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Chemical Reactions & Stoichiometry

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Introduction: The Language of Chemical Change

Chemistry is the science of transformation. Every time you cook an egg, a battery powers a device, or rust forms on iron, chemistry is happening—matter is rearranging at the atomic level. But these rearrangements follow strict mathematical rules. You can't create matter or destroy it; atoms are simply recombined into new substances.

In this chapter, you'll master the language of chemical reactions: how to write and balance equations, how to classify reactions, and most importantly, how to calculate exact quantities using stoichiometry. You'll learn why recipes work in specific proportions, how to identify the limiting reagent (the ingredient that runs out first), and how to predict yields with precision. These concepts are foundational not just for exams, but for understanding industrial chemistry and everyday life.

Part 1: Chemical Equations and Balancing

Understanding Chemical Equations

A chemical equation represents a reaction using chemical formulas and coefficients.

Unbalanced equation (skeleton): H₂ + O₂ → H₂O

This says: "Hydrogen gas reacts with oxygen gas to form water."

Problem: The atoms don't balance!

Left side: 2 H atoms + 2 O atoms
Right side: 2 H atoms + 1 O atom

Balanced equation: 2H₂ + O₂ → 2H₂O

Now atoms balance:

Left: 4 H atoms + 2 O atoms
Right: 4 H atoms + 2 O atoms

[Memory Hook] Balancing = ensuring equal numbers of each atom type on both sides

Method 1: Traditional Balancing (Hit and Trial)

Steps:

Write unbalanced skeleton equation
Count atoms of each element on both sides
Adjust coefficients (not subscripts!) until balanced
Check again

Example: Combustion of Methane (CH₄)

Skeleton: CH₄ + O₂ → CO₂ + H₂O

Attempt 1:
    C: 1 left, 1 right ✓
    H: 4 left, 2 right ✗ (need 2 H₂O)
    O: 2 left, 4 right ✗

Attempt 2: CH₄ + O₂ → CO₂ + 2H₂O
    C: 1 left, 1 right ✓
    H: 4 left, 4 right ✓
    O: 2 left, 4 right ✗ (need 2 O₂)

Attempt 3: CH₄ + 2O₂ → CO₂ + 2H₂O
    C: 1 left, 1 right ✓
    H: 4 left, 4 right ✓
    O: 4 left, 4 right ✓

Balanced!

Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O

Method 2: Algebraic Balancing

For complex equations, assign coefficients a, b, c, d, ... and solve simultaneously.

Example: Photosynthesis

Skeleton: CO₂ + H₂O → C₆H₁₂O₆ + O₂

Assign coefficients: aCO₂ + bH₂O → cC₆H₁₂O₆ + dO₂

Balance each element:

Carbon: a = 6c
Hydrogen: 2b = 12c
Oxygen: 2a + b = 6c + 2d

Set c = 1 (simplest):

a = 6
b = 6
12 + 6 = 6 + 2d → d = 6

Balanced: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

[Exam Trap] Never change subscripts to balance equations. Only adjust coefficients (numbers in front of formulas).

Part 2: Classification of Reactions

Type 1: Synthesis (Combination) Reactions

Definition: Two or more reactants combine to form a single product.

General form: A + B → AB

Examples:

2Na + Cl₂ → 2NaCl (metal + nonmetal)
2H₂ + O₂ → 2H₂O (nonmetals)
CaO + H₂O → Ca(OH)₂ (oxides)

Type 2: Decomposition Reactions

Definition: One reactant breaks into two or more products.

General form: AB → A + B

Examples:

2H₂O₂ → 2H₂O + O₂ (hydrogen peroxide decomposes when exposed to light or MnO₂ catalyst)
CaCO₃ → CaO + CO₂ (limestone decomposes when heated; used in industries)
2NaCl → 2Na + Cl₂ (electrolysis; produces sodium metal and chlorine gas)

Type 3: Combustion Reactions

Definition: A substance burns in oxygen, releasing energy (usually heat and light).

General form: CₓHᵧ + O₂ → CO₂ + H₂O (for hydrocarbons)

Examples:

CH₄ + 2O₂ → CO₂ + 2H₂O (methane burning)
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O (octane burning in gasoline engines)
S + O₂ → SO₂ (sulfur burning; produces acid rain precursor)

Always exothermic (releases energy). This is why combustion is used for heating and power generation.

