Final Exam Study Guide
Short Answer
1. How many representative particles are in 1.45 g of a molecular compound with a molar mass of 237 g?
2. Find the mass in grams of 3.10
10
molecules of F .
3. Find the number of moles of argon in 607 g of argon.
4. Find the mass, in grams, of 1.40
10
molecules of N .
5. What is the percent composition of NiO, if a sample of NiO with a mass of 41.9 g contains 33.1 g Ni and 8.8 g O?
6. What is the percent by mass of hydrogen in aspirin, C H O ?
7. Calculate the molecular formulas of the compounds having the following empirical formulas and molar masses: C H ,
58 g/mol; CH, 78 g/mol; and HgCl, 236.1 g/mol.
8. Complete and balance the following equation.
Al Cl
9. Complete and balance the following equation.
CH
O
CO
10. Complete and balance the following equation.
Fe (SO )
Ba(OH)
11. Balance the following equation. Indicate whether combustion is complete or incomplete.
C H
O
CO H O
12. Write a balanced net ionic equation for the following reaction.
H PO (aq) Ca(OH) (aq) Ca (PO ) (aq) H O(l)
13. Complete and balance the following equation:
K PO
BaCl
14. If a total of 13.5 mol of NaHCO and 4.5 mol of C H O react, how many moles of CO and Na C H O will be
produced?
3NaHCO (aq) + C H O (aq) 3CO (g) + 3H O(s) +Na C H O (aq)
15. If 8.00 mol of NH reacted with 14.0 mol of O , how many moles of H O will be produced?
4NH (g) + 7O (g) 4NO + 6H O(g)
16. If 8.6 L of H reacted with 4.3 L of O at STP, what is the volume of the gaseous water collected (assuming that none of
it condenses)?
2H (g) + O (g) 2H O(g)
17. Assuming STP and a stoichiometric amount of NH and NO in an expandable container originally at 15 L, what is the
final volume if the reaction goes to completion?
4NH (g) + 6NO(g) 5N (g) + 6H O(g)
18. How many grams of CO are needed to react with an excess of Fe O to produce 209.7 g Fe?
Fe O (s) + 3CO(g) 3CO (g) + 2Fe(s)
19. How many liters of O are needed to react completely with 45.0 L of H S at STP?
2H S(g) + 3O (g)
2SO (g) + 2H O(g)
20. If 5.0 g of H are reacted with excess CO, how many grams of CH OH are produced, based on a yield of 86%?
CO(g) + 2H (g) CH OH(l)
21. For the reaction 2Na(s) + Cl (g)
L of Cl (at STP)?
2NaCl(s), how many grams of NaCl could be produced from 103.0 g of Na and 13.0
22. Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of
hydrogen gas are formed when 48.7 g of sodium are added to water?
2Na + 2H O 2NaOH + H
23. The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%, how many grams of KClO are needed
to produce 10.0 L of O ?
2KClO (s) 2KCl(s) + 3O (g)
24. Consider the following reaction:
2H S(g) + 3O (g)
2SO (g) + 2H O(g)
If O was the excess reagent, 8.3 mol of H S were consumed, and 137.1 g of water were collected after the reaction has
gone to completion, what is the percent yield of the reaction?
25. The volume of a gas is 250 mL at 340.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa,
assuming the temperature remains constant?
26. A balloon filled with helium has a volume of 30.0 L at a pressure of 100 kPa and a temperature of 15.0 C. What will the
volume of the balloon be if the temperature is increased to 80.0 C and the pressure remains constant?
27. A gas has a volume of 590 mL at a temperature of –55.0 C. What volume will the gas occupy at 30.0 C?
28. A rigid container of O has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K?
29. A 10-g mass of krypton occupies 15.0 L at a pressure of 210 kPa. Find the volume of the krypton when the pressure is
increased to 790 kPa.
30. \A gas has a pressure of 710 kPa at 227 C. What will its pressure be at 27 C, if the volume does not change?
31. A gas occupies a volume of 140 mL at 35.0 C and 97 kPa. What is the volume of the gas at STP?
32. A gas storage tank has a volume of 3.5 10 m when the temperature is 27 C and the pressure is 101 kPa. What is the
new volume of the tank if the temperature drops to –10 C and the pressure drops to 95 kPa?
