Prentice Hall Chemistry
(c) 2005
Section Assessment Answers
Chapter
12
By Daniel R. Barnes
Init: 12/17/2008
WARNING: some images
and content in this
presentation may have
been taken without
permission from the world
wide web. It is intended for
use only by Mr. Barnes and
his students. It is not
meant to be copied or
distributed.
SWBAT . . .
. . . determine the mass of one substance
involved in a chemical reaction if given the
mass of any one of the other substances
involved in the reaction.
in other words . . .
. . . convert grams of the known into grams
of the unknown.
12.1 Section Assessment
5. How is a balanced chemical equation similar to a recipe?
A recipe says what ingredients you need to make some kind of
food. A balanced chemical equation says what reactants you
need to make some kind of product.
As the teacher’s edition says, “Both a balanced equation and a
recipe give quantitative information about the starting and end
materials.
12.1 Section Assessment
6. How do chemists use balanced equations?
Chemists use balanced equations as a basis to calculate how
much reactant is needed or how much product is formed in a
reaction.
For instance . . .
If you run a factory that makes chemicals, knowing how much of
each reactant you’re going to use lets you know how much it’s
going to cost you, and knowing how much product will be formed
tells you how much money you can expect to make when you sell
the product.
Gross sales – fabrication costs =
PROFIT
12.1 Section Assessment
7. Chemical reactions can be described in terms of what
quantities?
Chemical reactions can be described in terms of numbers atoms,
molecules, or moles; in terms of mass; or in terms of volume.
12.1 Section Assessment
8. What quantities are always conserved in chemical reactions?
Mass and atoms are always conserved in chemical reactions.
For example . . .
If you start with 18 kg of reactants, you should end up with . . .
. . .18 kg of products. Chemical reactions do not create or destroy
matter.
If 3 million carbon atoms react with 6 million oxygen atoms to form
carbon dioxide, the carbon dioxide will consist of . . .
3 million carbon atoms and 6 million oxygen atoms.
Volume is NOT conserved. (Example: a small handful of nitrogen
triiodide can explode to form several liters of nitrogen gas and
iodine vapors.)
Molecules are NOT conserved. (Example: two million hydrogen
molecules and one million oxygen molecules become two million
water molecules. 3 million molecules  2 million molecules.)
12.1 Section Assessment
9. Interpret the given equation in terms of relative numbers of
representative particles , numbers of moles, and masses of
No coefficient = an invisible coefficient of “1”
reactants and products.
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
In terms of “representative particles”:
Two atoms of K react with 2 molecules of H2O to form 2 formula
units of KOH and one molecule of H2.
In terms of mass:
78.2 g K + 36.0 g H2O  112.2 g KOH + 2.0 g H2
Two moles Two moles
of K
of H2O
Two moles
of KOH
One mole
of H2
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
Show that the balanced equation obeys the law of conservation of
mass.
C2H5OH(l) +
C= 2
H= 6
O= 37
3 O2(g) 
2 CO2(g) +
C= 1 2
H= 2 6
O= 3 5 7
3 H2O(g)
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
46 g
Show that the balanced equation obeys the law of conservation of
mass.
First, I’d like to show how the unbalanced equation disobeys the
law of conservation of mass
To do this, I’ll need to calculate the molar mass of each reactant
and each product . . .
C2H5OH:
C: 2 x 12 = 24
H: 6 x 1 = 6
O: 1 x 16 = 16
----------------------46 g/mol
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
46 g
32 g
Show that the balanced equation obeys the law of conservation of
mass.
First, I’d like to show how the unbalanced equation disobeys the
law of conservation of mass
To do this, I’ll need to calculate the molar mass of each reactant
and each product . . .
O 2:
O: 2 x 16 = 32
----------------------32 g/mol
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
46 g
32 g
44 g
Show that the balanced equation obeys the law of conservation of
mass.
First, I’d like to show how the unbalanced equation disobeys the
law of conservation of mass
To do this, I’ll need to calculate the molar mass of each reactant
and each product . . .
CO2:
C: 1 x 12 = 12
O: 2 x 16 = 32
----------------------44 g/mol
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
46 g
32 g
44 g
18 g
Show that the balanced equation obeys the law of conservation of
mass.
First, I’d like to show how the unbalanced equation disobeys the
law of conservation of mass
To do this, I’ll need to calculate the molar mass of each reactant
and each product . . .
