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How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0139 | Chemistry: Balance Equations Without Changing Formulae — Count Atoms, Then Check States

G3 SEC Chemistry K326/K328 | Learner’s Guide Vol 0139. Balance the numbers of reacting particles without changing the formula of any substance. Count atoms first, check ionic charge when appropriate, and read state symbols meaningfully.

Chemical equations turn a reaction into a precise statement about what changes and what remains conserved. This original guide builds that skill from the first atom count through ionic equations and mole ratios, with twenty-nine worked cases and a practical revision route.

Chapter index

1. Why a balanced equation tells a chemical story

A pupil writes magnesium + oxygen → magnesium oxide. That word equation describes the starting substances and product, but it does not yet tell us how many units of each substance participate in the reaction. The symbolic equation Mg + O₂ → MgO makes the substances clearer, but its oxygen atoms do not balance. The complete balanced version is 2Mg + O₂ → 2MgO. Both sides now contain two magnesium atoms and two oxygen atoms.

This G3 SEC Science Chemistry Learner’s Guide Vol 0139 teaches the habit that makes equation writing reliable: fix the chemical formulae first, adjust only the numbers in front, and count every type of atom before accepting an answer. The same equation can then support observations, state symbols and chemical calculations. A balanced equation is more than neat handwriting; it is a compact statement of the amounts in which substances react.

It is easy to make a small change that silently changes the substance. Writing MgO₂ instead of MgO does not simply “add enough oxygen”. It gives a different formula from the intended magnesium oxide. That is why balancing uses coefficients, not arbitrary changes to the subscripts inside chemical formulae. The formula of a substance and the amount of that substance are different questions.

This guide moves deliberately from word equations and atom counts to polyatomic groups, equations with state symbols, simple ionic equations and mole ratios. It also includes original exercises, complete answers and a calm seven-day revision route. The first step is always to understand what is reacting before calculating how much reacts.

2. Which 2027 SEC Science syllabus covers this

The official SEAB 2027 G3 syllabus directory lists K326 Science (Physics, Chemistry) and K328 Science (Chemistry, Biology). These two combined-Science combinations contain Chemistry. K327 Science (Physics, Biology) does not contain the Chemistry section. G3 Pure Chemistry is a separate subject, K324, with its own full syllabus and assessment scheme.

The official 2027 K326/K327/K328 combined Science syllabus includes Chemistry Section II on chemical reactions, Topic 4.1 Formulae and Equation Writing and Topic 4.2 Mole Concept and Stoichiometry. It explicitly expects learners to interpret equations with state symbols and construct chemical equations, including ionic equations. The syllabus explains that chemical equations represent conservation of matter and molar ratios.

For the combined Science examination, Paper 1 is a one-hour, 40-mark multiple-choice paper, worth 20 percent. Each selected Science has a one-hour-fifteen-minute structured/free-response paper carrying 65 marks and 32.5 percent. The Chemistry paper is Paper 3 for K326 and K328. Paper 5 is a practical assessment lasting one hour and thirty minutes, worth 30 marks and 15 percent. Always read the exact subject combination to know the papers required.

This guide provides original teaching exercises rather than past-paper questions or predictions about the specific reactions that will appear. The balanced equations are intended to develop the syllabus skill of constructing and interpreting chemical equations; they are not a replacement for learning the required substances, practical evidence and reaction patterns in the published syllabus.

3. A chemical formula is not an adjustable label

The formula H₂O represents water: each molecule contains two hydrogen atoms and one oxygen atom. The formula H₂O₂ represents a different substance, hydrogen peroxide, with two hydrogen atoms and two oxygen atoms. When balancing a chemical reaction involving water, you cannot change H₂O into H₂O₂ just to satisfy an atom count.

A coefficient placed before a formula tells how many units of that substance are represented. Thus 3H₂O denotes three water molecules in the molecular picture: altogether six hydrogen atoms and three oxygen atoms. In mole calculations, the same coefficient represents a three-mole proportion rather than exactly three separate molecules in a beaker.

