CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY -
Stoichiometry
Question Bank - Set 1
Liberty University
Question 1
Question
A student performs an experiment in the laboratory to determine the molar
mass of an unknown metal carbonate. The student takes a 2.50 g sample of
the metal carbonate and reacts it with excess hydrochloric acid. The student
collects the carbon dioxide gas produced and finds that it has a volume of 0.896
L at a pressure of 1.00 atm and a temperature of 25
°
C. Given that the reaction
is:
Metal Carbonate(s)+2HCl(aq)→Metal Chloride(aq)+Carbon Dioxide(g)+Water(l)
Calculate the molar mass of the metal carbonate.
(Relative atomic masses: H = 1.01 u, C = 12.01 u, O = 16.00 u, Cl = 35.45
u)
Solution
Step 1: Calculate the moles of carbon dioxide produced. Given that the volume
of carbon dioxide gas is 0.896 L, the pressure is 1.00 atm, and the temperature
is 25
°
C, we can use the ideal gas law to calculate the moles of carbon dioxide:
P V =nRT
n=P V
RT
n=(1.00 atm)(0.896 L)
0.0821 L ·atm/K ·mol ×(25 + 273) K
n=0.896
24.75
n≈0.036 mol
Step 2: Determine the moles of metal carbonate in the sample. From the
balanced chemical equation, we can see that one mole of metal carbonate pro-
duces one mole of carbon dioxide. Therefore, the moles of metal carbonate
present in the sample is also 0.036 mol.
Step 3: Calculate the molar mass of the metal carbonate. The molar mass
can be calculated by dividing the mass of the sample by the moles present:
Molar mass = Mass of sample
Moles of metal carbonate
Molar mass = 2.50 g
0.036 mol
Molar mass ≈69.44 g/mol
Therefore, the molar mass of the metal carbonate is approximately 69.44
g/mol.
Question 2
Question
A mixture of sodium sulfide (Na2S) and barium sulfide (BaS) weighing 2.00 g
was dissolved in water. Precipitation of barium sulfide (BaSO4) occurred when
barium nitrate solution was added which resulted in the formation of 3.32 g of
BaSO4. What is the weight percentage of Na2S in the mixture?
Solution
Step 1: Find the moles of BaSO4formed. The molar mass of BaSO4is:
Ba = 137.33 g/mol,S = 32.06 g/mol,O = 15.99 g/mol
So, the molar mass of BaSO4is:
137.33 + 32.06 + 4(15.99) = 137.33 + 32.06 + 63.96 = 233.35 g/mol
Given that 3.32 g of BaSO4is formed, we can find the moles of BaSO4:
Moles = Mass
Molar mass =3.32 g
233.35 g/mol ≈0.0142 mol
Step 2: Find the moles of BaS that reacted. From the balanced chemical
equation:
BaS + Ba(NO3)2→2BaSO4
2
it is clear that 1 mole of BaS reacts to form 1 mole of BaSO4. Therefore, the
moles of BaS that reacted is also approximately 0.0142 mol.
Step 3: Find the moles of Na2S in the mixture. From the balanced chemical
equation:
Na2S + Ba(NO3)2→2NaNO3+ BaSO4
we see that 1 mole of Na2S reacts to form 1 mole of BaSO4. Thus, the moles of
Na2S is also approximately 0.0142 mol.
Step 4: Find the weight of Na2S in the mixture. The molar mass of Na2S is:
2(Na) + S = 2(22.99) + 32.06 = 45.98 g/mol
Using the moles of Na2S and its molar mass, we can calculate the weight of
Na2S in the mixture:
Weight = Moles ×Molar mass = 0.0142 mol ×45.98 g/mol ≈0.652 g
Step 5: Find the weight percentage of Na2S in the mixture. The weight
percentage of Na2S in the mixture is:
Weight of Na2S
Total weight of mixture ×100%
Substitute the values and calculate:
Weight % = 0.652 g
2.00 g ×100% = 0.652
2.00 ×100% ≈32.6%
Therefore, the weight percentage of Na2S in the mixture is approximately
32.6
Question 3
Question
Calculate the mass of iron(III) oxide (Fe2O3) that can be produced from the
reaction of 5.00 grams of iron and excess oxygen gas according to the following
balanced chemical equation:
4Fe(s) + 3O2(g)→2Fe2O3(s)
Solution
Step 1: Calculate the moles of iron used. Given mass of iron: 5.00 grams
Molar mass of iron (Fe): 55.85 g/mol
Moles of iron = Mass
Molar mass =5.00 g
55.85 g/mol ≈0.0894 mol
3
Step 2: Determine the limiting reagent. Using the balanced chemical equa-
tion, we can see that the mole ratio of Fe : O2is 4:3.
Moles of oxygen required = 3
4×0.0894 mol ≈0.0670 mol
Since oxygen is in excess, iron is the limiting reagent.
Step 3: Calculate the moles of Fe2O3produced. From the balanced chemical
equation, we see that 4 moles of iron produce 2 moles of Fe2O3.
Moles of Fe2O3=2
4×0.0894 mol = 0.0447 mol
Step 4: Calculate the mass of Fe2O3produced. Molar mass of iron(III) oxide
(Fe2O3): 2 ×55.85 g/mol + 3 ×16.00 g/mol = 159.69 g/mol
Mass of Fe2O3= Moles ×Molar mass = 0.0447 mol ×159.69 g/mol ≈7.15 g
Therefore, the mass of iron(III) oxide that can be produced is approximately
7.15 grams.
Question 4
Question
In the reaction:
2 Al(s) + 3 Cl2(g)→2 AlCl3(s)
how many moles of aluminum chloride can be produced from 5.0 moles of chlo-
rine gas?
Solution
Step 1: Write out the balanced chemical equation for the reaction. The balanced
chemical equation is:
2 Al(s) + 3 Cl2(g)→2 AlCl3(s)
Step 2: Determine the mole ratio between chlorine gas and aluminum chlo-
ride from the balanced equation. From the balanced equation, we see that the
mole ratio between Cl2and AlCl3is 3:2.
Step 3: Use the mole ratio to determine the moles of aluminum chloride
produced. Given that we have 5.0 moles of Cl2, we can set up the following
proportion: 5.0 moles Cl2
3=xmoles AlCl3
2
Solving for x:
x=5.0 moles Cl2×2
3= 3.33 moles AlCl3
4
Therefore, 3.33 moles of aluminum chloride can be produced from 5.0 moles
of chlorine gas.
Question 5
Question
A chemist wants to determine the percentage composition of sulfur in a sample
of iron(II) sulfide (FeS). If a 2.50 g sample of iron(II) sulfide is reacted with
excess hydrochloric acid and the resulting reaction produces 0.445 g of hydrogen
sulfide gas (H2S), what is the percentage composition of sulfur in the sample?
(Note: The balanced chemical equation for the reaction is FeS + 2HCl →
FeCl2+ H2S)
Solution
Step 1: Find the moles of H2S produced. Given: Mass of H2S=0.445 g
To find the moles of H2S, we use the molar mass of H2S.
Molar mass of H2S = 1 ×2 + 32 = 34 g/mol
Moles of H2S = mass
molar mass =0.445
34 = 0.0131 mol
Step 2: Use the mole ratio to find the moles of FeS. From the balanced
chemical equation, the mole ratio of FeS to H2S is 1:1. So, moles of FeS = 0.0131
mol.
Step 3: Find the molar mass of FeS.
Molar mass of FeS = 56 + 32 = 88 g/mol
Step 4: Calculate the mass percentage of sulfur in FeS.
Percentage of sulfur in FeS = Molar mass of sulfur
Molar mass of FeS ×100%
Percentage of sulfur in FeS = 32
88 ×100% = 36.4%
Therefore, the percentage composition of sulfur in the sample of iron(II)
sulfide is 36.4%.
Question 6
Question
A sample of iron pyrite (FeS2), also known as fool’s gold, contains 5.00 moles
of sulfur. How many moles of iron are present in the sample?
Given: Molecular weight of FeS2= 119.98 g/mol
5
Solution
Step 1: Write the balanced chemical equation for the reaction involving iron
pyrite:
FeS2→Fe + S2
Step 2: Determine the molar ratio of sulfur to iron in the reaction. 1 mole
of FeS2produces 1 mole of iron (Fe) and 2 moles of sulfur (S).
Step 3: Calculate the number of moles of iron in the sample: Given that the
sample contains 5.00 moles of sulfur, and using the molar ratio from step 2, we
find: 5.00 mol S
2×1 mol Fe
1 mol S = 2.50 mol Fe
Therefore, there are 2.50 moles of iron present in the sample of iron pyrite.
Question 7
Question
A chemist is working with a reaction involving silver nitrate (AgNO3) and
sodium chloride (NaCl). If 10.0 grams of silver nitrate reacts completely with
sodium chloride to produce silver chloride (AgCl) and sodium nitrate (NaNO3),
what mass of sodium chloride is required for the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride. The balanced chemical equation is:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Calculate the molar mass of each compound involved in the reaction.
The molar mass of AgNO3can be calculated as:
(1 ×Ag) + (1 ×N) + (3 ×O) = 107.87 g/mol
The molar mass of NaCl can be calculated as:
(1 ×Na) + (1 ×Cl) = 58.44 g/mol
The molar mass of AgCl can be calculated as:
(1 ×Ag) + (1 ×Cl) = 143.32 g/mol
The molar mass of NaNO3can be calculated as:
(1 ×Na) + (1 ×N) + (3 ×O) = 84.99 g/mol
6
Step 3: Calculate the number of moles of AgNO3given 10.0 grams.
Moles of AgNO3=Mass
Molar mass =10.0 g
107.87 g/mol ≈0.093 mol
Step 4: Determine the mole ratio between AgNO3and NaCl from the bal-
anced chemical equation. The mole ratio is 1:1, which means 1 mole of AgNO3
reacts with 1 mole of NaCl.
