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first test
Student: ___________________________________________________________________________
1. What is the mass, in grams, of one arsenic atom?
A. 5.48  10-23 g
B. 33.0 g
C. 74.9 g
D. 1.24  10-22 g
E. 8.04  1021 g
2. The mass of 1.21  1020 atoms of sulfur is
A. 3.88  1021 g.
B. 2.00 mg.
C. 32.06 g.
D. 6.44 mg.
E. 2.00  10-4 g.
3. Which of these quantities does not represent 1.00 mol of the indicated substance?
A. 6.02  1023 C atoms
B. 26.0 g Fe
C. 12.01 g C
D. 65.4 g Zn
E. 6.02  1023 Fe atoms
4. One nanogram doesn't seem like a very large number. How many magnesium atoms are there in 1.00 ng of
magnesium?
A. 4.11  10-11 atoms
B. 2.48  1013 atoms
C. 6.83  10-35 atoms
D. 6.02  1014 atoms
E. 1.46  1034 atoms
5. How many atoms are in 0.0728 g of PCl3?
A. 1.28  1021 atoms
B. 4.38  1022 atoms
C. 4.39  1021 atoms
D. 3.19  1020 atoms
E. 6.02  1024 atoms
6. Calculate the number of moles of cesium in 50.0 g of cesium.
A. 0.376 mol
B. 0.357 mol
C. 2.66 mol
D. 2.80 mol
E. 0.0200 mol
7. Calculate the molecular mass of potassium permanganate, KMnO4.
A. 52 amu
B. 70 amu
C. 110 amu
D. 158 amu
E. 176 amu
8. What is the molar mass of nicotine, C10H14N2?
A. 134 g/mol
B. 148 g/mol
C. 158 g/mol
D. 210 g/mol
E. 162 g/mol
9. The molecular formula of aspirin is C9H8O4. How many aspirin molecules are present in one 500-milligram
tablet?
A. 2.77 molecules
B. 2.77  10-3 molecules
C. 1.67  1024 molecules
D. 1.67  1021 molecules
E. none of these is correct.
10. How many moles of oxygen atoms are there in 10 moles of KClO3?
A. 3 mol
B. 3.3 mol
C. 10 mol
D. 30 mol
E. 6.02  1024 mol
11. How many fluorine atoms are there in 65 g of CF4?
A. 0.74 atoms
B. 3.0 atoms
C. 4.5  1023 atoms
D. 1.8  1024 atoms
E. 2.4  1023 atoms
12. The percent composition by mass of a compound is 76.0% C, 12.8% H, and 11.2% O. The molar mass of
this compound is 284.5 g/mol. What is the molecular formula of the compound?
A. C10H6O
B. C9H18O
C. C16H28O4
D. C20H12O2
E. C18H36O2
13. A compound was discovered whose composition by mass is 85.6% C and 14.4% H. Which of these choices
could be the molecular formula of this compound?
A. CH4
B. C2H4
C. C3H4
D. C2H6
E. C3H8
14. When a chemical equation is balanced, it will have a set of whole number coefficients that cannot be
reduced to smaller whole numbers. What is the coefficient for O2 when the following combustion reaction of a
hydrocarbon is balanced?
___ C7H14 + ___ O2  ___ CO2 + ___ H2O
A. 42
B. 21
C. 11
D. 10
E. none of these
15. Balance the following equation using the smallest set of whole numbers, then add together the coefficients.
Don't forget to count coefficients of one.
___ Al + ___ H2SO4  ___ Al2(SO4)3 + ___ H2
The sum of the coefficients is
A. 3.
B. 5.
C. 6.
D. 9.
E. 12.
16. Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3 + 5O2  4NO + 6H2O
When 40.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent?
A. NH3
B. O2
C. Neither reagent is limiting.
17. How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according
to the following chemical equation?
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
A. 0.82 g
B. 5.8 g
C. 13.0 g
D. 14.6 g
E. 58.4 g
18. How many grams of Cr can be produced by the reaction of 44.1 g of Cr2O3 with 35.0 g of Al according to
the following chemical equation?
2Al + Cr2O3  Al2O3 + 2Cr
A.
B.
C.
D.
E.
7.56 g
30.2 g
67.4 g
104 g
60.4 g
19. The first step in the Ostwald process for producing nitric acid is
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g).
If the reaction of 150. g of ammonia with 150. g of oxygen gas yields 87. g of nitric oxide (NO), what is the
percent yield of this reaction?
A. 100%
B. 49%
C. 77%
D. 33%
E. 62%
20. One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona,
Na5(CO3)2(HCO3)·2H2O,
Na5(CO3)2(HCO3)·2H2O(s)  5Na2CO3(s) + CO2(g) + 3H2O(g)
When 1.00 metric ton (1  103 kg) of trona is decomposed, 0.74 metric ton of Na2CO3 is recovered. What is the
percent yield of this reaction?
A. 93%
B. 43%
C. 22%
D. 83%
E. 17%
first test Key
1. What is the mass, in grams, of one arsenic atom?
a. 5.48  10-23 g
b. 33.0 g
c. 74.9 g
D. 1.24  10-22 g
e. 8.04  1021 g
Chang - 003 Mass... #4
Difficulty: Easy
2. The mass of 1.21  1020 atoms of sulfur is
a. 3.88  1021 g.
b. 2.00 mg.
c. 32.06 g.
D. 6.44 mg.
e. 2.00  10-4 g.
