Chemical Equations
1. What This Chapter Covers
The chapter opens with six everyday processes and asks you to classify them: digestion, burning crackers, respiration, powdered lime added to water, a mango ripening, and iron nails left in moist air.
In each, the nature of the original substance changes. The book's working definition follows from that: if new substances are formed with properties completely unlike those of the original substances, a chemical change has taken place.
That raises the question the first half of the chapter answers — how do we actually know a reaction has happened? — and then the rest builds the notation for recording it. The chapter is allotted 5 periods in June and runs from textbook page 44 to page 68.
2. How Do We Know a Reaction Has Taken Place?
Three activities are used to collect evidence before any notation is introduced.
Activity 1 — quick lime and water
Take about 1 g of quick lime (calcium oxide) in a beaker and add 10 ml of water. Touch the beaker.
The beaker is hot. Calcium oxide reacts with water and liberates heat energy, dissolving to give a colourless solution.
Now test the solution with litmus. Red litmus turns blue; blue litmus does not change. So the solution is basic.
Activity 2 — sodium sulphate and barium chloride
Dissolve a small quantity of sodium sulphate in about 100 ml of water in one beaker, and a small quantity of barium chloride in 100 ml of water in another. Note the colour of each solution.
Add the Na₂SO₄ solution to the BaCl₂ solution. A white barium sulphate precipitate forms (figure 1) — an insoluble solid appearing out of two clear solutions.
Activity 3 — zinc and dilute hydrochloric acid
Take some zinc granules in a conical flask and add about 20 ml of dilute hydrochloric acid. Observe the changes.
Hold a burning matchstick near the mouth of the flask, then touch the bottom of the flask with your fingers. The gas responds to the flame, and the flask is hot (figure 2).
The four signs, collected
From these activities the book draws its list of what can happen during a chemical change:
- The original substances lose their characteristic properties, so products may differ in physical state and colour.
- The change may be exothermic or endothermic — heat may be liberated or absorbed.
- An insoluble substance, a precipitate, may form.
- A gas may be evolved.
Notice that all four are things you can observe without any equipment beyond your hands and eyes. That is deliberate: the notation introduced next has to be able to record each of them, and section 2.2 does exactly that.
3. Word Equations (Textbook 2.1)
Describing activity 1 in a sentence is accurate but long. The same information compresses into a word equation:
calcium oxide + water -> calcium hydroxide
The vocabulary comes with it:
- The substances that undergo the chemical change are the reactants.
- The new substances formed are the products.
The conventions are equally simple. Reactants go on the left of the arrow, products on the right, and the arrowhead points towards the products to show the direction of the reaction. Where there is more than one reactant or product, they are separated by a plus sign.
4. From Words to Formulae (Textbook 2.1.1)
Word equations can be made more precise and more useful by replacing words with chemical formulae.
A compound's formula lists the symbols of its constituent elements and uses a subscript to show how many atoms of each are present. If no subscript is written, 1 is understood. So calcium oxide is CaO, water is H₂O, and the product of activity 1 is calcium hydroxide, Ca(OH)₂:
CaO + H₂O -> Ca(OH)₂
The book then asks you to count the atoms of each element on each side — and to keep asking it. Two more reactions from the activities are set up the same way:
Zn + HCl -> ZnCl₂ + H₂
Na₂SO₄ + BaCl₂ -> BaSO₄ + NaCl
For each, the questions are: are the atoms of each element equal on both sides, and do all the elements present in the reactants also appear in the products? Working through them shows the first two are not balanced, which is what motivates the next section.
5. Balancing Chemical Equations (Textbook 2.1.2)
Why balancing is compulsory
By the law of conservation of mass, the total mass of products must equal the total mass of reactants consumed. An atom is the smallest particle of an element that takes part in a reaction, and it is the atom that accounts for the mass of a substance.
So the number of atoms of each element must be the same before and after. Atoms are neither created nor destroyed.
A chemical equation in which the number of atoms of each element on the reactant side equals the number on the product side is called a balanced equation.
