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Title: Stoichiometry Practice Exercises with detailed solutions
Description: This material contains worked-out practice exercises on stoichiometry. The solutions are given in a stepwise manner and are easy to follow.
Description: This material contains worked-out practice exercises on stoichiometry. The solutions are given in a stepwise manner and are easy to follow.
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Stoichiometry Practice Exercises | Reymon T
...
1
...
C8 H18 + O2 → CO2 + H2O
a
...
b
...
What mass of Oxygen is needed to completely consume 100 grams of Octane?
2
...
The
chemical reaction equation is shown below
...
After the experiment, the total mass of CaCl2 produced is 120g
...
Based on the amount of CaCl2 produced in the reaction, how much CO2 in grams
effervesced out of the Calcium carbonate used?
3
...
A by-product of
the process is Calcium sulfate
...
Write the balanced chemical reaction equation for the production of Hydrogen fluoride
...
How many kilograms of Calcium fluoride should be used to produce 1
...
What is the minimum amount, in grams, of Sulfuric acid needed to completely consume
140g of Calcium fluoride?
4
...
a
...
b
...
Complete the table below
...
What is the theoretical yield of the reaction?
e
...
Dela Cruz
Stoichiometry Practice Exercises | Reymon T
...
The complete combustion of Octane results to the production of Carbon dioxide and
water as shown in the unbalanced chemical equation below
...
Write the balanced chemical reaction equation
...
1
...
We do this to determine the stoichiometric
ratio of the reactants and products with each other in the chemical reaction
...
The mass of the system does not change
before and after a chemical reaction
...
Never change the subscript
...
The coefficient is a multiplier
...
Step 1
...
C8 H18 + O2 → CO2 + H2O
Reactant
Carbon
8
Hydrogen
18
Oxygen
2
Product
Carbon
1
Hydrogen
2
Oxygen
3
So, we multiply CO2 by placing the number 8 as a coefficient
...
The carbon atoms are balanced
...
There are 18
hydrogen atoms on the left side of the equation and only 2 on the right side
...
Dela Cruz
Stoichiometry Practice Exercises | Reymon T
...
C8 H18 + O2 → 8CO2 + 9H2O
Reactant
Carbon
8
Hydrogen
18
Oxygen
2
Product
Carbon
8
Hydrogen
18
Oxygen
25
Step 3
...
We now only need to balance
the number of Oxygen atoms
...
Also, since the Oxygen on the reactant is diatomic, we have to multiply it by
25
2
to equal the number of Oxygen atoms on the product side
...
We can see that the equation is already balanced
...
However, a proper, balanced chemical equation
should be expressed in small, whole number ratios
...
2(C8 H18 + 25
O2 → 8CO2 + 9H2O)
2
2C8 H18 + 25O2 → 16CO2 + 18H2O
Reactant
Carbon
16
Hydrogen
36
Oxygen
50
Product
Carbon
16
Hydrogen
36
Oxygen
50
If we look at the table, we see that we have equal number of atoms on both sides of the
equation and at the same time, the coefficients are expressed as small whole numbers
...
b
...
Determine the given quantity and the unknown quantity that we will solve for
...
Dela Cruz
Stoichiometry Practice Exercises | Reymon T
...
Convert the given quantity to moles using the substance’s molar mass
...
To simplify the solution, we will express the molar mass to the
nearest whole number
...
263 mol C8H18
Step 3
...
263 moles of C8 H18
...
(
...
104 mol CO2
Step 4
...
This means that we have to multiply the number
of moles by the molar mass to get our final answer
...
104 l CO2) (
44𝑔 𝐶𝑂2
1 𝑚𝑜𝑙𝑒 𝐶𝑂2
)=
92
...
What mass of Oxygen is needed to completely consume 100 grams of Octane?
Step 1
...
1 𝑚𝑜𝑙𝑒 𝐶8 𝐻18
(100g C8 H18) ((
) =
...
Use the balanced chemical equation to determine the number of moles of Oxygen needed
to completely consume
...
2C8H18 + 25O2 → 16CO2 + 18H2O
We can see from the balanced chemical equation that for every 2 moles of C8H18, we need 25
moles of O2
...
877 mol C8H18) (
25 𝑚𝑜𝑙 𝑂2
2 𝑚𝑜𝑙 𝐶8 𝐻18
) = 10
...
The question asks for the mass of Oxygen, not the number of moles
...
963 moles of O 2
(10
...
816g O2
4
Reymon T
...
Dela Cruz
Problem 2
...
The chemical reaction equation is shown below
...
After the experiment, the total mass of CaCl2 produced is 120g
...
Balance the chemical equation
...
Reactant
Product
Hydrogen
1
Hydrogen
2
Chlorine
1
Chlorine
2
Calcium
1
Calcium
1
Carbon
1
Carbon
1
Oxygen
3
Oxygen
3
We only need to balance the number of hydrogen and chlorine atoms so we put a
coefficient of 2 on HCl
...
Product
Hydrogen
Chlorine
Calcium
Carbon
Oxygen
2
2
1
1
3
2HCl + CaCO3 → CaCl2 + H2O + CO2
Step 2
...
(120g CaCl2) (
1𝑚𝑜𝑙𝑒 𝐶𝑎𝐶𝑙2
110𝑔 𝐶𝑎𝐶𝑙2
) = 1
...
Use the stoichiometric ratio based on the balanced chemical equation to solve for the
number of moles of CaCO3 that was used in the experiment
...
(1
...
091mol CaCO3
The question asks us for the number of moles of CaCO3 used in the experiment
...
091mol CaCO3
5
Reymon T
...
