Chemistry 10th Edition International Edition by Kenneth W – Test Bank
To Purchase this Complete Test Bank with Answers Click the link Below
https://tbzuiqe.com/product/chemistry-10th-edition-international-edition-by-kenneth-w-test-bank/
If face any problem or
Further information contact us At tbzuiqe@gmail.com
Sample Test
Chapter 3—Chemical Equations and Reaction Stoichiometry
MULTIPLE CHOICE
1. Which
of the following is not a consequence of the Law of Conservation of Matter?
2. It provides
a basis for balancing chemical equations.
3. It
means that there will be no observable change in the quantity of matter during
a chemical reaction.
4. As a
result, there will be the same number of moles on both the reactant and the
product side of a balanced equation.
5. It
can be stated as “matter is neither created nor destroyed during a chemical
reaction.”
6. All
of these are a consequence of the Law of Conservation of Mass.
ANS: C
OBJ: Understand the
implications of the Law of Conservation of Matter.
TOP: Chemical Equations
2. Balancing
a chemical equation so that it obeys the law of conservation of matter
requires:
3. Adjusting
the coefficients in front of the formulas so there are the same number and type
of atom on both sides of the equation.
4. Making
sure the reactants and products are in the same phase.
5. Keeping
the total charge the same on both sides of the equation.
6. Changing
the formulas of the products and reactants.
7. Keeping
the same number of molecules on both sides of the equation.
ANS: A
OBJ: Understand the
implications of the Law of Conservation of Matter.
TOP: Chemical Equations
3. What
scientific law requires that subscripts in formulas should never be changed
while balancing a chemical equation?
4. Law
of Multiple Proportions
5. Law
of Definite Proportions
6. Law
of Conservation of Matter
7. Law
of Conservation of Matter and Energy
8. Law
of Conservation of Energy
ANS: B
OBJ: Understand the
implications of the Law of Definite Proportions.
TOP: Chemical Equations
4. Balance
the following equation with the smallest whole number coefficients. What is the
coefficient for O2 in the balanced equation?
C4H10 + O2 ® CO2
+ H2O
1. 9
2. 5
3. 15
4. 6
5. 13
ANS: E
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
5. Balance
the following equation with the smallest whole number coefficients. What is the
coefficient for O2 in the balanced equation?
C4H9SO + O2 ® CO2
+ SO2 + H2O
1. 54
2. 29
3. 23
4. 32
5. 27
ANS: E
DIF: Harder Question
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
6. What
is the coefficient for carbon dioxide when the following equation showing the
combustion of isopropyl alcohol is balanced with the smallest whole number
coefficients?
C3H8O + O2 ® CO2
+ H2O
1. 3
2. 6
3. 13
4. 4
5. 1
ANS: B
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
7. Balance
the following equation with the smallest whole number coefficients. What is the
coefficient for H2O in the balanced equation?
Al(OH)3 + HCl ®
AlCl3 + H2O
1. 1
2. 2
3. 3
4. 14
5. 5
ANS: C
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
8. Balance
the following equation with the smallest whole number coefficients. What is the
coefficient for H2O in the balanced equation?
LiBF4 + H2O ®
H3BO3 + HF + LiF
1. 3
2. 2
3. 5
4. 6
5. 8
ANS: A
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
9. What
is the coefficient for HBr when the following equation is balanced with the
smallest whole number coefficients?
Br2 + H2O ® HBr
+ HBrO3
1. 5
2. 7
3. 8
4. 3
5. 6
ANS: A
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
10.
Balance the following equation with the smallest whole number
coefficients. What is the coefficient for NH3 in the balanced equation?
Fe(NO3)3 + NH3 + H2O ®
Fe(OH)3 + NH4NO3
1. 1
2. 3
3. 2
4. 6
5. 4
ANS: B
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
11.
Elemental phosphorus is produced from calcium phosphate in the
following reaction. What is the coefficient for C when this equation is
balanced with the smallest whole number coefficients?
Ca3(PO4)2 + SiO2 + C ® P4 +
CO + CaSiO3
1. 10
2. 3
3. 1
4. 6
5. 4
ANS: A
OBJ: Balance a chemical
equation.
TOP: Chemical Equations
12.
When heated lead nitrate decomposes according to the following
equation. What is the coefficient for NO2 when the this equation is balanced
with the smallest whole number coefficients?
Pb(NO3)2 ® PbO
+ O2 + NO2
1. 1
2. 2
3. 3
4. 4
5. 5
ANS: D
OBJ: Balance a chemical equation.
TOP: Chemical Equations
13.
The reaction of a hydrocarbon with oxygen to produce CO2 and
water is called complete combustion. Write the balanced reaction for the
complete combustion of heptane, C7H16, with the smallest whole number
coefficients. Then choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
14.
23
15.
27
16.
29
17.
30
18.
32
ANS: B
DIF: Harder Question
OBJ: Write a chemical
equation for the combustion of a hydrocarbon in oxygen, producing water and
carbon dioxide. | Balance a chemical equation. | Sum the coefficients.
TOP: Chemical Equations
14.
The discharge reaction that occurs in a nickel-cadmium (nicad)
battery is shown below. Choose the answer that represents the sum of all the
coefficients when this equation is balanced using the smallest whole number
coefficients.
Cd + NiO2 + H2O ®
Cd(OH)2 + Ni(OH)2
1. 5
2. 6
3. 7
4. 10
5. 5
ANS: B
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
15.
One of the reactions that take place inside a blast furnace to
produce iron is the reduction of iron ore by carbon monoxide. Choose the answer
that represents the sum of all the coefficients when this equation is balanced
using the smallest whole number coefficients.
Fe2O3 + CO ® CO2
+ Fe
1. 5
2. 9
3. 4
4. 7
5. 6
ANS: B
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
16.
Ultra-pure Si is required to produce semi-conductors. The first
step in this process is combining silicon dioxide with carbon and chlorine.
Choose the answer that represents the sum of all the coefficients when this
equation is balanced using the smallest whole number coefficients.
SiO2 + C + Cl2 ® CO +
SiCl4
1. 5
2. 6
3. 16
4. 7
5. 8
ANS: E
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
17.
Balance the following equation with the smallest whole number
coefficients. Choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
Cr + H2SO4 ®
Cr2(SO4)3 + H2
1. 7
2. 9
3. 11
4. 13
5. 15
ANS: B
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
18.
