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CHAPTER 3: STOICHIOMETRY
CHAPTER 3
STOICHIOMETRY
MULTIPLE CHOICE QUESTIONS
3.1
M
What is the largest mass of Cl2 that could be prepared from the reaction of 15.0 g of MnO 2 and
30.0 g of HCl according to the following equation?
 MnCl2 + Cl2 + 2H2O
MnO2 + 4HCl
A. 0.82 g
3.2
M
B. 5.8 g
A. NH3
and 50.0 g O2 are allowed to react, which is the limiting reagent?
B. O2
C. neither
When 22.0 g NaCl and 21.0 g H2SO4 are mixed and react according to the equation below,
which
is the limiting reagent?
2NaCl + H2SO4
A. NaCl
3.4
M
E. 58.4 g
 4NO + 6H2O
When 40.0 g NH3
M
D. 14.6 g
Ammonia reacts with diatomic oxygen to form nitrogen oxide and water vapor:
4NH3 + 5O2
3.3
C. 12.2 g
 Na2SO4 + 2HCl
B. H2SO4
C. Na2SO4
D. HCl
E. Neither reagent is limiting.
Hydrochloric acid can be prepared by the following reaction:
2NaCl(s) + H2SO4(l)
 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H 2SO4 and 150 g NaCl?
A. 7.30 g
3.5
M
B. 93.5 g
C. 146 g
D. 150 g
E. 196 g
Calculate the mass of FeS formed when 9.42 g of Fe are allowed to react with
8.50 g of S.
Fe(s) + S(s)  FeS(s)
A. 17.9 g
B. 87.9 g
C. 26.0 g
D. 14.8 g
E. 1.91  10 ─
3
g
1
CHAPTER 3: STOICHIOMETRY
3.6
H
What is the theoretical yield of chromium that can be produced by the reaction of
40.0 g of Cr2O3 with 8.00 g of aluminum according to the following reaction?
2Al + Cr2O3
A. 7.7 g
3.7
H
B. 15.4 g
C. 27.3 g
D. 30.8 g
E. 49.9 g
Calculate the mass of excess reagent remaining at the end of the reaction in which
90.0 g of SO2 are mixed with 100.0 g of O2.
2SO2 + O2
A. 11.5 g
3.8
M
 Al2O3 + 2Cr
 2SO3
B. 22.5 g
C. 67.5 g
D. 77.5 g
E. 400 g
What is the maximum number of grams of ammonia, NH 3 that can be obtained
from 10.0 g of H2 and 80.0 g of N2?
N2 + 3H2  2NH3
A. 28.4 g
3.9
M
B. 48 g
D. 90.0 g
E. 97.1 g
C. 54 g
D. 96 g
Write
E. 108 g
When 10.0 mol of V2O5 are mixed with 10.0 mol of Ca, which is the limiting
reagent according to the following reaction?
V2O5(s) + 5Ca(l)
A. V2O5
3.11
M
C. 56.7 g
How many grams of water could be made from 3.0 mol H 2 and 3.0 mol O2?
and balance the equation for the chemical reaction.
A. 6.0 g
3.10
M
B. 48.6 g
 2V(l) + 5CaO(s)
B. Ca
C. V
D. CaO
E. none of these
Ammonia reacts with diatomic oxygen to form nitrogen oxide and water vapor:
4NH3 + 5O2
 4NO + 6H2O
When 40.0 g NH3
A. NH3
and 50.0 g O2 are allowed to react, which is the limiting reagent?
B. O2
C. NO
D. H2O
E. none of these
2
CHAPTER 3: STOICHIOMETRY
3.12
M
Chlorine gas can be made by the reaction of manganese dioxide with hydrochloric
acid. Which is the limiting reagent when 28 g of MnO 2 are mixed with 42 g of HCl?
MnO2(s) + 4HCl(aq)
A.
B.
C.
D.
E.
3.13
M
 MnCl2(aq) + 2H2O(l) + Cl2(g)
MnO2
HCl
MnCl2
H2O
Cl2
Ammonia reacts with diatomic oxygen to form nitrogen oxide and water vapor:
4NH3 + 5O2
 4NO + 6H2O
What is the theoretical yield of water, in moles, when 40.0 g NH 3 and 50.0 g O2 are mixed and
allowed to react?