Step 5: Calculate the mass of NaCl required using the mole ratio and molar
mass of NaCl.
Mass of NaCl = Moles of AgNO3×Molar mass of NaCl
Mass of NaCl = 0.093 mol ×58.44 g/mol = 5.43 g
Therefore, 5.43 grams of sodium chloride is required for the reaction.
Question 8
Question
A student is performing a stoichiometry calculation involving the reaction of
iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. If
150.0 g of iron(III) oxide reacts with excess carbon monoxide, how many grams
of iron can be produced? (Molar masses: Fe = 55.85 g/mol, O = 16.00 g/mol,
C = 12.01 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between iron(III)
oxide and carbon monoxide.
F e2O3+ 3CO →2F e + 3CO2
Step 2: Calculate the molar mass of FeO.
Molar mass of F e2O3= 2(F e)+3(O) = 2(55.85 g/mol)+3(16.00 g/mol) = 159.70 g/mol
Step 3: Determine the number of moles of FeO present in 150.0 g.
Moles of F e2O3=Mass
Molar mass =150.0g
159.70 g/mol = 0.940 mol
Step 4: Use the stoichiometry of the balanced chemical equation to find the
number of moles of Fe produced.
Moles of F e =Moles of F e2O3
1×2mol F e
1mol F e2O3
= 0.940 mol ×2=1.880 mol
Step 5: Calculate the mass of iron produced.
Mass of F e =M oles of F e×M olar mass of F e = 1.880 mol×55.85 g/mol = 105.26 g
Therefore, 105.26 grams of iron can be produced from the reaction of 150.0
grams of iron(III) oxide with excess carbon monoxide.
7
Question 9
Question
A chemist wants to synthesize 1.50 g of ammonia (NH3) using the following
balanced chemical equation:
4NH3(g) + 5O2(g)→4NO(g) + 6H2O(g)
If the chemist has an excess of oxygen gas, what mass of oxygen gas is required
for the synthesis of 1.50 g of ammonia?
Solution
Step 1: Calculate the molar mass of NH3. The molar mass of NH3is 17.03
g/mol.
Step 2: Calculate the moles of NH3. Given mass of NH3= 1.50 g Number
of moles of NH3=1.50 g
17.03 g/mol = 0.0881 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the
moles of O2required. From the balanced chemical equation, we see that 4 moles
of NH3react with 5 moles of O2. Therefore, 0.0881 mol NH3×5
4= 0.110mol of O2
is required.
Step 4: Calculate the mass of O2required. Molar mass of O2= 32.00 g/mol
Mass of O2required = 0.110 mol ×32.00 g/mol = 3.52 g
Therefore, the mass of oxygen gas required for the synthesis of 1.50 g of
ammonia is 3.52 g.
Question 10
Question
A chemist is performing a reaction using 3.50 moles of iron(III) oxide and 4.00
moles of carbon monoxide according to the following balanced equation:
F e2O3(s)+3CO(g)→2F e(s)+3CO2(g)
What is the limiting reactant in this reaction? How many moles of iron are
produced?
Solution
Step 1: Calculate the moles of iron that can be produced by each reactant. -
For iron(III) oxide:
3.50 mol Fe2O3
1×2 mol Fe
1 mol Fe2O3
= 7.00 mol Fe
8
- For carbon monoxide:
4.00 mol CO
1×2 mol Fe
3 mol CO =8
3mol Fe = 2.67 mol Fe
Step 2: Identify the limiting reactant. - Since the carbon monoxide produces
fewer moles of iron, it is the limiting reactant in this reaction.
Step 3: Calculate the moles of iron produced using the limiting reactant. -
Using 4.00 moles of CO as the limiting reactant:
4.00 mol CO
1×2 mol Fe
3 mol CO =8
3mol Fe = 2.67 mol Fe
Therefore, the limiting reactant is carbon monoxide, and 2.67 moles of iron
are produced in this reaction.
Question 11
Question
Consider the reaction:
2KClO3→2KCl + 3O2
If 25.0 grams of KClO3 decomposes, what mass of oxygen is produced?
Solution
Step 1: Calculate the molar mass of KClO3, O2, and KCl.
The molar mass of KClO3 is:
2(39.10 g/mol K) + 35.45 g/mol Cl + 3(16.00 g/mol O) = 122.55 g/mol
The molar mass of O2 is:
2(16.00 g/mol O) = 32.00 g/mol
The molar mass of KCl is:
39.10 g/mol K + 35.45 g/mol Cl = 74.55 g/mol
Step 2: Find the number of moles of KClO3 using its molar mass.
Number of moles of KClO3:
25.0 g
122.55 g/mol = 0.204 mol
9
Step 3: Use the stoichiometry of the reaction to find the number of moles of
O2 produced.
From the balanced equation, we see that 2 moles of KClO3 produce 3
moles of O2.
0.204 mol KClO3 ×3 mol O2
2 mol KClO3 = 0.306 mol O2
Step 4: Convert the number of moles of O2 to grams.
Mass of O2 produced:
0.306 mol O2 ×32.00 g/mol O2 = 9.79 g O2
Therefore, 9.79 grams of oxygen is produced when 25.0 grams of KClO3
decomposes.
Question 12
Question
A reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) produces
silver chloride (AgCl) and sodium nitrate (NaNO3). If 10.0 grams of silver
nitrate reacts with an excess of sodium chloride to produce 12.0 grams of silver
chloride, what is the percent yield of silver chloride?
Solution
Step 1: Write the balanced chemical equation for the reaction.
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Calculate the molar mass of each compound: - AgNO3:
107.87 g/mol + 14.01 g/mol + (3 ×16.00 g/mol) = 169.87 g/mol
- AgCl:
107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Step 3: Determine the theoretical yield of silver chloride: - Using the molar
mass of AgNO3, we find the number of moles of AgNO3:
10.0 g
169.87 g/mol = 0.05885 mol
- According to the balanced equation, 1 mole of AgNO3produces 1 mole of
AgCl. - Therefore, the theoretical yield of AgCl is 0.05885 mol.
10
Step 4: Calculate the percent yield of AgCl: - The actual yield of AgCl is
given as 12.0 grams. - Convert the actual yield to moles:
12.0 g
143.32 g/mol = 0.08370 mol
- The percent yield is given by:
Percent Yield = Actual Yield
Theoretical Yield×100%
Percent Yield = 0.08370 mol
0.05885 mol×100% = 142.17%
Step 5: Since a percent yield greater than 100
Question 13
Question
A chemist wants to produce 25.0 g of potassium sulfate (K2SO4) by the reaction
of potassium hydroxide (KOH) and sulfuric acid (H2SO4). If the chemist has an
unlimited supply of potassium hydroxide but only 95.0 g of sulfuric acid, what
is the limiting reactant? Calculate the theoretical yield of potassium sulfate in
grams.
Solution
Step 1: Write the balanced chemical equation for the reaction:
2 KOH + H2SO4→K2SO4+ 2 H2O
Step 2: Calculate the molar mass of each substance: - KOH: K = 39.10 g/mol,
O = 16.00 g/mol, H = 1.01 g/mol Molar mass of KOH = 39.10 + 16.00 + 1.01 =
56.11 g/mol
- H2SO4: H = 1.01 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass of
H2SO4= 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol
- K2SO4: K = 39.10 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass
of K2SO4= 2(39.10) + 32.07 + 4(16.00) = 174.26 g/mol
Step 3: Calculate the number of moles of each reactant: - Moles of KOH =
25.0 g
56.11 g/mol ≈0.446 mol - Moles of H2SO4=95.0 g
98.09 g/mol ≈0.969 mol
Step 4: Determine the limiting reactant: The stoichiometry of the reaction
indicates that 1 mole of H2SO4reacts with 2 moles of KOH. Therefore, for
0.969 moles of H2SO4, we would need 0.969/2 = 0.484 moles of KOH. Since we
have less than 0.484 moles of KOH available, H2SO4is the limiting reactant.
Step 5: Calculate the theoretical yield of K2SO4: - Theoretical yield of
K2SO4= moles of limiting reactant×molar ratio from balanced equation
1×molar mass of K2SO4
Theoretical yield of K2SO4= 0.969 mol ×1
2×174.26 g/mol ≈84.35 g
Therefore, the limiting reactant is H2SO4and the theoretical yield of K2SO4
is 84.35 g.
11
Question 14
Question
A reaction between solid iron (III) oxide and carbon monoxide produces solid
iron and carbon dioxide gas. If 50.0 grams of solid iron (III) oxide react with
an excess of carbon monoxide and 35.0 grams of solid iron are produced, what
is the percent yield of the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The reaction is
between solid iron (III) oxide (F e2O3) and carbon monoxide (CO), producing
solid iron (F e) and carbon dioxide (CO2). The balanced equation is:
F e2O3+ 3CO →2F e + 3CO2
Step 2: Calculate the molar mass of F e2O3and F e. - Molar mass of F e2O3:
2×55.845 (molar mass of iron) + 3 ×16.00 (molar mass of oxygen) = 159.69
g/mol
- Molar mass of F e: 55.845 g/mol
Step 3: Calculate the number of moles of F e2O3used.
Moles of F e2O3=50.0 g
159.69 g/mol = 0.313 mol
Step 4: Using the balanced equation, calculate the theoretical yield of F e.
Since 1 mol of F e2O3produces 2 mol of F e, the theoretical yield of F e is:
0.313 mol F e2O3×2 mol F e
1 mol F e2O3
= 0.626 mol F e
Step 5: Calculate the theoretical mass of F e that should be produced.
Theoretical mass of F e = 0.626 mol ×55.845 g/mol = 35.0 g
Step 6: Calculate the percent yield of the reaction.