Balance the following equation with the smallest whole number coefficients.
Choose the answer that is the sum of the coefficients in the balanced equation.
Do not forget coefficients of “one”.
P4 + Cl2 ® PCl5
1. 7
2. 9
3. 11
4. 13
5. 15
ANS: E
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
19.
What is the sum of all coefficients when the following equation
is balanced, using the smallest whole number coefficients? Do not forget
coefficients of “one”.
CS2 + Cl2 ® CCl4
+ S2Cl2
1. 6
2. 8
3. 10
4. 12
5. 14
ANS: A
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
20.
Balance the following equation with the smallest whole number
coefficients. Choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
RbOH + H3PO4 ®
Rb3PO4 + H2O
1. 8
2. 10
3. 12
4. 4
5. 6
ANS: A
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
21.
What is the sum of all coefficients when the following equation
is balanced, using the smallest whole number coefficients? Do not forget
coefficients of “one”.
Cl2O7 + Ca(OH)2 ®
Ca(ClO4)2 + H2O
1. 6
2. 8
3. 10
4. 4
5. 5
ANS: D
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
22.
Balance the following equation with the smallest whole number
coefficients. Choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
XeF2 + H2O ® Xe +
HF + O2
1. 10
2. 11
3. 13
4. 15
5. 16
ANS: B
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
23.
What is the sum of all coefficients when the following equation
is balanced, using the smallest whole number coefficients? Do not forget coefficients
of “one”.
SiCl4 + H2O ®
H4SiO4 + HCl
1. 6
2. 8
3. 10
4. 12
5. 14
ANS: C
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
24.
What is the sum of all coefficients when the following equation
is balanced, using the smallest whole number coefficients? Do not forget
coefficients of “one”.
Ag + H2S + O2 ® Ag2S
+ H2O
1. 9
2. 10
3. 11
4. 12
5. 14
ANS: C
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
25.
Balance the following equation with the smallest whole number
coefficients. Choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
ZrO2 + CCl4 ®
ZrCl4 + COCl2
1. 6
2. 8
3. 10
4. 12
5. 14
ANS: A
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
26.
Balance the following equation with the smallest whole number
coefficients. Choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
Na2O + P4O10 ®
Na3PO4
1. 5
2. 8
3. 9
4. 10
5. 11
ANS: E
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
27.
What is the sum of all coefficients when the following equation
is balanced, using the smallest whole number coefficients? Do not forget
coefficients of “one”.
PtCl4 + XeF2 ® PtF6
+ ClF + Xe
1. 16
2. 22
3. 24
4. 26
5. 32
ANS: A
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
28.
Balance the following equation with the smallest whole number
coefficients. Choose the answer that is the sum of the coefficients in the
balanced equation. Do not forget coefficients of “one”.
CuSO4 + NH3 + H2O ®
(NH4)2SO4 + Cu(NH3)4(OH)2
1. 8
2. 9
3. 11
4. 12
5. 14
ANS: C
OBJ: Balance a chemical
equation. | Sum the coefficients.
TOP: Chemical Equations
29.
Consider the following balanced equation.
2H2 + O2 ® 2H2O
Which one of the following statements is false?
2. One
molecule of O2 will react with 2 molecules of H2.
3. One
mole of O2 will react with 2 moles of H2.
4. The
complete reaction of 32.0 g of O2 will produce 2 moles of H2O.
5. The
complete reaction of 2.0 g of H2 will produce 36.0 g of H2O.
6. The
amount of reaction that consumes 32.0 g of O2 produces 36.0 g of H2O.
ANS: D
OBJ: Translate a balanced
chemical equation into words.
TOP: Calculations Based on
Chemical Equations
30.
How many molecules of O2 would react with 56 C2H6 molecules
according to the following balanced equation?
2C2H6 + 7O2 ® 4CO2
+ 6H2O
1. 196
2. 392
3. 112
4. 50
5. 784
ANS: A
OBJ: Use the balanced
chemical reaction to convert molecules organic reactant to molecules dioxygen.
TOP: Calculations Based on
Chemical Equations
31.
How many moles of O2 are required to react with 23.5 moles of
methanol?
2CH3OH + 3O2 ® 2CO2
+ 4H2O
47.
47.0
48.
35.3
49.
11.8
50.
40
51.
23.5
ANS: B
OBJ: Use the balanced
chemical reaction to convert moles organic reactant to moles dioxygen.
TOP: Calculations Based on Chemical
Equations
32.
Ammonium nitrate fertilizer is sometimes used as an explosive.
How many moles of water can be formed from the decomposition of 13.2 moles of
ammonium nitrate?
2NH4NO3 ® 2N2
+ O2 + 4H2O
6. 6.60
7. 14.0
8. 26.4
9. 13.2
10.
18.0
ANS: C
OBJ: Use the balanced
chemical reaction to convert moles reactant to moles product.
TOP: Calculations Based on
Chemical Equations
33.
How many moles of O2 are required to burn completely 63.5 g of
C6H6, according to the following equation?
2C6H6 + 15O2 ® 12CO2
+ 6H2O
1. 0.814
2. 12.2
3. 6.1
4. 0.109
5. 9.21
ANS: C
OBJ: Use molecular weights
and the balanced chemical reaction to convert grams organic reactant to moles
dioxygen.
TOP: Calculations Based on
Chemical Equations
34.
Propane (C3H8) burns in oxygen to form CO2 and H2O according to
the following equation. How many grams of O2 are required to burn 3.01 ´ 1023 propane
molecules?
C3H8 + 5O2 ® 3CO2
+ 4H2O
80.
80.0 g
81.
40.0 g
82.
160 g
83.
16.0 g
84.
64.0 g
ANS: A
OBJ: Use Avogadro’s number,
molecular weights, and the balanced chemical reaction to convert molecules
organic reactant to grams dioxygen.
TOP: Calculations Based on
Chemical Equations
35.
How many moles of H2O will be produced from the complete combustion
of 2.4 grams of CH4?
CH4 + 2O2 ® CO2
+ 2H2O
1. 0.15
2. 0.30
3. 1.5
4. 3.0
5. 6.0
ANS: B
OBJ: Use molecular weights
and the balanced chemical reaction to convert grams organic reactant to moles
product.