A. 1.30 mol
3.14
M
B. 1.57 mol
A. 1.0 mol
E. none of these
2V(l) + 5CaO(s)
B. 1.6 mol
C. 2.0 mol
D. 2.4 mol
E. 4.0 mol
What is the theoretical yield of vanadium, in moles, that can be produced by the
reaction of 1.0 mol of V2O5 with 4.0 mol of calcium according to the following equation?
V2O5(s) + 5Ca(l)
A. 1.0 mol
3.16
M
D. 3.53 mol
What is the theoretical yield of vanadium, in moles, that can be produced by the
reaction of 2.0 mol of V2O5 with 6.0 mol of calcium according to the following equation?
V2O5(s) + 5Ca(l) 
3.15
M
C. 1.87 mol
 2V(l) + 5CaO(s)
B. 1.6 mol
C. 2.0 mol
D. 0.80 mol
E. none of these
What is the theoretical yield of vanadium that can be produced by the reaction of
40.0 g of V2O5 with 40.0 g of calcium according to the following equation?
V2O5(s) + 5Ca(l)
A. 11.2 g
 2V(l) + 5CaO(s)
B. 20.3 g
C. 22.4 g
D. 40.0 g
E. 5.6 g
3
CHAPTER 3: STOICHIOMETRY
3.17
M
3.18
M
How many grams of Cr can be produced by the reaction of 44.2 g of Cr2O3 with
35.0 g of Al according to the reaction?
2Al + Cr2O3
 Al2O3 + 2Cr
A. 7.56 g
B. 30.2 g
A. 30.0 g
E. none of these
 2Al + 3CO
B. 7.9 g
C. 101.2 g
D. 45.0 g
E. 31.8 g
An atom of bromine has a mass about four times greater than the mass of an atom
of neon. Which choice makes the correct comparison of the relative numbers of bromine and
neon atoms in 1000 g of each element?
A.
B.
C.
D.
E.
3.20
M
D. 104 g
What is the theoretical yield of aluminum that can be produced by the reaction of
60.0 g of aluminum oxide with 30.0 g of carbon according to the following equation?
Al2O3 + 3C
3.19
E
C. 67.4 g
The number of bromine and neon atoms is the same.
There are 1000 times as many bromine atoms as neon atoms.
There are 1000 times as many neon atoms as bromine atoms.
There are 4 times as many neon atoms as bromine atoms.
There are 4 times as many bromine atoms as neon atoms.
An atom of bromine has a mass about four times greater than the mass of an atom
of neon. How many grams of neon will contain the same number of atoms as 1000 g of
bromine?
A.
4 g Ne
B.
250 g Ne
C.
400 g Ne
D. 1000 g Ne
E. 4000 g Ne
3.21
M
What is the average mass, in grams, of one atom of iron?
A.
B.
C.
D.
E.
3.22
M
6.02  1023 g
1.66  10─ 24 g
9.28  10─ 23 g
55.85 g
55.85  10 ─ 23 g
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
4
CHAPTER 3: STOICHIOMETRY
3.23
M
What is the mass, in grams, of one copper atom?
A.
B.
C.
D.
E.
3.24
M
1.055  10─ 22 g
63.55 g
1 amu
1.66  10─ 24 g
9.476  1021 g
The mass of 1.21  1020 atoms of sulfur is:
A. 3.88  1021 g
3.25
M
E. 2.00  10─ 4 g
What is the mass of 7.80  1018 carbon atoms?
B. 6.43  103 g
C. 7.80  1018 g
0.274 moles of a substance weighs 62.5 g.
substance in units of g/mol?
A.
B.
C.
D.
E.
3.28
E
D. 6.45 mg
2.71  10─ 23 g
4.58  1022 g
28.08 g
1.04  104 g
7.60  10─ 2 g
A. 1.30  10 ─ 5 g
E. 12.01 g
3.27
M
C. 32.06 g
The mass of 1.63  1021 silicon atoms is:
A.
B.
C.
D.
E.
3.26
M
B. 2.00 mg
2.28  102 g/mol
1.71  101 g/mol
4.38  10 ─ 3 g/mol
2.17  102 g/mol
none of these
One mole of iron:
A.
B.
C.
D.
E.