[Memory Hook] Combustion = burning in oxygen; always exothermic; products are CO₂ and H₂O (for hydrocarbons)

Type 4: Redox (Oxidation-Reduction) Reactions

Definition: Reactions where electrons are transferred between atoms, changing their oxidation states.

Oxidation: Loss of electrons (or gain of oxygen, or loss of hydrogen) Reduction: Gain of electrons (or loss of oxygen, or gain of hydrogen)

Example: Rusting of Iron

4Fe + 3O₂ → 2Fe₂O₃

Fe: 0 → +3 (loses 3 electrons each; oxidized)
O: 0 → -2 (gains 2 electrons each; reduced)

Mnemonic: "OIL RIG" = Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons)

Type 5: Acid-Base Reactions

Definition: A reaction between an acid and a base producing a salt and water.

General form: H⁺ + OH⁻ → H₂O (ionic equation)

Example: HCl + NaOH → NaCl + H₂O

HCl (acid) provides H⁺
NaOH (base) provides OH⁻
Combine to form water; Na⁺ and Cl⁻ form salt

Type 6: Precipitation Reactions

Definition: Two solutions react to form an insoluble solid (precipitate).

Example: AgNO₃ + NaCl → AgCl↓ + NaNO₃

AgCl is insoluble; forms white precipitate
↓ symbol indicates precipitation

Common insoluble salts:

AgCl, AgBr, AgI (silver halides)
PbSO₄ (lead sulfate)
BaSO₄ (barium sulfate)
Mg(OH)₂, Al(OH)₃ (metal hydroxides)

Type 7: Displacement Reactions

Definition: A more reactive element displaces a less reactive element from a compound.

Single displacement: AB + C → AC + B (if C is more reactive than A)

Example: Zn + CuSO₄ → ZnSO₄ + Cu

Zinc (more reactive) displaces copper (less reactive) from copper sulfate
Zinc sulfate forms (soluble), copper metal precipitates

Double displacement: AB + CD → AD + CB

Example: AgNO₃ + NaCl → AgCl↓ + NaNO₃

Part 3: The Mole Concept

Avogadro's Number

Definition: The number of atoms/molecules in one mole of a substance.

Value: Nₐ = 6.022 × 10²³ (atoms/molecules per mole)

[Memory Hook] Avogadro's number = 6.022 × 10²³; represents huge quantities at atomic scale

Why this number? It's defined such that 12 grams of pure carbon-12 contains exactly one mole (6.022 × 10²³ atoms).

Molar Mass

Definition: Mass of one mole of a substance (grams per mole, g/mol).

Calculation: Add atomic masses of all atoms in the formula.

Example: Glucose (C₆H₁₂O₆)

C: 6 × 12 = 72 g/mol
H: 12 × 1 = 12 g/mol
O: 6 × 16 = 96 g/mol
Total: 72 + 12 + 96 = 180 g/mol

So one mole of glucose = 180 grams = 6.022 × 10²³ molecules

Mole Conversions

The central relationship:

Moles ←→ Grams ←→ Particles (atoms/molecules)

        molar mass
Moles ←-------→ Grams
        (g/mol)

                Avogadro's number
Moles ←-------------------------→ Particles
        (6.022 × 10²³)

Example: How many grams in 2.5 moles of NaCl?

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = moles × molar mass = 2.5 mol × 58.5 g/mol = 146.25 g

Example: How many molecules in 18 grams of water (H₂O)?

Molar mass of H₂O = 2(1) + 16 = 18 g/mol
Moles = mass / molar mass = 18 g / 18 g/mol = 1 mol
Molecules = moles × Nₐ = 1 × 6.022 × 10²³ = 6.022 × 10²³ molecules

[Memory Hook] Mole = bridge between macroscopic (grams) and atomic (molecules) worlds

Part 4: Stoichiometry

Mole Ratios from Balanced Equations

A balanced equation provides mole ratios between reactants and products.

Example: Combustion of methane CH₄ + 2O₂ → CO₂ + 2H₂O

Mole ratios:

1 mole CH₄ : 2 moles O₂
1 mole CH₄ : 1 mole CO₂
2 moles O₂ : 1 mole CO₂
1 mole CO₂ : 2 moles H₂O

Stoichiometric Calculations

Goal: Given quantity of one substance, find quantity of another.