The formula Na₂O contains two sodium atoms for every oxygen atom in the formula unit. Writing 2Na₂O represents two such formula units, with a total of four sodium and two oxygen atoms. The coefficient multiplies the entire formula; it does not attach only to the first element.

If a learner repeatedly rewrites subscripts to fix a count, pause and return to the earlier skill of finding a correct chemical formula. The Vol 0132 Ionic Formula Workshop explains how charges determine ionic formulae. The present volume begins after that decision and asks how correctly identified substances can be balanced within a reaction.

4. Conservation: where did the atoms go?

During an ordinary chemical reaction, atoms are rearranged into different combinations; they are not simply created or destroyed. That conservation is the central test of an equation. Write a small inventory of the elements on the reactant side, then write the inventory on the product side. Every element must have the same number of atoms on both sides.

For H₂ + O₂ → H₂O, the unbalanced inventory gives two hydrogen atoms on each side but two oxygen atoms on the left and one on the right. Place 2 before H₂O to make two oxygen atoms on the right. That now creates four hydrogen atoms on the product side, so place 2 before H₂. The result is 2H₂ + O₂ → 2H₂O.

Check it independently: left side has four H and two O; right side has four H and two O. The smallest whole-number coefficients are 2:1:2. You have balanced a reaction by changing the numbers of participating molecules, not the chemical identities of the substances.

In an experiment, a gas may leave an open container, making the measured mass remaining inside seem lower. That does not invalidate conservation of matter. It means the measurement does not include all products that crossed the chosen boundary. This distinction is explored in Vol 0104 on system boundaries.

5. A five-step balancing method

Step one: write the correct reactant and product formulae. Identify the elements present and read each subscript as part of the formula. Step two: count the atoms of every element on each side. Step three: choose an element whose imbalance is easy to repair and place a coefficient before a complete formula.

Step four: recount all elements because changing a coefficient may have altered several atom totals. Work through the remaining imbalances, leaving common diatomic substances such as O₂ or H₂ until the appropriate stage when that makes the arithmetic simpler. Step five: reduce any common coefficient factor and check every element again.

Take Fe + O₂ → Fe₂O₃. Each unit of iron(III) oxide contains two Fe and three O atoms. Three oxygen molecules supply six oxygen atoms; two Fe₂O₃ units contain six oxygen atoms. Those two product units also require four Fe atoms. Thus 4Fe + 3O₂ → 2Fe₂O₃.

There is no single magical element order that works in every equation. What matters is deliberate counting and a final independent check. If two attempts produce different coefficient sets, expand both into atom totals before concluding one is wrong. Sometimes a learner has simply used a larger equivalent set that should be reduced.

6. Coefficient versus subscript: the most useful contrast

Compare 2CO₂ with CO₄. The first denotes two units of carbon dioxide, containing two carbon atoms and four oxygen atoms. The second is a different formula and does not represent two carbon dioxide molecules. You may count the atoms and notice the same totals for certain elements, but that coincidence does not make the substances identical.

In Mg + O₂ → MgO, writing MgO₂ would change the product. Instead, put 2 before MgO, then 2 before Mg. The final equation is 2Mg + O₂ → 2MgO. The magnesium-oxygen compound remains MgO throughout.

The same discipline applies to NaCl. You cannot change sodium chloride to NaCl₂ to balance a chlorine count in an ordinary reaction when sodium chloride is the intended product. Put coefficients before the appropriate complete substances and verify formulae using bonding and charges.

Practise reading “3CaCl₂” aloud: three formula units of calcium chloride, each with one calcium and two chlorine units, give three calcium and six chlorine units in the equation count. This spoken expansion makes the multiplier easier to understand than memorising a visual trick.

7. Keep whole groups together when possible

Some formulae contain groups of atoms in parentheses. Calcium hydroxide is Ca(OH)₂. Each formula unit has one Ca, two O and two H. If a group such as OH appears unchanged on both sides of an equation, it can be convenient to count the whole group first, provided that doing so remains chemically and arithmetically valid.

Consider Ca(OH)₂ + HCl → CaCl₂ + H₂O. The product CaCl₂ needs two chloride units, so place 2 before HCl. The two hydroxide groups combine with the two hydrogen ions supplied by the acid to form two water molecules. The balanced equation is Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O.