Chang - 003 Mass... #6
Difficulty: Medium
3. Which of these quantities does not represent 1.00 mol of the indicated substance?
a. 6.02  1023 C atoms
B. 26.0 g Fe
c. 12.01 g C
d. 65.4 g Zn
e. 6.02  1023 Fe atoms
Chang - 003 Mass... #11
Difficulty: Medium
4. One nanogram doesn't seem like a very large number. How many magnesium atoms are there in 1.00 ng of
magnesium?
a. 4.11  10-11 atoms
B. 2.48  1013 atoms
c. 6.83  10-35 atoms
d. 6.02  1014 atoms
e. 1.46  1034 atoms
Chang - 003 Mass... #12
Difficulty: Medium
5. How many atoms are in 0.0728 g of PCl3?
A. 1.28  1021 atoms
b. 4.38  1022 atoms
c. 4.39  1021 atoms
d. 3.19  1020 atoms
e. 6.02  1024 atoms
Chang - 003 Mass... #16
Difficulty: Medium
6. Calculate the number of moles of cesium in 50.0 g of cesium.
A. 0.376 mol
b. 0.357 mol
c. 2.66 mol
d. 2.80 mol
e. 0.0200 mol
Chang - 003 Mass... #21
Difficulty: Easy
7. Calculate the molecular mass of potassium permanganate, KMnO4.
a. 52 amu
b. 70 amu
c. 110 amu
D. 158 amu
e. 176 amu
Chang - 003 Mass... #23
Difficulty: Easy
8. What is the molar mass of nicotine, C10H14N2?
a. 134 g/mol
b. 148 g/mol
c. 158 g/mol
d. 210 g/mol
E. 162 g/mol
Chang - 003 Mass... #26
Difficulty: Easy
9. The molecular formula of aspirin is C9H8O4. How many aspirin molecules are present in one 500-milligram
tablet?
a. 2.77 molecules
b. 2.77  10-3 molecules
c. 1.67  1024 molecules
D. 1.67  1021 molecules
e. none of these is correct.
Chang - 003 Mass... #29
Difficulty: Medium
10. How many moles of oxygen atoms are there in 10 moles of KClO3?
a. 3 mol
b. 3.3 mol
c. 10 mol
D. 30 mol
e. 6.02  1024 mol
Chang - 003 Mass... #34
Difficulty: Easy
11. How many fluorine atoms are there in 65 g of CF4?
a. 0.74 atoms
b. 3.0 atoms
c. 4.5  1023 atoms
D. 1.8  1024 atoms
e. 2.4  1023 atoms
Chang - 003 Mass... #37
Difficulty: Medium
12. The percent composition by mass of a compound is 76.0% C, 12.8% H, and 11.2% O. The molar mass of
this compound is 284.5 g/mol. What is the molecular formula of the compound?
a. C10H6O
b. C9H18O
c. C16H28O4
d. C20H12O2
E. C18H36O2
Chang - 003 Mass... #54
Difficulty: Medium
13. A compound was discovered whose composition by mass is 85.6% C and 14.4% H. Which of these choices
could be the molecular formula of this compound?
a. CH4
B. C2H4
c. C3H4
d. C2H6
e. C3H8
Chang - 003 Mass... #55
Difficulty: Medium
14. When a chemical equation is balanced, it will have a set of whole number coefficients that cannot be
reduced to smaller whole numbers. What is the coefficient for O2 when the following combustion reaction of a
hydrocarbon is balanced?
___ C7H14 + ___ O2  ___ CO2 + ___ H2O
a. 42
B. 21
c. 11
d. 10
e. none of these
Chang - 003 Mass... #59
Difficulty: Medium
15. Balance the following equation using the smallest set of whole numbers, then add together the coefficients.
Don't forget to count coefficients of one.
___ Al + ___ H2SO4  ___ Al2(SO4)3 + ___ H2
The sum of the coefficients is
a. 3.
b. 5.
c. 6.
D. 9.
e. 12.
Chang - 003 Mass... #66
Difficulty: Medium
16. Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3 + 5O2  4NO + 6H2O
When 40.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent?
a. NH3
B. O2
c. Neither reagent is limiting.
Chang - 003 Mass... #68
Difficulty: Medium
17. How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according
to the following chemical equation?
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
a. 0.82 g
b. 5.8 g
C. 13.0 g
d. 14.6 g
e. 58.4 g
Chang - 003 Mass... #73
Difficulty: Medium
18. How many grams of Cr can be produced by the reaction of 44.1 g of Cr2O3 with 35.0 g of Al according to
the following chemical equation?
2Al + Cr2O3  Al2O3 + 2Cr
a. 7.56 g
B. 30.2 g
c. 67.4 g
d. 104 g
e. 60.4 g
Chang - 003 Mass... #84
Difficulty: Medium
19. The first step in the Ostwald process for producing nitric acid is
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g).
If the reaction of 150. g of ammonia with 150. g of oxygen gas yields 87. g of nitric oxide (NO), what is the
percent yield of this reaction?
a. 100%
b. 49%
c. 77%
D. 33%
e. 62%
Chang - 003 Mass... #88
Difficulty: Difficult
20. One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona,
Na5(CO3)2(HCO3)·2H2O,
Na5(CO3)2(HCO3)·2H2O(s)  5Na2CO3(s) + CO2(g) + 3H2O(g)
When 1.00 metric ton (1  103 kg) of trona is decomposed, 0.74 metric ton of Na2CO3 is recovered. What is the
percent yield of this reaction?
A. 93%
b. 43%
c. 22%
d. 83%
e. 17%
Chang - 003 Mass... #89
Difficulty: Difficult
# of Questions
Category
20
Chang - 003 Mass...
2
Difficulty: Difficult
5
Difficulty: Easy
13
Difficulty: Medium