CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY -
Stoichiometry
Question Bank - Set 3
Liberty University
Question 1
Question
Iron reacts with hydrochloric acid to produce iron (II) chloride and hydrogen
gas according to the balanced chemical equation:
Fe + 2HCl →FeCl2+ H2
If 5.00 grams of iron reacts with excess hydrochloric acid, what mass of
hydrogen gas is produced?
Solution
Step 1: Find the molar mass of iron (Fe) and hydrogen (H2). The molar mass
of Fe is 55.85 g/mol, and the molar mass of H2is 2.02 g/mol.
Step 2: Calculate the number of moles of Fe reacted.
Moles of Fe = Mass
Molar mass =5.00 g
55.85 g/mol = 0.0894 mol Fe
Step 3: Use the balanced chemical equation to determine the mole ratio
between Fe and H2. From the balanced equation, 1 mol of Fe produces 1 mol
of H2. This means 0.0894 mol of Fe will produce 0.0894 mol of H2.
Step 4: Calculate the mass of hydrogen gas produced.
Mass of H2= Moles ×Molar mass = 0.0894 mol ×2.02 g/mol = 0.181 g
Therefore, 0.181 grams of hydrogen gas is produced when 5.00 grams of iron
reacts with excess hydrochloric acid.
Question 2
Question
Calcium hydroxide, Ca(OH)2, is produced by the reaction of calcium oxide,
CaO, with water. If 25.0 g of CaO reacts with excess water to produce 12.0 g
of Ca(OH)2, what is the percent yield of the reaction?
Solution
Step 1: Find the theoretical yield of calcium hydroxide produced.
Molar mass of CaO = 40.08 g/mol
Molar mass of Ca(OH2) = 74.10 g/mol
Step 1:
Moles of CaO = 25.0 g
40.08 g/mol = 0.6237 mol
Moles of Ca(OH2) = 12.0 g
74.10 g/mol = 0.1617 mol
From the chemical reaction:
CaO + H2O→Ca(OH2)
1 mol →1 mol
0.6237 mol →0.6237 mol Ca(OH2)
Step 2: Calculate the percent yield of the reaction.
Step 2:
Theoretical yield = 0.6237 mol ×74.10 g/mol = 46.35 g
Percent yield = 12.0 g
46.35 g×100% = 25.88%
Answer: The percent yield of the reaction is 25.88%.
Question 3
Question
Calculate the mass of nitrogen dioxide that can be produced from the reaction of
100.0 g of nitric oxide with excess oxygen gas. The balanced chemical equation
is:
2NO(g)+O2(g)→2NO2(g)
2
Solution
Step 1: Write down the balanced chemical equation.
2NO(g)+O2(g)→2NO2(g)
Step 2: Calculate the molar mass of the substances involved. - Molar mass
of NO: 30.01 g/mol - Molar mass of NO2: 46.01 g/mol
Step 3: Calculate the number of moles of NO present. Given: Mass of NO
= 100.0 g Molar mass of NO = 30.01 g/mol
Number of moles of NO = 100.0 g
30.01 g/mol = 3.33 mol
Step 4: Determine the limiting reactant. According to the balanced chemical
equation, 1 mol of NO reacts with 1/2 mol of O2 to produce 1 mol of NO2.
Therefore, the ratio of NO to O2 is 1:0.5.
Moles of O2 needed for complete reaction = 3.33 mol NO×1 mol O2
2 mol NO = 1.665
mol
Since the reaction is taking place with excess oxygen, NO is the limiting
reactant.
Step 5: Calculate the mass of NO2 produced. According to the balanced
chemical equation, 2 mol of NO produce 2 mol of NO2. Molar ratio of NO to
NO2 is 1:1.
Number of moles of NO2 produced = 3.33 mol
Mass of NO2 produced = 3.33 mol ×46.01 g/mol = 153.34 g
Therefore, 153.34 grams of nitrogen dioxide can be produced from the reac-
tion.
Question 4
Question
A compound contains only carbon, hydrogen, oxygen, and nitrogen. A 2.00 g
sample of the compound is burned in oxygen to yield 3.20 g of CO2and 1.44 g
of H2O. Another sample of the compound with a mass of 1.80 g is burned and
produces 1.08 g of CO2and 0.352 g of H2O. Determine the empirical formula
of the compound.
Solution
Step 1: Calculate the moles of CO2and H2O produced in the first combustion
reaction. Step 2: Calculate the moles of carbon and hydrogen in the compound.
Step 3: Calculate the mass of nitrogen in the compound. Step 4: Determine
the empirical formula of the compound.
Step 1: Calculate the moles of CO2and H2O produced in the first combus-
tion reaction.
Given: Mass of CO2= 3.20 g Molar mass of CO2= 44.01 g/mol
Number of moles of CO2=3.20 g
44.01 g/mol = 0.0727 mol
3
Mass of H2O = 1.44 g Molar mass of H2O = 18.02 g/mol
Number of moles of H2O = 1.44 g
18.02 g/mol = 0.0799 mol
Step 2: Calculate the moles of carbon and hydrogen in the compound.
From the combustion reactions:
1 mol of CO2contains 1 mol of carbon 0.0727 mol of CO2contains 0.0727
mol of carbon
1 mol of H2O contains 2 moles of hydrogen 0.0799 mol of H2O contains
2(0.0799) mol of hydrogen
Step 3: Calculate the mass of nitrogen in the compound.
The total mass of the compound = 2.00 g Mass of Carbon = 0.0727 mol *
12.01 g/mol = 0.872 g Mass of Hydrogen = 0.0799 mol * 1.01 g/mol = 0.0807
g Mass of Nitrogen = 2.00 g - (0.872 g + 0.0807 g) = 1.0473 g
Step 4: Determine the empirical formula of the compound.
The mass percent composition of the compound is approximately: Carbon:
0.872 g
2.00 g ×100% Hydrogen: 0.0807 g
2.00 g ×100% Nitrogen: 1.0473 g
2.00 g ×100%
After calculating the values, determine the empirical formula by finding the
lowest whole number ratio of these elements.
Question 5
Question
Given the following balanced chemical equation:
2C4H10(g) + 13O2(g)→8CO2(g) + 10H2O(g)
If 25.0 g of C4H10arereactedwithexcessoxygen, whatisthemaximumamountofCO2
that can be produced? (Molar masses: C = 12.01 g/mol, H = 1.008 g/mol,
O = 16.00 g/mol)
Solution
Step 1: Calculate the molar mass of C4H10.
Molar mass of C4H10 = (4 ×Molar mass of C) + (10 ×Molar mass of H)
= (4 ×12.01 g/mol) + (10 ×1.008 g/mol)
= 48.04 g/mol + 10.08 g/mol
= 58.12 g/mol
Step 2: Calculate the number of moles of C4H10.
Moles of C4H10 =Mass
Molar mass
=25.0 g
58.12 g/mol
≈0.430 mol
4
Step 3: Use the mole ratio from the balanced chemical equation to determine
the moles of CO2produced. From the balanced chemical equation, 2 moles of
C4H10 produces 8 moles of CO2.
Moles of CO2=0.430 mol ×8 mol CO2
2 mol C4H10
= 1.72 mol
Step 4: Convert the moles of CO2to grams using the molar mass of CO2.
Mass of CO2= Moles ×Molar mass
= 1.72 mol ×(12.01 g/mol + 2 ×16.00 g/mol)
= 1.72 mol ×44.01 g/mol
= 75.61 g
Therefore, the maximum amount of CO2that can be produced is 75.61 g.
Question 6
Question
In a chemical reaction, 5.00 grams of aluminum (Al) reacts with excess hy-
drochloric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas
(H2). If 1.50 grams of aluminum chloride is produced, what is the percentage
yield of the reaction?
Solution
Step 1: Calculate the molar mass of AlCl3: The molar mass of AlCl3is the
sum of the atomic masses of aluminum and chlorine:
Molar mass of AlCl3= (1 ×Atomic mass of Al) + (3 ×Atomic mass of Cl)
Step 2: Determine the number of moles of AlCl3produced: Given that the
mass of AlCl3produced is 1.50 grams, we can calculate the number of moles
using the formula:
moles =mass
molar mass
Step 3: Calculate the theoretical yield of AlCl3: The balanced chemical
equation for the reaction is:
2Al + 6HCl →2AlCl3+ 3H2
From the equation, we see that two moles of aluminum produce two moles of
AlCl3. Therefore, the molar ratio of AlCl3to aluminum is 1:1.
5
Step 4: Determine the theoretical mass of AlCl3that should be produced:
Using the molar mass of AlCl3, we can calculate the theoretical yield of alu-
minum chloride if all the aluminum reacted:
T heoretical yield =moles of Al ×Molar mass of AlCl3
Step 5: Calculate the percentage yield of the reaction: The percentage yield
is given by the formula:
P ercentage yield =Actual yield
T heoretical yield ×100
Question 7
Question
A chemist is performing a reaction in the laboratory and observes that 5.00 moles
of sulfur dioxide (SO2) react with excess oxygen gas, producing sulfur trioxide
(SO3) as the product. Assuming the reaction goes to completion, calculate the
mass of sulfur trioxide that would be produced.
Given: Molar mass of SO2= 64.07 g/mol Molar mass of SO3= 80.06 g/mol
Solution
Step 1: Write down the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between sulfur dioxide and
oxygen to produce sulfur trioxide is:
2 SO2+ O2→2 SO3
Step 2: Determine the limiting reactant.
Since sulfur dioxide is provided in a specific quantity, we need to determine the
limiting reactant between sulfur dioxide and oxygen.
Given: - Moles of sulfur dioxide (SO2) = 5.00 mol - Stoichiometry of the
reaction: - 1 mol of SO2reacts to produce 1 mol of SO3- 2 mol of SO2reacts
to produce 2 mol of SO3
Calculate the mole ratio between sulfur dioxide and sulfur trioxide:
5.00 mol SO2
1×2 mol SO3
2 mol SO2
= 5.00 mol SO3
Since the mole ratio is 5.00 mol of SO3, the limiting reactant is sulfur dioxide
(SO2).
Step 3: Calculate the mass of sulfur trioxide produced.
Given: Molar mass of SO3= 80.06 g/mol
Calculate the mass of sulfur trioxide produced using the mole amount ob-
tained from step 2:
Mass of SO3= Moles of SO3×Molar mass of SO3
6
Mass of SO3= 5.00 mol ×80.06 g/mol = 400.30 g
Therefore, the mass of sulfur trioxide that would be produced is 400.30
grams.
