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Prentice Hall Chemistry(c) 2005
Section Assessment Answers
Chapter 12By Daniel R. BarnesInit: 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 Assessment5. 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 Assessment6. 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.
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
For instance . . .
12.1 Section Assessment7. 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 Assessment8. What quantities are always conserved in chemical reactions?Mass and atoms are always conserved in chemical reactions.
If you start with 18 kg of reactants, you should end up with . . .
For example . . .
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.)
. . .18 kg of products. Chemical reactions do not create or destroy matter.
12.1 Section Assessment9. 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)
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 of K
Two moles of H2O
Two moles of KOH
One mole of H2
No coefficient = an invisible coefficient of “1”
12.1 Section Assessment10. 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)
C = C =H = H = O = O =
263
123
2
2
3
65 7
3
7
12.1 Section Assessment10. Balance this equation:C2H5OH(l) + O2(g) CO2(g) + H2O(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
46 g
12.1 Section Assessment10. Balance this equation:C2H5OH(l) + O2(g) CO2(g) + H2O(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 . . .
O2: O: 2 x 16 = 32----------------------- 32 g/mol
46 g 32 g
12.1 Section Assessment10. Balance this equation:C2H5OH(l) + O2(g) CO2(g) + H2O(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
46 g 32 g 44 g
12.1 Section Assessment10. Balance this equation:C2H5OH(l) + O2(g) CO2(g) + H2O(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
46 g 32 g 44 g 18 g
12.1 Section Assessment10. Balance this equation:C2H5OH(l) + O2(g) CO2(g) + H2O(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 g 32 g 44 g 18 g
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 Assessment10. 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)2 33
46 g 32 g 44 g 18 gx 1
46 g
x 3
96 g
x 2
88 g
x 3
54 g
46 g + 96 g = 142 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 Problems13. 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?
5 g CaC2
1x
g CaC2
mol CaC2
CaC2: Ca: 1 x 40 = 40 C: 2 x 12 = 24
64 g/mol
64
1x
mol CaC2
mol C2H2
1
1x
mol C2H2
g C2H2
126
C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2
26 g/mol
5 g g
26x 5
130 ) 130 64
2.03
2.03
12.2 Practice Problems14. 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 H2O1
xg H2O
mol H2O18
1x
mol H2O
mol CaC22
1
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
= 4.82 x 1022 molecules O2
12.2 Practice Problems15. 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
1x
122 g KClO3
1 mol KClO3 x
KClO3:
K: 1 x 39 = 39 Cl: 1 x 35 = 35 O: 3 x 16 = 48 ---------------------- 122 g/mol
x2 mol KClO3
3 mol O21 mol O2
6 x 1023
molecules O2
6.54 x 3 x 6 x 1023
122 x 2117.72 x 1023
244= = 0.482459 x 1023
= 4.82459 x 1022
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?
6 x 1023 molecules NO
12.2 Practice
Pro
blem
s
1
5 x 3 x 46 x 1022
1 mol NO
NO: N: 1 x 14 = 14 O: 2 x 16 = 32 ---------------------- 46 g/mol
5 x 1022 molecules NO 3 mol NO2
1 mol NO46 g NO2
1 mol NO2
6 x 1023= 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
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 . . .
12.2 Practice Problems
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 Assessment21. 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 Assessment22. 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 Assessment23.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
9 mol O2
2 mol C3H7OH
6 mol CO2
2 mol C3H7OH
8 mol H2O
2 mol C3H7OH
9 mol O2
6 mol CO2 8 mol H2O
9 mol O2 9 mol O2
2 mol C3H7OH
6 mol CO2
8 mol H2O9 mol O2
6 mol CO2 6 mol CO2
2 mol C3H7OH
8 mol H2O
9 mol O2 6 mol CO2
8 mol H2O 8 mol H2O
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?
12.2 Section Assessment
52.0 g C2H2
1x
g C2H2
mol C2H2x
mol C2H2
mol CO2 xg CO2
mol CO2
C2H2:
C: 2 x 12 = 24 H: 2 x 1 = 2---------------------- 26 g/mol
26
1
2
4
CO2:
C: 1 x 12 = 12 O: 2 x 16 = 32---------------------- 44 g/mol
1
44
= 176 g CO2
17652.0 g g
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?
52.0 g C2H2
1x
g C2H2
mol C2H2x
mol C2H2
mol H2O xg H2O
mol H2O
C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2---------------------- 26 g/mol
26
1
2
2
H2O: H: 2 x 1 = 2 O: 1 x 16 = 16---------------------- 18 g/mol
1
18
= 36 g H2O
176 g 36 g
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
2 mol Al
3 mol Cu
1 mol Cu
63.55 g Cu= 6.6 g Cu
27 g Al
1 mol Al
% yield =theoretical yield
actual yield =6.6 g Cu
4.65 g Cu = 0.705 = 70.5%
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?
5F2 + 2NH3 N2F4 + 6HF
66.6 gg
66.6 g NH3
1x
17 g NH3
1 mol NH3
NH3: N: 1 x 14 = 14 H: 3 x 1 = 3 17 g/mol
x2 mol NH3
5 mol F2 x1 mol F2
38 g F2
F2: F: 2 x 19 = 38 38 g/mol
= 372 g F2
372
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?
5F2 + 2NH3 N2F4 + 6HF
4.65 gg
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?
5F2 + 2NH3 N2F4 + 6HF
4.65 gg
4.65 g HF
1x
20 g HF
1 mol HF
HF: H: 1 x 1 = 1 F: 1 x 19 = 19 20 g/mol
x6 mol HF
2 mol NH3 x1 mol NH3
17 g NH3 = 1.32 g NH3
1.32
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?
5F2 + 2NH3 N2F4 + 6HF
225 g g?
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?
5F2 + 2NH3 N2F4 + 6HF
225 g g
225 g F21
x38 g F2
1 mol F2 x5 mol HF2
1 mol N2F4 x1 mol N2F4
104 g N2F4 = 123 g N2F4
123
N2F4:
N: 2 x 14 = 28 F: 4 x 19 = 76 104 g/mol
F2:
F: 2 x 19 = 38 38 g/mol
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?
Li3N + 3H2O NH3 + 3LiOH
32.9 g g?
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?
Li3N + 3H2O NH3 + 3LiOH
32.9 g g
32.9 g Li3N1
x35 g Li3N
1 mol Li3N x1 mol Li3N
3 mol H2O x1 mol H2O
18 g H2O= 51 g H2O
51
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
This Power Point may soon again be . . .
10Q + “E”1,000,000,000,000,000,000,000,000,000,000,000,000,000,000
I’m working on the honors stuff, mostly . . .
Title Page
SWBATS
12.1 Section Assessment
12.2 PracticeProblems
12.2 SectionAssessment
12.3 PracticeProblems
Ch 12Assessment
12.3 Section Assessment