Percent yield = Actual yield
Theoretical yield ×100%
Percent yield = 35.0 g
35.0 g ×100% = 100%
So, the percent yield of the reaction is 100
12
Question 15
Question
A student is tasked with determining the amount of magnesium oxide that can
be produced when 15.0 grams of magnesium metal reacts with excess oxygen
gas. The balanced chemical equation for the reaction is:
2Mg + O2→2MgO
Assuming 100
Solution
Step 1: Determine the molar mass of each compound involved. The molar mass
of magnesium (Mg) is 24.305 g/mol, oxygen (O) is 16.00 g/mol, and magnesium
oxide (MgO) is 40.30 g/mol.
Step 2: Calculate the number of moles of magnesium. Given: Mass of
magnesium = 15.0 g Using the molar mass of magnesium:
Moles of Mg = 15.0 g
24.305 g/mol = 0.617 mol Mg
Step 3: Use the stoichiometry of the balanced equation to convert moles of
magnesium to moles of magnesium oxide. From the balanced chemical equation:
2 Mg + O2→2 MgO
This means that 2 moles of magnesium produce 2 moles of magnesium oxide.
1 mol Mg = 1 mol MgO
Step 4: Calculate the theoretical yield of magnesium oxide in grams.
Moles of MgO = 0.617 mol Mg ×1 mol MgO
1 mol Mg = 0.617 mol MgO
Mass of MgO = 0.617 mol MgO ×40.30 g/mol = 24.88 g
Therefore, the theoretical yield of magnesium oxide in grams is 24.88g.
Question 16
Question
A reaction takes place according to the following balanced chemical equation:
C3H8(g) + 5O2(g)→3CO2(g) + 4H2O(g)
If 32.0 g of C3H8and64.0gofO2areallowedtoreact, whatisthelimitingreactant?Howmanygramsof waterareproduced?
13
Solution
Step 1: Calculate the number of moles of each reactant. Step 2: Determine
the limiting reactant and the amount of product formed. Step 3: Calculate the
mass of water produced.
Step 1: Calculate the number of moles of each reactant. The molar mass
of C3H8is44.1g/mol.T hemolarmassof O2is32.0g/mol.
Number of moles of C3H8: moles = mass
molar mass =32.0 g
44.1 g/mol = 0.726 mol
Number of moles of O2: moles = mass
molar mass =64.0 g
32.0 g/mol = 2.00 mol
Step 2: Determine the limiting reactant and the amount of product formed.
From the balanced chemical equation, 1 mol of C3H8reactswith5molofO2.T heref ore, thelimitingreactantisC3H8.
Amount of H2Oproducedfrom32.0gofC3H8: moles of H2O = 1
3×moles of C3H8=
1
3×0.726 = 0.242 mol
Step 3: Calculate the mass of water produced. The molar mass of H2Ois18.0g/mol.mass =
moles ×molar mass = 0.242 mol ×18.0 g/mol = 4.36 g
Therefore, the limiting reactant is C3H8and4.36gofH2Oareproduced.
Question 17
Question
A chemist wants to synthesize 10.0 grams of a compound. The molar mass of
the compound is 100 g/mol. If the chemist has a 90
Solution
Step 1: Calculate the number of moles of the desired compound: Given: Mass
of compound = 10.0 g, Molar mass of compound = 100 g/mol
Moles of compound = Mass of compound
Molar mass of compound =10.0 g
100 g/mol = 0.1 mol
Step 2: Account for the 90Since the yield is 90Let x be the number of moles
of reactant needed to achieve the desired amount of product.
Expected yield = 0.9×Actual yield
0.9×x= 0.1
x=0.1
0.9=1
9= 0.111 mol
Therefore, the chemist should start with 0.111 moles of reactant to achieve
the desired 10.0 grams of product, accounting for a 90
14
Question 18
Question
A student is conducting a reaction between 2.50 moles of magnesium (Mg) and
3.00 moles of hydrochloric acid (HCl) to produce magnesium chloride (MgCl2)
and hydrogen gas (H2). Write the balanced chemical equation for this reaction
and determine: a) The limiting reactant, b) The theoretical yield of magnesium
chloride in grams.
Given: Molar mass of Mg = 24.31 g/mol Molar mass of HCl = 36.46 g/mol
Molar mass of MgCl2= 95.21 g/mol
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and hydrochloric acid.
The balanced equation is:
Mg + 2HCl →MgCl2+ H2
Step 2: Calculate the number of moles of magnesium chloride that can be
produced from each reactant.
From the balanced equation, we see that 1 mole of magnesium reacts with
2 moles of hydrochloric acid to produce 1 mole of magnesium chloride.
For magnesium: Number of moles of MgCl2= 2.50 moles Mg ×1 mol MgCl2
1 mol Mg =
2.50 moles
For hydrochloric acid: Number of moles of MgCl2= 3.00 moles HCl ×1 mol MgCl2
2 mol HCl =
1.50 moles
Step 3: Determine the limiting reactant. The limiting reactant is the one
that produces the least amount of product. Since hydrochloric acid produces
less magnesium chloride (1.50 moles) compared to magnesium (2.50 moles),
hydrochloric acid is the limiting reactant.
Step 4: Calculate the theoretical yield of magnesium chloride in grams.
To find the theoretical yield of magnesium chloride in grams, we need to
convert the number of moles of MgCl2produced by hydrochloric acid to grams.
Theoretical yield of MgCl2= 1.50 moles MgCl2×95.21 g/mol = 142.81
grams
Therefore, the theoretical yield of magnesium chloride is 142.81 grams.
Question 19
Question
A chemist is trying to determine the mass of a sample that contains an unknown
metal carbonate. The chemist first reacts the metal carbonate with hydrochloric
acid to produce carbon dioxide gas, water, and a chloride salt of the metal. If
15
3.71 grams of the metal carbonate produces 0.431 grams of carbon dioxide gas,
what is the molar mass of the metal carbonate?
Solution
Step 1: Write the balanced chemical equation for the reaction between the metal
carbonate and hydrochloric acid. The reaction can be written as:
Metal Carbonate+Hydrochloric Acid →Carbon Dioxide+Water+Chloride Salt of Metal
Step 2: Calculate the molar mass of carbon dioxide (CO2). The molar mass
of carbon dioxide is 44.01 g/mol.
Step 3: Calculate the number of moles of carbon dioxide produced. Given
that 0.431 grams of carbon dioxide is produced and the molar mass of carbon
dioxide is 44.01 g/mol:
Moles of CO2=0.431 g
44.01 g/mol = 0.0098 mol
Step 4: Determine the molar ratio between the metal carbonate and carbon
dioxide in the chemical equation. From the balanced chemical equation, we can
see that 1 mole of metal carbonate produces 1 mole of carbon dioxide.
Step 5: Calculate the number of moles of metal carbonate used. Since the
number of moles of carbon dioxide is the same as the number of moles of metal
carbonate:
Moles of Metal Carbonate = 0.0098 mol
Step 6: Calculate the molar mass of the metal carbonate. Given that 3.71
grams of the metal carbonate is used:
Molar mass of Metal Carbonate = 3.71 g
0.0098 mol = 377.81 g/mol
Therefore, the molar mass of the metal carbonate is 377.81 g/mol.
Question 20
Question
A compound X contains only elements A and B. When 2.00 g of X is analyzed,
it is found to contain 0.571 g of A and 1.43 g of B. Calculate the atomic masses
of A and B if the formula of the compound is AB2.
Solution
Step 1: Determine the number of moles of A and B present in the compound.
Given that the compound is AB2, we can set up the following equations:
Moles of A = 0.571 g A
Atomic mass of A
16
Moles of B = 1.43 g B
Atomic mass of B
Step 2: Calculate the molar ratios of A and B in the compound. From the
formula AB2, we can determine the molar ratio between A and B:
Molar ratio of A:B = 1 : 2
Step 3: Use the molar ratios to find the atomic masses of A and B. Since
the molar ratio of A:B is 1:2, the moles of A and B must also be in the ratio of
1:2. This means that:
Moles of A
Atomic mass of A =Moles of B
2×Atomic mass of B
Step 4: Solve for the atomic masses of A and B. Substitute the moles of A
and B from Step 1 into the ratio equation from Step 3:
0.571
Atomic mass of A =1.43
2×Atomic mass of B
Atomic mass of A = 2×0.571
1.43
Atomic mass of B = 1.43
2×0.571
Step 5: Calculate the atomic masses of A and B.
Atomic mass of A = 2×0.571
1.43 = 1.59 g/mol
Atomic mass of B = 1.43
2×0.571 = 1.25 g/mol
Therefore, the atomic masses of elements A and B in the compound AB2
are 1.59 g/mol and 1.25 g/mol, respectively.
Question 21
Question
Calculate the mass of phosphorus in grams that can be produced from the
reaction of 125 grams of calcium phosphate (Ca3(PO4)2) with an excess of
sodium. The balanced chemical equation is:
Ca3(PO4)2+ 6Na →3Ca + 2Na3PO4
17
Solution
Step 1: Calculate the molar mass of Ca3(PO4)2.
Molar mass of Ca3(PO4)2= 3×molar mass of Ca+2×molar mass of P+8×molar mass of O
= 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 310.18 g/mol
Step 2: Calculate the number of moles of Ca3(PO4)2.
Moles of Ca3(PO4)2=125 g
310.18 g/mol = 0.403 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the
moles of P4produced. From the balanced chemical equation, we see that 1 mole
of Ca3(PO4)2produces 2 moles of P4.
Moles of P4= 0.403 mol ×2 mol P4
1 mol Ca3(PO4)2
= 0.806 mol P4
Step 4: Calculate the mass of P4produced in grams.
