UNIT 4

Chemical Reactions

Learn how to represent chemical changes, identify reaction types, calculate reacting quantities, and explain reactions using particles, ions, and electrons.

Topics

Representing Chemical Reactions

A chemical reaction rearranges atoms to form different substances. Chemical equations represent the identities and relative amounts of the reactants and products.

Chemical reactions rearrange atoms, but they do not create or destroy atoms. A balanced equation must contain the same number of each type of atom on both sides.

Parts of a Chemical Equation

Reactants → Products

Reactants are the substances present before the reaction. Products are the substances formed by the reaction. An arrow indicates the direction of the chemical change.

State Symbol Meaning
(s) Solid
(l) Liquid
(g) Gas
(aq) Dissolved in water

Coefficients and Subscripts

A coefficient is written before a chemical formula and indicates the relative number of particles or moles. A subscript is part of the chemical formula and indicates the ratio of atoms within a particle.

Change coefficients when balancing an equation. Never change a chemical formula’s subscripts because doing so changes the identity of the substance.

Example: Coefficient Versus Subscript

2H2O

The coefficient 2 represents two water molecules. Each molecule contains two hydrogen atoms and one oxygen atom.

Total: 4 H atoms and 2 O atoms

Steps for Balancing an Equation

  1. Write the correct chemical formulas for all reactants and products.
  2. Count the number of atoms of each element on both sides.
  3. Add coefficients to make the number of each type of atom equal.
  4. Reduce the coefficients to the smallest whole-number ratio when possible.
  5. Check the final atom counts and, for ionic equations, the total charge on both sides.

Example: Formation of Water

The unbalanced equation is:

H2 + O2 → H2O

Place a coefficient of 2 before H₂O to balance oxygen. Then place a coefficient of 2 before H₂ to balance hydrogen.

2H2(g) + O2(g) → 2H2O(g)
Element Reactant Side Product Side
Hydrogen 4 atoms 4 atoms
Oxygen 2 atoms 2 atoms
Particle diagram showing two hydrogen molecules reacting with one oxygen molecule to produce two water molecules
The coefficients in the balanced equation match the relative numbers of particles in the diagram. Every atom present before the reaction remains present after the reaction.

Physical and Chemical Changes

A physical change alters a substance’s state, shape, or arrangement without changing the identities of its particles. A chemical change rearranges atoms and forms substances with different chemical identities.

Physical Change Chemical Change
Melting or freezing Formation of a precipitate
Boiling or condensation Formation of a gas through reaction
Cutting or changing shape Formation of substances with new chemical identities
Dissolving without reaction Transfer of electrons or rearrangement of bonds

Evidence of a Chemical Reaction

Observations that may provide evidence of a chemical reaction include:

An observation alone does not always prove that a chemical reaction occurred. The strongest evidence shows that substances with new chemical identities were formed.

Representing Reactions Practice

Try each question before opening its solution.

Question 1: Balancing an Equation

Balance the following equation using the smallest whole-number coefficients:

Al + O2 → Al2O3
Show solution

Begin by balancing oxygen. The least common multiple of 2 and 3 is 6, so use 3O₂ and 2Al₂O₃.

Al + 3O2 → 2Al2O3

The product side now contains four aluminum atoms, so place a coefficient of 4 before aluminum.

4Al + 3O2 → 2Al2O3

Question 2: Interpreting Coefficients

Consider the balanced equation:

N2 + 3H2 → 2NH3

If two N₂ molecules react with six H₂ molecules, how many NH₃ molecules can form?

Show solution

The amounts given are twice the coefficients in the balanced equation:

2N2 + 6H2 → 4NH3

Therefore, four NH₃ molecules can form.

Question 3: Checking Atom Conservation

A student proposes the equation:

CH4 + O2 → CO2 + H2O

Is the equation balanced? If not, provide the balanced equation.

Show solution

The original equation is not balanced. It contains four hydrogen atoms on the reactant side but only two on the product side.

Place a coefficient of 2 before water, then balance oxygen with a coefficient of 2 before O₂:

CH4 + 2O2 → CO2 + 2H2O

Question 4: Physical or Chemical Change?