33. How many moles of N are in a flask with a volume of 250 mL at a pressure of 300.0 kPa and a temperature of 300.0 K?
34. The gaseous product of a reaction is collected in a 25.0-L container at 27 C. The pressure in the container is 300.0 kPa
and the gas has a mass of 96.0 g. How many moles of the gas are in the container?
35. What is the pressure exerted by 32 g of O in a 22.0-L container at 30.0 C?
36. A mixture of gases at a total pressure of 95 kPa contains N , CO , and O . The partial pressure of the CO is 24 kPa
and the partial pressure of the N is 48 kPa. What is the partial pressure of the O ?
37. Use Graham’s law to calculate how much faster fluorine gas, F , will effuse than chlorine gas, Cl , will. The molar
mass of F = 38.0; the molar mass of Cl = 70.9.
38. If the solubility of a gas is 7.5 g/L at 404 kPa pressure, what is the solubility of the gas when the pressure is 202 kPa?
39. What is the molarity of a solution containing 1.2 grams of solute in 450 mL of solution? (molar mass of solute = 24 g)
40. If 1.0 mL of 6.0M HCl is added to 499 mL of water to give exactly a 500-mL solution, what is the molarity of the dilute
solution?
41. What is the molality of a solution containing 15 grams of solute in 0.50 kilograms of solvent? (molar mass of solute = 24
g)
42. Calculate the molality of a solution prepared by dissolving 175 g of KNO in 750 g of water.
43. What is the number of moles of solute in a 0.3 molal solution containing 0.10 kg of solvent?
44. What is the mole fraction of KCl in a 0.20 molal solution of KCl? (molar mass of KCl = 75 g; molar mass of water = 18
g)
Problem
45. Write the noble-gas electron configuration for silicon.
46. Draw the orbital diagram for phosphorus.
47. Draw the orbital diagram for argon.
48. Write the noble-gas electron configuration represented in the orbital diagram below.
49. Draw a Lewis structure for carbon disulfide, CS2.
50. The molar mass of aluminum is 26.98 g/mol and the molar mass of fluorine is 19.00 g/mol. Calculate the molar mass of
aluminum trifluoride, AlF3.
51. The molar mass of copper is 63.55 g/mol, the molar mass of sulfur is 32.07 g/mol, and the molar mass of oxygen is
16.00 g/mol. Calculate the molar mass of copper(II) sulfate, CuSO 4.
52. The molar mass of iron is 55.85 g/mol, the molar mass of silicon is 28.08 g/mol, and the molar mass of oxygen is 16.00
g/mol. Calculate the molar mass of iron(II) silicate, Fe2SiO4.
53. The molar mass of aluminum is 26.98 g/mol and the molar mass of oxygen is 16.00 g/mol. Determine the molar mass of
Al2O3.
Essay
54. Consider the description: copper(II) oxide reacts with sulfuric acid to produce copper(II) sulfate and water. List the steps
required to write its balanced chemical equation. State how to determine which of the five general types of reaction this
represents.
55. Predict the precipitate that forms when aqueous solutions of silver nitrate and potassium chloride react to form products
in a double-replacement reaction. Include a discussion of how to write the complete chemical equation describing this
reaction.