TOP: Calculations Based on Chemical
Equations
36.
How many moles of CaCO3 would have to be decomposed to produce
129 grams of CaO?
CaCO3 ® CaO
+ CO2
1. 1.75
2. 1.86
3. 2.00
4. 2.25
5. 2.30
ANS: E
OBJ: Use formula weights and
the balanced chemical reaction to convert grams product to moles reactant.
TOP: Calculations Based on
Chemical Equations
37.
Acrylonitrile, C3H3N, is a molecule used to produce a plastic
called Orlon. How many grams of acrylonitrile could be produced by reacting 583
g of propene, C3H6 with excess ammonia, NH3 and oxygen?
2C3H6 + 2NH3 + 3O2 ®
2C3H3N + 6H2O
1. 368 g
2. 1470
g
3. 462 g
4. 735 g
5. 583 g
ANS: D
OBJ: Use molecular weights
and the balanced chemical reaction to convert grams organic reactant to grams
product.
TOP: Calculations Based on
Chemical Equations
38.
A secondary step in the process to produce ultra-pure silicon is
to combine silicon tetrachloride with magnesium. How many grams of Si could be
produced by reacting 2.00 kg of SiCl4 with excess Mg?
SiCl4 + 2Mg ® Si +
2MgCl2
454.
454.1 g
455.
33.07 g
456.
377.5 g
457.
12,100 g
458.
331 g
ANS: E
OBJ: Convert kilograms to
grams. | Use molecular weights, atomic weights, and the balanced chemical
reaction to convert grams reactant to grams product.
TOP: Calculations Based on
Chemical Equations
39.
How many grams of magnesium are required to produce 5.000 kg of
Si?
SiCl4 + 2Mg ® Si +
2MgCl2
1. 8654
g
2. 7581
g
3. 4327
g
4. 9999
g
5. 2164
g
ANS: A
OBJ: Convert kilograms to
grams. | Use molecular weights, atomic weights, and the balanced chemical
reaction to convert grams product to grams reactant.
TOP: Calculations Based on
Chemical Equations
40.
How many grams of oxygen are required to burn 0.10 mole of C3H8?
C3H8 + 5O2 ® 3CO2
+ 4H2O
8. 8.0 g
9. 12 g
10.
16 g
11.
32 g
12.
64 g
ANS: C
OBJ: Use molecular weights
and the balanced chemical reaction to convert moles organic reactant to grams
dioxygen.
TOP: Calculations Based on
Chemical Equations
41.
If sufficient acid is used to react completely with 72.9 g of
magnesium, how much hydrogen will be produced?
2HCl + Mg ®
MgCl2 + H2
4. 4.5 g
5. 3.0
mol
6. 1.5
mol
7. 9.0 g
8. 6.0
mol
ANS: B
OBJ: Use molecular weights,
atomic weights, and the balanced chemical reaction to convert grams reactant to
grams or moles product.
TOP: Calculations Based on
Chemical Equations
42.
How many grams of O2 are required to burn 46.0 grams of C5H12?
C5H12 + 8O2 ® 5CO2
+ 6H2O
1. 1.78
g
2. 828 g
3. 164 g
4. 104 g
5. 12.9
g
ANS: C
OBJ: Use molecular weights
and the balanced chemical reaction to convert grams organic reactant to grams
dioxygen.
TOP: Calculations Based on
Chemical Equations
43.
What mass of phosphoric acid, H3PO4, would actually react with
7.17 grams of LiOH?
3LiOH + H3PO4 ®
Li3PO4 + 3H2O
3. 3.27
g
4. 6.53
g
5. 9.80
g
6. 19.6
g
7. 29.4
g
ANS: C
OBJ: Use formula weights,
molecular weights, and the balanced chemical reaction to convert grams of one
reactant to grams of another.
TOP: Calculations Based on
Chemical Equations
44.
The single largest use of phosphoric acid is in the production
of fertilizer. How many grams of phosphoric acid (with excess NH3) are required
to produce 800 g of (NH4)2HPO4, a common fertilizer?
H3PO4 + 2NH3 ®
(NH4)2HPO4
1. 1186
g
2. 593 g
3. 391 g
4. 686 g
5. 1080
g
ANS: B
OBJ: Use molecular weights,
formula weights, and the balanced chemical reaction to convert grams product to
grams reactant.
TOP: Calculations Based on
Chemical Equations
45.
What mass of SiF4 could be produced by the reaction of 15 g of
SiO2 with an excess of
HF? The equation for the reaction is:
SiO2 + 4HF ® SiF4
+ 2H2O
1. 1.04
g
2. 12 g
3. 26 g
4. 104 g
5. 52 g
ANS: C
OBJ: Use molecular weights,
formula weights, and the balanced chemical reaction to convert grams reactant
to grams product.
TOP: Calculations Based on
Chemical Equations
46.
What mass of SrF2 can be prepared from the reaction of 10.0 g of
Sr(OH)2 with excess
HF?
Sr(OH)2 + 2HF ® SrF2
+ 2H2O
9. 9.67
g
10.
9.82 g
11.
10.0 g
12.
11.6 g
13.
10.3 g
ANS: E
OBJ: Use formula weights and
the balanced chemical reaction to convert grams reactant to grams product.
TOP: Calculations Based on
Chemical Equations
47.
What mass of Li3PO4 can be prepared from the complete reaction
of 7.17 grams of LiOH with a stoichiometric amount of H3PO4?
3LiOH + H3PO4 ®
Li3PO4 + 3H2O
9. 9.80
g
10.
9.34 g
11.
9.61 g
12.
10.4 g
13.
11.6 g
ANS: E
OBJ: Use formula weights and
the balanced chemical reaction to convert grams reactant to grams product.
TOP: Calculations Based on Chemical
Equations
48.
What is the total mass of products formed when 3.2 grams of CH4
is burned in air?
CH4 + 2O2 ® CO2
+ 2H2O
1. 16 g
2. 36 g
3. 44 g
4. 80 g
5. 32 g
ANS: A
DIF: Harder Question
OBJ: Use molecular weights
and the balanced chemical reaction to convert grams organic reactant to grams
product.| Sum the mass of all products.
TOP: Calculations Based on
Chemical Equations
49.