Formula units
Balancing means finding how many formula units of each substance take part. A formula unit is one unit — atom, ion or molecule — corresponding to a given formula.
| Substance | One formula unit is |
|---|---|
| NaCl | one Na⁺ ion and one Cl⁻ ion |
| MgBr₂ | one Mg²⁺ ion and two Br⁻ ions |
| H₂O | one H₂O molecule |
The four-step method, on hydrogen and oxygen
Step 1. Write the equation with correct chemical formulae for every reactant and product:
H₂ + O₂ -> H₂O
Step 2. Balance it. The critical rule: do not touch the ratio of atoms inside a molecule. You may only assign suitable numbers, called coefficients, in front of the formulae. Writing 2 before H₂ and 2 before H₂O gives
2H₂ + O₂ -> 2H₂O
Counting now: hydrogen 4 and 4, oxygen 2 and 2.
Step 3. Check whether all the coefficients share a common factor. Since the lowest whole-number ratio is wanted, divide through if they do. Here they do not, so nothing changes.
Step 4. Verify by counting the atoms of each element on both sides once more.
The method the chapter applies to all three of its balancing examples. The book notes this is a trial and error method, and that some equations need more care.
6. Worked Example 1 — Combustion of Propane
Propane, C₃H₈, is a colourless, odourless gas used as a heating and cooking fuel. The reactants are propane and oxygen; the products are carbon dioxide and water.
Step 1. Write the unbalanced equation:
C₃H₈ + O₂ -> CO₂ + H₂O
The book names this a skeleton equation: an unbalanced chemical equation containing the molecular formulae of the substances.
Step 2. Compare atoms and find coefficients. The advice given is to start with the most complex substance, here C₃H₈.
There are 3 carbon atoms on the left but only 1 on the right. Adding a coefficient of 3 to CO₂ balances carbon:
C₃H₈ + O₂ -> 3CO₂ + H₂O
There are 8 hydrogens on the left but only 2 on the right. Adding a coefficient of 4 to H₂O balances hydrogen:
C₃H₈ + O₂ -> 3CO₂ + 4H₂O
Now oxygen: 2 on the left, but 10 on the right (6 in 3CO₂ and 4 in 4H₂O). Adding a coefficient of 5 to O₂ balances it:
C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O
Step 3. The coefficients are already the smallest whole numbers. To show why this step exists, the book offers a counter-example:
2C₃H₈ + 10O₂ -> 6CO₂ + 8H₂O
This equation is balanced — count and check. But the coefficients share a factor of 2, so dividing throughout returns the accepted final form.
Step 4. Count the numbers and kinds of atoms on both sides to confirm.
7. Worked Example 2 — Iron Oxide and Aluminium
Iron oxide reacts with aluminium to form iron and aluminium oxide.
Step 1. Fe₂O₃ + Al -> Fe + Al₂O₃
Step 2. Examine the atoms. Oxygen is already equal on both sides (3 and 3), so only the metals need work.
There are 2 Fe atoms on the left and 1 on the right. Multiply Fe on the product side by 2:
Fe₂O₃ + Al -> 2Fe + Al₂O₃
Aluminium is still unbalanced: 1 on the left, 2 on the right. Multiply Al on the left by 2:
Fe₂O₃ + 2Al -> 2Fe + Al₂O₃
Step 3 and 4. The coefficients are already smallest whole numbers, and a final count confirms the balance:
| Element | Atoms in reactants | Atoms in products |
|---|---|---|
| Fe | 2 (in Fe₂O₃) | 2 (in 2Fe) |
| O | 3 (in Fe₂O₃) | 3 (in Al₂O₃) |
| Al | 2 (in 2Al) | 2 (in Al₂O₃) |
The book adds an honest caveat: this is a trial and error method, and some equations need more care than these.
Back to the two unbalanced activity equations
Section 4 left two equations from the activities unbalanced. The method now settles both.