(complete)
is heavier than one mole of lead (Pb).
is equivalent to 77.0 g of iron.
is equivalent to 26.0 g of iron.
has the same mass as one mole of lead .
none of these
D. 1.56  10 ─ 4 g
What is the molar mass of the
5
CHAPTER 3: STOICHIOMETRY
3.29
E
Which one of the following is not equivalent to 1.00 mol of the substance?
A.
B.
C.
D.
E.
3.30
E
One nanogram doesn't seem like a very large number.
present
in 1.00 ng of magnesium?
A.
B.
D.
C.
E.
3.31
M
6.02  1023 C atoms
0.146 atoms
0.292 atoms
8.78  1022 atoms
1.76  1023 atoms
0.279 mol
3.58 mol
7.43 mol
4.21 mol
6.02  1023 mol
Calculate the number of moles of phosphorus represented by 12.0 g of phosphorus.
A.
B.
C.
D.
E.
3.34
M
4.11  10 ─11 atoms
2.48  1013 atoms
6.02  1014 atoms
6.83  10 ─35 atoms
1.46  1034 atoms
Determine the number of moles of aluminum represented by 96.7 g of Al.
A.
B.
C.
D.
E.
3.33
E
How many magnesium atoms are
How many atoms are present in 5.54 g F2?
A.
B.
C.
D.
E.
3.32
E
6.02  1023 C atoms
26.0 g Fe
12.01 g C
65.38 g Zn
6.02  1023 Fe atoms
0.387 mol
0.80 mol
2.58 mol
12.0 mol
6.02  1023 mol
Which of the following contains the greatest number of atoms?
A.
B.
C.
D.
E.
100 g of Pb
2.0 mol of Ar
0.1 mol of Fe
5 g of He
20 million O2 molecules
6
CHAPTER 3: STOICHIOMETRY
3.35
E
How many phosphorus atoms are present in 1.0 g of phosphorus?
A.
B.
C.
D.
E.
3.36
E
Calculate the formula mass of potassium permanganate KMnO4.
A.
B.
C.
D.
E.
3.37
E
1 atom
0.032 atoms
4.0  1022 atoms
1.94  1022 atoms
6.0  1023 atoms
52 amu
70 amu
110 amu
158 amu
176 amu
Which is the molar mass of acetaminophen, C8H9NO2?
A. 151 g/mol
3.38
E
E. 125 g/mol
How many molecules of aspirin are
2.77 molecules
2.77  10 ─ 3 molecules
1.67  1024 molecules
1.67 1021 molecules
None of these
What is the mass of 0.0250 mol P2O5?
A. 35.5 g
3.41
M
D. 162 g/mol
0.51 mol
1.94 mol
4.07 mol
88.0 mol
171 mol
The molar mass of aspirin is 180.2 g/mol.
present in one 500-milligram tablet?
A.
B.
C.
D.
E.
3.40
E
C. 43 g/mol
How many moles of CF4 are there in 171 g of CF4?
A.
B.
C.
D.
E.
3.39
M
B. 76 g/mol
B. 5676 g
Formaldehyde has the formula CH2O.
formaldehyde?
A. 6.1  10 ─
27
D. 1.51  1022 g
C. 0.0250 g
B. 3.7  10 ─
3
E. 3.55 g
How many molecules are in 0.11 g
C. 4
D. 2.2  1021
E. 6.6  1022
7
CHAPTER 3: STOICHIOMETRY
3.42
M
How many ozone (O3) molecules are in 8.0 g of ozone?
A.
B.
C.
D.
E.
3.43
M
3 molecules
3.6  1024 molecules
1.0  1023 molecules
3.0  1023 molecules
6.0  1023 molecules
How many moles of HCl are represented by 1.0  1019 HCl molecules?
A. 1.7  10 ─ 5 mol
E. 6.02  104 mol
3.44
E
B. 174 g
D. 36.5 mol
C. 363 g
D. 1.81  1024 g
E. 40.3 g
How many moles of NH3 (ammonia) are represented by in 77.3 g of NH3?
B. 4.54 mol
C. 14.0 mol
D. 1.31  103 mol
How many moles of oxygen atoms are contained in 10 moles of KClO 3 (potassium
chlorate)?
A.
B.
C.
D.
E.
3.48
M
C. 1.0  1019 mol
3.6  1024 atoms
4.6  1022 atoms
1.3  1023 atoms
0.217 atoms
0.072 atoms
A. 0.220 mol
E. none of these
3.47
E
mol
How many sodium atoms are in 6.0 g of Na3N?