How many grams of KOH would have to be reacted with excess CO2
to produce 69.1 grams of K2CO3?
2KOH + CO2 ®
K2CO3 + H2O
56.
56.1 g
57.
112 g
58.
28.1 g
59.
222 g
60.
84.2 g
ANS: A
OBJ: Use formula weights and
the balanced chemical reaction to convert grams product to grams reactant.
TOP: Calculations Based on
Chemical Equations
50.
A mixture of calcium oxide, CaO, and calcium carbonate, CaCO3,
that had a mass of 3.454 g was heated until all the calcium carbonate was
decomposed according to the following equation. After heating, the sample had a
mass of 3.102 g. Calculate the mass of CaCO3 present in the original sample.
CaCO3 (solid) ® CaO
(solid) + CO2 (gas)
1. 0.400
g
2. 0.800
g
3. 1.00
g
4. 1.60
g
5. 0.200
g
ANS: B
DIF: Harder Question
OBJ: Determine the grams of
carbon dioxide released. | Use formula weights, molecular weights, and the balanced
chemical reaction to convert grams carbon dioxide to grams reactant.
TOP: Calculations Based on
Chemical Equations
51.
The following statements apply to the interpretation of chemical
equations. Not all of the statements are true. Which statement is false?
52.
The number of grams of the reactants must equal the number of
grams of products in a balanced equation.
53.
There will always be some of the excess reagent left at the
completion of the reaction.
54.
The following equation for the reaction involving hypothetical
substances, A, B, C, and D, implies that the products C and D are always
produced in a three to one mole ratio.
A + 2B ® 3C +
D
1. The
equation shown in c implies that in any reaction involving A and B as
reactants, A must be the limiting reactant.
2. The
total number of atoms in the reactants that react must always equal the total
number of atoms in the products produced by the reaction.
ANS: D
OBJ: Understand the
implications of a balanced chemical equation. | Understand the term limiting
reactant. | Understand the term excess reagent.
TOP: The Limiting Reactant Concept
52.
If 25.0 g of each reactant were used in performing the following
reaction, which would be the limiting reactant?
3PbO2 + Cr2(SO4)3 + K2SO4 + H2O ® 3PbSO4 + K2Cr2O7 + H2SO4
1. PbO2
2. H2O
3. K2SO4
4. PbSO4
5. Cr2(SO4)3
ANS: A
OBJ: Use molecular weights,
formula weights, and the balanced chemical equation to determine the limiting
reactant.
TOP: The Limiting Reactant Concept
53.
If 20.0 g of each reactant were used in performing the following
reaction, which would be the limiting reactant?
2MnO2 + 4KOH + O2 + Cl2 ® 2KMnO4
+ 2KCl + 2H2O
1. MnO2
2. KOH
3. O2
4. Cl2
5. KMnO4
ANS: B
OBJ: Use molecular weights,
formula weights, and the balanced chemical equation to determine the limiting
reactant.
TOP: The Limiting Reactant Concept
54.
How many moles of carbon dioxide could be produced if 10 moles
of octane, C8H18, are combined with 20 moles of oxygen?
C8H18 + 25O2 ®
16CO2 + 18H2O
1. 40
mol
2. 8.0
mol
3. 12.8
mol
4. 62.5
mol
5. 20
mol
ANS: C
OBJ: Use the balanced
chemical equation to determine the limiting reactant.
TOP: The Limiting Reactant Concept
55.
If 58 moles of NH3 are combined with 32 moles of sulfuric acid,
what is the limiting reactant and how much of the excess reactant is left over?
2NH3 + H2SO4 ®
(NH4)2SO4
1. H2SO4;
29 mol
2. NH3;
1.0 mol
3. NH3;
29 mol
4. NH3;
3.0 mol
5. H2SO4;
3.0 mol
ANS: D
OBJ: Use the balanced
chemical equation to determine the limiting reactant and the remaining moles of
excess reactant.
TOP: The Limiting Reactant Concept
56.
If 25 grams of methane, CH4, and 30 g of ammonia, NH3, are
combined w/excess oxygen, how much methane or ammonia will be left when the reaction
is finished?
CH4 + NH3 + O2 ® 2HCN
+ 6H2O
1. 0.10g
NH3
2. 0.10
mol CH4
3. 0.10
g CH4
4. 10 g
NH3
5. 0.10
mol NH3
ANS: E
OBJ: Use molecular weights
and the balanced chemical equation to determine the limiting reactant and the
remaining moles of limited reactant in excess.
TOP: The Limiting Reactant Concept
57.
The thermite reaction is performed using 8.6 g Fe2O3 and 1.8 g
powdered Al metal. Which reactant is in excess and by how much?
Fe2O3 + 2 Al ®
Al2O3 + 2Fe
1. Al;
0.3 g
2. Fe2O3;
2.0 g
3. Al;
2.1 g
4. Al;
1.1 g
5. Fe2O3;
3.3 g
ANS: E
OBJ: Use formula weights,
atomic weights, and the balanced chemical equation to determine the limiting
reactant and the grams remaining of the reactant in excess.
TOP: The Limiting Reactant Concept
58.
What mass of Cu(NO3)2 can be prepared from the reaction of 3.65
grams of copper with 5.16 grams of HNO3?
Cu + 2HNO3 ®
Cu(NO3)2 + H2
15.
15.4 g
16.
10.8 g
17.
5.14 g
18.
10.3 g
19.
7.68 g
ANS: E
OBJ: Use molecular weights,
atomic weights, and the balanced chemical equation to determine the limiting
reactant. | Use the limiting reactant,the balanced chemical equation, and
molecular weights, formula weights, and atomic weights as needed, to determine
grams product.
TOP: The Limiting Reactant Concept
59.
What mass of SrF2 can be prepared from the reaction of 8.05 g of
Sr(OH)2 with 3.88 g of HF?
Sr(OH)2 + 2HF ® SrF2
+ 2H2O
11.
11.7 g
12.
12.2 g
13.
10.5 g
14.
8.31 g
15.
8.62 g
ANS: D
OBJ: Use molecular weights,
formula weights, and the balanced chemical equation to determine the limiting
reactant. | Use the limiting reactant, formula weights and molecular weights as
needed, and the balanced chemical equation to determine grams product.