A sample of solid water melts into liquid water. Is this a physical or chemical change? Explain using particle identity.

Show solution

Melting is a physical change. The arrangement and movement of the water molecules change, but every particle remains an H₂O molecule.

No substance with a new chemical identity is formed.

Question 5: Evaluating Evidence

Bubbles appear when a liquid is heated. Does this observation alone prove that a chemical reaction occurred?

Show solution

No. The bubbles could result from a physical change, such as the liquid boiling and entering the gas phase.

More evidence would be needed to show that a gas with a new chemical identity was produced.

VIDEO EXPLANATION

Watch: Representing Chemical Reactions

Watch an explanation of balancing equations, coefficients, particle representations, and physical versus chemical changes.

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Net Ionic Equations

A net ionic equation shows only the particles that undergo a chemical change during an aqueous reaction. Ions that remain unchanged are called spectator ions and are omitted.

Strong Electrolytes in Ionic Equations

Soluble ionic compounds and strong acids or bases are represented as separated ions when they are dissolved in water.

NaCl(aq) → Na+(aq) + Cl(aq)

Substances that are solids, liquids, gases, weak electrolytes, or insoluble compounds are generally kept together in ionic equations.

Separate a substance into ions only when it is present as a strong electrolyte in aqueous solution. Do not separate a precipitate, pure liquid, gas, or weak electrolyte.

Spectator Ions

Spectator ions appear in the same form on both sides of a complete ionic equation. They remain dissolved and do not undergo the chemical change represented by the net ionic equation.

Steps for Writing a Net Ionic Equation

  1. Write and balance the molecular equation.
  2. Separate strong aqueous electrolytes into their individual ions.
  3. Keep solids, liquids, gases, and weak electrolytes together.
  4. Cancel spectator ions that appear unchanged on both sides.
  5. Confirm that both atoms and total charge are balanced in the final net ionic equation.

Example: Formation of AgCl

Aqueous silver nitrate and sodium chloride react to form solid silver chloride.

Molecular Equation

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Complete Ionic Equation

Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)



AgCl(s) + Na+(aq) + NO3(aq)

Na+ and NO3 appear unchanged on both sides, so they are spectator ions.

Net Ionic Equation

Ag+(aq) + Cl(aq) → AgCl(s)
Particle diagram showing silver and chloride ions forming a solid precipitate while sodium and nitrate remain dissolved
Ag⁺ and Cl⁻ ions form solid AgCl. Na⁺ and NO₃⁻ remain separated in solution and act as spectator ions.

Checking a Net Ionic Equation

A correct net ionic equation must conserve both atoms and electrical charge. For the formation of AgCl, the total charge on the reactant side is zero, matching the neutral solid product.

(+1) + (−1) = 0

Helpful Solubility Patterns

Solubility patterns help predict whether ions remain separated in aqueous solution or form an insoluble precipitate. These patterns include exceptions, so they should be applied carefully.

Ion or Compound Type General Solubility Pattern
Group 1 metal ions Compounds are soluble
NH4+ Compounds are soluble
NO3 Compounds are soluble
Cl, Br, and I Usually soluble; important exceptions include compounds containing Ag+, Pb2+, or Hg22+
SO42− Usually soluble; important exceptions include compounds containing Ba2+, Sr2+, or Pb2+
CO32− and PO43− Usually insoluble except with Group 1 ions or NH4+
OH Usually insoluble, with important exceptions involving Group 1 ions and some larger Group 2 ions
Soluble strong electrolytes are separated into ions in a complete ionic equation. An insoluble product is written together as a solid.

Strong Acid–Strong Base Reactions

Strong acids and strong bases are represented as separated ions in aqueous solution. Their neutralization reaction commonly reduces to the formation of liquid water.

Example: HCl and NaOH

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Complete ionic equation:

H+(aq) + Cl(aq) + Na+(aq) + OH(aq)



Na+(aq) + Cl(aq) + H2O(l)

Na+ and Cl are spectator ions.