Atom audit: Ca 1:1, Cl 2:2, O 2:2, and H 4:4. You can reason in terms of the unchanged hydroxide groups at first and then confirm using individual atoms. Group shortcuts do not replace final verification.

For reactions where a polyatomic ion changes chemically, counting it as a permanent block can become misleading. Use the group as a shortcut only while it remains intact in the relevant formulae. If you are unsure, return to the individual elements. Conservation applies to atoms, not to an imagined promise that every group survives.

8. Diatomic elements and the oxygen trap

Several common elemental substances are represented as diatomic molecules in equations, including H₂, N₂, O₂ and Cl₂. The subscript belongs to the molecular formula. If an equation involves oxygen gas, the reactant is usually O₂, not a solitary O atom written for convenience.

Consider burning carbon to form carbon dioxide: C + O₂ → CO₂ is already balanced with coefficients 1:1:1. Both sides contain one carbon and two oxygen atoms. If the product were carbon monoxide, a different balanced equation would be 2C + O₂ → 2CO. The products are distinct substances, so you cannot swap them to make arithmetic easier.

Hydrogen reacting with chlorine to form hydrogen chloride is H₂ + Cl₂ → 2HCl. Each side has two H and two Cl atoms. The two molecules of hydrogen chloride are represented by a coefficient of 2, not by changing HCl to H₂Cl₂ as a supposed replacement formula.

Learning familiar elemental formulae saves time because it prevents an error at the first stage of equation writing. But do not make the opposite mistake of turning every element into a diatomic molecule. The appropriate formula comes from chemical knowledge and the question context.

9. Metal and acid: combine meaning with balancing

A common reaction type is a suitable metal reacting with dilute acid to produce a salt and hydrogen gas. For magnesium and hydrochloric acid, the chemical equation is Mg + 2HCl → MgCl₂ + H₂. Its coefficients are determined by conservation, while the product formula MgCl₂ comes from the charges on magnesium and chloride ions.

Check Mg 1:1, Cl 2:2 and H 2:2. The equation does not say every possible metal behaves identically with every acid, and it does not guarantee any reaction without considering the metal’s reactivity. It describes the specified reaction under suitable conditions.

For zinc with sulfuric acid, Zn + H₂SO₄ → ZnSO₄ + H₂ is balanced as written. The sulfate group remains intact. Writing ZnSO₄ with coefficient 2 would unnecessarily double the product and require corresponding changes to the reactants; do not change anything once the simplest correct coefficients work.

A strong examination response may combine a word equation, the formulae, an equation with state symbols and an observation. Those are separate pieces of the explanation. Balancing the symbols does not independently prove that the reaction occurs; the chemistry and experimental conditions must make sense.

10. Acids, alkalis and neutralisation

Neutralisation can be represented by an acid reacting with a suitable alkali to produce salt and water. For hydrochloric acid and sodium hydroxide, HCl + NaOH → NaCl + H₂O is already balanced: one Na, one Cl, two H in total and one O on each side.

For sulfuric acid and sodium hydroxide, H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The sodium sulfate formula needs two sodium ions per sulfate ion, so the reactant side needs two sodium hydroxide formula units. The resulting two water molecules then balance hydrogen and oxygen.

This comparison teaches an important habit. Do not mechanically copy coefficients from one neutralisation question to another. The formula and quantity of the salt depend on the ions involved. A correct balancing method begins with the right products and formulae.

The simplified ionic equation for neutralisation between hydrogen ions and hydroxide ions is H⁺(aq) + OH⁻(aq) → H₂O(l). The atoms and net charge balance. It highlights the essential ionic change without listing spectator ions, a distinction addressed later in this guide.

11. Carbonates and gases: follow what leaves the vessel

A typical acid-carbonate reaction produces a salt, water and carbon dioxide. For sodium carbonate and hydrochloric acid, Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. The two sodium atoms require two sodium chloride units, and the two hydrogen atoms yield one water molecule.