Question 8
Question
In a chemical reaction, 10.0 grams of magnesium (Mg) react with excess hy-
drochloric acid (HCl) to produce magnesium chloride (MgCl2) and hydrogen
gas (H2). How many moles of magnesium chloride are produced in the reac-
tion?
Solution
Step 1: Write the balanced chemical equation for the reaction:
Mg + 2HCl →MgCl2+ H2
Step 2: Calculate the molar mass of magnesium chloride (MgCl2):
Mg : 24.305 g/mol,Cl : 35.453 g/mol
MgCl2: 24.305 + 2(35.453) = 95.211 g/mol
Step 3: Convert the given mass of magnesium to moles:
Moles of Mg = 10.0 g
24.305 g/mol = 0.411 mol
Step 4: Determine the stoichiometry ratio between magnesium and magne-
sium chloride: From the balanced chemical equation, 1 mole of Mg produces
1 mole of MgCl2. Therefore, 0.411 moles of Mg will produce 0.411 moles of
MgCl2.
Step 5: Determine the mass of magnesium chloride produced:
Mass of MgCl2= 0.411 mol ×95.211 g/mol = 39.126 g
Step 6: Convert the mass of magnesium chloride to moles:
Moles of MgCl2=39.126 g
95.211 g/mol = 0.411 mol
Therefore, 0.411 moles of magnesium chloride are produced in the reaction.
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Question 9
Question
A student performs a reaction in the laboratory in which 5.00 grams of solid
aluminum reacts with excess hydrochloric acid to produce aluminum chloride
and hydrogen gas. If the student collects 2.50 grams of hydrogen gas, what is
the percent yield of hydrogen gas in this reaction? (Molar masses: Al = 27.0
g/mol, Cl = 35.5 g/mol, H = 1.00 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2 Al + 6 HCl →2 AlCl3+ 3 H2
Step 2: Calculate the theoretical yield of hydrogen gas: 1. Determine the
number of moles of aluminum used:
Moles of Al = Mass of Al
Molar mass of Al =5.00 g
27.0 g/mol = 0.185 mol Al
2. According to the balanced chemical equation, 3 moles of hydrogen gas are
produced for every 2 moles of aluminum:
Moles of H2=3
2×Moles of Al = 3
2×0.185 mol = 0.278 mol H2
3. Calculate the mass of hydrogen gas produced:
Mass of H2= Moles of H2×Molar mass of H = 0.278 mol×1.00 g/mol = 0.278 g
Step 3: Calculate the percent yield of hydrogen gas:
Percent Yield (%) = Actual Yield
Theoretical Yield ×100
Given that the student collects 2.50 grams of hydrogen gas, the actual yield is
2.50 g. Therefore, the percent yield is:
Percent Yield (%) = 2.50 g
0.278 g ×100 ≈900%
Therefore, the percent yield of hydrogen gas in this reaction is approximately
900%.
Question 10
Question
Calculate the mass of sulfuric acid (H2SO4) required to react with 25.0 g of
magnesium hydroxide (Mg(OH)2) according to the following balanced chemical
8
equation:
2 H2SO4(aq) + Mg(OH)2(s) →MgSO4(aq) + 2 H2O (l)
(Molar mass: H2SO4= 98.08 g/mol, Mg(OH)2= 58.32 g/mol)
Solution
Step 1: Calculate the number of moles of magnesium hydroxide (Mg(OH)2)
used.
Moles of Mg(OH)2=Mass
Molar mass =25.0 g
58.32 g/mol = 0.429 mol
Step 2: Use the mole ratio from the balanced chemical equation to find the
moles of sulfuric acid (H2SO4) required. From the balanced equation, we see
that 2 moles of H2SO4reacts with 1 mole of Mg(OH)2. So, the moles of H2SO4
needed is:
0.429 mol Mg(OH)2×2 mol H2SO4
1 mol Mg(OH)2
= 0.858 mol H2SO4
Step 3: Calculate the mass of sulfuric acid (H2SO4) required.
Mass of H2SO4= Moles ×Molar mass = 0.858 mol ×98.08 g/mol = 84.12 g
Therefore, 84.12 g of sulfuric acid is required to react with 25.0 g of magne-
sium hydroxide.
Question 11
Question
A student performs a reaction between 40.0 g of magnesium (Mg) and excess
sulfuric acid (H2SO4) to produce magnesium sulfate (MgSO4) and hydrogen
gas (H2). Calculate the mass of magnesium sulfate that can be produced in this
reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and sulfuric acid:
Mg + H2SO4→MgSO4+ H2
Step 2: Calculate the molar mass of each compound: - Mg (Magnesium):
24.305 g/mol - H2SO4 (Sulfuric acid): 1(2.016 g/mol) + 32.065 g/mol + 4(15.999
g/mol) = 98.079 g/mol - MgSO4 (Magnesium sulfate): 24.305 g/mol + 1(32.065
g/mol) + 4(15.999 g/mol) = 120.366 g/mol
9
Step 3: Determine the molar quantity of magnesium from the given mass:
Moles of Mg = 40.0 g
24.305 g/mol ≈1.645 mol
Step 4: Use the balanced equation to find the mole ratio between magnesium
and magnesium sulfate. According to the balanced equation, 1 mol of Mg
produces 1 mol of MgSO4.
Moles of MgSO4= 1.645 mol
Step 5: Calculate the mass of magnesium sulfate produced:
Mass of MgSO4= Moles of MgSO4×Molar mass of MgSO4
Mass of MgSO4= 1.645 mol ×120.366 g/mol ≈197.79 g
Therefore, the mass of magnesium sulfate that can be produced in this re-
action is approximately 197.79 g.
Question 12
Question
When 50.0 g of aluminum oxide (Al2O3) reacts with excess hydrochloric acid
(HCl), how many moles of aluminum chloride (AlCl3) are produced?
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
oxide and hydrochloric acid:
Al2O3+ 6HCl →2AlCl3+ 3H2O
Step 2: Calculate the molar mass of aluminum oxide (Al2O3) and aluminum
chloride (AlCl3). The molar mass of Al2O3:
(2×Molar mass of Al)+(3×Molar mass of O) = (2×26.98 g/mol)+(3×16.00 g/mol) = 101.96 g/mol
The molar mass of AlCl3:
(Molar mass of Al)+(3×Molar mass of Cl) = 26.98 g/mol+(3×35.45 g/mol) = 133.33 g/mol
Step 3: Calculate the number of moles of aluminum oxide:
moles = mass
molar mass =50.0 g
101.96 g/mol = 0.490 mol
10
Step 4: Determine the stoichiometry relationship between aluminum oxide
and aluminum chloride: From the balanced equation, 1 mole of Al2O3produces
2 moles of AlCl3. Therefore, 0.490 mol of Al2O3will produce:
0.490 mol ×2 mol AlCl3
1 mol Al2O3
= 0.980 mol AlCl3
Step 5: Answer Therefore, when 50.0 g of aluminum oxide reacts with excess
hydrochloric acid, 0.980 moles of aluminum chloride are produced.
Question 13
Question
Silver nitrate reacts with sodium chloride to produce silver chloride and sodium
nitrate. If 10.0 g of silver nitrate reacts with an excess of sodium chloride, what
mass of silver chloride is produced? (Molar masses: Ag = 107.87 g/mol, N =
14.01 g/mol, O = 16.00 g/mol, Na = 22.99 g/mol, Cl= 35.45 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between silver nitrate (AgNO3) and sodium
chloride (NaCl) is:
AgNO3+NaCl →AgCl +NaNO3
Step 2: Calculate the molar mass of AgNO3.
Molar mass of AgN O3= 107.87 + 14.01 + (3 ×16.00) = 169.88 g/mol
Step 3: Calculate the number of moles of AgNO3.
Moles of AgNO3=Mass
Molar mass =10.0g
169.88 g/mol ≈0.059 mol
Step 4: Determine the stoichiometry between AgNO3and AgCl from the
balanced chemical equation. From the balanced equation, 1 mol of AgNO3
produces 1 mol of AgCl.
Step 5: Calculate the mass of AgCl.
Moles of AgCl =Moles of AgN O3
Mass of AgCl =Moles of AgCl×M olar mass of AgCl = 0.059 mol×(107.87+35.45) = 7.11 g
Therefore, 7.11 g of silver chloride is produced during the reaction.
11
Question 14
Question
A sample of iron sulfide, FeS, has a mass of 5.00 grams. When this sample is
heated in air, the iron reacts with oxygen to form iron(III) oxide, Fe2O3, and
sulfur dioxide, SO2. Calculate the mass of iron(III) oxide that can be formed.
(Note: The balanced chemical equation for this reaction is: 4FeS + 7O2→
2Fe2O3+ 4SO2)
Solution
Step 1: Determine the molar masses of FeS and Fe2O3. The molar mass of FeS
can be calculated as:
Molar mass of FeS = (atomic mass of Fe)+(atomic mass of S) = 55.85 g/mol+32.07 g/mol = 87.92 g/mol
The molar mass of Fe2O3can be calculated as:
Molar mass of Fe2O3= 2(atomic mass of Fe)+3(atomic mass of O) = 2(55.85 g/mol)+3(16.00 g/mol) = 159.70 g/mol
Step 2: Determine the number of moles of FeS in the sample.
Number of moles of FeS = mass of FeS
molar mass of FeS =5.00 g
87.92 g/mol = 0.057 mol
Step 3: Use the stoichiometry of the balanced chemical equation to determine
the number of moles of Fe2O3that can be formed. From the balanced chemical
equation, 4 moles of FeS react to form 2 moles of Fe2O3.
Number of moles of Fe2O3=0.057 mol FeS ×2 mol Fe2O3
4 mol FeS = 0.0285 mol Fe2O3
Step 4: Calculate the mass of Fe2O3that can be formed.
Mass of Fe2O3= Number of moles of Fe2O3×Molar mass of Fe2O3
= 0.0285 mol ×159.70 g/mol = 4.55 g
Therefore, the mass of iron(III) oxide that can be formed is 4.55 grams.
Question 15
Question
A student is conducting an experiment in the lab that involves the reaction
between solid iron(III) oxide and solid aluminum to produce iron metal and
aluminum oxide according to the following balanced equation:
2F e2O3+ 4Al →6F e + 4Al2O3
If the student starts with 30.0 grams of iron(III) oxide and 20.0 grams of alu-
minum, determine: a) the limiting reactant b) the theoretical yield of iron (in
grams) c) the percentage yield if the actual yield of iron obtained is 45.0 grams.