TOP: The Limiting Reactant Concept
60.
How many grams of nitric acid can be prepared from the reaction
of 69.0 grams of nitrogen dioxide with 36.0 grams of water?
3NO2 + H2O ®
2HNO3 + NO
1. 252 g
2. 63.0
g
3. 116 g
4. 84.0
g
5. 76.0
g
ANS: B
OBJ: Use molecular weights
and the balanced chemical equation to determine the limiting reactant. | Use
the limiting reactant, molecular weights as needed, and the balanced chemical
equation to determine grams product.
TOP: The Limiting Reactant Concept
61.
Salicylic acid reacts with acetic anhydride to form aspirin,
acetylsalicylic acid. How much aspirin can be produced from the reaction of
100. g of salicylic acid and 100. g of acetic anhydride?
2
C7H6O3
+ C4H6O3 ®
2
C9H8O4
+ H2O
salicylic acid acetic
anhydride
aspirin
88.
88.2 g
89.
44.1 g
90.
176 g
91.
130 g
92.
65.2 g
ANS: A
OBJ: Use molecular weights
and the balanced chemical equation to determine the limiting reactant. | Use
the limiting reactant, molecular weights as needed, and the balanced chemical
equation to determine grams product.
TOP: The Limiting Reactant Concept
62.
How many grams of MnSO4 can be prepared from the reaction of
15.0 g of KMnO4 with 22.6 g of H2SO3?
2KMnO4 + 5H2SO3 ®
2MnSO4 + K2SO4 + 2H2SO4
14.
14.3 g
15.
15.4 g
16.
15.9 g
17.
16.6 g
18.
17.4 g
ANS: A
OBJ: Use molecular weights,
formula weights, and the balanced chemical equation to determine the limiting
reactant. | Use the limiting reactant, molecular weights and formula weights as
needed, and the balanced chemical equation to determine grams product.
TOP: The Limiting Reactant Concept
63.
A mixture of 13.1 g Zn and 22.0 g I2 is reacted to completion in
a closed, evacuated container. What are the contents of the container after this
reaction?
Zn + I2 ® ZnI2
27.
27.7 g of ZnI2 and 5.7 g of Zn
28.
63.9 g of ZnI2 and 3.4 g of I2
29.
63.9 g of ZnI2
30.
27.7 g of ZnI2 and 7.4 g of Zn
31.
31.2 g of ZnI2 and 3.9 g of I2
ANS: D
OBJ: Use molecular weights,
formula weights, and the balanced chemical equation to determine the limiting
reactant. | Use the limiting reactant, molecular weights and formula weights as
needed, and the balanced chemical equation to determine grams product and grams
of reactant in excess.
TOP: The Limiting Reactant Concept
64.
What is the percent yield of CO2 if the reaction of 10.0 grams
of CO with excess O2
produces 12.8 grams of CO2?
2CO(g) + O2(g) ®
2CO2(g)
76.
76.4%
77.
78.1%
78.
81.5%
79.
84.4%
80.
88.9%
ANS: C
OBJ: Use molecular weights
and the balanced chemical reaction to determine the theoretical yield. |
Calculate the percent yield given the actual yield.
TOP: Percent Yields from Chemical
Reactions
65.
Salicylic acid reacts with acetic anhydride to form aspirin,
acetylsalicylic acid. If the percent yield in this reaction is 78.5%, what mass
of salicylic acid is required to produce 150. g aspirin?
2
C7H6O3
+ C4H6O3 ®
2
C9H8O4
+ H2O
salicylic acid acetic
anhydride
aspirin
1. 292 g
2. 146 g
3. 90.0
g
4. 73.0
g
5. 191 g
ANS: B
OBJ: Determine the
theoretical yield of product required to produce the desired mass of product
given percent yield. | Use molecular weights and the balanced chemical
reaction to convert theoretical grams product to grams reactant.
TOP: Percent Yields from Chemical
Reactions
66.
The reaction of 5.0 g of CuSO4 with excess sodium hydroxide
produced 2.6 g of Cu(OH)2. What percent yield of Cu(OH)2 was obtained?
CuSO4 + 2NaOH ®
Cu(OH)2 + Na2SO4
1. 52%
2. 61%
3. 32%
4. 85%
5. 92%
ANS: D
OBJ: Use formula weights and
the balanced chemical reaction to determine the theoretical yield. | Calculate
the percent yield given the actual yield.
TOP: Percent Yields from Chemical
Reactions
67.
What is the percent yield of elemental sulfur if 7.54 grams of
sulfur are obtained from the reaction of 6.16 grams of SO2 with an excess of
H2S?
2H2S + SO2 ® 2H2O
+ 3S
72.
72.6%
73.
40.8%
74.
81.5%
75.
88.4%
76.
91.4%
ANS: C
OBJ: Use molecular weights,
atomic weights, and the balanced chemical reaction to determine the theoretical
yield. | Calculate the percent yield given the actual yield.
TOP: Percent Yields from Chemical
Reactions
68.
What is the percent yield if 28.50 of FeO reacts with excess CO
and produces 17.841 g Fe?
FeO + CO ® Fe +
CO2
62.
62.62%
63.
81%
64.
80.55%
65.
124%
66.
77.72%
ANS: C
OBJ: Use formula weights,
atomic weights, and the balanced chemical reaction to determine the theoretical
yield. | Calculate the percent yield given the actual yield.
TOP: Percent Yields from Chemical
Reactions
69.
How many grams of PI3 could be produced from 250. g of I2 and
excess phosphorus if the reaction gives a 98.5% yield?
P4 + 6I2 ® 4PI3
1. 246 g
2. 254 g
3. 266 g
4. 270 g
5. 286 g
ANS: C
OBJ: Use molecular weights
and the balanced chemical reaction to determine the theoretical yield. |
Determine actual yield from percent yield and theoretical yield.
TOP: Percent Yields from Chemical
Reactions
70.
Suppose 600. g of P4 reacts with 1300. g of S8. How many grams
of P4S10 can be produced, assuming 80.0% yield based on the limiting reactant?
4P4 + 5S8 ®
4P4S10
4. 4.62 ´ 102 g
5. 1.44 ´ 103 g
6. 2.16 ´ 103 g
7. 4.92 ´ 103 g
8. 6.50 ´ 102 g
ANS: B
OBJ: Use molecular weights
and the balanced chemical reaction to determine the theoretical yield. |
Determine actual yield from percent yield and theoretical yield.