H2O:
H: 2 x 1 = 2
O: 1 x 16 = 16
----------------------18 g/mol
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
46 g
32 g
44 g
18 g
Show that the balanced equation obeys the law of conservation of
mass.
First, I’d like to show how the unbalanced equation disobeys the
law of conservation of mass
To do this, I’ll need to calculate the molar mass of each reactant
and each product . . .
46 + 32 =
78 grams of reactants
44 + 18 =
62 grams of products
You can’t have 16 grams of matter just disappear like that.
In its unbalanced state, the equation violates the law of
conservation of matter. It needs to be balanced!
12.1 Section Assessment
10. Balance this equation:
C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
Show that the balanced equation obeys the law of conservation of
mass.
C2H5OH(l) +
46 g
x1
46 g
3 O2(g) 
32 g
x3
96 g
46 g + 96 g = 142 g
2 CO2(g) +
44 g
x2
88 g
3 H2O(g)
18 g
x3
54 g
88 g + 54 g = 142 g
(in terms of mass, anyway)
Don’t forget that one “mole” of a material =
6.022 x 1023 molecules*
of that material.
*If the material is a noble gas or a metal, subsitute the word “atoms” for “molecules”. If the material is ionic,
substitute “forumula units” for “molecules”.
12.2 Practice Problems
13. Acetylene gas (C2H2) is produced by adding water to calcium
carbide (CaC2).
2.03 g
5g
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
How many grams of acetylene are produced by adding water to
5.00 g CaC2?
5 g CaC2
1
x
1 mol CaC2
64 g CaC2
x
1 mol C2H2
26 g C2H2
x
1 mol CaC2 1 mol C2H2
CaC2:
C2H2:
Ca: 1 x 40 = 40
C: 2 x 12 = 24
C: 2 x 12 = 24
H: 2 x 1 = 2
64 g/mol
26 g/mol
26
2.03
x5
64 ) 130
130
12.2 Practice Problems
14. Using the same equation, determine how many moles of
CaC2 are needed to react completely with 49.0 g H2O.
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
49 g H2O
x
1
1 mol H2O
x
18 g H2O
1 mol CaC2
2 mol H2O
H2O:
H: 2 x 1 = 2
O: 1 x 16 = 16
---------------------18 g/mol
49
18 x 2
mol CaC2
=
49
36
mol CaC2 = 1.36 mol CaC2
12.2 Practice Problems
15. How many molecules of oxygen are produced by the
decomposition of 6.54 g of potassium chlorate (KClO3)?
2KClO3  2KCl + 3O2
6.54 g KClO3
1
x
1 mol KClO3
122 g KClO3
x
3 mol O2
2 mol KClO3
6.54 x 3 x 6 x 1023
117.72 x 1023
=
122 x 2
244
KClO3:
K: 1 x 39 = 39
Cl: 1 x 35 = 35
O: 3 x 16 = 48
---------------------122 g/mol
x
6 x 1023
molecules O2
1 mol O2
= 0.482459 x 1023
= 4.82459 x 1022
= 4.82 x 1022 molecules O
2
16. The last step in the production of nitric acid is the reaction
of nitrogen dioxide with water:
3NO2 + H2O  2HNO3 + NO
How many grams of nitrogen dioxide must react with water to
produce 5.00 x 1022 molecules of nitrogen monoxide?
5 x 1022
molecules NO
1
1 mol NO
3 mol NO2 46 g NO2
1 mol NO
6 x 1023
molecules NO
6 x 1023
NO:
N: 1 x 14 = 14
O: 2 x 16 = 32
---------------------46 g/mol
690 x 1022
=
6 x 1023
= 115 x 10-1 g NO2
= 1.15 x 101 g NO2
= 11.5 g NO2
12.2 Practice
Problems
5 x 3 x 46 x 1022
1 mol NO2
12.2 Practice Problems
12.2 Practice Problems
Notice how the 22.4’s just ended up canceling each other out? It
almost makes you wonder why I bothered putting them there in
the first place . . .
Look back at my work-out for #18 and notice again how the 22.4’s
just ended up canceling each other out. If I’d put those two
fractions in my work-out for this one, the same thing would have
happened, so I just didn’t bother putting them in. I could have
also put in two fractions to turn mL to L and then L back to mL
again, but the 1000’s would have also have just canceled each
other out, so I didn’t bother.
12.2 Practice Problems
12.2 Practice Problems
12.2 Section Assessment
21. How are mole ratios used in chemical calculations?
Mole ratios are written using the coefficients from a balanced
chemical equation. They are used to relate moles of reactants
and products in stoichiometric calculations.