Method:

Write balanced equation
Convert given quantity to moles
Use mole ratio from equation
Convert result to desired units

Example: How much CO₂ is produced from 64g of CH₄?

Step 1: Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O

Step 2: Convert 64g CH₄ to moles

Molar mass CH₄ = 12 + 4 = 16 g/mol
Moles = 64 g / 16 g/mol = 4 moles

Step 3: Use mole ratio

From equation: 1 mole CH₄ produces 1 mole CO₂
4 moles CH₄ produces 4 moles CO₂

Step 4: Convert to grams

Molar mass CO₂ = 12 + 2(16) = 44 g/mol
Mass = 4 mol × 44 g/mol = 176 g

Answer: 176 g CO₂ is produced

Limiting Reagent

Definition: The reactant that is completely consumed, limiting the amount of product formed.

Method to find limiting reagent:

Calculate moles of each reactant
Divide by stoichiometric coefficient from equation
The reactant with smallest ratio is limiting

Example: Reaction of 56g Fe and 32g O₂

Balanced: 4Fe + 3O₂ → 2Fe₂O₃

Step 1: Calculate moles

Fe: 56 g / 56 g/mol = 1 mole
O₂: 32 g / 32 g/mol = 1 mole

Step 2: Divide by coefficients

Fe: 1 / 4 = 0.25
O₂: 1 / 3 = 0.33

Step 3: O₂ has smallest ratio (0.25 < 0.33), so O₂ is limiting

Answer: O₂ is the limiting reagent; Fe is in excess

[Memory Hook] Limiting reagent = runs out first; determines how much product forms

[Exam Trap] Students often confuse "smallest coefficient" with "limiting reagent." The limiting reagent is determined by mole ratio divided by stoichiometric coefficient, not by the coefficient itself.

Percent Yield

Definition: Actual yield (experimental) as percentage of theoretical yield (calculated).

Formula: Percent Yield = (Actual Yield / Theoretical Yield) × 100%

Example:

Theoretical yield from stoichiometry: 100 g product
Actual yield from lab experiment: 75 g
Percent yield = (75 / 100) × 100% = 75%

Why < 100%?

Side reactions (unwanted products form)
Incomplete reactions (equilibrium prevents 100% completion)
Losses during transfer (product stuck to containers)
Evaporation or decomposition

Part 5: Real-World Stoichiometry Applications

Industrial Examples

Haber Process (ammonia synthesis): N₂ + 3H₂ ⇌ 2NH₃

This reaction feeds the world through fertilizers. Exact stoichiometry is critical: if you don't have the right ratio of N₂ to H₂, one reactant is wasted.

Steel Production: 4Fe + 3O₂ → 2Fe₂O₃ (oxidation) Fe₂O₃ + 3CO → 2Fe + 3CO₂ (reduction)

Iron ore (Fe₂O₃) is reduced by carbon monoxide to produce pure iron. Stoichiometry determines how much CO is needed and how much iron is produced.

Combustion in Engines: C₈H₁₈ + 12.5O₂ → 8CO₂ + 9H₂O (gasoline combustion)

Complete combustion requires a precise air-to-fuel ratio. Too much fuel (rich mixture) produces CO and smoke; too little (lean mixture) reduces power.

Percent Composition

Definition: Percentage by mass of each element in a compound.

Formula: % of element = (mass of element in formula / molar mass of compound) × 100%

Example: Find percent composition of CO₂

C: (12 / 44) × 100% = 27.3%
O: (32 / 44) × 100% = 72.7%

Check: 27.3% + 72.7% = 100% ✓

Conclusion

Stoichiometry is the mathematics of chemistry. It transforms abstract equations into concrete calculations: how many grams of product, how much reactant is needed, which ingredient runs out first. Master this chapter, and you'll navigate both exam problems and real-world applications with confidence.

23 MCQ Questions

Q1: In a balanced chemical equation, which of the following must be equal on both sides?

A) Coefficients of reactants and products
B) Number of molecules of reactants and products
C) Number of atoms of each element
D) Masses of reactants and products

Q2: When balancing a chemical equation, which of the following should NOT be changed?