Mass of P4= 0.806 mol P4×molar mass of P = 0.806 mol×30.97 g/mol = 24.91 g
Therefore, the mass of phosphorus produced in this reaction is 24.91 grams.
Question 22
Question
A reaction between iron (III) oxide and carbon monoxide produces iron and
carbon dioxide according to the following balanced chemical equation:
F e2O3(s)+3CO(g)→2F e(s)+3CO2(g)
If 25.0 grams of iron (III) oxide reacts with an excess of carbon monoxide, how
many grams of iron are produced?
Solution
Step 1: Calculate the molar mass of F e2O3: The molar mass of iron is approxi-
mately 55.85 g/mol and the molar mass of oxygen is approximately 16.00 g/mol.
Therefore, the molar mass of F e2O3is:
2×55.85 + 3 ×16.00 = 159.70 g/mol
Step 2: Calculate the number of moles of F e2O3with 25.0 grams:
Moles of F e2O3=25.0 g
159.70 g/mol = 0.157 mol
18
Step 3: Use the mole ratio from the balanced equation to find the moles of
iron produced: From the balanced chemical equation, 1 mole of F e2O3produces
2 moles of iron. Therefore, the moles of iron produced is:
0.157 mol F e2O3×2 mol Fe
1 mol F e2O3
= 0.314 mol Fe
Step 4: Calculate the mass of iron produced: The molar mass of iron is
approximately 55.85 g/mol. Therefore, the mass of iron produced is:
0.314 mol Fe ×55.85 g/mol = 17.6 g
Therefore, 17.6 grams of iron are produced when 25.0 grams of iron (III)
oxide reacts with an excess of carbon monoxide.
Question 23
Question
A chemist has 50.0 grams of aluminum (Al) and reacts it with excess hydrochlo-
ric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas (H2).
Calculate the theoretical yield of aluminum chloride that can be produced in
grams. (Molar masses: Al = 26.98 g/mol, Cl = 35.45 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2Al + 6HCl →2AlCl3+ 3H2
Step 2: Calculate the number of moles of aluminum. Given: mass of alu-
minum (Al) = 50.0 g
Molar mass of aluminum (Al) = 26.98 g/mol
Number of moles of Al = Mass of Al
Molar mass of Al =50.0 g
26.98 g/mol
Number of moles of Al ≈1.855 mol
Step 3: Use the mole ratio from the balanced chemical equation to find the
number of moles of AlCl3. From the balanced chemical equation: 1 mol of Al
produces 2 mol of AlCl3
Number of moles of AlCl3= 1.855 mol ×2 mol AlCl3
1 mol Al
Number of moles of AlCl3≈3.71 mol
19
Step 4: Calculate the mass of aluminum chloride produced. Molar mass of
aluminum chloride (AlCl3) = 26.98 g/mol + 3 ×35.45 g/mol = 133.33 g/mol
Mass of AlCl3= Number of moles of AlCl3×Molar mass of AlCl3≈3.71 mol×133.33 g/mol
Mass of AlCl3≈494.64 g
Therefore, the theoretical yield of aluminum chloride that can be produced
is 494.64 grams.
Question 24
Question
A 2.50 g sample of magnesium powder reacts with excess hydrochloric acid
to produce 51.3 mL of hydrogen gas at 25.0
°
C and 1.00 atm. Determine the
theoretical yield of hydrogen gas in grams. (Molar volume at STP = 22.4 L/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and hydrochloric acid:
Mg + 2HCl →MgCl2+ H2
Step 2: Calculate the number of moles of hydrogen gas produced using the
ideal gas law equation:
P V =nRT
Given: - Pressure, P= 1.00 atm - Volume, V= 51.3 mL = 0.0513 L -
Temperature, T= 25.0+273 = 298 K - Gas constant, R= 0.0821 atm·L/mol·K
Calculating moles of hydrogen:
n=P V
RT =(1.00 atm)(0.0513 L)
0.0821 atm ·L/mol ·K·298 K ≈0.0021 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the
theoretical yield of hydrogen gas in grams:
Molar mass of H2= 2.02 g/mol
Theoretical yield of H2= 0.0021 mol ×1 mol H2
1 mol ×2.02 g/mol ≈0.0042 g
Therefore, the theoretical yield of hydrogen gas in this reaction is 0.0042
grams.
20
Question 25
Question
A sample of iron(III) oxide (F e2O3) is decomposed by heating in a current of
hydrogen gas to produce metallic iron and water vapor. If 50.0 grams of iron(III)
oxide is decomposed, what mass of water vapor is produced?
Solution
Step 1: Write the balanced chemical equation for the reaction. The decomposi-
tion of iron(III) oxide can be represented by the following chemical equation:
F e2O3(s)+3H2(g)→2F e(s)+3H2O(g)
Step 2: Calculate the molar mass of F e2O3and H2O. - Molar mass of
F e2O3:F e: 55.85 g/mol
O: 16.00 g/mol ×3 = 48.00 g/mol
Total: 55.85 + 48.00 = 103.85 g/mol
- Molar mass of H2O:H: 1.01 g/mol ×2=2.02 g/mol
O: 16.00 g/mol
Total: 2.02 + 16.00 = 18.02 g/mol
Step 3: Calculate the number of moles of F e2O3in the sample.
Moles of F e2O3=mass
molar mass =50.0 g
103.85 g/mol = 0.481 mol
Step 4: Use the stoichiometry of the balanced chemical equation to determine
the moles of water produced. From the balanced chemical equation, 1 mole of
F e2O3produces 3 moles of H2O. Therefore, 0.481 moles of F e2O3will produce
0.481 ×3 = 1.443 moles of H2O.
Step 5: Calculate the mass of water vapor produced.
Mass of H2O= moles ×molar mass = 1.443 mol ×18.02 g/mol = 25.97 g
Answer: The mass of water vapor produced is 25.97 grams.
Question 26
Question
When 45.0 g of aluminum react with excess oxygen, how many grams of alu-
minum oxide are produced? [Molar mass of aluminum = 26.98 g/mol, molar
mass of oxygen = 16.00 g/mol]
21
Solution
Step 1: Write the balanced chemical equation for the reaction between alu-
minum and oxygen. Step 2: Calculate the number of moles of aluminum using
its molar mass. Step 3: Use the mole ratio from the balanced equation to find
the number of moles of aluminum oxide produced. Step 4: Convert the number
of moles of aluminum oxide to grams using its molar mass.
Step 1: The balanced chemical equation for the reaction is:
4Al + 3O2→2Al2O3
Step 2: Calculate the number of moles of aluminum:
Moles of Al = 45.0 g Al
26.98 g/mol Al = 1.67 mol Al
Step 3: Use the mole ratio from the balanced equation to find the number
of moles of aluminum oxide produced:
Moles of Al2O3=2 mol Al2O3
4 mol Al ×1.67 mol Al = 0.835 mol Al2O3
Step 4: Convert the number of moles of aluminum oxide to grams using its
molar mass:
Mass of Al2O3= 0.835 mol Al2O3×101.96 g/mol Al2O3= 85.1 g Al2O3
Therefore, 85.1 grams of aluminum oxide are produced when 45.0 grams of
aluminum react with excess oxygen.
Question 27
Question
A mixture of potassium perchlorate (KClO4) and powdered aluminum is often
used as a rocket propellant. When ignited, the aluminum reduces the per-
chlorate to produce aluminum oxide, potassium chloride, and chlorine gas. If
20.0 g of potassium perchlorate reacts with excess aluminum to produce 8.60
g of potassium chloride, what is the percent yield of potassium chloride in this
reaction? (Molar masses: KClO4= 138.55 g/mol, KCl = 74.55 g/mol)
Solution
Step 1: Calculate the moles of KCl produced. Given: - Mass of KCl produced:
8.60 g - Molar mass of KCl: 74.55 g/mol
We can use the formula:
moles = mass
molar mass
22
Substitute the values:
moles of KCl = 8.60 g
74.55 g/mol = 0.115 mol
Step 2: Write and balance the chemical equation for the reaction. The
reaction between potassium perchlorate and aluminum can be represented as:
3 KClO4+ 8 Al →3 KCl + 4 Al2O3+ 6 Cl2
Step 3: Calculate the theoretical yield of KCl. From the balanced equation,
we see that 3 moles of KCl are produced for every 3 moles of KClO4consumed.
Therefore, the molar ratio of KCl to KClO4is 1:1.
Since 20.0 g of KClO4is reacted, we first determine the moles of KClO4:
moles of KClO4=20.0 g
138.55 g/mol = 0.144 mol
The moles of KCl produced should also be 0.144 mol.
Therefore, the theoretical yield of KCl is:
moles of KCl ×molar mass of KCl = 0.144 mol ×74.55 g/mol = 10.746 g
Step 4: Calculate the percent yield of KCl. The percent yield is given by
the formula:
Percent yield = actual yield
theoretical yield×100%
Substitute the values:
Percent yield = 8.60 g
10.746 g×100% = 79.99%
Therefore, the percent yield of potassium chloride in this reaction is 79.99
Question 28
Question
A student wants to synthesize 10.0 grams of silver chloride (AgCl) from silver
nitrate and hydrochloric acid. If the reaction is:
AgNO3+HCl −→ AgCl +HNO3
and the student has 23.0 grams of silver nitrate (AgNO3) and an excess of
hydrochloric acid (HCl), how many grams of silver chloride can be synthesized?
23
Solution
Step 1: Write the balanced chemical equation for the reaction.
AgNO3+HCl −→ AgCl +HNO3
Step 2: Calculate the molar mass of each substance: The molar mass of
AgNO3 (silver nitrate) is:
107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol) = 169.87 g/mol
The molar mass of AgCl (silver chloride) is:
107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Step 3: Calculate the number of moles of AgNO3.