12
Solution
a) To determine the limiting reactant, we need to calculate the amount of moles
of each reactant and then use the stoichiometric coefficients of the balanced
reaction to determine which reactant limits the amount of product that can be
formed.
Step 1: Calculate the moles of iron(III) oxide (Fe2O3) Given mass
of iron(III) oxide: 30.0 grams Molar mass of Fe2O3: 2(55.85 g/mol) + 3(16.00
g/mol) = 159.70 g/mol Number of moles of Fe2O3: 30.0 g
159.70 g/mol ≈0.188 mol
Step 2: Calculate the moles of aluminum (Al) Given mass of alu-
minum: 20.0 grams Molar mass of Al: 26.98 g/mol Number of moles of Al:
20.0 g
26.98 g/mol ≈0.741 mol
Step 3: Determine the limiting reactant From the balanced chemical
equation, the stoichiometry between Fe2O3 and Al is 1:2. Therefore, we need
2 moles of Al for every 1 mole of Fe2O3. For 0.188 moles of Fe2O3, we need
0.376 moles of Al. Since the student has only 0.741 moles of Al, aluminum is in
excess and iron(III) oxide is the limiting reactant.
b) Theoretical yield of iron (in grams) Step 1: Calculate the moles
of iron (Fe) From the balanced chemical equation, 2 moles of Fe2O3 produce
6 moles of Fe. Number of moles of Fe produced: 0.188 mol ×6
2= 0.564 mol
Step 2: Calculate the theoretical yield of iron Molar mass of Fe: 55.85
g/mol Theoretical yield of Fe: 0.564 mol ×55.85 g/mol = 31.54 grams
Therefore, the theoretical yield of iron is 31.54 grams.
c) Percentage yield Given actual yield of iron: 45.0 grams
Step 1: Calculate the percentage yield Percentage yield = Actual yield
Theoretical yield ×
100% Percentage yield = 45.0 g
31.54 g ×100% ≈142.6%
Therefore, the percentage yield is approximately 142.6
Question 16
Question
When 4.00 g of aluminum metal reacts with excess hydrochloric acid, hydrogen
gas is produced. Calculate the volume of hydrogen gas (at STP) produced in
the reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
and hydrochloric acid:
2Al + 6HCl →2AlCl3+3H2
Step 2: Calculate the moles of aluminum used: Given mass of aluminum =
4.00 g Molar mass of aluminum (Al) = 26.98 g/mol
13
Number of moles of aluminum = 4.00 g
26.98 g/mol
Step 3: Calculate the moles of hydrogen gas produced: From the balanced
chemical equation, 2 moles of aluminum produce 3 moles of hydrogen gas. So,
the number of moles of hydrogen gas = 3
2×(moles of aluminum used)
Step 4: Calculate the volume of hydrogen gas at STP: 1 mole of any gas at
STP occupies 22.4 L of volume. Therefore, volume of hydrogen gas produced
= (moles of hydrogen gas) ×22.4 L/mol
Step 5: Substitute the calculated values and solve for the volume of hydrogen
gas.
Question 17
Question
A sample of carbon disulfide (CS2) has a mass of 8.96 g. When it is burned
in excess oxygen in a combustion reaction, it produces 22.7 g of carbon dioxide
(CO2). Calculate the mass of sulfur dioxide (SO2) that would be produced in
this reaction.
Solution
Step 1: Write the balanced chemical equation for the combustion of carbon
disulfide:
CS2+ 3O2→CO2+ 2SO2
Step 2: Calculate the molar mass of CS2:
Molar mass of CS2= 12.01 ×1 + 32.07 ×2 = 76.15 g/mol
Step 3: Calculate the number of moles of CS2using its mass:
Number of moles of CS2=8.96 g
76.15 g/mol = 0.118 mol
Step 4: Use the mole ratio from the balanced equation to find the moles of
SO2produced:
Moles of SO2= 0.118 mol ×2 mol SO2
1 mol CS2
= 0.236 mol
Step 5: Calculate the molar mass of SO2:
Molar mass of SO2= 32.07 + 2 ×16.00 = 64.07 g/mol
Step 6: Calculate the mass of SO2produced:
Mass of SO2= 0.236 mol ×64.07 g/mol = 15.14 g
Therefore, 15.14 g of sulfur dioxide (SO2) would be produced in this reaction.
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Question 18
Question
A student is performing a stoichiometry calculation involving the reaction:
2H2O2(aq)→2H2O(l)+O2(g)
If 25.0 grams of H2O2reacts, how many grams of O2will be produced? (Assume
the reaction goes to completion.)
Solution
Step 1: Convert the given mass of H2O2to moles using the molar mass.
Molar mass of H2O2= 2(1.01 g/mol) + 2(16.00 g/mol)
= 2.02 g/mol + 32.00 g/mol
= 34.02 g/mol
Moles of H2O2=25.0 g
34.02 g/mol
= 0.7357 mol
Step 2: Use the coefficients of the balanced chemical equation to relate the
moles of H2O2to the moles of O2. Since the balanced equation has a 1:1 mole
ratio of H2O2to O2, the moles of O2produced will also be 0.7357 moles.
Step 3: Convert the moles of O2to grams using the molar mass.
Molar mass of O2= 2(16.00 g/mol)
= 32.00 g/mol
Mass of O2= 0.7357 mol ×32.00 g/mol
= 23.46 g
Therefore, 23.46 grams of O2will be produced when 25.0 grams of H2O2
reacts.
Question 19
Question
Sulfuric acid can be produced through the contact process, where sulfur dioxide
gas reacts with oxygen gas to produce sulfur trioxide gas, which then reacts
15
with water to produce sulfuric acid. If 20.0 g of sulfur dioxide reacts with
excess oxygen gas according to the following balanced chemical equation:
2SO2(g)+O2(g)→2SO3(g)
and then the sulfur trioxide gas produced reacts with water to form sulfuric acid
according to the following balanced chemical equation:
SO3(g)+H2O(l)→H2SO4(l)
what mass of sulfuric acid is produced?
Solution
Step 1: Calculate the moles of sulfur dioxide.
Given: mass of sulfur dioxide = 20.0 g Molar mass of sulfur dioxide (SO2)
= 32.07 g/mol + 2(16.00 g/mol) = 64.07 g/mol
Number of moles of sulfur dioxide:
moles of SO2=mass
molar mass =20.0 g
64.07 g/mol ≈0.3127 mol
Step 2: Determine the limiting reagent.
From the balanced chemical equation: 1 mol of SO2reacts with 0.5 mol of
O2
Number of moles of oxygen needed:
moles of O2needed = 0.5×moles of SO2= 0.5×0.3127 ≈0.1563 mol
Since oxygen is in excess, it is not the limiting reagent. Therefore, sulfur
dioxide is the limiting reagent.
Step 3: Calculate the moles of sulfuric acid produced.
From the balanced chemical equation: 1 mol of SO3produces 1 mol of H2SO4
Number of moles of sulfuric acid produced:
moles of H2SO4= moles of SO3= moles of limiting reagent = 0.3127 mol
Step 4: Calculate the mass of sulfuric acid produced.
Molar mass of sulfuric acid (H2SO4) = 1(1.01 g/mol) + 2(15.999 g/mol) +
4(16.00 g/mol) = 98.09 g/mol
Mass of sulfuric acid produced:
mass = moles ×molar mass = 0.3127 mol ×98.09 g/mol ≈30.70 g
Therefore, approximately 30.70 g of sulfuric acid is produced.
Question 20
Question
In the reaction between solid lead(II) oxide and hydrogen gas, lead metal and
water are produced. If 35.0 grams of lead(II) oxide and 4.0 moles of hydrogen
gas are allowed to react, what mass of lead will be produced?
16
Solution
Step 1: Write the balanced chemical equation for the reaction, then calculate
the molar masses of the compounds involved.
PbO →Pb + H2O
Lead(II) oxide →Lead + Water
The molar masses are:
PbO: 207.2 g/mol
Pb: 207.2 g/mol
H2O: 18.02 g/mol
Step 2: Determine the limiting reactant by calculating the number of moles
of each reactant.
Moles of PbO: 35.0 g
207.2 g/mol = 0.169mol
Moles of H2:
4.0 mol
Since PbO produces 1 mol of Pb for every 1 mol of PbO, the limiting reactant
is PbO.
Step 3: Calculate the theoretical yield of lead using the moles of PbO.
0.169 mol PbO ×1 mol Pb
1 mol PbO ×207.2 g/mol Pb = 35.0 g Pb
Final Answer
The mass of lead produced is 35.0 grams.
Question 21
Question
A compound contains only carbon, hydrogen, and nitrogen. When 0.275 g of
the compound is burned in excess oxygen, 0.275 g of carbon dioxide, 0.224 g
of water, and 0.569 g of nitrogen gas are produced. Determine the empirical
formula of the compound.
17
Solution
Step 1: Calculate the moles of carbon dioxide, water, and nitrogen gas produced.
Moles of CO2=0.275 g
44.01 g/mol = 0.006251 mol
Moles of H2O = 0.224 g
18.015 g/mol = 0.012436 mol
Moles of N2=0.569 g
28.02 g/mol = 0.020313 mol
Step 2: Determine the moles of each element present in the compound.
Moles of C = 0.006251 mol CO2
1 mol CO2
= 0.006251 mol
Moles of H = 0.012436 mol H2O×2 mol H
1 mol H2O= 0.024872 mol
Moles of N = 0.020313 mol N2
1 mol N2
= 0.020313 mol
Step 3: Determine the simplest molar ratio of the elements in the compound.
Moles of C : Moles of H : Moles of N
0.006251 : 0.024872 : 0.020313
0.006251 : 0.024872 : 0.020313
1:3.98 : 3.25
Step 4: Find the empirical formula by multiplying the subscripts by a com-
mon factor to obtain whole numbers. The closest whole number ratio is 1:4:3,
so the empirical formula of the compound is CH4N3.