Final Exam Study Guide
Answer Section
SHORT ANSWER
1. ANS:
1.45 g
1.00 mol/237 g
= 3.68
10
10
molecules/1.00 mol
molecules
PTS: 1
2. ANS:
3.10 10
= 19.6 g F
6.02
DIF:
molecules
L2
1 mol F /6.02
REF:
10
p. 292
molecules
OBJ:
10.1.2
38.0 g F /1 mol F
PTS: 1
DIF: L2
3. ANS:
607 g Ar 1 mol Ar/39.9 g Ar = 15.2 mol Ar
REF:
p. 297
OBJ:
10.2.1
PTS: 1
4. ANS:
REF:
p. 299
OBJ:
10.2.1
1.40 10
= 6.51 g
DIF:
molecules N
L2
(1.00 mol N /6.02
PTS: 1
DIF: L3
OBJ: 10.1.2 | 10.2.1
5. ANS:
33.1 g Ni/41.9 g NiO 100% = 79% Ni
8.8 g O/41.9 g NiO 100% = 21% O
10
REF:
molecules N )
(28.0 g N /1 mol N )
p. 291 | p. 297
PTS: 1
DIF: L2
REF:
6. ANS:
8.00 g H /180 g C H O
100% = 4.44% H
p. 307
OBJ:
10.3.1
PTS: 1
DIF: L3
7. ANS:
58 g/mol/29 g/efm = 2 efm/mol; C H
78 g/mol/13 g/efm = 6 efm/mol; C H
236.1 g/mol/236.1 g/efm = 1 efm/mol; HgCl
REF:
p. 307
OBJ:
10.3.1
PTS: 1
8. ANS:
2Al 3Cl
2O
PTS:
1
L3
REF:
p. 312
OBJ:
10.3.3
DIF:
L3
REF:
p. 327
OBJ:
11.1.3 | 11.2.2
REF:
p. 327
OBJ:
11.1.3 | 11.2.2
2AlCl
PTS: 1
STA: SC.HS.1.1.8
9. ANS:
CH
DIF:
CO
2H O
DIF:
L3
STA: SC.HS.1.1.8
10. ANS:
Fe (SO )
3Ba(OH)
2Fe(OH)
3BaSO
PTS: 1
DIF: L3
STA: SC.HS.1.1.8
11. ANS:
2C H
7O
6CO 8H O (incomplete)
REF:
p. 327
PTS: 1
DIF:
OBJ: 11.1.3 | 11.2.2
12. ANS:
L3
REF:
STA:
p. 327 | p. 336
SC.HS.1.1.8
L3
SC.HS.1.1.8
REF:
p. 342 | p. 343
PTS: 1
DIF: L2
STA: SC.HS.1.1.8
14. ANS:
13.5 mol of CO ; 4.5 mol of Na C H O
REF:
PTS: 1
15. ANS:
12.0 mol of H O
REF:
H (aq)
OH (aq)
PTS: 1
OBJ: 11.3.1
13. ANS:
2K PO
3BaCl
OBJ:
11.1.3 | 11.2.2
p. 327
OBJ:
11.3.2
p. 359
OBJ:
12.2.1
H O(l)
DIF:
STA:
Ba (PO ) s
DIF:
L1
6KCl
PTS: 1
DIF: L2
REF: p. 359
OBJ: 12.2.1
16. ANS:
8.6 L H / (22.4 L/1 mol) 2 mol H O/2 mol H 22.4 L/ 1 mol = 8.6 L H O
PTS: 1
17. ANS:
17 L
DIF:
L2
REF:
p. 363
OBJ:
12.2.2
PTS: 1
18. ANS:
209.7 g Fe
DIF:
L3
REF:
p. 363
OBJ:
12.2.2
PTS: 1
19. ANS:
45.0 L H S
1 mol Fe/55.85 g Fe
DIF:
3 mol CO/2 mol Fe
L2
1 mol H S/22.4 L H S
REF:
28.01 g CO/1 mol CO = 157.8 g CO
p. 371
3 mol O /2 mol H S
OBJ:
12.3.1
22.4 L O /1 mol O = 67.5 L O
PTS: 1
DIF: L3
REF: p. 371
OBJ: 12.3.1
20. ANS:
Theoretical yield:
5.0 g H
1 mol H /2.0 g H
1 mol CH OH/2 mol H
32 g CH OH/1 mol CH OH
= 40 g CH OH
40 g CH OH 86% = 34 g CH OH
PTS: 1
DIF: L3
REF: p. 371
OBJ: 12.3.1
21. ANS:
13.0 L Cl
1 mol Cl /22.4 L Cl = 0.580 mol Cl
103.0 g Na 1 mol Na/23 g Na = 4.48 mol Na
Cl is limiting reagent: 0.580 mol Cl
2 mol NaCl/1 mol Cl = 1.16 mol NaCl