Moles of AgNO3 = 23.0 g
169.87 g/mol = 0.1355 mol
Step 4: Use the balanced chemical equation to determine the moles of AgCl
that can be produced. From the balanced chemical equation: 1 mole of AgNO3
produces 1 mole of AgCl.
So, 0.1355 moles of AgNO3 will produce 0.1355 moles of AgCl.
Step 5: Calculate the mass of silver chloride that can be synthesized.
Mass of AgCl = 0.1355 mol ×143.32 g/mol = 19.41 g
Therefore, the student can synthesize 19.41 grams of silver chloride.
Question 29
Question
When 1.00 g of aluminum reacts with excess hydrochloric acid, hydrogen gas
is produced. Calculate the volume of hydrogen gas produced at 25
°
C and 1.00
atm pressure. Assume the reaction goes to completion and that the gas behaves
ideally. (Molar mass of aluminum = 26.98 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
and hydrochloric acid. The balanced chemical equation for the reaction is:
2Al + 6HCl →2AlCl3+ 3H2
Step 2: Calculate the number of moles of aluminum based on its mass.
Given: Mass of aluminum = 1.00 g Molar mass of aluminum = 26.98 g/mol
Number of moles of aluminum = 1.00 g
26.98 g/mol = 0.0371 mol
24
Step 3: Determine the number of moles of hydrogen gas produced using the
mole ratio from the balanced chemical equation. From the balanced chemical
equation, 2 moles of Al produce 3 moles of H2. So, 0.0371 mol of Al will produce
0.0371 mol×3
2= 0.0557 mol of H2 gas.
Step 4: Apply the ideal gas law to find the volume of hydrogen gas produced.
Given: Pressure (P) = 1.00 atm, Temperature (T) = 25
°
C = 298 K, Volume of
gas (V) = ?
Using the ideal gas law equation P V =nRT , where R is the ideal gas
constant:
V=nRT
P=(0.0557 mol)(0.0821 L atm/mol K)(298 K)
1.00 atm
V= 1.30 L
Therefore, the volume of hydrogen gas produced is 1.30 L.
Question 30
Question
Balance the following chemical equation and determine the mass of magnesium
oxide (MgO) produced when 20.0 grams of magnesium (Mg) reacts with excess
oxygen gas (O2).
Mg + O2→MgO
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
2Mg + O2→2MgO
Step 2: Calculate the molar mass of each substance. - Molar mass of Mg:
24.31 g/mol - Molar mass of O2: 32.00 g/mol - Molar mass of MgO: 40.31 g/mol
Step 3: Find the number of moles of magnesium reacting.
20.0 g
24.31 g/mol = 0.823 mol Mg
Step 4: Determine the limiting reactant. Using the balanced equation, we
can see that 2 moles of Mg react with 1 mole of O2to produce 2 moles of MgO.
Therefore, for the 0.823 moles of Mg to react completely, we would need:
0.823 mol Mg
2 mol Mg ×1 mol O2= 0.411 mol O2
Since we have excess O2, magnesium is the limiting reactant.
25
Therefore, 3.33 moles of aluminum chloride can be produced from 5.0 moles
of chlorine gas.
Question 5
Question
A chemist wants to determine the percentage composition of sulfur in a sample
of iron(II) sulfide (FeS). If a 2.50 g sample of iron(II) sulfide is reacted with
excess hydrochloric acid and the resulting reaction produces 0.445 g of hydrogen
sulfide gas (H2S), what is the percentage composition of sulfur in the sample?
(Note: The balanced chemical equation for the reaction is FeS + 2HCl →
FeCl2+ H2S)
Solution
Step 1: Find the moles of H2S produced. Given: Mass of H2S=0.445 g
To find the moles of H2S, we use the molar mass of H2S.
Molar mass of H2S = 1 ×2 + 32 = 34 g/mol
Moles of H2S = mass
molar mass =0.445
34 = 0.0131 mol
Step 2: Use the mole ratio to find the moles of FeS. From the balanced
chemical equation, the mole ratio of FeS to H2S is 1:1. So, moles of FeS = 0.0131
mol.
Step 3: Find the molar mass of FeS.
Molar mass of FeS = 56 + 32 = 88 g/mol
Step 4: Calculate the mass percentage of sulfur in FeS.
Percentage of sulfur in FeS = Molar mass of sulfur
Molar mass of FeS ×100%
Percentage of sulfur in FeS = 32
88 ×100% = 36.4%
Therefore, the percentage composition of sulfur in the sample of iron(II)
sulfide is 36.4%.
Question 6
Question
A sample of iron pyrite (FeS2), also known as fool’s gold, contains 5.00 moles
of sulfur. How many moles of iron are present in the sample?
Given: Molecular weight of FeS2= 119.98 g/mol
5
Solution
Step 1: Write the balanced chemical equation for the reaction involving iron
pyrite:
FeS2→Fe + S2
Step 2: Determine the molar ratio of sulfur to iron in the reaction. 1 mole
of FeS2produces 1 mole of iron (Fe) and 2 moles of sulfur (S).
Step 3: Calculate the number of moles of iron in the sample: Given that the
sample contains 5.00 moles of sulfur, and using the molar ratio from step 2, we
find: 5.00 mol S
2×1 mol Fe
1 mol S = 2.50 mol Fe
Therefore, there are 2.50 moles of iron present in the sample of iron pyrite.
Question 7
Question
A chemist is working with a reaction involving silver nitrate (AgNO3) and
sodium chloride (NaCl). If 10.0 grams of silver nitrate reacts completely with
sodium chloride to produce silver chloride (AgCl) and sodium nitrate (NaNO3),
what mass of sodium chloride is required for the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride. The balanced chemical equation is:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Calculate the molar mass of each compound involved in the reaction.
The molar mass of AgNO3can be calculated as:
(1 ×Ag) + (1 ×N) + (3 ×O) = 107.87 g/mol
The molar mass of NaCl can be calculated as:
(1 ×Na) + (1 ×Cl) = 58.44 g/mol
The molar mass of AgCl can be calculated as:
(1 ×Ag) + (1 ×Cl) = 143.32 g/mol
The molar mass of NaNO3can be calculated as:
(1 ×Na) + (1 ×N) + (3 ×O) = 84.99 g/mol
6
Step 3: Calculate the number of moles of AgNO3given 10.0 grams.
Moles of AgNO3=Mass
Molar mass =10.0 g
107.87 g/mol ≈0.093 mol
Step 4: Determine the mole ratio between AgNO3and NaCl from the bal-
anced chemical equation. The mole ratio is 1:1, which means 1 mole of AgNO3
reacts with 1 mole of NaCl.
Step 5: Calculate the mass of NaCl required using the mole ratio and molar
mass of NaCl.
Mass of NaCl = Moles of AgNO3×Molar mass of NaCl
Mass of NaCl = 0.093 mol ×58.44 g/mol = 5.43 g
Therefore, 5.43 grams of sodium chloride is required for the reaction.
Question 8
Question
A student is performing a stoichiometry calculation involving the reaction of
iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. If
150.0 g of iron(III) oxide reacts with excess carbon monoxide, how many grams
of iron can be produced? (Molar masses: Fe = 55.85 g/mol, O = 16.00 g/mol,
C = 12.01 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between iron(III)
oxide and carbon monoxide.
F e2O3+ 3CO →2F e + 3CO2
Step 2: Calculate the molar mass of FeO.
Molar mass of F e2O3= 2(F e)+3(O) = 2(55.85 g/mol)+3(16.00 g/mol) = 159.70 g/mol
Step 3: Determine the number of moles of FeO present in 150.0 g.
Moles of F e2O3=Mass
Molar mass =150.0g
159.70 g/mol = 0.940 mol
Step 4: Use the stoichiometry of the balanced chemical equation to find the
number of moles of Fe produced.
Moles of F e =Moles of F e2O3
1×2mol F e
1mol F e2O3
= 0.940 mol ×2=1.880 mol
Step 5: Calculate the mass of iron produced.
Mass of F e =M oles of F e×M olar mass of F e = 1.880 mol×55.85 g/mol = 105.26 g
Therefore, 105.26 grams of iron can be produced from the reaction of 150.0
grams of iron(III) oxide with excess carbon monoxide.
7
Question 9
Question
A chemist wants to synthesize 1.50 g of ammonia (NH3) using the following
balanced chemical equation:
4NH3(g) + 5O2(g)→4NO(g) + 6H2O(g)
If the chemist has an excess of oxygen gas, what mass of oxygen gas is required
for the synthesis of 1.50 g of ammonia?
Solution
Step 1: Calculate the molar mass of NH3. The molar mass of NH3is 17.03
g/mol.
Step 2: Calculate the moles of NH3. Given mass of NH3= 1.50 g Number
of moles of NH3=1.50 g
17.03 g/mol = 0.0881 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the
moles of O2required. From the balanced chemical equation, we see that 4 moles
of NH3react with 5 moles of O2. Therefore, 0.0881 mol NH3×5
4= 0.110mol of O2
is required.
Step 4: Calculate the mass of O2required. Molar mass of O2= 32.00 g/mol
Mass of O2required = 0.110 mol ×32.00 g/mol = 3.52 g
Therefore, the mass of oxygen gas required for the synthesis of 1.50 g of
ammonia is 3.52 g.
Question 10
Question
A chemist is performing a reaction using 3.50 moles of iron(III) oxide and 4.00
moles of carbon monoxide according to the following balanced equation:
F e2O3(s)+3CO(g)→2F e(s)+3CO2(g)
What is the limiting reactant in this reaction? How many moles of iron are
produced?
Solution
Step 1: Calculate the moles of iron that can be produced by each reactant. -
For iron(III) oxide:
3.50 mol Fe2O3
1×2 mol Fe
1 mol Fe2O3
= 7.00 mol Fe
8
- For carbon monoxide:
4.00 mol CO
1×2 mol Fe
3 mol CO =8
3mol Fe = 2.67 mol Fe
Step 2: Identify the limiting reactant. - Since the carbon monoxide produces
fewer moles of iron, it is the limiting reactant in this reaction.