Question 22
Question
A sample of iron(III) oxide, Fe2O3, is decomposed by heating it in a stream of
hydrogen gas to produce metallic iron and water. If 100.0 g of iron(III) oxide
produces 88.0 g of iron, what is the percent yield of the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the decomposition of iron(III) oxide is:
Fe2O3(s) + 3H2(g)−→ 2Fe(s) + 3H2O(g)
18
Step 2: Calculate the molar mass of Fe2O3and Fe. The molar mass of Fe2O3
is:
2×atomic mass of Fe+3×atomic mass of O = 2(55.85)+3(16.00) = 159.70 g/mol
The molar mass of Fe is:
atomic mass of Fe = 55.85 g/mol
Step 3: Calculate the number of moles of Fe produced. First, calculate the
number of moles of Fe2O3:
100.0 g Fe2O3
159.70 g/mol Fe2O3
= 0.626 mol Fe2O3
According to the balanced chemical equation, 1 mole of Fe2O3will produce 2
moles of Fe. Therefore, the number of moles of Fe produced is:
0.626 mol Fe2O3×2 mol Fe
1 mol Fe2O3
= 1.25 mol Fe
Step 4: Calculate the theoretical yield of Fe. The theoretical yield of Fe can
be calculated using the number of moles of Fe produced:
Theoretical yield = 1.25 mol Fe ×55.85 g/mol Fe = 69.8 g Fe
Step 5: Calculate the percent yield. The percent yield is given by the for-
mula:
Percent yield = Actual yield
Theoretical yield ×100%
Given that the actual yield is 88.0 g of Fe, the percent yield is:
Percent yield = 88.0 g
69.8 g ×100% = 126.3%
Therefore, the percent yield of the reaction is 126.3
Question 23
Question
For the reaction:
2 Al + 3 Cl2→2 AlCl3
If 1.50 grams of aluminum (Al) reacts with excess chlorine (Cl2), what mass of
aluminum chloride (AlCl3) will be produced?
19
Solution
Step 1: Calculate the molar mass of aluminum (Al):
Molar mass of Al = 26.98 g/mol
Step 2: Determine the number of moles of aluminum present:
Number of moles of Al = Given mass
Molar mass =1.50 g
26.98 g/mol
Step 3: Calculate the molar mass of aluminum chloride (AlCl3):
Molar mass of AlCl3= (1 ×Molar mass of Al) + (3 ×Molar mass of Cl)
Step 4: Determine the mass of aluminum chloride (AlCl3) produced using
the mole ratio from the balanced chemical equation:
2 Al : 2 AlCl3
Number of moles of AlCl3=Number of moles of Al ×2
2
Step 5: Convert the number of moles of aluminum chloride to mass using
the molar mass of AlCl3:
Mass of AlCl3= Number of moles of AlCl3×Molar mass of AlCl3
Now, we can perform the calculations to find the mass of aluminum chloride
produced.
Question 24
Question
In a chemical reaction, 10 moles of hydrogen gas react with excess nitrogen gas
to produce ammonia. Calculate the maximum mass of ammonia that can be
produced.
Solution
Step 1: Write the balanced chemical equation for the reaction.
3H2+N2→2NH3
Step 2: Calculate the molar mass of ammonia (NH).
Molar mass of NH3= 1(14) + 3(1) = 17 g/mol
20
Step 3: Determine the number of moles of ammonia that can be produced
using the mole ratio from the balanced equation.
Moles of NH3=10 mol H2
3×2
1=20
3mol NH3
Step 4: Calculate the mass of ammonia that can be produced.
Mass of NH3= Moles of NH3×Molar mass of NH3
Mass of NH3=20
3×17 g = 340
3g = 113.3 g
Therefore, the maximum mass of ammonia that can be produced is approx-
imately 113.33 grams.
Question 25
Question
Suppose you have a chemical reaction represented by the following balanced
equation:
2H2O2(l)→2H2O(l)+O2(g)
If 25.0 grams of H2O2are decomposed, what mass of O2gas will be produced?
Solution
Step 1: Calculate the molar mass of H2O2. The molar mass of H2O2can be
calculated as follows:
2(H) + 2(O) = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
Step 2: Convert the given mass of H2O2to moles.
Moles of H2O2=25.0 g
34.02 g/mol = 0.736 mol
Step 3: Determine the mole ratio between H2O2and O2from the balanced
chemical equation. From the balanced chemical equation, we can see that for
every 2 moles of H2O2decomposed, 1 mole of O2gas is produced.
Step 4: Calculate the moles of O2gas produced.
Moles of O2= 0.736 mol ×1 mol O2
2 mol H2O2
= 0.368 mol
Step 5: Convert moles of O2gas to mass.
Mass of O2= 0.368 mol ×32.00 g/mol = 11.74 g
Therefore, if 25.0 grams of H2O2are decomposed, 11.74 grams of O2gas
will be produced.
21
Question 26
Question
A chemist wants to determine the amount of product that can be formed from
a reaction between 5.0 g of calcium chloride (CaCl2) and an excess of silver
nitrate (AgNO3) according to the following balanced chemical equation:
CaCl2+ 2AgNO3→Ca(NO3)2+ 2AgCl
What is the maximum amount of silver chloride (AgCl) that can be produced
from this reaction?
Solution
Step 1: Calculate the molar mass of CaCl2and AgCl. The molar mass of CaCl2
is:
Ca : 40.08 g/mol,Cl : 35.45 g/mol
Molar mass of CaCl2= 40.08 + 2(35.45) = 110.98 g/mol
The molar mass of AgCl is:
Ag : 107.87 g/mol,Cl : 35.45 g/mol
Molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol
Step 2: Calculate the number of moles of CaCl2.
Moles of CaCl2=5.0 g
110.98 g/mol = 0.045 mol
Step 3: Use the mole ratio from the balanced chemical equation to find the
moles of AgCl produced. From the balanced equation, we see that 1 mole of
CaCl2produces 2 moles of AgCl.
Moles of AgCl = 0.045 mol ×2 mol AgCl
1 mol CaCl2
= 0.09 mol
Step 4: Calculate the mass of AgCl produced.
Mass of AgCl = Moles of AgCl ×Molar mass of AgCl
Mass of AgCl = 0.09 mol ×143.32 g/mol = 12.899 g
Therefore, the maximum amount of silver chloride that can be produced
from the reaction is 12.899 grams.
22
Question 27
Question
A compound contains only nitrogen and oxygen elements. When a 0.750 g
sample of the compound is burned in excess oxygen gas, 1.460 g of nitrogen
dioxide gas is produced. Determine the empirical formula of the compound.
Solution
Step 1: Find the moles of nitrogen dioxide produced. Given that the molar
mass of nitrogen dioxide (NO2) is 46.0055 g/mol, we can calculate the moles of
nitrogen dioxide produced using the given mass.
Moles of NO2=1.460 g
46.0055 g/mol
Moles of NO2= 0.03175 mol
Step 2: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the combustion of the compound is:
2NxOy+ 2xO2→2xNO2
Step 3: Calculate the moles of nitrogen in the original compound. From the
balanced chemical equation, we can see that 2 moles of nitrogen produce 2 moles
of nitrogen dioxide. Therefore, the moles of nitrogen in the original compound
is equal to the moles of nitrogen dioxide produced.
Moles of nitrogen in the compound = 0.03175 mol
Step 4: Calculate the moles of oxygen in the original compound. Using
the Law of Conservation of Mass, we can calculate the moles of oxygen in the
original compound by subtracting the moles of nitrogen from the total moles in
the nitrogen dioxide produced.
Moles of oxygen in the compound = 2 moles of nitrogen dioxide−moles of nitrogen
Moles of oxygen in the compound = 2(0.03175) −0.03175
Moles of oxygen in the compound = 0.03175 mol
Step 5: Determine the empirical formula of the compound. The empirical
formula of the compound is the simplest whole-number ratio of nitrogen to
oxygen. Since the moles of nitrogen and oxygen are equal (0.03175 mol), the
empirical formula is NO.
23
Question 28
Question
A chemical reaction produces 4.50 moles of product A. If the reaction has a 75
2A + 3B →4C
Solution
Step 1: First, we need to determine the theoretical yield of product A if the
reaction had 100
4.50 moles A ×3 moles B
2 moles A = 6.75 moles B
Step 2: Since the reaction only had a 75
4.50 moles A ×0.75 = 3.375 moles A
Step 3: We know that product A was the limiting reactant, which means all
of B was used up in the reaction. Therefore, the moles of B used is equal to the
moles of B required to produce 3.375 moles of A:
3.375 moles A ×3 moles B
2 moles A = 5.0625 moles B
So, 5.0625 moles of the limiting reactant (B) were used in the reaction.
Question 29
Question
A reaction between solid aluminum and chlorine gas produces solid aluminum
chloride. If 5.00 grams of aluminum is reacted with an excess of chlorine gas
to produce 25.0 grams of aluminum chloride, what is the percent yield of the
reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between aluminum and chlorine is:
2Al + 3Cl2→2AlCl3
Step 2: Calculate the molar mass of aluminum chloride (AlCl3). The molar
mass of Al: 26.98 g/mol The molar mass of Cl: 35.45 g/mol
Molar mass of AlCl3= (3 x molar mass of Cl) + molar mass of Al Molar
mass of AlCl3= (3 x 35.45 g/mol) + 26.98 g/mol Molar mass of AlCl3= 106.35
g/mol + 26.98 g/mol = 133.33 g/mol
24
Step 3: Calculate the theoretical yield of aluminum chloride. To find the
theoretical yield, we first need to determine the number of moles of aluminum
used in the reaction. Given mass of aluminum = 5.00 g Molar mass of Al =
26.98 g/mol
Number of moles of Al = (Given mass of Al) / (Molar mass of Al) Number
of moles of Al = 5.00 g / 26.98 g/mol 0.1852 mol
Using the stoichiometry from the balanced equation, we can see that 2 moles
of Al will produce 2 moles of AlCl3. Therefore, 0.1852 mol of Al will produce
(0.1852 mol / 2) x 133.33 g/mol = 12.96 g of AlCl3.
Step 4: Calculate the percent yield of the reaction. The actual yield of
the reaction is given as 25.0 grams of AlCl3. Percent yield = (Actual yield /
Theoretical yield) x 100Percent yield = (25.0 g / 12.96 g) x 100
Therefore, the percent yield of the reaction is approximately 192.89
Question 30
Question
A 2.00 g sample of impure potassium chloride was dissolved in water. The
solution was then treated with excess silver nitrate to precipitate 3.70 g of silver
chloride. Calculate the percentage of potassium chloride in the impure sample.
Solution
Step 1: Find the moles of silver chloride precipitated. Given that the molar mass
of silver chloride is 143.32 g/mol, we can calculate the moles of silver chloride
precipitated:
Moles of AgCl = Mass of AgCl
Molar mass of AgCl =3.70 g
143.32 g/mol = 0.0258 mol
Step 2: Determine the moles of potassium chloride in the impure sample.
Since silver chloride and potassium chloride react in a 1:1 ratio, the moles of
potassium chloride in the impure sample will be the same as the moles of silver
chloride precipitated.