For zinc and hydrochloric acid, chlorine is 1 on the left and 2 on the right, and hydrogen is 1 on the left and 2 on the right. A coefficient of 2 on HCl fixes both at once:
Zn + 2HCl -> ZnCl₂ + H₂
Counting: Zn 1 and 1, H 2 and 2, Cl 2 and 2.
For sodium sulphate and barium chloride, barium and sulphate are already matched, but sodium is 2 on the left against 1 on the right, and chlorine is 2 against 1. A coefficient of 2 on NaCl balances both:
Na₂SO₄ + BaCl₂ -> BaSO₄ + 2NaCl
Counting: Na 2 and 2, S 1 and 1, O 4 and 4, Ba 1 and 1, Cl 2 and 2.
A question the book leaves open
A box titled THINK AND WRITE poses this: you brush a wall with an aqueous suspension of Ca(OH)₂, and after two days the wall has turned white. Write the balanced chemical reaction using appropriate symbols and formulae.
The textbook sets this as an exercise and does not print the answer, so none is asserted here. What it is testing is worth naming, though: the wall changes over two days with nothing added to it, so whatever is reacting with the calcium hydroxide must be coming from the air. Identifying that reactant is the whole question, and the four-step method then does the rest.
8. Making Equations More Informative (Textbook 2.2)
A balanced equation can carry four further pieces of information — matching, one for one, the four observations from the opening activities.
Physical state (2.2.1)
Gaseous, liquid and solid states are written as (g), (l) and (s). A substance dissolved in water is aqueous, written (aq). Adding states to the iron oxide equation:
Fe₂O₃(s) + 2Al(s) -> 2Fe(s) + Al₂O₃(s)
The symbol Δ above the arrow represents heating.
Heat changes (2.2.2)
Heat is written as Q on the product side, with a plus sign for exothermic and a minus sign for endothermic reactions:
C(s) + O₂(g) -> CO₂(g) + Q (exothermic)
N₂(g) + O₂(g) -> 2NO(g) - Q (endothermic)
Gas evolved (2.2.3)
A gas evolved is marked with an upward arrow after the formula, or by the state symbol (g):
Zn(s) + H₂SO₄(aq) -> ZnSO₄(aq) + H₂(g)
Precipitate formed (2.2.4)
A precipitate is marked with a downward arrow:
AgNO₃(aq) + NaCl(aq) -> AgCl(s) + NaNO₃(aq)
where AgCl is the precipitate. Reaction conditions such as temperature, pressure or a catalyst are written above or below the arrow:
2AgCl(s) -> 2Ag(s) + Cl₂(g) in sunlight
6CO₂(g) + 6H₂O(l) -> C₆H₁₂O₆(s) + 6O₂(g) in sunlight, with chlorophyll
That last one is photosynthesis, written in the same notation as everything else in the chapter.
9. What a Balanced Equation Tells You (Textbook 2.3)
The chapter lists six things:
- It identifies the reactants and products through symbols and formulae.
- It gives the ratio of molecules of reactants and products.
- Since molecular masses are expressed in unified masses (U), it gives the relative masses of reactants and products.
- If those masses are expressed in grams, it gives the molar ratios.
- Where gases are involved, masses can be converted to volumes using the molar mass and molar volume relationship.
- Using molar mass and Avogadro's number, the number of molecules and atoms can be calculated.
Taken together these give four kinds of relationship: mass-mass, mass-volume, volume-volume, and mass-volume-number of molecules.
10. Stoichiometry — the Textbook's Three Examples
Example 1 — mass to mass
2Al(s) + Fe₂O₃(s) -> Al₂O₃(s) + 2Fe(s), with Al = 27 U, Fe = 56 U, O = 16 U.
Reading the masses off the balanced equation:
(2 × 27) + (2 × 56 + 3 × 16) -> (2 × 27 + 3 × 16) + (2 × 56)
54 U + 160 U -> 102 U + 112 U, or 2 mol + 1 mol giving 1 mol + 2 mol.
Problem. Calculate the aluminium required to obtain 1120 kg of iron.
54 g of Al gives 112 g of Fe, so
x = (1120 × 1000 × 54) / 112 = 540000 g = 540 kg
So 540 kg of aluminium yields 1120 kg of iron.