Step 3: Calculate the moles of iron produced using the limiting reactant. -
Using 4.00 moles of CO as the limiting reactant:
4.00 mol CO
1×2 mol Fe
3 mol CO =8
3mol Fe = 2.67 mol Fe
Therefore, the limiting reactant is carbon monoxide, and 2.67 moles of iron
are produced in this reaction.
Question 11
Question
Consider the reaction:
2KClO3→2KCl + 3O2
If 25.0 grams of KClO3 decomposes, what mass of oxygen is produced?
Solution
Step 1: Calculate the molar mass of KClO3, O2, and KCl.
The molar mass of KClO3 is:
2(39.10 g/mol K) + 35.45 g/mol Cl + 3(16.00 g/mol O) = 122.55 g/mol
The molar mass of O2 is:
2(16.00 g/mol O) = 32.00 g/mol
The molar mass of KCl is:
39.10 g/mol K + 35.45 g/mol Cl = 74.55 g/mol
Step 2: Find the number of moles of KClO3 using its molar mass.
Number of moles of KClO3:
25.0 g
122.55 g/mol = 0.204 mol
9
Step 3: Use the stoichiometry of the reaction to find the number of moles of
O2 produced.
From the balanced equation, we see that 2 moles of KClO3 produce 3
moles of O2.
0.204 mol KClO3 ×3 mol O2
2 mol KClO3 = 0.306 mol O2
Step 4: Convert the number of moles of O2 to grams.
Mass of O2 produced:
0.306 mol O2 ×32.00 g/mol O2 = 9.79 g O2
Therefore, 9.79 grams of oxygen is produced when 25.0 grams of KClO3
decomposes.
Question 12
Question
A reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) produces
silver chloride (AgCl) and sodium nitrate (NaNO3). If 10.0 grams of silver
nitrate reacts with an excess of sodium chloride to produce 12.0 grams of silver
chloride, what is the percent yield of silver chloride?
Solution
Step 1: Write the balanced chemical equation for the reaction.
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Calculate the molar mass of each compound: - AgNO3:
107.87 g/mol + 14.01 g/mol + (3 ×16.00 g/mol) = 169.87 g/mol
- AgCl:
107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Step 3: Determine the theoretical yield of silver chloride: - Using the molar
mass of AgNO3, we find the number of moles of AgNO3:
10.0 g
169.87 g/mol = 0.05885 mol
- According to the balanced equation, 1 mole of AgNO3produces 1 mole of
AgCl. - Therefore, the theoretical yield of AgCl is 0.05885 mol.
10
Step 4: Calculate the percent yield of AgCl: - The actual yield of AgCl is
given as 12.0 grams. - Convert the actual yield to moles:
12.0 g
143.32 g/mol = 0.08370 mol
- The percent yield is given by:
Percent Yield = Actual Yield
Theoretical Yield×100%
Percent Yield = 0.08370 mol
0.05885 mol×100% = 142.17%
Step 5: Since a percent yield greater than 100
Question 13
Question
A chemist wants to produce 25.0 g of potassium sulfate (K2SO4) by the reaction
of potassium hydroxide (KOH) and sulfuric acid (H2SO4). If the chemist has an
unlimited supply of potassium hydroxide but only 95.0 g of sulfuric acid, what
is the limiting reactant? Calculate the theoretical yield of potassium sulfate in
grams.
Solution
Step 1: Write the balanced chemical equation for the reaction:
2 KOH + H2SO4→K2SO4+ 2 H2O
Step 2: Calculate the molar mass of each substance: - KOH: K = 39.10 g/mol,
O = 16.00 g/mol, H = 1.01 g/mol Molar mass of KOH = 39.10 + 16.00 + 1.01 =
56.11 g/mol
- H2SO4: H = 1.01 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass of
H2SO4= 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol
- K2SO4: K = 39.10 g/mol, S = 32.07 g/mol, O = 16.00 g/mol Molar mass
of K2SO4= 2(39.10) + 32.07 + 4(16.00) = 174.26 g/mol
Step 3: Calculate the number of moles of each reactant: - Moles of KOH =
25.0 g
56.11 g/mol ≈0.446 mol - Moles of H2SO4=95.0 g
98.09 g/mol ≈0.969 mol
Step 4: Determine the limiting reactant: The stoichiometry of the reaction
indicates that 1 mole of H2SO4reacts with 2 moles of KOH. Therefore, for
0.969 moles of H2SO4, we would need 0.969/2 = 0.484 moles of KOH. Since we
have less than 0.484 moles of KOH available, H2SO4is the limiting reactant.
Step 5: Calculate the theoretical yield of K2SO4: - Theoretical yield of
K2SO4= moles of limiting reactant×molar ratio from balanced equation
1×molar mass of K2SO4
Theoretical yield of K2SO4= 0.969 mol ×1
2×174.26 g/mol ≈84.35 g
Therefore, the limiting reactant is H2SO4and the theoretical yield of K2SO4
is 84.35 g.
11
Question 14
Question
A reaction between solid iron (III) oxide and carbon monoxide produces solid
iron and carbon dioxide gas. If 50.0 grams of solid iron (III) oxide react with
an excess of carbon monoxide and 35.0 grams of solid iron are produced, what
is the percent yield of the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The reaction is
between solid iron (III) oxide (F e2O3) and carbon monoxide (CO), producing
solid iron (F e) and carbon dioxide (CO2). The balanced equation is:
F e2O3+ 3CO →2F e + 3CO2
Step 2: Calculate the molar mass of F e2O3and F e. - Molar mass of F e2O3:
2×55.845 (molar mass of iron) + 3 ×16.00 (molar mass of oxygen) = 159.69
g/mol
- Molar mass of F e: 55.845 g/mol
Step 3: Calculate the number of moles of F e2O3used.
Moles of F e2O3=50.0 g
159.69 g/mol = 0.313 mol
Step 4: Using the balanced equation, calculate the theoretical yield of F e.
Since 1 mol of F e2O3produces 2 mol of F e, the theoretical yield of F e is:
0.313 mol F e2O3×2 mol F e
1 mol F e2O3
= 0.626 mol F e
Step 5: Calculate the theoretical mass of F e that should be produced.
Theoretical mass of F e = 0.626 mol ×55.845 g/mol = 35.0 g
Step 6: Calculate the percent yield of the reaction.
Percent yield = Actual yield
Theoretical yield ×100%
Percent yield = 35.0 g
35.0 g ×100% = 100%
So, the percent yield of the reaction is 100
12
Question 15
Question
A student is tasked with determining the amount of magnesium oxide that can
be produced when 15.0 grams of magnesium metal reacts with excess oxygen
gas. The balanced chemical equation for the reaction is:
2Mg + O2→2MgO
Assuming 100
Solution
Step 1: Determine the molar mass of each compound involved. The molar mass
of magnesium (Mg) is 24.305 g/mol, oxygen (O) is 16.00 g/mol, and magnesium
oxide (MgO) is 40.30 g/mol.
Step 2: Calculate the number of moles of magnesium. Given: Mass of
magnesium = 15.0 g Using the molar mass of magnesium:
Moles of Mg = 15.0 g
24.305 g/mol = 0.617 mol Mg
Step 3: Use the stoichiometry of the balanced equation to convert moles of
magnesium to moles of magnesium oxide. From the balanced chemical equation:
2 Mg + O2→2 MgO
This means that 2 moles of magnesium produce 2 moles of magnesium oxide.
1 mol Mg = 1 mol MgO
Step 4: Calculate the theoretical yield of magnesium oxide in grams.
Moles of MgO = 0.617 mol Mg ×1 mol MgO
1 mol Mg = 0.617 mol MgO
Mass of MgO = 0.617 mol MgO ×40.30 g/mol = 24.88 g
Therefore, the theoretical yield of magnesium oxide in grams is 24.88g.
Question 16
Question
A reaction takes place according to the following balanced chemical equation:
C3H8(g) + 5O2(g)→3CO2(g) + 4H2O(g)
If 32.0 g of C3H8and64.0gofO2areallowedtoreact, whatisthelimitingreactant?Howmanygramsof waterareproduced?
13
Solution
Step 1: Calculate the number of moles of each reactant. Step 2: Determine
the limiting reactant and the amount of product formed. Step 3: Calculate the
mass of water produced.
Step 1: Calculate the number of moles of each reactant. The molar mass
of C3H8is44.1g/mol.T hemolarmassof O2is32.0g/mol.
Number of moles of C3H8: moles = mass
molar mass =32.0 g
44.1 g/mol = 0.726 mol
Number of moles of O2: moles = mass
molar mass =64.0 g
32.0 g/mol = 2.00 mol
Step 2: Determine the limiting reactant and the amount of product formed.
From the balanced chemical equation, 1 mol of C3H8reactswith5molofO2.T heref ore, thelimitingreactantisC3H8.
Amount of H2Oproducedfrom32.0gofC3H8: moles of H2O = 1
3×moles of C3H8=
1
3×0.726 = 0.242 mol
Step 3: Calculate the mass of water produced. The molar mass of H2Ois18.0g/mol.mass =
moles ×molar mass = 0.242 mol ×18.0 g/mol = 4.36 g
Therefore, the limiting reactant is C3H8and4.36gofH2Oareproduced.
Question 17
Question
A chemist wants to synthesize 10.0 grams of a compound. The molar mass of
the compound is 100 g/mol. If the chemist has a 90
Solution
Step 1: Calculate the number of moles of the desired compound: Given: Mass
of compound = 10.0 g, Molar mass of compound = 100 g/mol
Moles of compound = Mass of compound
Molar mass of compound =10.0 g
100 g/mol = 0.1 mol
Step 2: Account for the 90Since the yield is 90Let x be the number of moles
of reactant needed to achieve the desired amount of product.