Moles of KCl = 0.0258 mol
Step 3: Calculate the mass of potassium chloride in the impure sample. The
molar mass of potassium chloride is 74.55 g/mol. Using the moles of potassium
chloride calculated above, we find the mass of KCl in the impure sample:
Mass of KCl = Moles of KCl×Molar mass of KCl = 0.0258 mol×74.55 g/mol = 1.92 g
Step 4: Calculate the percentage of potassium chloride in the impure sample.
Finally, we can determine the percentage of potassium chloride in the impure
25
Question 2
Question
Calcium hydroxide, Ca(OH)2, is produced by the reaction of calcium oxide,
CaO, with water. If 25.0 g of CaO reacts with excess water to produce 12.0 g
of Ca(OH)2, what is the percent yield of the reaction?
Solution
Step 1: Find the theoretical yield of calcium hydroxide produced.
Molar mass of CaO = 40.08 g/mol
Molar mass of Ca(OH2) = 74.10 g/mol
Step 1:
Moles of CaO = 25.0 g
40.08 g/mol = 0.6237 mol
Moles of Ca(OH2) = 12.0 g
74.10 g/mol = 0.1617 mol
From the chemical reaction:
CaO + H2O→Ca(OH2)
1 mol →1 mol
0.6237 mol →0.6237 mol Ca(OH2)
Step 2: Calculate the percent yield of the reaction.
Step 2:
Theoretical yield = 0.6237 mol ×74.10 g/mol = 46.35 g
Percent yield = 12.0 g
46.35 g×100% = 25.88%
Answer: The percent yield of the reaction is 25.88%.
Question 3
Question
Calculate the mass of nitrogen dioxide that can be produced from the reaction of
100.0 g of nitric oxide with excess oxygen gas. The balanced chemical equation
is:
2NO(g)+O2(g)→2NO2(g)
2
Solution
Step 1: Write down the balanced chemical equation.
2NO(g)+O2(g)→2NO2(g)
Step 2: Calculate the molar mass of the substances involved. - Molar mass
of NO: 30.01 g/mol - Molar mass of NO2: 46.01 g/mol
Step 3: Calculate the number of moles of NO present. Given: Mass of NO
= 100.0 g Molar mass of NO = 30.01 g/mol
Number of moles of NO = 100.0 g
30.01 g/mol = 3.33 mol
Step 4: Determine the limiting reactant. According to the balanced chemical
equation, 1 mol of NO reacts with 1/2 mol of O2 to produce 1 mol of NO2.
Therefore, the ratio of NO to O2 is 1:0.5.
Moles of O2 needed for complete reaction = 3.33 mol NO×1 mol O2
2 mol NO = 1.665
mol
Since the reaction is taking place with excess oxygen, NO is the limiting
reactant.
Step 5: Calculate the mass of NO2 produced. According to the balanced
chemical equation, 2 mol of NO produce 2 mol of NO2. Molar ratio of NO to
NO2 is 1:1.
Number of moles of NO2 produced = 3.33 mol
Mass of NO2 produced = 3.33 mol ×46.01 g/mol = 153.34 g
Therefore, 153.34 grams of nitrogen dioxide can be produced from the reac-
tion.
Question 4
Question
A compound contains only carbon, hydrogen, oxygen, and nitrogen. A 2.00 g
sample of the compound is burned in oxygen to yield 3.20 g of CO2and 1.44 g
of H2O. Another sample of the compound with a mass of 1.80 g is burned and
produces 1.08 g of CO2and 0.352 g of H2O. Determine the empirical formula
of the compound.
Solution
Step 1: Calculate the moles of CO2and H2O produced in the first combustion
reaction. Step 2: Calculate the moles of carbon and hydrogen in the compound.
Step 3: Calculate the mass of nitrogen in the compound. Step 4: Determine
the empirical formula of the compound.
Step 1: Calculate the moles of CO2and H2O produced in the first combus-
tion reaction.
Given: Mass of CO2= 3.20 g Molar mass of CO2= 44.01 g/mol
Number of moles of CO2=3.20 g
44.01 g/mol = 0.0727 mol
3
Mass of H2O = 1.44 g Molar mass of H2O = 18.02 g/mol
Number of moles of H2O = 1.44 g
18.02 g/mol = 0.0799 mol
Step 2: Calculate the moles of carbon and hydrogen in the compound.
From the combustion reactions:
1 mol of CO2contains 1 mol of carbon 0.0727 mol of CO2contains 0.0727
mol of carbon
1 mol of H2O contains 2 moles of hydrogen 0.0799 mol of H2O contains
2(0.0799) mol of hydrogen
Step 3: Calculate the mass of nitrogen in the compound.
The total mass of the compound = 2.00 g Mass of Carbon = 0.0727 mol *
12.01 g/mol = 0.872 g Mass of Hydrogen = 0.0799 mol * 1.01 g/mol = 0.0807
g Mass of Nitrogen = 2.00 g - (0.872 g + 0.0807 g) = 1.0473 g
Step 4: Determine the empirical formula of the compound.
The mass percent composition of the compound is approximately: Carbon:
0.872 g
2.00 g ×100% Hydrogen: 0.0807 g
2.00 g ×100% Nitrogen: 1.0473 g
2.00 g ×100%
After calculating the values, determine the empirical formula by finding the
lowest whole number ratio of these elements.
Question 5
Question
Given the following balanced chemical equation:
2C4H10(g) + 13O2(g)→8CO2(g) + 10H2O(g)
If 25.0 g of C4H10arereactedwithexcessoxygen, whatisthemaximumamountofCO2
that can be produced? (Molar masses: C = 12.01 g/mol, H = 1.008 g/mol,
O = 16.00 g/mol)
Solution
Step 1: Calculate the molar mass of C4H10.
Molar mass of C4H10 = (4 ×Molar mass of C) + (10 ×Molar mass of H)
= (4 ×12.01 g/mol) + (10 ×1.008 g/mol)
= 48.04 g/mol + 10.08 g/mol
= 58.12 g/mol
Step 2: Calculate the number of moles of C4H10.
Moles of C4H10 =Mass
Molar mass
=25.0 g
58.12 g/mol
≈0.430 mol
4
Step 3: Use the mole ratio from the balanced chemical equation to determine
the moles of CO2produced. From the balanced chemical equation, 2 moles of
C4H10 produces 8 moles of CO2.
Moles of CO2=0.430 mol ×8 mol CO2
2 mol C4H10
= 1.72 mol
Step 4: Convert the moles of CO2to grams using the molar mass of CO2.
Mass of CO2= Moles ×Molar mass
= 1.72 mol ×(12.01 g/mol + 2 ×16.00 g/mol)
= 1.72 mol ×44.01 g/mol
= 75.61 g
Therefore, the maximum amount of CO2that can be produced is 75.61 g.
Question 6
Question
In a chemical reaction, 5.00 grams of aluminum (Al) reacts with excess hy-
drochloric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas
(H2). If 1.50 grams of aluminum chloride is produced, what is the percentage
yield of the reaction?
Solution
Step 1: Calculate the molar mass of AlCl3: The molar mass of AlCl3is the
sum of the atomic masses of aluminum and chlorine:
Molar mass of AlCl3= (1 ×Atomic mass of Al) + (3 ×Atomic mass of Cl)
Step 2: Determine the number of moles of AlCl3produced: Given that the
mass of AlCl3produced is 1.50 grams, we can calculate the number of moles
using the formula:
moles =mass
molar mass
Step 3: Calculate the theoretical yield of AlCl3: The balanced chemical
equation for the reaction is:
2Al + 6HCl →2AlCl3+ 3H2
From the equation, we see that two moles of aluminum produce two moles of
AlCl3. Therefore, the molar ratio of AlCl3to aluminum is 1:1.
5
Step 4: Determine the theoretical mass of AlCl3that should be produced:
Using the molar mass of AlCl3, we can calculate the theoretical yield of alu-
minum chloride if all the aluminum reacted:
T heoretical yield =moles of Al ×Molar mass of AlCl3
Step 5: Calculate the percentage yield of the reaction: The percentage yield
is given by the formula:
P ercentage yield =Actual yield
T heoretical yield ×100
Question 7
Question
A chemist is performing a reaction in the laboratory and observes that 5.00 moles
of sulfur dioxide (SO2) react with excess oxygen gas, producing sulfur trioxide
(SO3) as the product. Assuming the reaction goes to completion, calculate the
mass of sulfur trioxide that would be produced.
Given: Molar mass of SO2= 64.07 g/mol Molar mass of SO3= 80.06 g/mol
Solution
Step 1: Write down the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between sulfur dioxide and
oxygen to produce sulfur trioxide is:
2 SO2+ O2→2 SO3
Step 2: Determine the limiting reactant.
Since sulfur dioxide is provided in a specific quantity, we need to determine the
limiting reactant between sulfur dioxide and oxygen.
Given: - Moles of sulfur dioxide (SO2) = 5.00 mol - Stoichiometry of the
reaction: - 1 mol of SO2reacts to produce 1 mol of SO3- 2 mol of SO2reacts
to produce 2 mol of SO3
Calculate the mole ratio between sulfur dioxide and sulfur trioxide:
5.00 mol SO2
1×2 mol SO3
2 mol SO2
= 5.00 mol SO3
Since the mole ratio is 5.00 mol of SO3, the limiting reactant is sulfur dioxide
(SO2).
Step 3: Calculate the mass of sulfur trioxide produced.
Given: Molar mass of SO3= 80.06 g/mol
Calculate the mass of sulfur trioxide produced using the mole amount ob-
tained from step 2:
Mass of SO3= Moles of SO3×Molar mass of SO3
6
Mass of SO3= 5.00 mol ×80.06 g/mol = 400.30 g
Therefore, the mass of sulfur trioxide that would be produced is 400.30
grams.
Question 8
Question
In a chemical reaction, 10.0 grams of magnesium (Mg) react with excess hy-
drochloric acid (HCl) to produce magnesium chloride (MgCl2) and hydrogen
gas (H2). How many moles of magnesium chloride are produced in the reac-
tion?
Solution
Step 1: Write the balanced chemical equation for the reaction:
Mg + 2HCl →MgCl2+ H2
Step 2: Calculate the molar mass of magnesium chloride (MgCl2):
Mg : 24.305 g/mol,Cl : 35.453 g/mol
MgCl2: 24.305 + 2(35.453) = 95.211 g/mol
Step 3: Convert the given mass of magnesium to moles:
Moles of Mg = 10.0 g
24.305 g/mol = 0.411 mol
Step 4: Determine the stoichiometry ratio between magnesium and magne-
sium chloride: From the balanced chemical equation, 1 mole of Mg produces
1 mole of MgCl2. Therefore, 0.411 moles of Mg will produce 0.411 moles of
MgCl2.