Net ionic equation:

H+(aq) + OH(aq) → H2O(l)

Weak Electrolytes in Net Ionic Equations

Weak acids and weak bases are not separated completely into ions because most of their particles remain in molecular form.

Example: HF and OH

HF is a weak acid, so it remains together on the reactant side.

HF(aq) + OH(aq) → F(aq) + H2O(l)

Net Ionic Equation Practice

Try each question before opening its solution.

Question 1: Identifying Spectator Ions

Consider the reaction:

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

Identify the spectator ions and write the net ionic equation.

Show solution

Na+ and Cl remain aqueous and unchanged, so they are spectator ions.

Ba2+(aq) + SO42−(aq) → BaSO4(s)

Question 2: Strong Acid and Strong Base

Write the net ionic equation for the reaction between aqueous HNO3 and aqueous KOH.

Show solution

HNO₃ and KOH are strong electrolytes. K+ and NO3 are spectator ions.

H+(aq) + OH(aq) → H2O(l)

Question 3: Weak Acid Reaction

A student writes HF as H+ and F in a net ionic equation. Explain why this representation is incorrect.

Show solution

HF is a weak acid and does not ionize completely in water. Most HF particles remain together, so HF should be written in molecular form in the net ionic equation.

Question 4: Predicting a Precipitate

A solution containing Ag+ is mixed with a solution containing Br. Predict whether a precipitate forms and write the net ionic equation.

Show solution

AgBr is insoluble, so a solid precipitate forms.

Ag+(aq) + Br(aq) → AgBr(s)

Question 5: Checking Charge Conservation

Show that charge is conserved in:

Ba2+(aq) + SO42−(aq) → BaSO4(s)
Show solution

The total reactant charge is:

(+2) + (−2) = 0

BaSO₄ is a neutral solid, so the product side also has a total charge of zero.

VIDEO EXPLANATION

Watch: Net Ionic Equations

Watch an explanation of complete ionic equations, spectator ions, precipitation reactions, and acid–base neutralization.

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Stoichiometry and Titration

Stoichiometry uses a balanced chemical equation to calculate the relative amounts of reactants consumed and products formed during a reaction.

Mole Ratios

The coefficients in a balanced equation provide mole ratios between every reactant and product.

N2 + 3H2 → 2NH3

This equation gives several possible mole ratios:

1 mol N2 : 3 mol H2 : 2 mol NH3
Mole ratios come from the coefficients—not the subscripts—of the balanced chemical equation.

Steps for a Stoichiometry Calculation

  1. Write and balance the chemical equation.
  2. Convert the given quantity to moles if it is not already expressed in moles.
  3. Use the coefficients to convert from moles of the given substance to moles of the desired substance.
  4. Convert the resulting moles into the requested unit, such as grams, particles, volume, or concentration.

Example: Mole-to-Mole Stoichiometry

Nitrogen is present in excess. How many moles of NH₃ can form from 4.50 mol of H₂?

4.50 mol H2 × 2 mol NH3 3 mol H2 = 3.00 mol NH3

Limiting Reactants

The limiting reactant is consumed first and determines the maximum amount of product that can form. Any reactant remaining after the limiting reactant is consumed is an excess reactant.

To identify the limiting reactant, calculate how much product each reactant could form. The reactant that produces less product is limiting.

Example: Identifying the Limiting Reactant

2H2 + O2 → 2H2O

A mixture contains 5.00 mol H₂ and 2.00 mol O₂. Determine the limiting reactant.

Product possible from H₂:

5.00 mol H2 × 2 mol H2O 2 mol H2 = 5.00 mol H2O

Product possible from O₂:

2.00 mol O2 × 2 mol H2O 1 mol O2 = 4.00 mol H2O

O₂ produces less water, so O₂ is the limiting reactant. A maximum of 4.00 mol H₂O can form.

Theoretical Yield and Percent Yield

The theoretical yield is the maximum amount of product predicted from the limiting reactant. The actual yield is the amount of product obtained experimentally.

Percent yield = actual yield theoretical yield × 100%

Titration

A titration uses a solution with a known concentration to determine the concentration of another solution. The titrant is gradually delivered to the analyte until the reaction reaches its equivalence point.