Expected yield = 0.9×Actual yield
0.9×x= 0.1
x=0.1
0.9=1
9= 0.111 mol
Therefore, the chemist should start with 0.111 moles of reactant to achieve
the desired 10.0 grams of product, accounting for a 90
14
Question 18
Question
A student is conducting a reaction between 2.50 moles of magnesium (Mg) and
3.00 moles of hydrochloric acid (HCl) to produce magnesium chloride (MgCl2)
and hydrogen gas (H2). Write the balanced chemical equation for this reaction
and determine: a) The limiting reactant, b) The theoretical yield of magnesium
chloride in grams.
Given: Molar mass of Mg = 24.31 g/mol Molar mass of HCl = 36.46 g/mol
Molar mass of MgCl2= 95.21 g/mol
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and hydrochloric acid.
The balanced equation is:
Mg + 2HCl →MgCl2+ H2
Step 2: Calculate the number of moles of magnesium chloride that can be
produced from each reactant.
From the balanced equation, we see that 1 mole of magnesium reacts with
2 moles of hydrochloric acid to produce 1 mole of magnesium chloride.
For magnesium: Number of moles of MgCl2= 2.50 moles Mg ×1 mol MgCl2
1 mol Mg =
2.50 moles
For hydrochloric acid: Number of moles of MgCl2= 3.00 moles HCl ×1 mol MgCl2
2 mol HCl =
1.50 moles
Step 3: Determine the limiting reactant. The limiting reactant is the one
that produces the least amount of product. Since hydrochloric acid produces
less magnesium chloride (1.50 moles) compared to magnesium (2.50 moles),
hydrochloric acid is the limiting reactant.
Step 4: Calculate the theoretical yield of magnesium chloride in grams.
To find the theoretical yield of magnesium chloride in grams, we need to
convert the number of moles of MgCl2produced by hydrochloric acid to grams.
Theoretical yield of MgCl2= 1.50 moles MgCl2×95.21 g/mol = 142.81
grams
Therefore, the theoretical yield of magnesium chloride is 142.81 grams.
Question 19
Question
A chemist is trying to determine the mass of a sample that contains an unknown
metal carbonate. The chemist first reacts the metal carbonate with hydrochloric
acid to produce carbon dioxide gas, water, and a chloride salt of the metal. If
15
3.71 grams of the metal carbonate produces 0.431 grams of carbon dioxide gas,
what is the molar mass of the metal carbonate?
Solution
Step 1: Write the balanced chemical equation for the reaction between the metal
carbonate and hydrochloric acid. The reaction can be written as:
Metal Carbonate+Hydrochloric Acid →Carbon Dioxide+Water+Chloride Salt of Metal
Step 2: Calculate the molar mass of carbon dioxide (CO2). The molar mass
of carbon dioxide is 44.01 g/mol.
Step 3: Calculate the number of moles of carbon dioxide produced. Given
that 0.431 grams of carbon dioxide is produced and the molar mass of carbon
dioxide is 44.01 g/mol:
Moles of CO2=0.431 g
44.01 g/mol = 0.0098 mol
Step 4: Determine the molar ratio between the metal carbonate and carbon
dioxide in the chemical equation. From the balanced chemical equation, we can
see that 1 mole of metal carbonate produces 1 mole of carbon dioxide.
Step 5: Calculate the number of moles of metal carbonate used. Since the
number of moles of carbon dioxide is the same as the number of moles of metal
carbonate:
Moles of Metal Carbonate = 0.0098 mol
Step 6: Calculate the molar mass of the metal carbonate. Given that 3.71
grams of the metal carbonate is used:
Molar mass of Metal Carbonate = 3.71 g
0.0098 mol = 377.81 g/mol
Therefore, the molar mass of the metal carbonate is 377.81 g/mol.
Question 20
Question
A compound X contains only elements A and B. When 2.00 g of X is analyzed,
it is found to contain 0.571 g of A and 1.43 g of B. Calculate the atomic masses
of A and B if the formula of the compound is AB2.
Solution
Step 1: Determine the number of moles of A and B present in the compound.
Given that the compound is AB2, we can set up the following equations:
Moles of A = 0.571 g A
Atomic mass of A
16
Moles of B = 1.43 g B
Atomic mass of B
Step 2: Calculate the molar ratios of A and B in the compound. From the
formula AB2, we can determine the molar ratio between A and B:
Molar ratio of A:B = 1 : 2
Step 3: Use the molar ratios to find the atomic masses of A and B. Since
the molar ratio of A:B is 1:2, the moles of A and B must also be in the ratio of
1:2. This means that:
Moles of A
Atomic mass of A =Moles of B
2×Atomic mass of B
Step 4: Solve for the atomic masses of A and B. Substitute the moles of A
and B from Step 1 into the ratio equation from Step 3:
0.571
Atomic mass of A =1.43
2×Atomic mass of B
Atomic mass of A = 2×0.571
1.43
Atomic mass of B = 1.43
2×0.571
Step 5: Calculate the atomic masses of A and B.
Atomic mass of A = 2×0.571
1.43 = 1.59 g/mol
Atomic mass of B = 1.43
2×0.571 = 1.25 g/mol
Therefore, the atomic masses of elements A and B in the compound AB2
are 1.59 g/mol and 1.25 g/mol, respectively.
Question 21
Question
Calculate the mass of phosphorus in grams that can be produced from the
reaction of 125 grams of calcium phosphate (Ca3(PO4)2) with an excess of
sodium. The balanced chemical equation is:
Ca3(PO4)2+ 6Na →3Ca + 2Na3PO4
17
Solution
Step 1: Calculate the molar mass of Ca3(PO4)2.
Molar mass of Ca3(PO4)2= 3×molar mass of Ca+2×molar mass of P+8×molar mass of O
= 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 310.18 g/mol
Step 2: Calculate the number of moles of Ca3(PO4)2.
Moles of Ca3(PO4)2=125 g
310.18 g/mol = 0.403 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the
moles of P4produced. From the balanced chemical equation, we see that 1 mole
of Ca3(PO4)2produces 2 moles of P4.
Moles of P4= 0.403 mol ×2 mol P4
1 mol Ca3(PO4)2
= 0.806 mol P4
Step 4: Calculate the mass of P4produced in grams.
Mass of P4= 0.806 mol P4×molar mass of P = 0.806 mol×30.97 g/mol = 24.91 g
Therefore, the mass of phosphorus produced in this reaction is 24.91 grams.
Question 22
Question
A reaction between iron (III) oxide and carbon monoxide produces iron and
carbon dioxide according to the following balanced chemical equation:
F e2O3(s)+3CO(g)→2F e(s)+3CO2(g)
If 25.0 grams of iron (III) oxide reacts with an excess of carbon monoxide, how
many grams of iron are produced?
Solution
Step 1: Calculate the molar mass of F e2O3: The molar mass of iron is approxi-
mately 55.85 g/mol and the molar mass of oxygen is approximately 16.00 g/mol.
Therefore, the molar mass of F e2O3is:
2×55.85 + 3 ×16.00 = 159.70 g/mol
Step 2: Calculate the number of moles of F e2O3with 25.0 grams:
Moles of F e2O3=25.0 g
159.70 g/mol = 0.157 mol
18
Step 3: Use the mole ratio from the balanced equation to find the moles of
iron produced: From the balanced chemical equation, 1 mole of F e2O3produces
2 moles of iron. Therefore, the moles of iron produced is:
0.157 mol F e2O3×2 mol Fe
1 mol F e2O3
= 0.314 mol Fe
Step 4: Calculate the mass of iron produced: The molar mass of iron is
approximately 55.85 g/mol. Therefore, the mass of iron produced is:
0.314 mol Fe ×55.85 g/mol = 17.6 g
Therefore, 17.6 grams of iron are produced when 25.0 grams of iron (III)
oxide reacts with an excess of carbon monoxide.
Question 23
Question
A chemist has 50.0 grams of aluminum (Al) and reacts it with excess hydrochlo-
ric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas (H2).
Calculate the theoretical yield of aluminum chloride that can be produced in
grams. (Molar masses: Al = 26.98 g/mol, Cl = 35.45 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2Al + 6HCl →2AlCl3+ 3H2
Step 2: Calculate the number of moles of aluminum. Given: mass of alu-
minum (Al) = 50.0 g
Molar mass of aluminum (Al) = 26.98 g/mol
Number of moles of Al = Mass of Al
Molar mass of Al =50.0 g
26.98 g/mol
Number of moles of Al ≈1.855 mol
Step 3: Use the mole ratio from the balanced chemical equation to find the
number of moles of AlCl3. From the balanced chemical equation: 1 mol of Al
produces 2 mol of AlCl3
Number of moles of AlCl3= 1.855 mol ×2 mol AlCl3
1 mol Al
Number of moles of AlCl3≈3.71 mol
19
Step 4: Calculate the mass of aluminum chloride produced. Molar mass of
aluminum chloride (AlCl3) = 26.98 g/mol + 3 ×35.45 g/mol = 133.33 g/mol
Mass of AlCl3= Number of moles of AlCl3×Molar mass of AlCl3≈3.71 mol×133.33 g/mol
Mass of AlCl3≈494.64 g
Therefore, the theoretical yield of aluminum chloride that can be produced
is 494.64 grams.