Step 5: Determine the mass of magnesium chloride produced:
Mass of MgCl2= 0.411 mol ×95.211 g/mol = 39.126 g
Step 6: Convert the mass of magnesium chloride to moles:
Moles of MgCl2=39.126 g
95.211 g/mol = 0.411 mol
Therefore, 0.411 moles of magnesium chloride are produced in the reaction.
7
Question 9
Question
A student performs a reaction in the laboratory in which 5.00 grams of solid
aluminum reacts with excess hydrochloric acid to produce aluminum chloride
and hydrogen gas. If the student collects 2.50 grams of hydrogen gas, what is
the percent yield of hydrogen gas in this reaction? (Molar masses: Al = 27.0
g/mol, Cl = 35.5 g/mol, H = 1.00 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2 Al + 6 HCl →2 AlCl3+ 3 H2
Step 2: Calculate the theoretical yield of hydrogen gas: 1. Determine the
number of moles of aluminum used:
Moles of Al = Mass of Al
Molar mass of Al =5.00 g
27.0 g/mol = 0.185 mol Al
2. According to the balanced chemical equation, 3 moles of hydrogen gas are
produced for every 2 moles of aluminum:
Moles of H2=3
2×Moles of Al = 3
2×0.185 mol = 0.278 mol H2
3. Calculate the mass of hydrogen gas produced:
Mass of H2= Moles of H2×Molar mass of H = 0.278 mol×1.00 g/mol = 0.278 g
Step 3: Calculate the percent yield of hydrogen gas:
Percent Yield (%) = Actual Yield
Theoretical Yield ×100
Given that the student collects 2.50 grams of hydrogen gas, the actual yield is
2.50 g. Therefore, the percent yield is:
Percent Yield (%) = 2.50 g
0.278 g ×100 ≈900%
Therefore, the percent yield of hydrogen gas in this reaction is approximately
900%.
Question 10
Question
Calculate the mass of sulfuric acid (H2SO4) required to react with 25.0 g of
magnesium hydroxide (Mg(OH)2) according to the following balanced chemical
8
equation:
2 H2SO4(aq) + Mg(OH)2(s) →MgSO4(aq) + 2 H2O (l)
(Molar mass: H2SO4= 98.08 g/mol, Mg(OH)2= 58.32 g/mol)
Solution
Step 1: Calculate the number of moles of magnesium hydroxide (Mg(OH)2)
used.
Moles of Mg(OH)2=Mass
Molar mass =25.0 g
58.32 g/mol = 0.429 mol
Step 2: Use the mole ratio from the balanced chemical equation to find the
moles of sulfuric acid (H2SO4) required. From the balanced equation, we see
that 2 moles of H2SO4reacts with 1 mole of Mg(OH)2. So, the moles of H2SO4
needed is:
0.429 mol Mg(OH)2×2 mol H2SO4
1 mol Mg(OH)2
= 0.858 mol H2SO4
Step 3: Calculate the mass of sulfuric acid (H2SO4) required.
Mass of H2SO4= Moles ×Molar mass = 0.858 mol ×98.08 g/mol = 84.12 g
Therefore, 84.12 g of sulfuric acid is required to react with 25.0 g of magne-
sium hydroxide.
Question 11
Question
A student performs a reaction between 40.0 g of magnesium (Mg) and excess
sulfuric acid (H2SO4) to produce magnesium sulfate (MgSO4) and hydrogen
gas (H2). Calculate the mass of magnesium sulfate that can be produced in this
reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and sulfuric acid:
Mg + H2SO4→MgSO4+ H2
Step 2: Calculate the molar mass of each compound: - Mg (Magnesium):
24.305 g/mol - H2SO4 (Sulfuric acid): 1(2.016 g/mol) + 32.065 g/mol + 4(15.999
g/mol) = 98.079 g/mol - MgSO4 (Magnesium sulfate): 24.305 g/mol + 1(32.065
g/mol) + 4(15.999 g/mol) = 120.366 g/mol
9
Step 3: Determine the molar quantity of magnesium from the given mass:
Moles of Mg = 40.0 g
24.305 g/mol ≈1.645 mol
Step 4: Use the balanced equation to find the mole ratio between magnesium
and magnesium sulfate. According to the balanced equation, 1 mol of Mg
produces 1 mol of MgSO4.
Moles of MgSO4= 1.645 mol
Step 5: Calculate the mass of magnesium sulfate produced:
Mass of MgSO4= Moles of MgSO4×Molar mass of MgSO4
Mass of MgSO4= 1.645 mol ×120.366 g/mol ≈197.79 g
Therefore, the mass of magnesium sulfate that can be produced in this re-
action is approximately 197.79 g.
Question 12
Question
When 50.0 g of aluminum oxide (Al2O3) reacts with excess hydrochloric acid
(HCl), how many moles of aluminum chloride (AlCl3) are produced?
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
oxide and hydrochloric acid:
Al2O3+ 6HCl →2AlCl3+ 3H2O
Step 2: Calculate the molar mass of aluminum oxide (Al2O3) and aluminum
chloride (AlCl3). The molar mass of Al2O3:
(2×Molar mass of Al)+(3×Molar mass of O) = (2×26.98 g/mol)+(3×16.00 g/mol) = 101.96 g/mol
The molar mass of AlCl3:
(Molar mass of Al)+(3×Molar mass of Cl) = 26.98 g/mol+(3×35.45 g/mol) = 133.33 g/mol
Step 3: Calculate the number of moles of aluminum oxide:
moles = mass
molar mass =50.0 g
101.96 g/mol = 0.490 mol
10
Step 4: Determine the stoichiometry relationship between aluminum oxide
and aluminum chloride: From the balanced equation, 1 mole of Al2O3produces
2 moles of AlCl3. Therefore, 0.490 mol of Al2O3will produce:
0.490 mol ×2 mol AlCl3
1 mol Al2O3
= 0.980 mol AlCl3
Step 5: Answer Therefore, when 50.0 g of aluminum oxide reacts with excess
hydrochloric acid, 0.980 moles of aluminum chloride are produced.
Question 13
Question
Silver nitrate reacts with sodium chloride to produce silver chloride and sodium
nitrate. If 10.0 g of silver nitrate reacts with an excess of sodium chloride, what
mass of silver chloride is produced? (Molar masses: Ag = 107.87 g/mol, N =
14.01 g/mol, O = 16.00 g/mol, Na = 22.99 g/mol, Cl= 35.45 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between silver nitrate (AgNO3) and sodium
chloride (NaCl) is:
AgNO3+NaCl →AgCl +NaNO3
Step 2: Calculate the molar mass of AgNO3.
Molar mass of AgN O3= 107.87 + 14.01 + (3 ×16.00) = 169.88 g/mol
Step 3: Calculate the number of moles of AgNO3.
Moles of AgNO3=Mass
Molar mass =10.0g
169.88 g/mol ≈0.059 mol
Step 4: Determine the stoichiometry between AgNO3and AgCl from the
balanced chemical equation. From the balanced equation, 1 mol of AgNO3
produces 1 mol of AgCl.
Step 5: Calculate the mass of AgCl.
Moles of AgCl =Moles of AgN O3
Mass of AgCl =Moles of AgCl×M olar mass of AgCl = 0.059 mol×(107.87+35.45) = 7.11 g
Therefore, 7.11 g of silver chloride is produced during the reaction.
11
Question 14
Question
A sample of iron sulfide, FeS, has a mass of 5.00 grams. When this sample is
heated in air, the iron reacts with oxygen to form iron(III) oxide, Fe2O3, and
sulfur dioxide, SO2. Calculate the mass of iron(III) oxide that can be formed.
(Note: The balanced chemical equation for this reaction is: 4FeS + 7O2→
2Fe2O3+ 4SO2)
Solution
Step 1: Determine the molar masses of FeS and Fe2O3. The molar mass of FeS
can be calculated as:
Molar mass of FeS = (atomic mass of Fe)+(atomic mass of S) = 55.85 g/mol+32.07 g/mol = 87.92 g/mol
The molar mass of Fe2O3can be calculated as:
Molar mass of Fe2O3= 2(atomic mass of Fe)+3(atomic mass of O) = 2(55.85 g/mol)+3(16.00 g/mol) = 159.70 g/mol
Step 2: Determine the number of moles of FeS in the sample.
Number of moles of FeS = mass of FeS
molar mass of FeS =5.00 g
87.92 g/mol = 0.057 mol
Step 3: Use the stoichiometry of the balanced chemical equation to determine
the number of moles of Fe2O3that can be formed. From the balanced chemical
equation, 4 moles of FeS react to form 2 moles of Fe2O3.
Number of moles of Fe2O3=0.057 mol FeS ×2 mol Fe2O3
4 mol FeS = 0.0285 mol Fe2O3
Step 4: Calculate the mass of Fe2O3that can be formed.
Mass of Fe2O3= Number of moles of Fe2O3×Molar mass of Fe2O3
= 0.0285 mol ×159.70 g/mol = 4.55 g
Therefore, the mass of iron(III) oxide that can be formed is 4.55 grams.
Question 15
Question
A student is conducting an experiment in the lab that involves the reaction
between solid iron(III) oxide and solid aluminum to produce iron metal and
aluminum oxide according to the following balanced equation:
2F e2O3+ 4Al →6F e + 4Al2O3
If the student starts with 30.0 grams of iron(III) oxide and 20.0 grams of alu-
minum, determine: a) the limiting reactant b) the theoretical yield of iron (in
grams) c) the percentage yield if the actual yield of iron obtained is 45.0 grams.
12
Solution
a) To determine the limiting reactant, we need to calculate the amount of moles
of each reactant and then use the stoichiometric coefficients of the balanced
reaction to determine which reactant limits the amount of product that can be
formed.