TOP: Percent Yields from Chemical
Reactions
71.
If 6.6 g of fluorine reacts with 5.6 g chlorine to produce 8.5 g
of chlorine trifluoride, what is the limiting reactant and the percent yield of
chlorine trifluoride?
Cl2 + 3F2 ®
2ClF3
1. F2;
45%
2. Cl2;
58%
3. Cl2;
53%
4. F2;
69%
5. F2;
79%
ANS: E
OBJ: Use molecular weights
and the balanced chemical equation to determine the limiting reactant. | Use
the limiting reactant, molecular weights, and the balanced chemical reaction to
determine the theoretical yield. | Calculate the percent yield given the actual
yield.
TOP: Percent Yields from Chemical
Reactions
72.
If a reaction of 5.0 g of hydrogen with 5.0 g of carbon monoxide
produced 4.5 g of methanol, what was the percent yield?
2H2 + CO ®
CH3OH
1. 11%
2. 79%
3. 96%
4. 24%
5. 63%
ANS: B
DIF: Harder Question
OBJ: Use molecular weights
and the balanced chemical equation to determine the limiting reactant. | Use
the limiting reactant, molecular weights, and the balanced chemical reaction to
determine the theoretical yield. | Calculate the percent yield given the actual
yield.
TOP: Percent Yields from Chemical
Reactions
73.
In the sequential reactions shown below, the first reaction has
an 80% yield and the second reaction has a 60% yield.
A + B ® C +
D
C + E ® F
What is the yield of F?
1. 60%
2. 80%
3. 20%
4. 48%
5. 70%
ANS: D
OBJ: Determine the overall
percent yield given the percent yield of each step.
TOP: Sequential Reactions
74.
Sulfuric acid is probably the most important industrial chemical
because it is used in so many industrial processes to produce or purify other
chemicals. It can be produced by a three step process. First, sulfur is burned
in air to give sulfur dioxide. Second, the sulfur dioxide is converted to
sulfur trioxide by passing the sulfur dioxide over a catalyst in the presence
of oxygen at a high temperature. Third, the sulfur trioxide is reacted with
water to form sulfuric acid. What mass of sulfuric acid would be formed from
1.00 ´ 103
kg of sulfur, if the three steps gave yields as listed below?
S + O2 ®
SO2 92.5%
2SO2 + O2 ®
2SO3 72.6%
SO3 + H2O ®
H2SO4 98.2%
4. 4.03 ´ 103 kg
5. 2.02 ´ 103 kg
6. 3.06 ´ 103 kg
7. 2.84 ´ 103 kg
8. 2.22 ´ 103 kg
ANS: B
DIF: Harder Question
OBJ: Determine the overall
percent yield given the percent yield of each step. | Use the sequence of
balanced reactions, and molecular weights and atomic weights as needed, to
convert grams reactant to grams final product. | Determine the actual yield of
product from the overall percent yield.
TOP: Sequential Reactions
75.
Gold is recovered from low-grade ore by first dissolving it with
KCN and then precipitating the solid gold by reacting the aqueous gold complex
ion with Zn. How much gold could be recovered with 250. g of KCN?
4 Au + 8KCN + O2 + 2 H2O ®
4K[Au(CN)2] + 4KOH
4 Zn + 8 K[Au(CN)2] ® 8 Au
+ 4 K2[Zn(CN)4]
1. 756 g
2. 126 g
3. 189 g
4. 378 g
5. 250 g
ANS: D
OBJ: Use molecular weights,
formula weights, and the balanced chemical reaction to determine the
theoretical yield of gold given grams limiting reactant.
TOP: Sequential Reactions
76.
Ethyl butyrate (C6H12O2), “oil of pineapple”, can be produced in
the laboratory from butyric acid (C4H8O2), a foul-smelling liquid, and ethyl
alcohol (C2H6O). First, butyric acid is reacted with PCl5 to produce butyryl
chloride (C4H7ClO) which is then reacted with the ethyl alcohol to produce the
ethyl butyrate. What mass of ethyl butyrate could be produced from 12.8 g of
butyric acid if the first reaction gave a 65.8% yield of butyryl chloride and
the second reaction gave a 88.4% yield of ethyl butyrate?
C4H8O2 + PCl5 ®
C4H7ClO + HCl + POCl3
C4H7ClO + C2H6O ®
C6H12O2 + HCl
8. 8.80
g
9. 9.01
g
10.
8.47 g
11.
16.9 g
12.
9.81 g
ANS: E
DIF: Harder Question
OBJ: Determine the overall
percent yield given the percent yield of each step. | Use the sequence of
balanced reactions, and molecular weights and atomic weights as needed, to
convert grams reactant to grams final product. | Determine the actual yield of
product from the overall percent yield.
TOP: Sequential Reactions
77.
What mass of silver nitrate, AgNO3, is required to prepare 800.
g of 3.50% solution of AgNO3?
78.
24.6 g
79.
26.7 g
80.
27.0 g
81.
25.5 g
82.
28.0 g
ANS: E
OBJ: Convert percent solute
and total grams solution to grams solute.
TOP: Concentrations of Solutions
78.
What mass of water is contained in 200. grams of 15.0% KCl
solution?
79.
11.3 g
80.
170. g
81.
174. g
82.
30.0 g
83.
25.5 g
ANS: B
OBJ: Convert percent solute
and total grams solution to grams solvent.
TOP: Concentrations of Solutions
79.
What mass of 25.0% Ba(NO3)2 solution contains 40.0 grams of
Ba(NO3)2?
80.
117 g
81.
160 g
82.
321 g
83.
10.0 g
84.
62.5 g
ANS: B
OBJ: Convert percent solute
and grams solute to grams solution.
TOP: Concentrations of Solutions
80.
What mass of 30.0% Ca(NO3)2 solution contains 60.0 grams of
water?
81.
42.0 g
82.
85.7 g
83.
58.0 g
84.
14.3 g
85.
62.4 g
ANS: B
OBJ: Convert percent solute
and grams solute to grams solution.
TOP: Concentrations of Solutions
81.