A) Coefficients
B) Subscripts in chemical formulas
C) Both coefficients and subscripts
D) The arrow direction

Q3: In the combustion of methane (CH₄ + 2O₂ → CO₂ + 2H₂O), what type of reaction is this?

A) Decomposition
B) Combustion
C) Precipitation
D) Acid-base

Q4: In a redox reaction, oxidation is defined as:

A) Gain of oxygen
B) Gain of electrons
C) Loss of electrons
D) Loss of oxygen

Q5: In the reaction 2H₂O₂ → 2H₂O + O₂, what type of reaction is this?

A) Synthesis
B) Combustion
C) Decomposition
D) Displacement

Q6: Avogadro's number (6.022 × 10²³) represents:

A) The number of atoms in 1 kilogram of any element
B) The number of atoms/molecules in one mole
C) The atomic mass of carbon-12
D) The density of a gas at STP

Q7: What is the molar mass of calcium carbonate (CaCO₃)?

A) 60 g/mol
B) 80 g/mol
C) 100 g/mol
D) 120 g/mol

Q8: How many moles are present in 36 grams of water (H₂O)? (Molar mass = 18 g/mol)

A) 1 mole
B) 2 moles
C) 3 moles
D) 4 moles

Q9: In the reaction 2Na + Cl₂ → 2NaCl, how many moles of NaCl are produced from 2 moles of Na?

A) 1 mole
B) 2 moles
C) 3 moles
D) 4 moles

Q10: The limiting reagent in a reaction is:

A) The reactant in largest amount
B) The reactant that is completely consumed
C) The reactant with highest molar mass
D) The reactant with lowest coefficient

Q11: In the reaction 4Fe + 3O₂ → 2Fe₂O₃, if 8 moles of Fe and 5 moles of O₂ are available, which is the limiting reagent?

A) Fe
B) O₂
C) Both are equal
D) Cannot be determined

Q12: Percent yield is calculated as:

A) (Theoretical yield / Actual yield) × 100%
B) (Actual yield / Theoretical yield) × 100%
C) (Reactants / Products) × 100%
D) (Moles reacted / Total moles) × 100%

Q13: A reaction has a theoretical yield of 50g but an actual yield of 40g. What is the percent yield?

A) 80%
B) 90%
C) 75%
D) 100%

Q14: In an acid-base neutralization reaction, what is always produced?

A) A salt only
B) Water and a gas
C) A salt and water
D) A base and an acid

Q15: In a precipitation reaction, AgNO₃ + NaCl → AgCl↓ + NaNO₃, the arrow with a downward sign (↓) indicates:

A) A gas is released
B) A solid precipitate forms
C) Heat is released
D) The reaction is reversible

Q16: How many molecules are in 2 moles of oxygen gas (O₂)?

A) 2 × 6.022 × 10²³
B) 2 × 6.022 × 10²³
C) 1.2044 × 10²⁴
D) All of the above

Q17: In the reaction C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, the mole ratio of glucose to oxygen is:

A) 1:1
B) 1:6
C) 6:1
D) 6:6

Q18: Stoichiometry is best described as:

A) The study of element properties
B) The calculation of quantities in chemical reactions
C) The balancing of equations only
D) The measurement of reaction rates

Q19: In a synthesis reaction:

A) One reactant breaks into multiple products
B) Multiple reactants combine into one product
C) A more reactive element displaces another
D) Acid and base neutralize each other

Q20: The percent composition of oxygen in CO₂ is approximately:

A) 27%
B) 45%
C) 73%
D) 100%

Q21: In the combustion reaction 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O, if 3 moles of octane react, how many moles of CO₂ are produced?

A) 8 moles
B) 16 moles
C) 24 moles
D) 48 moles

Q22: When 64g of O₂ reacts according to 2H₂ + O₂ → 2H₂O, how many grams of H₂O are produced?

A) 36g
B) 72g
C) 144g
D) 180g

Q23: In an incomplete combustion of hydrocarbons, the product is:

A) Only CO₂ and H₂O
B) CO, CO₂, H₂O, and possibly soot
C) Only CO
D) Only soot

Answer Key: 1-C, 2-B, 3-B, 4-C, 5-C, 6-B, 7-C, 8-B, 9-B, 10-B, 11-B, 12-B, 13-A, 14-C, 15-B, 16-D, 17-B, 18-B, 19-B, 20-C, 21-C, 22-B, 23-B

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