1. 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.
Stoichiometry

2. 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.

3. 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.

4. PROFIT 12.1 Section Assessment Gross sales – fabrication costs =
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

5. 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.

6. 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 . . .
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.)

7. 12.1 Section Assessment
Interpret the given equation in terms of relative numbers of representative particles , numbers of moles, and masses of reactants and products.
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
No coefficient = an invisible coefficient of “1”
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 g H2O  g KOH g H2
Two moles of K
Two moles of H2O
Two moles of KOH
One mole of H2

8. 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) O2(g)  CO2(g) H2O(g) 3 2 3
C = C =
H = H =
O = O = 2 1 2 6 2 6 3 7 3 5 7

9. 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.
46 g
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

10. 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.
46 g
32 g
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 . . .
O2:
O: 2 x 16 = 32
32 g/mol

11. 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.
46 g
32 g
44 g
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. 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.
46 g
32 g
44 g
18 g
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

13. 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.
46 g
32 g
44 g
18 g
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 . . . =
62 grams of products =
78 grams of reactants
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!

14. REACTANTS PRODUCTS = 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) O2(g)  CO2(g) H2O(g) 3 2 3
46 g
32 g
44 g
18 g
x 1
x 3
x 2
x 3
46 g
96 g
88 g
54 g
46 g + 96 g = 142 g
88 g + 54 g = 142 g
REACTANTS
PRODUCTS =
(in terms of mass, anyway)

15. REALITY CHECK 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”.

16. 12.2 Practice Problems
13. Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2).
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?
2.03 g
5 g
5 g CaC2 1
mol CaC2 1
mol C2H2 26
g C2H2 x x x 1 64
g CaC2 1
mol CaC2 1
mol C2H2
CaC2:
Ca: 1 x 40 = 40
C: 2 x 12 = 24
C2H2:
C: 2 x 12 = 24
H: 2 x 1 = 2
64 g/mol
26 g/mol 26
x 5
2.03 ) 64
130
130

17. CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
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 1
mol H2O 1
mol CaC2 x x 1 18
g H2O 2
mol H2O
H2O:
H: 2 x 1 = 2
O: 1 x 16 = 16
18 g/mol 49 49
mol CaC2 =
mol CaC2 =
1.36 mol CaC2
18 x 2 36

18. 12.2 Practice Problems
How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO3)?
2KClO3  2KCl + 3O2
6 x 1023
molecules O2
6.54 g KClO3
1 mol KClO3
3 mol O2 x x x
122 g KClO3
2 mol KClO3
1 mol O2 1
6.54 x 3 x 6 x 1023
x 1023 =
= x 1023
122 x 2
244
= x 1022
KClO3:
K: 1 x 39 = 39
Cl: 1 x 35 = 35
O: 3 x 16 = 48
122 g/mol
= 4.82 x 1022 molecules O2

19. 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 mol NO
3 mol NO2
46 g NO2 1
1 mol NO
1 mol NO2
6 x 1023
molecules NO
5 x 3 x 46 x 1022
690 x 1022 =
6 x 1023
6 x 1023
12.2 Practice Problems
NO:
N: 1 x 14 = 14
O: 2 x 16 = 32
46 g/mol
= 115 x 10-1 g NO2
= 1.15 x 101 g NO2
= 11.5 g NO2

20. 12.2 Practice Problems

21. 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 . . .

22. Look back at my work-out for #18 and notice again how the 22
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

23. 12.2 Practice Problems

24. 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.

25. 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.

26. 12.2 Section Assessment
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
9 mol O2
2 mol C3H7OH
6 mol CO2
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
8 mol H2O
2 mol C3H7OH
9 mol O2
6 mol CO2

27. 12.2 Section Assessment
24. The combustion of acetylene gas is represented by this equation:
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?
176 g
52.0 g 1
mol C2H2
52.0 g C2H2 4
mol CO2 44
g CO2 x x x 1 26
g C2H2 2
mol C2H2 1
mol CO2
C2H2:
C: 2 x 12 = 24
H: 2 x 1 = 2
26 g/mol
CO2:
C: 1 x 12 = 12
O: 2 x 16 = 32
44 g/mol
= 176 g CO2

28. 12.2 Section Assessment
24. The combustion of acetylene gas is represented by this equation:
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?
176 g
36 g 1
mol C2H2
52.0 g C2H2 2
mol H2O 18
g H2O x x x 1 26
g C2H2 2
mol C2H2 1
mol H2O
C2H2:
C: 2 x 12 = 24
H: 2 x 1 = 2
26 g/mol
H2O:
H: 2 x 1 = 2
O: 1 x 16 = 16
18 g/mol
= 36 g H2O

29. 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.

30. 12.3 Practice Problems

31. 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.)

32. 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.

33. 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”.

34. 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.

35. 12.3 Practice Problems
. . .

36. 12.3 Practice Problems
#32 = evil (no formulas or equation given AND 2 stoich calcs)

37. 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.”

38. 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
3 mol Cu
63.55 g Cu
= 6.6 g Cu
27 g Al
2 mol Al
1 mol Cu
actual yield
4.65 g Cu
% yield = =
= = 70.5%
theoretical yield
6.6 g Cu

39. 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
Chapter 12 Assessment
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?

40. 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
Chapter 12 Assessment
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?
372 g
66.6 g
5F NH3  N2F HF
66.6 g NH3
1 mol NH3
5 mol F2
38 g F2 x x x
= 372 g F2 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

41. 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
Chapter 12 Assessment
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
4.65 g
5F NH3  N2F HF

42. 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
Chapter 12 Assessment
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
4.65 g
5F NH3  N2F HF
4.65 g HF
1 mol HF
2 mol NH3
17 g NH3 x x x
= 1.32 g NH3 1
20 g HF
6 mol HF
1 mol NH3
HF:
H: 1 x 1 = 1
F: 1 x 19 = 19
20 g/mol
NH3:
N: 1 x 14 = 14
H: 3 x 1 = 3
17 g/mol

43. 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
Chapter 12 Assessment
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 ? g
5F NH3  N2F HF

44. 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g)
Chapter 12 Assessment
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
123 g
5F NH3  N2F HF
225 g F2
1 mol F2
1 mol N2F4
104 g N2F4 x x x
= 123 g N2F4 1
38 g F2
5 mol HF2
1 mol N2F4
N2F4:
N: 2 x 14 = 28
F: 4 x 19 = 76
104 g/mol
F2:
F: 2 x 19 = 38
38 g/mol

45. 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 ? g
Li3N + 3H2O  NH3 + 3LiOH

46. 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 mol Li3N
3 mol H2O
18 g H2O x x x
= 51 g H2O 1
35 g Li3N
1 mol Li3N
1 mol H2O
Li3N:
Li: 3 x 7 = 21
N: 1 x 14 = 14
35 g/mol
H2O:
H: 2 x 1 = 2
O: 1 x 16 = 16
18 g/mol

47. 10Q + “E” UNDER CONSTRUCTION This Power Point may soon again be . . .
I’m working on the honors stuff, mostly . . .
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000
10Q
+ “E”

48. hyperTABLE of CONTENTS
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