The density of a 7.50% solution of ammonium sulfate, (NH4)2SO4,
is 1.04 g/mL. What mass of (NH4)2SO4 would be required to prepare 750. mL of
this solution?
82.
45.8 g
83.
54.0 g
84.
58.5 g
85.
62.4 g
86.
65.7 g
ANS: C
OBJ: Convert volume solution
to grams solute given percent solute, and solution density.
TOP: Concentrations of Solutions
82.
What mass of NaF and water are required to produce 680. g of
15.0 wt % NaF solution?
83.
170 g NaF; 510 g H2O
84.
11 g NaF; 669 g H2O
85.
578 g NaF; 102 g H2O
86.
15 g NaF; 665 g H2O
87.
102 g NaF; 578 g H2O
ANS: E
OBJ: Convert percent solute
and total grams solution to grams solvent. | Convert percent solute and total
grams solution to grams solute.
TOP: Concentrations of Solutions
83.
What volume of 40.0% NaNO3 solution contains 0.15 mole of NaNO3?
Density = 1.32 g/mL.
84.
42.0 mL
85.
3.86 mL
86.
9.60 mL
87.
24.1 mL
88.
38.2 mL
ANS: D
OBJ: Convert moles solute to
volume solution given percent solute and solution density.
TOP: Concentrations of Solutions
84.
The molarity of a solution is defined as
85.
the number of moles of solute per kilogram of solvent.
86.
the number of moles of solute per liter of solution.
87.
the number of equivalent weights of solute per liter of
solution.
88.
the number of moles of solute per kilogram of solution.
89.
the number of moles of solute per liter of solvent.
ANS: B
OBJ: Define molarity.
TOP: Concentrations of Solutions
85.
Calculate the molarity of a solution that contains 70.0 g of
H2SO4 in 280. mL of solution.
86.
2.55 M
87.
6.84 M
88.
8.62 M
89.
9.78 M
90.
11.84 M
ANS: A
OBJ: Calculate solution
molarity given grams solute and volume solution.
TOP: Concentrations of Solutions
86.
What is the molarity of 850. mL of a solution containing 46.2
grams of NaBr?
87.
0.495 M
88.
0.506 M
89.
0.516 M
90.
0.528 M
91.
0.545 M
ANS: D
OBJ: Calculate solution
molarity given solution volume and grams solute.
TOP: Concentrations of Solutions
87.
What volume of 0.365 M NaOH solution contains 53.4 g NaOH?
88.
3.66 L
89.
2.05 L
90.
146 L
91.
19.5 L
92.
14.6 L
ANS: A
OBJ: Calculate volume of
solution given grams solute and solution molarity.
TOP: Concentrations of Solutions
88.
Which of the following statements does not apply to an aqueous
solution of sodium chloride?
89.
Water is the solvent
90.
The % by mass would be calculated as (mass sodium chloride/mass
water).
91.
Sodium chloride is the solute.
92.
The solution is a homogeneous mixture.
93.
All of these statements apply to an aqueous sodium chloride
solution.
ANS: B
OBJ: Understand the term
aqueous. | Define percent solute by mass for a two component solution. |
Understand the term solute. | Recognize a solution is a homogeneous mixture.
TOP: Concentrations of Solutions
89.
What is the molarity of a barium chloride solution prepared by
dissolving 2.50 g of BaCl2•2H2O in enough water to make 400. mL of solution?
90.
2.56 ´ 10-2 M
91.
4.97 ´ 10-3 M
92.
4.09 ´ 10-2 M
93.
7.31 ´ 10-3 M
94.
5.20 ´ 10-2 M
ANS: A
OBJ: Calculate solution
molarity given grams solute and volume solution.
TOP: Concentrations of Solutions
90.
What mass of glucose (mw = 180 g/mol) must be dissolved in
enough water to produce 1000. mL of 0.55 M glucose solution?
91.
99 g
92.
327 g
93.
0.099 g
94.
235 g
95.
99.0 g
ANS: A
OBJ: Calculate grams solute
given solution volume and solution molarity.
TOP: Concentrations of Solutions
91.
What mass of Na2SO4 is required to prepare 400. mL of 1.50 M
Na2SO4 solution?
92.
213 g
93.
56.8 g
94.
71.4 g
95.
85.2 g
96.
8.52 ´ 104
g
ANS: D
OBJ: Calculate grams solute
given solution volume and solution molarity.
TOP: Concentrations of Solutions
92.
What volume of 0.250 M KOH solution contains 6.31 grams of KOH?
93.
631 mL
94.
28.1 mL
95.
450 mL
96.
2.22 mL
97.
0.44 mL
ANS: C
OBJ: Calculate volume of
solution given grams solute and solution molarity.
TOP: Concentrations of Solutions
93.
The specific gravity of commercial nitric acid solution is 1.42
and it is 70.0% HNO3 by mass. Calculate its molarity.
94.
18.0 M
95.
15.8 M
96.
12.8 M
97.
99.4 M
98.
26.2 M
ANS: B
OBJ: Calculate solution molarity
given the percent by mass and specific gravity.
TOP: Concentrations of Solutions
94.
Concentrated hydrochloric acid is about 12.1 M. What volume of
concentrated HCl is required to produce 5500 mL of 0.250 M?
95.
980 mL
96.
0.114 L
97.
98 mL
98.
0.211 L
99.
1114 mL
ANS: B
OBJ: Calculate the volume of
concentrated solution to be diluted given molarity concentrated solution,
volume dilute solution, and molarity dilute solution. | Convert mL to L.
TOP: Dilution of Solutions
95.
A laboratory stock solution is 1.50 M NaOH. Calculate the volume
of this stock solution that would be needed to prepare 300. mL of 0.200 M NaOH.
96.
2.25 mL
97.
10.0 mL
98.
40.0 mL
99.
1.00 mL
100.
0.100 mL
ANS: C
OBJ: Calculate the volume of
concentrated solution to be diluted given molarity concentrated solution,
volume dilute solution, and molarity dilute solution.
TOP: Dilution of Solutions
96.
When a solution is diluted, what is the relationship of the
number of moles of solute in the more concentrated initial volume of solution
to the number of moles of solute in the less concentrated final volume of
solution?
97.
The ratio of the numbers is directly proportional to the two
volumes.
98.
The ratio of the numbers is inversely proportional to the two
volumes.
99.