A.
B.
C.
D.
E.
3.46
M
─3
Calculate the mass of 3.00 moles of CF2Cl2.
A. 3.00 g
3.45
M
B. 1.5  10
3 mol
3.3 mol
10 mol
30 mol
6.02  1024 mol
How many sulfur atoms are there in 21.0 g of Al2S3?
A.
B.
C.
D.
E.
8.42  1022 atoms
2.53  1023 atoms
2.14  1023 atoms
6.02  1023 atoms
6.3  1026 atoms
8
CHAPTER 3: STOICHIOMETRY
3.49
M
How many sulfur atoms are present in 25.6 g of Al 2(S2O3)3?
A. 0.393
3.50
M
How many moles of oxygen atoms are in 25.7 g CaSO 4?
How many moles of Cl atoms are in 65.2 g CHCl3?
3.3  1023 mol
D. 1.64 mol
How many grams of sulfur are contained in 6.0 g of Fe 2(SO4)3?
B. 0.48 g
C. 6.00 g
D. 0.92 g
E. 1.44 g
How many grams of sodium are contained in 10 g of sodium sulfate, Na 2SO4?
A. 0.16 g
3.57
M
C.
9.6  1025 atoms
8.0  1024 atoms
159 atoms
4.21 atoms
342 atoms
A. 2.40 g
3.56
M
B. 1.09 mol
How many carbon atoms are in 10 lbs of sugar, C12H22O11?
A.
B.
C.
D.
E.
3.55
M
D. 1.14  1023 mol
4 O atoms
2.40  1024 O atoms
1.13 O atoms
9.09  1023 O atoms
2.28  1023 O atoms
A. 0.548 mol
3.54
H
C. 4.00 mol
How many oxygen atoms are in 51.4 g CaSO4?
A.
B.
C.
D.
E.
3.53
M
E. 2.37  1023
0.74 atoms
3.0 atoms
4.5  1023 atoms
1.8  1024 atoms
2.4  1023 atoms
A. 0.189 mol
B. 0.755 mol
E. 4.55  1023 mol
3.52
M
D. 7.90  1022
How many fluorine atoms are there in 65 g CF4?
A.
B.
C.
D.
E.
3.51
M
C. 3.95  1022
B. 6
B. 0.32 g
C. 3.2 g
D. 1.6 g
E. 142 g
How many grams of nitrogen are contained in 7.5 g of Ca(NO3)2?
A. 0.64 g
B. 1.3 g
C. 0.15 g
D. 1.15 g
E. 2.3 g
E. 3.0 mol
9
CHAPTER 3: STOICHIOMETRY
3.58
M
The mass of four moles of molecular bromine (Br 2) is:
A.
3.59
M
80 g
B.
320 g
E.
24  1023 g
B 4.5 g
C. 15.7 g
D. 160 g
E. 319 g
B. 46 g
C. 23 g
D. 4.6 g
E. none of these
What is the mass of 8.25  1019 UF6 molecules?
B. 0.0482 g
C. 1.37  10
─4
g
D. 2.90  1022 g
E. 8.25  1019 g
On the average an atom of uranium (U) is approximately how many times heavier
than an atom of potassium?
A.
B.
C.
D.
E.
3.64
M
140 g
What is the mass of 0.20 mol of C2H6O (ethyl alcohol)?
A. 352 g
3.63
E
D.
4.99  10 ─ 24 g
138 g
6.52  10 ─ 2 g
50 g
1.81  1024 g
A. 230 g
3.62
M
639 g
What is the mass of 3.00 mol of ethanol, C2H6O?
A.
B.
C.
D.
E.
3.61
E
C.
Calculate the mass of 4.50 mol of chlorine gas, Cl2.
A. 6.34  10 ─ 2 g
3.60
E
6.1 times
4.8 times
2.4 times
12.5 times
7.7 times
Boron obtained from borax deposits in Death Valley consists of two isotopes. They
are boron-10 and boron-11 with atomic masses of 10.013 amu and 11.009 amu, respectively.
The atomic mass of boron is 10.81 amu. See periodic table. Which isotope of boron is more
abundant, boron-10 or boron-11?
A. cannot be determined from data given
B. neither, their abundances are the same
C. boron-10
D. boron-11
3.65
M
10
The element oxygen consists of three naturally occurring isotopes: 16O, 17O, and 18O.