At the equivalence point, the reactants have been combined in the stoichiometric ratio given by the balanced equation. Their moles are equal only when the coefficients form a 1:1 ratio.
Diagram of a burette delivering titrant into an Erlenmeyer flask containing the analyte
The burette measures the volume of titrant delivered to the analyte in the flask.

Example: Acid–Base Titration

A 25.00 mL sample of HCl is neutralized by 20.00 mL of 0.1500 M NaOH.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

First, calculate the moles of NaOH delivered:

0.1500 mol/L × 0.02000 L = 0.003000 mol NaOH

The equation has a 1:1 ratio, so the original sample contained 0.003000 mol HCl.

MHCl = 0.003000 mol 0.02500 L = 0.1200 M

Practice Questions

Question 1: Mass Stoichiometry

Nitrogen and hydrogen react according to the following balanced equation:

N2 + 3H2 → 2NH3

What mass of NH3 can be produced from 14.0 g of N2 if H2 is available in excess?

Show solution

First, convert the mass of nitrogen into moles. The molar mass of N2 is 28.02 g/mol.

14.0 g N2 × 1 mol N2 28.02 g N2 = 0.500 mol N2

Use the coefficients in the balanced equation to convert moles of N2 into moles of NH3.

0.500 mol N2 × 2 mol NH3 1 mol N2 = 1.00 mol NH3

Finally, convert moles of NH3 into grams. Its molar mass is 17.03 g/mol.

1.00 mol NH3 × 17.03 g NH3 1 mol NH3 = 17.0 g NH3

Therefore, approximately 17.0 g of NH3 can be produced.

Question 2: Limiting Reactant

Carbon monoxide reacts with oxygen according to the following equation:

2CO + O2 → 2CO2

If 3.00 mol of CO reacts with 2.00 mol of O2, identify the limiting reactant and determine the number of moles of CO2 produced.

Show solution

Calculate how much CO2 each reactant could produce.

3.00 mol CO × 2 mol CO2 2 mol CO = 3.00 mol CO2
2.00 mol O2 × 2 mol CO2 1 mol O2 = 4.00 mol CO2

CO produces the smaller amount of product, so CO is the limiting reactant. The reaction produces 3.00 mol of CO2.

Question 3: Percent Yield

A reaction has a theoretical yield of 12.5 g, but an experiment produces only 10.0 g of product. Calculate the percent yield.

Show solution
Percent yield = Actual yield Theoretical yield × 100
Percent yield = 10.0 g 12.5 g × 100 = 80.0%

The percent yield is 80.0%.

Question 4: Titration with a Two-to-One Ratio

A 25.00 mL sample of H2SO4 is neutralized by 30.00 mL of 0.2000 M NaOH.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

What is the concentration of the H2SO4 solution?

Show solution

First, convert the NaOH volume to liters and calculate its number of moles.

0.2000 mol/L × 0.03000 L = 0.006000 mol NaOH

The balanced equation shows that two moles of NaOH react with one mole of H2SO4.

0.006000 mol NaOH × 1 mol H2SO4 2 mol NaOH = 0.003000 mol H2SO4

Divide the acid’s moles by its volume in liters.

M = 0.003000 mol 0.02500 L = 0.1200 M

The concentration of the acid is 0.1200 M H2SO4.

Question 5: Titration Error Analysis

During a titration, a student accidentally adds more titrant after reaching the endpoint. The student uses this larger volume to calculate the concentration of the unknown solution.

Will the calculated concentration of the unknown be too high or too low? Explain.

Show solution

The calculated concentration will be too high.

The recorded titrant volume is greater than the volume actually needed to reach the endpoint. This makes the calculated number of moles of titrant—and therefore the calculated number of moles of the unknown—too large.

VIDEO EXPLANATION

Watch: Stoichiometry and Titration

Watch a worked example involving mole ratios, limiting reactants, or titration calculations.

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Reaction Types

Chemical reactions can be classified by examining which substances react, which products form, and whether electrons are transferred. Recognizing a reaction type can help predict its products and determine how it should be analyzed.