The number of moles of solute in the more concentrated initial
volume is always greater.
100.
The number of moles of solute in the less concentrated final
volume is always greater.
101.
The number of moles of solute in both solutions is the same.
ANS: E
OBJ: Compare moles of solute
in a dilute solution to moles of solute in the volume of concentrated stock
solution used to make the dilute solution.
TOP: Dilution of Solutions
97.
Calculate the molarity of the resulting solution if 25.0 mL of
1.50 M HCl solution is diluted to 500. mL.
98.
13.3 M
99.
1.67 M
100.
0.0333 M
101.
0.00417 M
102.
0.0750 M
ANS: E
OBJ: Calculate the molarity
of a dilution given volume and molarity of concentrated stock and the final
volume of the dilute solution.
TOP: Dilution of Solutions
98.
Calculate the molarity of the resulting solution if enough water
is added to 50.0 mL of 4.20 M NaCl solution to make a solution with a volume of
2.80 L.
99.
75.0 M
100.
0.043 M
101.
33.1 M
102.
0.067 M
103.
0.0750 M
ANS: E
OBJ: Calculate the molarity
of a dilution given volume and molarity of concentrated stock and the final
volume of the dilute solution.
TOP: Dilution of Solutions
99.
Calculate the resulting molarity of a solution prepared by
mixing 25.0 mL of 0.160 M NaBr and 55.0 mL of 0.0320 M NaBr.
100.
0.522 M
101.
0.272 M
102.
0.230 M
103.
0.0658 M
104.
0.0720 M
ANS: E
DIF: Harder Question
OBJ: Determine the molarity
of a solution made by mixing two known volumes of solution of differing solute
concentration.
TOP: Dilution of Solutions
100.
Calculate the molarity of the resulting solution prepared by
diluting 25.0 mL of 18.0% ammonium chloride, NH4Cl, (density = 1.05 g/mL) to a
final volume of 80.0 mL.
101.
0.292 M
102.
0.059 M
103.
1.10 M
104.
0.0536 M
105.
0.00105 M
ANS: C
DIF: Harder Question
OBJ: Calculate the molarity
of a dilution given volume, percent solute, and solution density of the
concentrated stock and the final volume of the dilute solution.
TOP: Dilution of Solutions
101.
A sample of commercial perchloric acid is 70.0% HClO4 by mass;
its density is 1.664 g/mL. How many milliliters of this concentrated HClO4
would be required to prepare 500. mL of 1.50 M HClO4 solution?
102.
33.0 mL
103.
45.3 mL
104.
54.1 mL
105.
64.7 mL
106.
78.6 mL
ANS: D
DIF: Harder Question
OBJ: Calculate the volume of
a concentrated stock required for a dilution given percent solute and solution
density of the concentrated stock and the final volume and molarity of the
dilute solution.
TOP: Dilution of Solutions
102.
How many grams of KOH are contained in 400. mL of 0.250 M KOH
solution?
103.
12.4 g
104.
5.61 g
105.
89.8 g
106.
35.1 g
107.
8.98 g
ANS: B
OBJ: Calculate grams solute
given solution volume and solution molarity.
TOP: Using Solutions in Chemical
Reactions
103.
Silver nitrate, AgNO3, reacts with sodium chloride as indicated
by the following equation. What mass of NaCl would be required to react with
200. mL of 0.200 M AgNO3 solution?
AgNO3 + NaCl ® AgCl
+ NaNO3
1. 0.117
g
2. 1.17
g
3. 2.34
g
4. 4.68
g
5. 3.06
g
ANS: C
OBJ: Use molarity, formula
weights, and the balanced chemical reaction to convert the volume of one
reactant to grams of another reactant.
TOP: Using Solutions in Chemical
Reactions
104.
If 45.0 mL of 0.250 M PbCl4 solution reacts with 20.0 mL of
(NH4)2SO4, what is the molarity of the (NH4)2SO4?
PbCl4 + 2(NH4)2SO4 ®
Pb(SO4)2 + 4NH4Cl
1. 0.502
M
2. 1.25
M
3. 2.25
M
4. 1.13
M
5. 0.563
M
ANS: D
OBJ: Use volume, molarity,
and the balanced chemical reaction to determine a reactant molarity.
TOP: Using Solutions in Chemical
Reactions
105.
What mass of calcium carbonate, CaCO3, is required to react with
100. mL of 2.00 M HCl solution?
CaCO3 + 2HCl ®
CaCl2 + CO2 + H2O
5. 5.00
g
6. 10.0
g
7. 15.0
g
8. 20.0
g
9. 23.0
g
ANS: B
OBJ: Use molarity, formula
weights, and the balanced chemical reaction to convert the volume of one
reactant to grams of another reactant.
TOP: Using Solutions in Chemical
Reactions
106.
What volume of 0.150 molar KOH is required to react with 1.259
grams of oxalic acid, (COOH)2?
2KOH + (COOH)2 ®
K2(COO)2 + 2H2O
1. 93 mL
2. 27.9
mL
3. 186
mL
4. 147
mL
5. 372
mL
ANS: C
OBJ: Use molarity, molecular
weight, and the balanced chemical reaction to convert the grams of one reactant
to volume of another reactant.
TOP: Using Solutions in Chemical
Reactions
107.
What volume of 0.130 M HCl solution will just react with 0.424
gram of Ba(OH)2?
2HCl + Ba(OH)2 ®
BaCl2 + 2H2O
38.
38.1 mL
39.
32.6 mL
40.
24.1 mL
41.
18.6 mL
42.
96.7 mL
ANS: A
OBJ: Use molarity, formula
weight, and the balanced chemical reaction to convert the grams of one reactant
to volume of another reactant.
TOP: Using Solutions in Chemical
Reactions
108.
How many grams of PbCl2 precipitate if 100. mL of 0.150 M LiCl
solution reacts with an excess of Pb(NO3)2 solution?
2 LiCl + Pb(NO3)2 ®
PbCl2 + 2 LiNO3
2. 2.09
g
3. 8.34
g
4. 13.9
g
5. 4.17
g
6. 92.7
g
ANS: A
OBJ: Use molarity, formula
weight, and the balanced chemical reaction to convert volume of reactant to
grams of product.
TOP: Using Solutions in Chemical
Reactions
Comments
Post a Comment