1.16 mol NaCl 58 g NaCl/1 mol NaCl = 67.3 g NaCl
PTS: 1
DIF: L3
22. ANS:
Assume the sodium is limiting:
48.7 g Na
= 6.37
10
1 mol Na/23.0 g Na
REF:
p. 371
1 mol H /2 mol Na
(6.02
OBJ:
10
12.3.1
molecules H )/1 mol H
molecules H
PTS: 1
DIF: L3
REF: p. 371
OBJ: 12.3.1
23. ANS:
10.0 L 100%/95% = 10.5 L theoretical yield
10.5 L O
1 mol O /22.4 L O
2 mol KClO /3 mol O
122.6 g KClO /1mol KClO
= 38.4 g KClO
PTS: 1
DIF: L3
REF: p. 374
OBJ: 12.3.2
24. ANS:
8.3 mol H S 2 mol H O/2 mol H S 18 g H O/1 mol = 149.4 g H O theoretical yield
percent yield = 137.1 g/149.4 g 100% = 92%
PTS: 1
25. ANS:
V =V
DIF:
= 250 mL
PTS: 1
STA: SC.HS.1.1.3
26. ANS:
V =V
DIF:
= 30.0 L
=V
P =P
p. 373
OBJ:
12.3.2
REF:
p. 419
OBJ:
14.2.1
REF:
p. 421
OBJ:
14.2.1
REF:
p. 421
OBJ:
14.2.1
= 1700 mL
L2
L2
= 590 mL
PTS: 1
STA: SC.HS.1.1.3
28. ANS:
REF:
= 34.3 L
PTS: 1
DIF:
STA: SC.HS.1.1.3
27. ANS:
T = –55 C + 273 = 218 K
T = 30.0 C + 273 = 303 K
V
L3
DIF:
= 340 kPa
= 820 mL
L2
= 140 kPa
PTS: 1
DIF: L2
STA: SC.HS.1.1.3
29. ANS:
P
V =P
V
210 kPa 15.0 L = 790 kPa V
REF:
p. 421
OBJ:
14.2.1
=V
V = 4.0 L
PTS: 1
STA: SC.HS.1.1.3
30. ANS:
227 C + 273 = 500 K
DIF:
L3
REF:
p. 419
OBJ:
14.2.1
PTS: 1
DIF:
STA: SC.HS.1.1.3
31. ANS:
T = 35.0 C + 273 = 308 K
T = 0.0 C + 273 = 273 K
L3
REF:
p. 421
OBJ:
14.2.1
p. 424
OBJ:
14.2.2
p. 424
OBJ:
14.2.2
OBJ:
14.3.1
27 C + 273 = 300 K
=
;
=
710 kPa
=P
P = 470 kPa
V =P
V
V = 97 kPa
140 mL
= 120 mL
PTS: 1
DIF: L3
STA: SC.HS.1.1.3
32. ANS:
T = 27 C + 273 = 300 K; P = 101 kPa
T = –10 C + 273 = 263 K; P = 95 kPa
V =P
V
V = 3.26
= (101 kP)
250 mL
PTS:
STA:
10 m )
10 m
PTS: 1
STA: SC.HS.1.1.3
33. ANS:
n=P
(3.5
REF:
DIF:
L3
REF:
= 0.25 L
=
1
SC.HS.1.1.3
= 0.030 mol
DIF:
L2
REF:
p. 427
34. ANS:
n=
=
= 3.0 mol
PTS: 1
STA: SC.HS.1.1.3
35. ANS:
32 g O
P=
DIF:
L2
REF:
p. 427
OBJ:
14.3.1
OBJ:
14.3.1
= 1 mol O
=
= 110 kPa
PTS: 1
STA: SC.HS.1.1.3
36. ANS:
=
–(
DIF:
L2
REF:
p. 427
+
) = 95 kPa – (48 kPa + 24 kP) = 23 kPa
PTS: 1
STA: SC.HS.1.1.3
37. ANS:
DIF:
L2
REF:
p. 434
OBJ:
14.4.1
REF:
p. 436
OBJ:
14.4.2
= 1.4
PTS: 1
STA: SC.HS.1.1.3
38. ANS:
S =
DIF:
L2
=
= 3.8 g/L
PTS: 1
OBJ: 16.1.3
39. ANS:
DIF:
STA:
L2
SC.HS.1.1.5
REF:
p. 476 | p. 477
PTS: 1
OBJ: 16.2.1
40. ANS:
DIF:
L2
REF:
p. 480 | p. 481
PTS: 1
OBJ: 16.2.2
41. ANS:
DIF:
L2
REF:
p. 483 | p. 484
PTS: 1
OBJ: 16.4.1
42. ANS:
molar mass KNO :
DIF:
STA:
L2
SC.HS.1.1.5
REF:
p. 491 | p. 492
K: 1 39.1 g = 39.1 g