Step 1: Calculate the moles of iron(III) oxide (Fe2O3) Given mass
of iron(III) oxide: 30.0 grams Molar mass of Fe2O3: 2(55.85 g/mol) + 3(16.00
g/mol) = 159.70 g/mol Number of moles of Fe2O3: 30.0 g
159.70 g/mol ≈0.188 mol
Step 2: Calculate the moles of aluminum (Al) Given mass of alu-
minum: 20.0 grams Molar mass of Al: 26.98 g/mol Number of moles of Al:
20.0 g
26.98 g/mol ≈0.741 mol
Step 3: Determine the limiting reactant From the balanced chemical
equation, the stoichiometry between Fe2O3 and Al is 1:2. Therefore, we need
2 moles of Al for every 1 mole of Fe2O3. For 0.188 moles of Fe2O3, we need
0.376 moles of Al. Since the student has only 0.741 moles of Al, aluminum is in
excess and iron(III) oxide is the limiting reactant.
b) Theoretical yield of iron (in grams) Step 1: Calculate the moles
of iron (Fe) From the balanced chemical equation, 2 moles of Fe2O3 produce
6 moles of Fe. Number of moles of Fe produced: 0.188 mol ×6
2= 0.564 mol
Step 2: Calculate the theoretical yield of iron Molar mass of Fe: 55.85
g/mol Theoretical yield of Fe: 0.564 mol ×55.85 g/mol = 31.54 grams
Therefore, the theoretical yield of iron is 31.54 grams.
c) Percentage yield Given actual yield of iron: 45.0 grams
Step 1: Calculate the percentage yield Percentage yield = Actual yield
Theoretical yield ×
100% Percentage yield = 45.0 g
31.54 g ×100% ≈142.6%
Therefore, the percentage yield is approximately 142.6
Question 16
Question
When 4.00 g of aluminum metal reacts with excess hydrochloric acid, hydrogen
gas is produced. Calculate the volume of hydrogen gas (at STP) produced in
the reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
and hydrochloric acid:
2Al + 6HCl →2AlCl3+3H2
Step 2: Calculate the moles of aluminum used: Given mass of aluminum =
4.00 g Molar mass of aluminum (Al) = 26.98 g/mol
13
Number of moles of aluminum = 4.00 g
26.98 g/mol
Step 3: Calculate the moles of hydrogen gas produced: From the balanced
chemical equation, 2 moles of aluminum produce 3 moles of hydrogen gas. So,
the number of moles of hydrogen gas = 3
2×(moles of aluminum used)
Step 4: Calculate the volume of hydrogen gas at STP: 1 mole of any gas at
STP occupies 22.4 L of volume. Therefore, volume of hydrogen gas produced
= (moles of hydrogen gas) ×22.4 L/mol
Step 5: Substitute the calculated values and solve for the volume of hydrogen
gas.
Question 17
Question
A sample of carbon disulfide (CS2) has a mass of 8.96 g. When it is burned
in excess oxygen in a combustion reaction, it produces 22.7 g of carbon dioxide
(CO2). Calculate the mass of sulfur dioxide (SO2) that would be produced in
this reaction.
Solution
Step 1: Write the balanced chemical equation for the combustion of carbon
disulfide:
CS2+ 3O2→CO2+ 2SO2
Step 2: Calculate the molar mass of CS2:
Molar mass of CS2= 12.01 ×1 + 32.07 ×2 = 76.15 g/mol
Step 3: Calculate the number of moles of CS2using its mass:
Number of moles of CS2=8.96 g
76.15 g/mol = 0.118 mol
Step 4: Use the mole ratio from the balanced equation to find the moles of
SO2produced:
Moles of SO2= 0.118 mol ×2 mol SO2
1 mol CS2
= 0.236 mol
Step 5: Calculate the molar mass of SO2:
Molar mass of SO2= 32.07 + 2 ×16.00 = 64.07 g/mol
Step 6: Calculate the mass of SO2produced:
Mass of SO2= 0.236 mol ×64.07 g/mol = 15.14 g
Therefore, 15.14 g of sulfur dioxide (SO2) would be produced in this reaction.
14
Question 18
Question
A student is performing a stoichiometry calculation involving the reaction:
2H2O2(aq)→2H2O(l)+O2(g)
If 25.0 grams of H2O2reacts, how many grams of O2will be produced? (Assume
the reaction goes to completion.)
Solution
Step 1: Convert the given mass of H2O2to moles using the molar mass.
Molar mass of H2O2= 2(1.01 g/mol) + 2(16.00 g/mol)
= 2.02 g/mol + 32.00 g/mol
= 34.02 g/mol
Moles of H2O2=25.0 g
34.02 g/mol
= 0.7357 mol
Step 2: Use the coefficients of the balanced chemical equation to relate the
moles of H2O2to the moles of O2. Since the balanced equation has a 1:1 mole
ratio of H2O2to O2, the moles of O2produced will also be 0.7357 moles.
Step 3: Convert the moles of O2to grams using the molar mass.
Molar mass of O2= 2(16.00 g/mol)
= 32.00 g/mol
Mass of O2= 0.7357 mol ×32.00 g/mol
= 23.46 g
Therefore, 23.46 grams of O2will be produced when 25.0 grams of H2O2
reacts.
Question 19
Question
Sulfuric acid can be produced through the contact process, where sulfur dioxide
gas reacts with oxygen gas to produce sulfur trioxide gas, which then reacts
15
with water to produce sulfuric acid. If 20.0 g of sulfur dioxide reacts with
excess oxygen gas according to the following balanced chemical equation:
2SO2(g)+O2(g)→2SO3(g)
and then the sulfur trioxide gas produced reacts with water to form sulfuric acid
according to the following balanced chemical equation:
SO3(g)+H2O(l)→H2SO4(l)
what mass of sulfuric acid is produced?
Solution
Step 1: Calculate the moles of sulfur dioxide.
Given: mass of sulfur dioxide = 20.0 g Molar mass of sulfur dioxide (SO2)
= 32.07 g/mol + 2(16.00 g/mol) = 64.07 g/mol
Number of moles of sulfur dioxide:
moles of SO2=mass
molar mass =20.0 g
64.07 g/mol ≈0.3127 mol
Step 2: Determine the limiting reagent.
From the balanced chemical equation: 1 mol of SO2reacts with 0.5 mol of
O2
Number of moles of oxygen needed:
moles of O2needed = 0.5×moles of SO2= 0.5×0.3127 ≈0.1563 mol
Since oxygen is in excess, it is not the limiting reagent. Therefore, sulfur
dioxide is the limiting reagent.
Step 3: Calculate the moles of sulfuric acid produced.
From the balanced chemical equation: 1 mol of SO3produces 1 mol of H2SO4
Number of moles of sulfuric acid produced:
moles of H2SO4= moles of SO3= moles of limiting reagent = 0.3127 mol
Step 4: Calculate the mass of sulfuric acid produced.
Molar mass of sulfuric acid (H2SO4) = 1(1.01 g/mol) + 2(15.999 g/mol) +
4(16.00 g/mol) = 98.09 g/mol
Mass of sulfuric acid produced:
mass = moles ×molar mass = 0.3127 mol ×98.09 g/mol ≈30.70 g
Therefore, approximately 30.70 g of sulfuric acid is produced.
Question 20
Question
In the reaction between solid lead(II) oxide and hydrogen gas, lead metal and
water are produced. If 35.0 grams of lead(II) oxide and 4.0 moles of hydrogen
gas are allowed to react, what mass of lead will be produced?
16
Solution
Step 1: Write the balanced chemical equation for the reaction, then calculate
the molar masses of the compounds involved.
PbO →Pb + H2O
Lead(II) oxide →Lead + Water
The molar masses are:
PbO: 207.2 g/mol
Pb: 207.2 g/mol
H2O: 18.02 g/mol
Step 2: Determine the limiting reactant by calculating the number of moles
of each reactant.
Moles of PbO: 35.0 g
207.2 g/mol = 0.169mol
Moles of H2:
4.0 mol
Since PbO produces 1 mol of Pb for every 1 mol of PbO, the limiting reactant
is PbO.
Step 3: Calculate the theoretical yield of lead using the moles of PbO.
0.169 mol PbO ×1 mol Pb
1 mol PbO ×207.2 g/mol Pb = 35.0 g Pb
Final Answer
The mass of lead produced is 35.0 grams.
Question 21
Question
A compound contains only carbon, hydrogen, and nitrogen. When 0.275 g of
the compound is burned in excess oxygen, 0.275 g of carbon dioxide, 0.224 g
of water, and 0.569 g of nitrogen gas are produced. Determine the empirical
formula of the compound.
17
Solution
Step 1: Calculate the moles of carbon dioxide, water, and nitrogen gas produced.
Moles of CO2=0.275 g
44.01 g/mol = 0.006251 mol
Moles of H2O = 0.224 g
18.015 g/mol = 0.012436 mol
Moles of N2=0.569 g
28.02 g/mol = 0.020313 mol
Step 2: Determine the moles of each element present in the compound.
Moles of C = 0.006251 mol CO2
1 mol CO2
= 0.006251 mol
Moles of H = 0.012436 mol H2O×2 mol H
1 mol H2O= 0.024872 mol
Moles of N = 0.020313 mol N2
1 mol N2
= 0.020313 mol
Step 3: Determine the simplest molar ratio of the elements in the compound.
Moles of C : Moles of H : Moles of N
0.006251 : 0.024872 : 0.020313
0.006251 : 0.024872 : 0.020313
1:3.98 : 3.25
Step 4: Find the empirical formula by multiplying the subscripts by a com-
mon factor to obtain whole numbers. The closest whole number ratio is 1:4:3,
so the empirical formula of the compound is CH4N3.
Question 22
Question
A sample of iron(III) oxide, Fe2O3, is decomposed by heating it in a stream of
hydrogen gas to produce metallic iron and water. If 100.0 g of iron(III) oxide
produces 88.0 g of iron, what is the percent yield of the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the decomposition of iron(III) oxide is:
Fe2O3(s) + 3H2(g)−→ 2Fe(s) + 3H2O(g)
18
Step 2: Calculate the molar mass of Fe2O3and Fe. The molar mass of Fe2O3
is:
2×atomic mass of Fe+3×atomic mass of O = 2(55.85)+3(16.00) = 159.70 g/mol
The molar mass of Fe is:
atomic mass of Fe = 55.85 g/mol
Step 3: Calculate the number of moles of Fe produced. First, calculate the
number of moles of Fe2O3:
100.0 g Fe2O3
159.70 g/mol Fe2O3
= 0.626 mol Fe2O3
According to the balanced chemical equation, 1 mole of Fe2O3will produce 2
moles of Fe. Therefore, the number of moles of Fe produced is:
0.626 mol Fe2O3×2 mol Fe
1 mol Fe2O3
= 1.25 mol Fe
Step 4: Calculate the theoretical yield of Fe. The theoretical yield of Fe can
be calculated using the number of moles of Fe produced:
Theoretical yield = 1.25 mol Fe ×55.85 g/mol Fe = 69.8 g Fe
Step 5: Calculate the percent yield. The percent yield is given by the for-
mula:
Percent yield = Actual yield
Theoretical yield ×100%
Given that the actual yield is 88.0 g of Fe, the percent yield is:
Percent yield = 88.0 g
69.8 g ×100% = 126.3%
Therefore, the percent yield of the reaction is 126.3
Question 23
Question
For the reaction:
2 Al + 3 Cl2→2 AlCl3
If 1.50 grams of aluminum (Al) reacts with excess chlorine (Cl2), what mass of
aluminum chloride (AlCl3) will be produced?