Question 24
Question
A 2.50 g sample of magnesium powder reacts with excess hydrochloric acid
to produce 51.3 mL of hydrogen gas at 25.0
°
C and 1.00 atm. Determine the
theoretical yield of hydrogen gas in grams. (Molar volume at STP = 22.4 L/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and hydrochloric acid:
Mg + 2HCl →MgCl2+ H2
Step 2: Calculate the number of moles of hydrogen gas produced using the
ideal gas law equation:
P V =nRT
Given: - Pressure, P= 1.00 atm - Volume, V= 51.3 mL = 0.0513 L -
Temperature, T= 25.0+273 = 298 K - Gas constant, R= 0.0821 atm·L/mol·K
Calculating moles of hydrogen:
n=P V
RT =(1.00 atm)(0.0513 L)
0.0821 atm ·L/mol ·K·298 K ≈0.0021 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the
theoretical yield of hydrogen gas in grams:
Molar mass of H2= 2.02 g/mol
Theoretical yield of H2= 0.0021 mol ×1 mol H2
1 mol ×2.02 g/mol ≈0.0042 g
Therefore, the theoretical yield of hydrogen gas in this reaction is 0.0042
grams.
20
Question 25
Question
A sample of iron(III) oxide (F e2O3) is decomposed by heating in a current of
hydrogen gas to produce metallic iron and water vapor. If 50.0 grams of iron(III)
oxide is decomposed, what mass of water vapor is produced?
Solution
Step 1: Write the balanced chemical equation for the reaction. The decomposi-
tion of iron(III) oxide can be represented by the following chemical equation:
F e2O3(s)+3H2(g)→2F e(s)+3H2O(g)
Step 2: Calculate the molar mass of F e2O3and H2O. - Molar mass of
F e2O3:F e: 55.85 g/mol
O: 16.00 g/mol ×3 = 48.00 g/mol
Total: 55.85 + 48.00 = 103.85 g/mol
- Molar mass of H2O:H: 1.01 g/mol ×2=2.02 g/mol
O: 16.00 g/mol
Total: 2.02 + 16.00 = 18.02 g/mol
Step 3: Calculate the number of moles of F e2O3in the sample.
Moles of F e2O3=mass
molar mass =50.0 g
103.85 g/mol = 0.481 mol
Step 4: Use the stoichiometry of the balanced chemical equation to determine
the moles of water produced. From the balanced chemical equation, 1 mole of
F e2O3produces 3 moles of H2O. Therefore, 0.481 moles of F e2O3will produce
0.481 ×3 = 1.443 moles of H2O.
Step 5: Calculate the mass of water vapor produced.
Mass of H2O= moles ×molar mass = 1.443 mol ×18.02 g/mol = 25.97 g
Answer: The mass of water vapor produced is 25.97 grams.
Question 26
Question
When 45.0 g of aluminum react with excess oxygen, how many grams of alu-
minum oxide are produced? [Molar mass of aluminum = 26.98 g/mol, molar
mass of oxygen = 16.00 g/mol]
21
Solution
Step 1: Write the balanced chemical equation for the reaction between alu-
minum and oxygen. Step 2: Calculate the number of moles of aluminum using
its molar mass. Step 3: Use the mole ratio from the balanced equation to find
the number of moles of aluminum oxide produced. Step 4: Convert the number
of moles of aluminum oxide to grams using its molar mass.
Step 1: The balanced chemical equation for the reaction is:
4Al + 3O2→2Al2O3
Step 2: Calculate the number of moles of aluminum:
Moles of Al = 45.0 g Al
26.98 g/mol Al = 1.67 mol Al
Step 3: Use the mole ratio from the balanced equation to find the number
of moles of aluminum oxide produced:
Moles of Al2O3=2 mol Al2O3
4 mol Al ×1.67 mol Al = 0.835 mol Al2O3
Step 4: Convert the number of moles of aluminum oxide to grams using its
molar mass:
Mass of Al2O3= 0.835 mol Al2O3×101.96 g/mol Al2O3= 85.1 g Al2O3
Therefore, 85.1 grams of aluminum oxide are produced when 45.0 grams of
aluminum react with excess oxygen.
Question 27
Question
A mixture of potassium perchlorate (KClO4) and powdered aluminum is often
used as a rocket propellant. When ignited, the aluminum reduces the per-
chlorate to produce aluminum oxide, potassium chloride, and chlorine gas. If
20.0 g of potassium perchlorate reacts with excess aluminum to produce 8.60
g of potassium chloride, what is the percent yield of potassium chloride in this
reaction? (Molar masses: KClO4= 138.55 g/mol, KCl = 74.55 g/mol)
Solution
Step 1: Calculate the moles of KCl produced. Given: - Mass of KCl produced:
8.60 g - Molar mass of KCl: 74.55 g/mol
We can use the formula:
moles = mass
molar mass
22
Substitute the values:
moles of KCl = 8.60 g
74.55 g/mol = 0.115 mol
Step 2: Write and balance the chemical equation for the reaction. The
reaction between potassium perchlorate and aluminum can be represented as:
3 KClO4+ 8 Al →3 KCl + 4 Al2O3+ 6 Cl2
Step 3: Calculate the theoretical yield of KCl. From the balanced equation,
we see that 3 moles of KCl are produced for every 3 moles of KClO4consumed.
Therefore, the molar ratio of KCl to KClO4is 1:1.
Since 20.0 g of KClO4is reacted, we first determine the moles of KClO4:
moles of KClO4=20.0 g
138.55 g/mol = 0.144 mol
The moles of KCl produced should also be 0.144 mol.
Therefore, the theoretical yield of KCl is:
moles of KCl ×molar mass of KCl = 0.144 mol ×74.55 g/mol = 10.746 g
Step 4: Calculate the percent yield of KCl. The percent yield is given by
the formula:
Percent yield = actual yield
theoretical yield×100%
Substitute the values:
Percent yield = 8.60 g
10.746 g×100% = 79.99%
Therefore, the percent yield of potassium chloride in this reaction is 79.99
Question 28
Question
A student wants to synthesize 10.0 grams of silver chloride (AgCl) from silver
nitrate and hydrochloric acid. If the reaction is:
AgNO3+HCl −→ AgCl +HNO3
and the student has 23.0 grams of silver nitrate (AgNO3) and an excess of
hydrochloric acid (HCl), how many grams of silver chloride can be synthesized?
23
Solution
Step 1: Write the balanced chemical equation for the reaction.
AgNO3+HCl −→ AgCl +HNO3
Step 2: Calculate the molar mass of each substance: The molar mass of
AgNO3 (silver nitrate) is:
107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol) = 169.87 g/mol
The molar mass of AgCl (silver chloride) is:
107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Step 3: Calculate the number of moles of AgNO3.
Moles of AgNO3 = 23.0 g
169.87 g/mol = 0.1355 mol
Step 4: Use the balanced chemical equation to determine the moles of AgCl
that can be produced. From the balanced chemical equation: 1 mole of AgNO3
produces 1 mole of AgCl.
So, 0.1355 moles of AgNO3 will produce 0.1355 moles of AgCl.
Step 5: Calculate the mass of silver chloride that can be synthesized.
Mass of AgCl = 0.1355 mol ×143.32 g/mol = 19.41 g
Therefore, the student can synthesize 19.41 grams of silver chloride.
Question 29
Question
When 1.00 g of aluminum reacts with excess hydrochloric acid, hydrogen gas
is produced. Calculate the volume of hydrogen gas produced at 25
°
C and 1.00
atm pressure. Assume the reaction goes to completion and that the gas behaves
ideally. (Molar mass of aluminum = 26.98 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
and hydrochloric acid. The balanced chemical equation for the reaction is:
2Al + 6HCl →2AlCl3+ 3H2
Step 2: Calculate the number of moles of aluminum based on its mass.
Given: Mass of aluminum = 1.00 g Molar mass of aluminum = 26.98 g/mol
Number of moles of aluminum = 1.00 g
26.98 g/mol = 0.0371 mol
24
Step 3: Determine the number of moles of hydrogen gas produced using the
mole ratio from the balanced chemical equation. From the balanced chemical
equation, 2 moles of Al produce 3 moles of H2. So, 0.0371 mol of Al will produce
0.0371 mol×3
2= 0.0557 mol of H2 gas.
Step 4: Apply the ideal gas law to find the volume of hydrogen gas produced.
Given: Pressure (P) = 1.00 atm, Temperature (T) = 25
°
C = 298 K, Volume of
gas (V) = ?
Using the ideal gas law equation P V =nRT , where R is the ideal gas
constant:
V=nRT
P=(0.0557 mol)(0.0821 L atm/mol K)(298 K)
1.00 atm
V= 1.30 L
Therefore, the volume of hydrogen gas produced is 1.30 L.
Question 30
Question
Balance the following chemical equation and determine the mass of magnesium
oxide (MgO) produced when 20.0 grams of magnesium (Mg) reacts with excess
oxygen gas (O2).
Mg + O2→MgO
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
2Mg + O2→2MgO
Step 2: Calculate the molar mass of each substance. - Molar mass of Mg:
24.31 g/mol - Molar mass of O2: 32.00 g/mol - Molar mass of MgO: 40.31 g/mol
Step 3: Find the number of moles of magnesium reacting.
20.0 g
24.31 g/mol = 0.823 mol Mg
Step 4: Determine the limiting reactant. Using the balanced equation, we
can see that 2 moles of Mg react with 1 mole of O2to produce 2 moles of MgO.
Therefore, for the 0.823 moles of Mg to react completely, we would need:
0.823 mol Mg
2 mol Mg ×1 mol O2= 0.411 mol O2
Since we have excess O2, magnesium is the limiting reactant.
25
Step 5: Calculate the mass of magnesium oxide produced.
Moles of MgO produced = 0.823 mol Mg ×2 mol MgO
2 mol Mg = 0.823 mol MgO
Mass = Moless of MgO ×Molar Mass of MgO
Mass = 0.823 mol ×40.31 g/mol = 33.2 g
Therefore, 33.2 grams of magnesium oxide are produced from 20.0 grams of
magnesium.
26
