Stoichiometry is the branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction.
That is, the relationship between the number of moles of reactants and products in a chemical reaction
In simple terms, stoichiometry helps us to calculate:
How much reactant is needed
How much product will be formed during a chemical reaction
Stoichiometry is based on the law of conservation of mass, which states that matter can neither be created nor destroyed in a chemical reaction.
Importance of Stoichiometry
Stoichiometry is used to:
Calculate masses of reactants and products
Determine the amount of substances in reactions
Predict product yield
Find limiting and excess reactants
Design industrial chemical processes
Basic Terms in Stoichiometry
1. Mole
A mole is the amount of substance that contains
6.02 × 10²³ particles (Avogadro’s number).
The mole is also the unit of measurement in chemistry.
2. Molar Mass
The molar mass is the mass of one mole of a substance in grams (g/mol).
Example:
Molar mass of H₂O = 2(1) + 16 = 18 g/mol
3. Chemical Equation
A chemical equation shows the relationship between reactants and products.
Example:
2H2 + O2 → 2H2O
2mols 1mol 2mol
or 2g (2x16)g (2x18)g
or 2g 32g 22.4dm3
This means:
2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.
Types of Stoichiometric Calculations
1. Mole-to-Mole Calculations
This involves using the ratio of moles in a balanced equation. That is, mole-mole relationship
Example 1:
How many moles of oxygen are needed to react with 4 moles of hydrogen?
2H2 + O2 → 2H2O
From equation:
2 moles H₂ react with 1 mole O₂
So, 4 moles H₂ will need:
1/2 x 4 = 2 moles of O2
2. Mass-to-Mole Calculations
Example 2:
What is the number of moles in 44 g of CO₂?
Molar mass of CO₂ = 12 + 2(16) = 44 g/mol
Moles = Mass =
Molar mass
44 = 1mole
44
3. Mass-to-Mass Calculations
Example 3:
What mass of CO₂ is produced when 10 g of CaCO₃ decomposes?
Equation:
CaCO3 →CaO + CO2
Step 1: Molar masses
CaCO₃ = 100 g/mol
CO₂ = 44 g/mol
From equation:
100 g CaCO₃ → 44 g CO₂
So,
10 g CaCO₃ →?
10 = 4.4 g of CO2
100
4. Volume-to-Volume (Gaseous Reactions)
At the same temperature and pressure, equal volumes of gases contain equal number of molecules.
Example 4:
What volume of oxygen is needed to react with 40 cm³ of hydrogen?
2H2 + O2 →2H2O
2 volumes H₂ react with 1 volume O₂
So,
40 cm³ H₂ will need:
1/2x 40 = 20 cm3 of O2
5. Limiting Reactant Calculations
The limiting reactant is the reactant that is completely used up first and stops the reaction.
Example 5:
If 2 g of hydrogen reacts with 16 g of oxygen, which is limiting?
2H2 + O2 →2H2O
Moles:
H₂ = 2 ÷ 2 = 1 mole
O₂ = 16 ÷ 32 = 0.5 mole
Required ratio:
2H₂ : 1O₂
Actual ratio:
1H₂ : 0.5 O₂ → correct ratio
So, no reactant is in excess — both are completely used up.
Percentage Yield
Not all reactions give maximum product.
Formula:
Percentage Yield = Actual Yield x 100
Theoretical Yield
Example 6:
If theoretical yield = 10 g and actual yield = 8 g
8 = 100 = 80%
10
Summary
Stoichiometry helps chemists:
Predict quantities in reactions
Save materials
Improve industrial efficiency
Avoid wastage
OBJECTIVE QUESTIONS (WAEC/NECO)
1. Stoichiometry deals with the
A. speed of reactions
B. colour of substances
C. quantitative relationship between reactants and products
D. energy changes in reactions
2. The number of particles in one mole of a substance is
A. 3.01 × 10²³
B. 6.02 × 10²³
C. 1.00 × 10²³
D. 12.00 × 10²³
3. The molar mass of CO₂ is
A. 12 g/mol
B. 16 g/mol
C. 28 g/mol
D. 44 g/mol
4. How many moles are present in 18 g of water?
A. 0.5
B. 1
C. 2
D. 18
5. In the equation
2H₂ + O₂ → 2H₂O
the mole ratio of H₂ to O₂ is
A. 1:1
B. 1:2
C. 2:1
D. 2:2
6. What mass of NaCl contains 1 mole of NaCl?
A. 23 g
B. 35.5 g
C. 58.5 g
D. 46 g
7. At the same temperature and pressure, equal volumes of gases contain
A. equal masses
B. equal densities
C. equal number of molecules
D. equal pressures
8. Which of the following is the limiting reactant?
A. The reactant in excess
B. The reactant completely used up
C. The product formed
D. The catalyst
9. The formula for calculating percentage yield is
A. Actual × Theoretical
B. Actual ÷ Theoretical × 100
C. Theoretical ÷ Actual × 100
D. Actual − Theoretical
10. How many moles are in 44 g of CO₂?
A. 0.5
B. 1
C. 2
D. 44
11. What volume of oxygen is required to react with 40 cm³ of hydrogen?
(2H₂ + O₂ → 2H₂O)
A. 10 cm³
B. 20 cm³
C. 30 cm³
D. 40 cm³
12. Which law is the basis of stoichiometry?
A. Law of definite proportion
B. Law of multiple proportions
C. Law of conservation of mass
D. Law of gaseous volumes
13. The molar mass of CaCO₃ is
A. 40
B. 56
C. 84
D. 100
14. How many moles are present in 32 g of O₂?
A. 0.5
B. 1
C. 2
D. 16
15. Which of the following is NOT used in stoichiometric calculations?
A. Balanced equation
B. Molar mass
C. Temperature only
D. Mole ratio
THEORY QUESTIONS (WAEC/NECO)
Short Answer Questions
1. Define stoichiometry.
2. What is a mole?
3. State Avogadro’s number.
4. Define molar mass.
5. What is a limiting reactant?
6. Calculate the number of moles in 22 g of CO₂.
7. What mass of CO₂ is produced when 50 g of CaCO₃ decomposes? according to the equation
CaCO₃ → CaO + CO₂
8. How many moles of oxygen are needed to react completely with 6 moles of hydrogen? given the equation below
2H₂ + O₂ → 2H₂O
9. Explain stoichiometry and state three of its applications.
10. Describe how to calculate the mass of a product formed from a given mass of reactant using a balanced chemical equation.
11. In a reaction, 10 g of calcium carbonate was heated and produced 3.5 g of carbon dioxide.
(a) Calculate the theoretical yield
(b) Calculate the percentage yield
CaCO₃ → CaO + CO₂
12. Explain the term “limiting reactant” and show with an example.
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