Example 2 — mass, volume and number of molecules
Problem. Calculate the volume, mass and number of molecules of hydrogen liberated when 230 g of sodium reacts with excess water at STP. (Na = 23 U, O = 16 U, H = 1 U)
2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)
Masses: 46 g + 36 g gives 80 g + 2 g.
Mass of hydrogen. 46 g of Na gives 2 g of H₂, so 230 g gives (230 × 2)/46 = 10 g.
Volume. One gram molar mass of any gas at STP — standard temperature 273 K and standard pressure 1 bar, or 760 mm of Hg — occupies 22.4 litres, the gram molar volume. So 2 g of hydrogen occupies 22.4 litres, and 10 g occupies (10 × 22.4)/2 = 112 litres.
Number of molecules. 2 g of hydrogen is 1 mole, containing 6.02 × 10²³ molecules. So 10 g contains (10 × 6.02 × 10²³)/2 = 3.01 × 10²⁴ molecules.
Example 3 — the limiting reagent
Problem. Calculate the volume and number of molecules of CO₂ liberated at STP when 50 g of CaCO₃ is treated with dilute hydrochloric acid containing 7.3 g of dissolved HCl gas.
CaCO₃(s) + 2HCl(aq) -> CaCl₂(aq) + H₂O(l) + CO₂(g)
From the equation, 100 g of CaCO₃ reacts with 73 g of HCl to liberate 44 g of CO₂.
Here is the catch. 100 g of CaCO₃ requires 73 g of HCl, so 50 g of CaCO₃ would require 36.5 g of HCl — but only 7.3 g is available.
So the CO₂ formed depends on the HCl, not the CaCO₃. This gives the chapter's definition:
The reactant available in the lesser amount is called the limiting reagent, as it limits the amount of product formed.
Mass of CO₂. 73 g of HCl releases 44 g of CO₂, so 7.3 g releases (7.3 × 44)/73 = 4.4 g.
Volume. 44 g of CO₂ occupies 22.4 L at STP, so 4.4 g occupies (4.4 × 22.4)/44 = 2.24 litres.
Number of molecules. 44 g of CO₂ contains 6.022 × 10²³ molecules, so 4.4 g contains (4.4 × 6.022 × 10²³)/44 = 6.022 × 10²² molecules.
The arithmetic is easy; the judgement is in recognising which reactant limits the reaction before any of it starts.
Key words from the chapter
Reactants, products, exothermic reaction, endothermic reaction, physical and chemical changes, primary or skeleton equation, formula unit, precipitation, coefficient, atomic mass, standard temperature and pressure (STP), molar mass, Avogadro's number, gram molar volume.
11. Summary
A chemical change is recognised by new substances with properties unlike the originals. The three opening activities supply four observable signs: a change in properties and state, heat liberated or absorbed, a precipitate formed, and a gas evolved.
A reaction is first written as a word equation, with reactants on the left, products on the right, and an arrow showing direction. Replacing words with chemical formulae makes it precise, and the unbalanced result is a skeleton equation.
Balancing is required by the law of conservation of mass: the atoms of each element must be equal on both sides. The four-step method is to write correct formulae, adjust coefficients only — never the subscripts inside a formula — reduce to the lowest whole-number ratio, and recount. The chapter applies it to hydrogen and oxygen, to propane combustion, and to iron oxide with aluminium, advising you to start from the most complex substance.
An equation is made more informative by adding physical states (s), (l), (g) and (aq), heat as +Q or −Q, an upward arrow for a gas evolved, a downward arrow for a precipitate, and reaction conditions above the arrow.
A balanced equation then supports quantitative work: molecular ratios, relative masses in unified mass units, molar ratios in grams, volumes through the gram molar volume of 22.4 litres at STP, and numbers of molecules through Avogadro's number. The third worked example adds the idea that decides many real calculations — the limiting reagent, the reactant present in the lesser amount, which fixes how much product can form regardless of how much of the other reactant is available.