Important: A reaction may fit more than one category. For example, a combustion reaction is also an oxidation–reduction reaction because electrons are transferred.

Precipitation Reactions

A precipitation reaction occurs when two aqueous ionic solutions combine and produce an insoluble ionic solid. The solid that forms is called a precipitate.

Example: Formation of Silver Chloride

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl is insoluble in water, so it forms a solid precipitate. Na+ and NO3 remain dissolved and act as spectator ions.

Ag+(aq) + Cl(aq) → AgCl(s)

Acid–Base Reactions

In a Brønsted–Lowry acid–base reaction, an acid donates a proton, H+, while a base accepts a proton.

Brønsted–Lowry definitions:
An acid is a proton donor.
A base is a proton acceptor.

When a strong acid reacts with a strong base, the important reaction is usually the formation of water from hydronium ions and hydroxide ions.

Example: Strong Acid–Strong Base Reaction

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Because HCl and NaOH are strong electrolytes, they separate into ions in water. Na+ and Cl are spectator ions.

H+(aq) + OH(aq) → H2O(l)

Weak acids and weak bases should generally remain together when writing net ionic equations because they do not ionize completely in water.

Example: Weak Acid Reacting with a Base

CH3COOH(aq) + OH(aq) → CH3COO(aq) + H2O(l)

Acetic acid, CH3COOH, donates a proton to the hydroxide ion. Because acetic acid is weak, it is written as a complete molecule rather than separated into ions.

Combustion Reactions

A combustion reaction occurs when a substance reacts with oxygen. When a hydrocarbon undergoes complete combustion, the products are carbon dioxide and water.

Example: Combustion of Methane

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

The carbon atoms in methane become part of carbon dioxide, while the hydrogen atoms become part of water.

Combustion pattern:

Hydrocarbon + O2 → CO2 + H2O

Oxidation–Reduction Reactions

An oxidation–reduction reaction, often called a redox reaction, involves the transfer of electrons. Oxidation and reduction always occur together.

Oxidation: loss of electrons and an increase in oxidation number.

Reduction: gain of electrons and a decrease in oxidation number.

The phrase “oxidation is loss, reduction is gain” can help you remember how electrons move.

Assigning Oxidation Numbers

Oxidation numbers are assigned values used to track how electrons are distributed during a reaction.

Rule Oxidation Number
An element by itself 0
A monatomic ion Equal to the ion’s charge
Oxygen in most compounds −2
Hydrogen in most compounds +1
Group 1 metals in compounds +1
Group 2 metals in compounds +2
Sum in a neutral compound 0
Sum in a polyatomic ion Equal to the ion’s charge

These are the most commonly used oxidation-number rules. Some elements have exceptions, which can be considered when the chemical formula requires them.

Example: Identifying Oxidation and Reduction

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Zn begins with an oxidation number of 0 and changes to +2. Its oxidation number increases, so Zn is oxidized.

Zn → Zn2+ + 2e

Cu begins at +2 and changes to 0. Its oxidation number decreases, so Cu2+ is reduced.

Cu2+ + 2e → Cu

Oxidizing and Reducing Agents

The substance that causes another substance to be oxidized is called the oxidizing agent. The oxidizing agent gains electrons and is itself reduced.

The substance that causes another substance to be reduced is called the reducing agent. The reducing agent loses electrons and is itself oxidized.

In the reaction between Zn and Cu2+:

Zn is oxidized, so Zn is the reducing agent.

Cu2+ is reduced, so Cu2+ is the oxidizing agent.

Recognizing a Redox Reaction

To determine whether a reaction is redox, assign oxidation numbers before and after the reaction. If at least one oxidation number increases and another decreases, electrons have been transferred and the reaction is redox.

Example: Is This a Redox Reaction?

2Mg(s) + O2(g) → 2MgO(s)

Mg changes from 0 to +2, so Mg is oxidized. Oxygen changes from 0 to −2, so oxygen is reduced. Because oxidation and reduction both occur, this is a redox reaction.

VIDEO EXPLANATION

Watch: Identifying Reaction Types

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