12.2 Section Assessment
22. Outline the sequence of steps needed to solve a typical
stoichiometric problem.
i. Convert the given quantity to moles using the molar mass of the
known chemical.
ii. Use the mole ratio (from the appropriate coefficients in the
balanced chemical equation for the reaction) to find moles of the
desired chemical.
iii. Using the molar mass of the desired chemical, convert moles
of the desired chemical into grams of the desired chemical.
12.2 Section Assessment
23.Write the 12 mole ratios that can be derived from the equation
for the combustion of isopropyl alcohol.
2C3H7OH(l) + 9O2(g)  6CO2(g) + 8H2O(g)
Holy crud. You ready? Here they come.
2 mol C3H7OH
2 mol C3H7OH
2 mol C3H7OH
9 mol O2
6 mol CO2
8 mol H2O
9 mol O2
9 mol O2
2 mol C3H7OH
6 mol CO2
9 mol O2
8 mol H2O
6 mol CO2
6 mol CO2
6 mol CO2
2 mol C3H7OH
9 mol O2
8 mol H2O
8 mol H2O
8 mol H2O
9 mol O2
6 mol CO2
8 mol H2O
2 mol C3H7OH
12.2 Section Assessment
24. The combustion of acetylene gas is represented by this
equation:
176 g
52.0 g
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)
How many grams of CO2 and H2O are produced when 52.0 g
C2H2 burns in oxygen?
1 mol C2H2
52.0 g C2H2
4 mol CO2
x
x
x
1
26 g C2H2
2 mol C2H2
C2H2:
C: 2 x 12 = 24
H: 2 x 1 = 2
---------------------26 g/mol
44 g CO2
1 mol CO2
CO2:
C: 1 x 12 = 12
= 176 g CO2
O: 2 x 16 = 32
---------------------44 g/mol
12.2 Section Assessment
24. The combustion of acetylene gas is represented by this
equation:
176 g
36 g
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)
How many grams of CO2 and H2O are produced when 52.0 g
C2H2 burns in oxygen?
1 mol C2H2
52.0 g C2H2
2 mol H2O
x
x
x
1
26 g C2H2
2 mol C2H2
C2H2:
C: 2 x 12 = 24
H: 2 x 1 = 2
---------------------26 g/mol
18 g H2O
1 mol H2O
H2O:
H: 2 x 1 = 2
= 36 g H2O
O: 1 x 16 = 16
---------------------18 g/mol
12.3 Practice Problems
The math shows that 2.7 moles of C2H4 can create 5.4 moles of
CO2, but 6.3 moles of O2 can only create 4.2 moles CO2. O2 is
the weakest link, so it’s the limiting reactant. In this reaction, all
the O2 would be used up, with leftover C2H4.
12.3 Practice Problems
12.3 Practice Problems
2.6 mol C2H4 can produce less H2O than 6.3 mol O2 can, so C2H4
is the limiting reactant. (I do this a little different from the book. I
see how much product each amount of reactant can make.
Whoever makes the least product is the limiting reactant.)
12.3 Practice Problems
As with #27, I calculated how much product each amount of
reactant could make. the 2.4 mol C2H2 produced the lesser
amount of H2O, so C2H2 is the limiting reactant.
12.3 Practice Problems
This is just an ordinary, g-this  g-that, stoichiometry problem.
The only difference is that now we’re referring to the results of
such calculations as “theoretical yield”.
12.3 Practice Problems
#30 is extra-evil because it doesn’t tell you what the formulas are
of the chemicals in the reaction, and, since you have to make the
equation from scratch, you have to balance it, too. You may have
to go back to chapter 9 if you forget how to figure out the formula
of a chemical when given only its name.
12.3 Practice Problems
...
12.3 Practice Problems
#32 = evil (no formulas or equation given AND 2 stoich calcs)
12.3 Section Assessment
33. “In a chemical equation, an insufficient quantity of any of the
reactants will limit the amount of product that forms.”
34. “The efficiency of a reaction carried out in a laboratory can be
measured by calculating the percent yield.”
12.3 Section Assessment
35. What is the percent yield if 4.65 g of copper is produced when
1.87 g of aluminum reacts with an excess of copper (II) sulfate?
