample Problems
105
5) 140.0 grams of sulfuric acid is added to 230.0 grams
of barium peroxide. Which reactant is in excess and by
how many grams?
H2SO4 + BaO2 -->
140. g
230. g
g excess
230. g BaO2
BaSO4 + H2O2
mole BaO2
mole H2SO4 98.1 g H2SO4
169.3 g BaO2 mole BaO2 mole H2SO4
= 133 g H2SO4 used
140. g - 133 = 7 g H2SO4 left