19
Solution
Step 1: Calculate the molar mass of aluminum (Al):
Molar mass of Al = 26.98 g/mol
Step 2: Determine the number of moles of aluminum present:
Number of moles of Al = Given mass
Molar mass =1.50 g
26.98 g/mol
Step 3: Calculate the molar mass of aluminum chloride (AlCl3):
Molar mass of AlCl3= (1 ×Molar mass of Al) + (3 ×Molar mass of Cl)
Step 4: Determine the mass of aluminum chloride (AlCl3) produced using
the mole ratio from the balanced chemical equation:
2 Al : 2 AlCl3
Number of moles of AlCl3=Number of moles of Al ×2
2
Step 5: Convert the number of moles of aluminum chloride to mass using
the molar mass of AlCl3:
Mass of AlCl3= Number of moles of AlCl3×Molar mass of AlCl3
Now, we can perform the calculations to find the mass of aluminum chloride
produced.
Question 24
Question
In a chemical reaction, 10 moles of hydrogen gas react with excess nitrogen gas
to produce ammonia. Calculate the maximum mass of ammonia that can be
produced.
Solution
Step 1: Write the balanced chemical equation for the reaction.
3H2+N2→2NH3
Step 2: Calculate the molar mass of ammonia (NH).
Molar mass of NH3= 1(14) + 3(1) = 17 g/mol
20
Step 3: Determine the number of moles of ammonia that can be produced
using the mole ratio from the balanced equation.
Moles of NH3=10 mol H2
3×2
1=20
3mol NH3
Step 4: Calculate the mass of ammonia that can be produced.
Mass of NH3= Moles of NH3×Molar mass of NH3
Mass of NH3=20
3×17 g = 340
3g = 113.3 g
Therefore, the maximum mass of ammonia that can be produced is approx-
imately 113.33 grams.
Question 25
Question
Suppose you have a chemical reaction represented by the following balanced
equation:
2H2O2(l)→2H2O(l)+O2(g)
If 25.0 grams of H2O2are decomposed, what mass of O2gas will be produced?
Solution
Step 1: Calculate the molar mass of H2O2. The molar mass of H2O2can be
calculated as follows:
2(H) + 2(O) = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
Step 2: Convert the given mass of H2O2to moles.
Moles of H2O2=25.0 g
34.02 g/mol = 0.736 mol
Step 3: Determine the mole ratio between H2O2and O2from the balanced
chemical equation. From the balanced chemical equation, we can see that for
every 2 moles of H2O2decomposed, 1 mole of O2gas is produced.
Step 4: Calculate the moles of O2gas produced.
Moles of O2= 0.736 mol ×1 mol O2
2 mol H2O2
= 0.368 mol
Step 5: Convert moles of O2gas to mass.
Mass of O2= 0.368 mol ×32.00 g/mol = 11.74 g
Therefore, if 25.0 grams of H2O2are decomposed, 11.74 grams of O2gas
will be produced.
21
Question 26
Question
A chemist wants to determine the amount of product that can be formed from
a reaction between 5.0 g of calcium chloride (CaCl2) and an excess of silver
nitrate (AgNO3) according to the following balanced chemical equation:
CaCl2+ 2AgNO3→Ca(NO3)2+ 2AgCl
What is the maximum amount of silver chloride (AgCl) that can be produced
from this reaction?
Solution
Step 1: Calculate the molar mass of CaCl2and AgCl. The molar mass of CaCl2
is:
Ca : 40.08 g/mol,Cl : 35.45 g/mol
Molar mass of CaCl2= 40.08 + 2(35.45) = 110.98 g/mol
The molar mass of AgCl is:
Ag : 107.87 g/mol,Cl : 35.45 g/mol
Molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol
Step 2: Calculate the number of moles of CaCl2.
Moles of CaCl2=5.0 g
110.98 g/mol = 0.045 mol
Step 3: Use the mole ratio from the balanced chemical equation to find the
moles of AgCl produced. From the balanced equation, we see that 1 mole of
CaCl2produces 2 moles of AgCl.
Moles of AgCl = 0.045 mol ×2 mol AgCl
1 mol CaCl2
= 0.09 mol
Step 4: Calculate the mass of AgCl produced.
Mass of AgCl = Moles of AgCl ×Molar mass of AgCl
Mass of AgCl = 0.09 mol ×143.32 g/mol = 12.899 g
Therefore, the maximum amount of silver chloride that can be produced
from the reaction is 12.899 grams.
22
Question 27
Question
A compound contains only nitrogen and oxygen elements. When a 0.750 g
sample of the compound is burned in excess oxygen gas, 1.460 g of nitrogen
dioxide gas is produced. Determine the empirical formula of the compound.
Solution
Step 1: Find the moles of nitrogen dioxide produced. Given that the molar
mass of nitrogen dioxide (NO2) is 46.0055 g/mol, we can calculate the moles of
nitrogen dioxide produced using the given mass.
Moles of NO2=1.460 g
46.0055 g/mol
Moles of NO2= 0.03175 mol
Step 2: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the combustion of the compound is:
2NxOy+ 2xO2→2xNO2
Step 3: Calculate the moles of nitrogen in the original compound. From the
balanced chemical equation, we can see that 2 moles of nitrogen produce 2 moles
of nitrogen dioxide. Therefore, the moles of nitrogen in the original compound
is equal to the moles of nitrogen dioxide produced.
Moles of nitrogen in the compound = 0.03175 mol
Step 4: Calculate the moles of oxygen in the original compound. Using
the Law of Conservation of Mass, we can calculate the moles of oxygen in the
original compound by subtracting the moles of nitrogen from the total moles in
the nitrogen dioxide produced.
Moles of oxygen in the compound = 2 moles of nitrogen dioxide−moles of nitrogen
Moles of oxygen in the compound = 2(0.03175) −0.03175
Moles of oxygen in the compound = 0.03175 mol
Step 5: Determine the empirical formula of the compound. The empirical
formula of the compound is the simplest whole-number ratio of nitrogen to
oxygen. Since the moles of nitrogen and oxygen are equal (0.03175 mol), the
empirical formula is NO.
23
Question 28
Question
A chemical reaction produces 4.50 moles of product A. If the reaction has a 75
2A + 3B →4C
Solution
Step 1: First, we need to determine the theoretical yield of product A if the
reaction had 100
4.50 moles A ×3 moles B
2 moles A = 6.75 moles B
Step 2: Since the reaction only had a 75
4.50 moles A ×0.75 = 3.375 moles A
Step 3: We know that product A was the limiting reactant, which means all
of B was used up in the reaction. Therefore, the moles of B used is equal to the
moles of B required to produce 3.375 moles of A:
3.375 moles A ×3 moles B
2 moles A = 5.0625 moles B
So, 5.0625 moles of the limiting reactant (B) were used in the reaction.
Question 29
Question
A reaction between solid aluminum and chlorine gas produces solid aluminum
chloride. If 5.00 grams of aluminum is reacted with an excess of chlorine gas
to produce 25.0 grams of aluminum chloride, what is the percent yield of the
reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between aluminum and chlorine is:
2Al + 3Cl2→2AlCl3
Step 2: Calculate the molar mass of aluminum chloride (AlCl3). The molar
mass of Al: 26.98 g/mol The molar mass of Cl: 35.45 g/mol
Molar mass of AlCl3= (3 x molar mass of Cl) + molar mass of Al Molar
mass of AlCl3= (3 x 35.45 g/mol) + 26.98 g/mol Molar mass of AlCl3= 106.35
g/mol + 26.98 g/mol = 133.33 g/mol
24
Step 3: Calculate the theoretical yield of aluminum chloride. To find the
theoretical yield, we first need to determine the number of moles of aluminum
used in the reaction. Given mass of aluminum = 5.00 g Molar mass of Al =
26.98 g/mol
Number of moles of Al = (Given mass of Al) / (Molar mass of Al) Number
of moles of Al = 5.00 g / 26.98 g/mol 0.1852 mol
Using the stoichiometry from the balanced equation, we can see that 2 moles
of Al will produce 2 moles of AlCl3. Therefore, 0.1852 mol of Al will produce
(0.1852 mol / 2) x 133.33 g/mol = 12.96 g of AlCl3.
Step 4: Calculate the percent yield of the reaction. The actual yield of
the reaction is given as 25.0 grams of AlCl3. Percent yield = (Actual yield /
Theoretical yield) x 100Percent yield = (25.0 g / 12.96 g) x 100
Therefore, the percent yield of the reaction is approximately 192.89
Question 30
Question
A 2.00 g sample of impure potassium chloride was dissolved in water. The
solution was then treated with excess silver nitrate to precipitate 3.70 g of silver
chloride. Calculate the percentage of potassium chloride in the impure sample.
Solution
Step 1: Find the moles of silver chloride precipitated. Given that the molar mass
of silver chloride is 143.32 g/mol, we can calculate the moles of silver chloride
precipitated:
Moles of AgCl = Mass of AgCl
Molar mass of AgCl =3.70 g
143.32 g/mol = 0.0258 mol
Step 2: Determine the moles of potassium chloride in the impure sample.
Since silver chloride and potassium chloride react in a 1:1 ratio, the moles of
potassium chloride in the impure sample will be the same as the moles of silver
chloride precipitated.
Moles of KCl = 0.0258 mol
Step 3: Calculate the mass of potassium chloride in the impure sample. The
molar mass of potassium chloride is 74.55 g/mol. Using the moles of potassium
chloride calculated above, we find the mass of KCl in the impure sample:
Mass of KCl = Moles of KCl×Molar mass of KCl = 0.0258 mol×74.55 g/mol = 1.92 g
Step 4: Calculate the percentage of potassium chloride in the impure sample.
Finally, we can determine the percentage of potassium chloride in the impure
25
sample by dividing the mass of KCl by the initial mass of the impure sample
and multiplying by 100
Percentage of KCl = Mass of KCl
Mass of impure sample ×100% = 1.92 g
2.00 g ×100% = 96%
Therefore, the impure sample contains 96
26