2Al + 3CuSO4  Al2(SO4)3 + 3Cu
1.87 g Al
1 mol Al
27 g Al
3 mol Cu 63.55 g Cu
= 6.6 g Cu
2 mol Al 1 mol Cu
% yield =
actual yield
4.65 g Cu
=
= 0.705 = 70.5%
theoretical yield
6.6 g Cu
Chapter 12 Assessment
42.The reaction of fluorine with ammonia produces dinitrogen
tetrafluoride and hydrogen fluoride.
5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
a. If you have 66.6 g of NH3, how many grams of F2 are required
for complete reaction?
Chapter 12 Assessment
42.The reaction of fluorine with ammonia produces dinitrogen
tetrafluoride and hydrogen fluoride.
5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
a. If you have 66.6 g of NH3, how many grams of F2 are required
for complete reaction?
66.6 g
372 g
5F2
+
2NH3

N2F4 +
6HF
66.6 g NH3 1 mol NH3
5 mol F2
38 g F2
x
x
x
1
17 g NH3
2 mol NH3
1 mol F2
NH3:
N: 1 x 14 = 14
H: 3 x 1 = 3
17 g/mol
F2:
F: 2 x 19 = 38
38 g/mol
= 372 g F2
Chapter 12 Assessment
42.The reaction of fluorine with ammonia produces dinitrogen
tetrafluoride and hydrogen fluoride.
5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
b. How many grams of NH3 are required to produce 4.65 g HF?
g
5F2
+
2NH3

4.65 g
N2F4 + 6HF
Chapter 12 Assessment
42.The reaction of fluorine with ammonia produces dinitrogen
tetrafluoride and hydrogen fluoride.
5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
b. How many grams of NH3 are required to produce 4.65 g HF?
1.32 g
5F2
+
4.65 g HF
1 mol HF
x
1
20 g HF
HF:
H: 1 x 1 = 1
F: 1 x 19 = 19
20 g/mol
2NH3

4.65 g
N2F4 + 6HF
2 mol NH3
17 g NH3
x
x
= 1.32 g NH3
6 mol HF
1 mol NH3
NH3:
N: 1 x 14 = 14
H: 3 x 1 = 3
17 g/mol
Chapter 12 Assessment
42.The reaction of fluorine with ammonia produces dinitrogen
tetrafluoride and hydrogen fluoride.
5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
c. How many grams of N2F4 can be produced from 225 g F2?
225 g
5F2 +
? g
2NH3

N2F4 +
6HF
Chapter 12 Assessment
42.The reaction of fluorine with ammonia produces dinitrogen
tetrafluoride and hydrogen fluoride.
5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
c. How many grams of N2F4 can be produced from 225 g F2?
123 g
225 g
5F2 +
225 g F2
1
x
1 mol F2
38 g F2
F2:
F: 2 x 19 = 38
38 g/mol
2NH3
x

1 mol N2F4
5 mol HF2
N2F4 +
x
6HF
104 g N2F4
1 mol N2F4
N2F4:
N: 2 x 14 = 28
F: 4 x 19 = 76
104 g/mol
= 123 g N2F4
Chapter 12 Assessment
44. Lithium nitride reacts with water to form ammonia and
aqueous lithium hydroxide.
Li3N + 3H2O  NH3 + 3LiOH
a. What mass of water is needed to react with 32.9 g Li3N?
? g
32.9 g
Li3N + 3H2O  NH3 + 3LiOH
Chapter 12 Assessment
44. Lithium nitride reacts with water to form ammonia and
aqueous lithium hydroxide.
Li3N + 3H2O  NH3 + 3LiOH
a. What mass of water is needed to react with 32.9 g Li3N?
32.9 g 51 g
Li3N + 3H2O  NH3 + 3LiOH
32.9 g Li3N
1
x
1 mol Li3N
35 g Li3N
Li3N:
Li: 3 x 7 = 21
N: 1 x 14 = 14
35 g/mol
x
3 mol H2O
1 mol Li3N
x
18 g H2O
1 mol H2O
= 51 g H2O
H2O:
H: 2 x 1 = 2
O: 1 x 16 = 16
18 g/mol
This Power Point may soon again be . . .
I’m working on the honors stuff, mostly . . .
10Q
1,000,000,000,000,0
00,000,000,000,000,
000,000,000,000,000
+ “E”
Title Page
12.3 Section
Assessment
SWBATS
Ch 12
Assessment
12.1 Section
Assessment
12.2 Practice
Problems
12.2 Section
Assessment
12.3 Practice
Problems