N: 1 14.0 g = 14.0 g
O: 3 16.0 g = 48.0 g
molar mass = 101.1 g
PTS: 1
DIF: L2
REF:
OBJ: 16.4.1
STA: SC.HS.1.1.5
43. ANS:
Number of moles = mass of solvent molality
p. 491 | p. 492
PTS: 1
OBJ: 16.4.1
44. ANS:
DIF:
STA:
L2
SC.HS.1.1.5
REF:
p. 491 | p. 492
DIF:
STA:
L2
SC.HS.1.1.5
REF:
p. 492 | p. 493
PTS: 1
46. ANS:
DIF:
III
REF:
3
OBJ:
3
PTS: 1
47. ANS:
DIF:
III
REF:
3
OBJ:
3
PTS:
OBJ:
1
16.4.1
PROBLEM
45. ANS:
[Ne] 3s2 3p2
PTS: 1
48. ANS:
[Ne] 3s2 3p4
DIF:
III
OBJ:
4
PTS: 1
49. ANS:
DIF:
III
REF:
3
OBJ:
3
PTS: 1
50. ANS:
83.98 g/mol AlF3
Solution:
26.98 g/mol Al + (3
DIF:
III
REF:
3
OBJ:
5
3
OBJ:
1
19.00 g/mol F) = 89.3 g/mol AlF3
PTS: 1
DIF: III
51. ANS:
159.62 g/mol CuSO4
Solution:
63.55 g/mol Cu + 32.07 g/mol S + (4
REF:
16.00 g/mol O) = 159.62 g/mol CuSO4
PTS: 1
DIF: III
52. ANS:
203.78 g/mol Fe2SiO4
Solution:
(2 55.85 g/mol Fe) + (28.08 g/mol Si) + (4
REF:
3
1
DIF:
III
1
16.00 g/mol O) = 203.78 g/mol Fe2SiO4
PTS: 1
DIF: III
REF: 3
53. ANS:
101.96 g/mol Al2O3
Solution:
(2 26.98 g/mol Al) + (3 16.00 g/mol O) = 101.96 g/mol Al 2O3
PTS:
OBJ:
REF:
3
OBJ:
1
OBJ:
1
ESSAY
54. ANS:
A first approach is to indicate the reactants and products in a word equation:
copper(II) oxide + sulfuric acid
copper(II) sulfate + water
Next, convert the word equation to a skeleton equation using chemical formulas:
CuO(s) + H SO (aq)
CuSO (aq) + H O(l)
Because the Cu
is exchanged in this reaction involving an aqueous solution, and the molecular compound H O is
formed, the reaction type is double-replacement. In this case, the skeleton equation already is balanced. Inspection of
substances on the left and right of the equation shows that the same relative amounts (number of atoms) of each element
are present in the reactants and the products.
PTS: 1
DIF: L3
REF: p. 334 | p. 335
OBJ: 11.1.3 | 11.2.1
STA: SC.HS.1.1.8
55. ANS:
Because the reaction is a double-replacement type, cations are exchanged among compounds during the reaction. The
first step is to write the equation in skeleton form:
AgNO + KCl
AgCl + KNO
Inspection of this equation shows that the insoluble precipitate silver chloride forms in an aqueous solution of potassium
nitrate. The relative amounts of elements are the same on either side of the equation, so the complete equation is:
AgNO (aq) + KCl(aq)
PTS:
OBJ:
1
11.3.2
AgCl(s) + KNO (aq) (balanced)
DIF:
STA:
L3
SC.HS.1.1.8
REF:
p. 334 | p. 335