Dela Cruz
b
...
We know that 120 grams of CaCl2 was produced in the reaction which is equivalent to
1
...
We go straight to the stoichiometric ratio to determine the number of moles
of CO2 that should have been produced given the amount of CaCl2 produced
...
This means that 1
...
Step 2
...
(1
...
004g CO2
Problem 3
...
A byproduct of the process is calcium sulfate
...
Write the balanced chemical reaction equation for the production of Hydrogen
fluoride
...
We first write the chemical formula for the reactants and products
...
Balance the chemical equation
...
Hence, we place a coefficient
of 2 before HF
...
Dela Cruz
1
2
2
1
4
Stoichiometry Practice Exercises | Reymon T
...
b
...
32 kg of
Hydrogen fluoride?
Step 1
...
We will also convert
the given mass to grams since molar masses are given in grams
...
(1
...
We determine the number of moles of CaF2 needed to produce 66 mol of HF
...
(66 mol HF) (
1𝑚𝑜𝑙 𝐶𝑎𝐹2
2 𝑚𝑜𝑙 𝐻𝐹
) = 33 mol CaF2
Step 3
...
(33 mol CaF2) (
The answer is
78𝑔 𝐶𝑎𝐹2
1 𝑚𝑜𝑙 𝐶𝑎𝐹2
)(
1𝑘𝑔
1000𝑔
) = 2
...
574 kg CaF2
c
...
Convert the mass of CaF2 to number of moles
...
795mol CaF2
Step 2
...
795 moles of
CaF2
...
795 mol CaF2) (
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
1 𝑚𝑜𝑙 𝐶𝑎𝐹2
) = 1
...
The question asks for the amount in grams so we will use the molar mass of H2SO4 to
calculate the mass
...
795 mol H2SO4) (
Therefore, it takes
98𝑔 𝐻2 𝑆𝑂4
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
) = 175
...
91g H2SO4
to completely consume 140g of CaF2
...
Dela Cruz
Stoichiometry Practice Exercises | Reymon T
...
In a halogenation reaction, 57g of Methane is made to react with 128 g of Bromine gas
to produce Carbon tetrabromide and Hydrogen bromide
...
Write the balanced chemical equation for the reaction
...
Which of the two reactants, Methane or Bromine, is the limiting reactant?
c
...
Amount
Reactants
Products
Methane
Bromine
Carbon tetrabromide Hydrogen bromide
Prior to reaction
Reaction
After reaction
d
...
What is the percent yield of the reaction if 47g of Hydrogen bromide was produced?
a
...
Step 1
...
Reactants
Name
Methane
Bromine
Products
Name
Formula
Carbon tetrabromide
CBr4
Hydrogen bromide
HBr
Formula
CH4
Br2
Step 2
...
The table below shows the unbalanced chemical equation
and the distribution of atoms in the reactant and product sides of the equation
...
CH4 + Br2 → CBr4 + 4HBr
Reactant
Carbon
1
Hydrogen
4
Bromine
2
Product
Carbon
1
Hydrogen
4
Bromine
8
Next, we multiply Br2 by 4 to balance the number of Bromine atoms
...
Dela Cruz
Stoichiometry Practice Exercises | Reymon T
...
Which of the two reactants, Methane or Bromine, is the limiting reactant?
Step 1
...
First, we convert the given mass for
each reactant to moles
...
563 mol CH4
) =
...
Test which reactant will produce less product if completely consumed
...
(3
...
8mol Br2) (
4𝑚𝑜𝑙 𝐻𝐵𝑟
1𝑚𝑜𝑙 𝐶𝐻4
) = 14
...
8 mol HBr
The limiting reactant in this case is Br2 because it can only produce, at a maximum,
...
252 mol of HBr that can be produced if 57g of Methane is completely consumed
...
Complete the table below
...
List the amounts of reactants and products before the reaction
...
Since the reaction has not taken place yet, the amounts
of products are zero
...
563 mol
...
The limiting reactant will be fully consumed, so the change in Br2 is -
...
The maximum amount of Hydrogen bromide that can be produced is
...
We still need to calculate the amount of
Methane that will be consumed in the process as well as the maximum amount of Carbon
tetrabromide that can be produced
...
8 mol Br2) (
1𝑚𝑜𝑙 𝐶𝐻4
4 𝑚𝑜𝑙 𝐵𝑟2
) =
...
Dela Cruz
Stoichiometry Practice Exercises | Reymon T
...
8 mol Br2) (
) =
...
Fill out the Change row of the table ensuring that you assign a negative sign to the change
in the amounts of reactants
...
563 mol
...
2mol
-
...
2mol
...
Add the amounts prior to reaction and the change
...
Amount
Reactants
Methane
Bromine
3
...
8mol
-
...
8mol
3
...
2mol
...
2 mol
...
What is the theoretical yield of the reaction?
The theoretical yield of the reaction is the maximum amount of product that can be produced in a
reaction assuming that the limiting reactant is fully consumed
...
2mol and
...
e
...
Since the actual yield for Hydrogen bromide is expressed in grams, we have to express our
theoretical yield in grams
...
8mol HBr) (
81𝑔 𝐻𝐵𝑟
) = 64
...
Apply the percent yield formula
...
8𝑔 𝐻𝐵𝑟
) X100
Percent yield = 75
...
Dela Cruz
Title: Stoichiometry Practice Exercises with detailed solutions
Description: This material contains worked-out practice exercises on stoichiometry. The solutions are given in a stepwise manner and are easy to follow.
Description: This material contains worked-out practice exercises on stoichiometry. The solutions are given in a stepwise manner and are easy to follow.