The atomic mass of oxygen is 16.0 amu. Which of the following can be implied about the
relative abundances of these isotopes?
A.
B.
C.
D.
E.
More than 50% of all O atoms are 17O.
Almost all O atoms are 18O.
Almost all O atoms are 17O.
The isotopes all have the same abundance, i.e. 33.3%.
The abundances of 17O and 18O are very small.
CHAPTER 3: STOICHIOMETRY
3.66
M
What is the coefficient for H2O when the following equation is balanced
with the smallest set of whole numbers?
 ___ NaOH + ___ H2
___ Na + ___ H2O
A. 1
3.67
M
B. 2
C. 3
A. 3
3.69
M
3.70
M
B. 4

C. 6
___ Al(OH)3 + ___ CH4
D. 12
E. 24
__ C2H4 + __ O2
 __ CO2 + __ H2O
A. 1
C. 3
B. 2
D. 4
E. 6
When a chemical equation is balanced, it will have a set of whole number
coefficients which cannot be reduced to smaller whole numbers. What is the smallest whole
number coefficient for O2 when the following combustion reaction of a hydrocarbon is
balanced?
___ C7H14 + ___ O2
 ___ CO2 + ___ H2O
A. 42
C. 11
B. 21
D. 10
E. none of these
What is the coefficient for O2 when the following combustion reaction of a
fatty acid is balanced?
A. 1
3.72
M
E. 5
When balanced, the coefficient of O2 in the following equation is:
 __ CO2 + __ H2O
__ C18H36O2 + __ O2
3.71
M
D. 4
What is the coefficient for H2O when the following equation is balanced
with the smallest set of whole numbers?
___ Al4C3 + ___ H2O
3.68
E
11
B. 8
C. 9
D. 26
E. 27
What is the coefficient of H2SO4 when the following equation is balanced
with the smallest set of whole numbers?
___ Ca3(PO4)2 + ___ H2SO4
 ___ CaSO4 + ___ H3PO4
A. 3
D. 11
B. 8
C. 10
E. none of these
Balance the equation using the smallest set of whole numbers.
___ PCl3(l) + ___ H2O(l)
A. 1
B. 2
C. 3
 ___ H3PO3(aq) + ___ HCl(aq)
D. 5
E. none of these
What is the coefficient for H 2O?
CHAPTER 3: STOICHIOMETRY
3.73
M
What is the coefficient for O2 when the following equation is balanced with the
smallest set of whole numbers?
 ___ CO2 + ___ H2O
___ CH3OH + ___ O2
A. 1
3.74
M
3.75
M
3.76
M
B. 2
C. 3
D. 7
E. none of these
When a chemical equation is balanced, it will have a set of whole-number
coefficients that cannot be reduced to smaller whole numbers. Balance the following equation.
Then add together the four coefficients, don't forget the 1's. The sum of the coefficients is:
__ SF4 + __ H2O
 __ H2SO3 + __ HF
A. 4
C. 7
B. 6
D. 9
E. none of these
Balance the equation using the smallest set of whole numbers. Then sum the
coefficients in the balanced equation. Don't forget coefficients of "one." The sum of the
coefficients is:
___ Cr + ___ H2SO4
 ___ Cr2(SO4)3 + ___ H2
A. 4
C. 11
B. 9
D. 13
E. 15
When a chemical equation is balanced, it will have a set of whole number
coefficients which cannot be reduced to smaller whole numbers. Balance the following
chemical equation. What is the sum of the coefficients (do not forget to add in the coefficient
one if it occurs)?
 ___ Al2(SO4)3 + ___ H2
___ Al + ___ H2SO4
A. 3
3.77
M
B. 5
C. 6
D. 9
E. 12
When a chemical equation is balanced, it will have a set of whole number
coefficients which cannot be reduced to smaller whole numbers. Balance the following chemical
equation with the smallest set of whole-number coefficients.
___ CH4 + ___ Cl2
 ___ CCl4 + ___ HCl
The sum of the coefficients is (do not forget to add in the coefficient one if it occurs):
A. 4
3.78
M
12
B. 6
C. 8
D. 10
E. 12
How many iron atoms are present in 5.00  102 g of iron?
A.
B.
C.
D.
E.
0.052 atoms
19.2 atoms
5.39  1024 atoms
1.15  1025 atoms
500 atoms