Check carefully: left Na 2, C 1, O 3, H 2 and Cl 2. Right Na 2, C 1, O 1 + 2 = 3, H 2 and Cl 2. No atom has disappeared. If the carbon dioxide escapes, the mass measured in an open reaction vessel can fall despite matter being conserved when the escaping gas is accounted for.

Calcium carbonate can decompose on heating to calcium oxide and carbon dioxide under suitable conditions: CaCO₃ → CaO + CO₂. The equation is already balanced. Sometimes no coefficient change is needed; adding unnecessary numbers is an error rather than a sign of extra effort.

Use observations such as effervescence only when relevant and supported. Bubbles can indicate a gas is evolving, but bubble formation alone does not identify the gas as carbon dioxide. Testing the gas and connecting the result to the expected reaction gives a stronger scientific answer.

12. State symbols: what exists as solid, liquid, gas or aqueous

A chemical equation can include state symbols: (s) solid, (l) liquid, (g) gas and (aq) an aqueous solution. These tell the reader the physical state of each substance under the described conditions. They are not extra coefficients, and they do not affect the numerical atom count.

For a familiar reaction in dilute aqueous conditions: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). The magnesium starts as a solid, the hydrochloric acid is aqueous, the soluble magnesium chloride is aqueous, and hydrogen is a gas. The equation and states work together to describe the experimental situation.

Do not assume a compound is aqueous merely because water is present. A substance may be insoluble and form a precipitate. Conversely, a soluble salt may remain in solution rather than appearing as a solid. Use solubility knowledge, the actual reagent conditions and stated observations.

State symbols can also clarify why an experiment changes appearance. For AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq), the silver chloride is a solid precipitate. Coefficients are 1:1:1:1, while the change from two aqueous solutions to a solid product supplies observable evidence of precipitation.

13. Ionic equations: retain the reacting particles

A full equation records the compounds in the reaction mixture. An ionic equation may isolate the species participating in the essential change. For precipitation between aqueous silver nitrate and sodium chloride, the full equation is AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).

The sodium and nitrate ions are present in solution before and after the reaction. They are spectator ions in this idealised representation. Removing them leaves Ag⁺(aq) + Cl⁻(aq) → AgCl(s). The net charge is zero on each side, and there is one Ag and one Cl in the product.

For acid-alkali neutralisation in the appropriate aqueous context, H⁺(aq) + OH⁻(aq) → H₂O(l) gives the essential change. It is balanced both for atoms and overall charge. The symbol “+” between ions means they are combined on the same side of the equation; the superscript charges identify the ions.

Do not cancel a species merely because the same element appears on both sides. Spectator-ion cancellation works only for identical species that remain unchanged in the same amount on the two sides of the ionic equation. If an ion changes into a precipitate, gas or different chemical species, it belongs in the essential equation.

14. The balancing check for charge as well as atoms

A molecular equation must conserve atoms. An ionic equation must also conserve overall electric charge. For Ag⁺ + Cl⁻ → AgCl, the total reactant charge is +1 − 1 = 0, and the product is neutral, so charge balances. For H⁺ + OH⁻ → H₂O, the same check gives zero on both sides.

Consider Mg²⁺ + 2OH⁻ → Mg(OH)₂. There is one magnesium and two hydroxide groups on either side. The left-hand total charge is +2 − 2 = 0, matching the neutral precipitate on the right. If you wrote only one hydroxide ion on the left, you would miss an atom count and a charge balance.

The superscript charge belongs to the ion formula; it is not a coefficient. Mg²⁺ means a magnesium ion with charge +2, not two magnesium ions. 2Mg²⁺ means two such magnesium ions, giving total charge +4. Both the number of species and the charge on each matter.

When the prompt asks for a net ionic equation, a final atom audit without a charge audit is incomplete. The extra check takes little time and catches mistakes that are invisible if you count only element symbols.

15. Balanced coefficients become mole ratios

Once the equation is balanced, coefficients provide the reaction’s mole ratios. For 2H₂ + O₂ → 2H₂O, the relationship is two moles of hydrogen to one mole of oxygen to two moles of water, under the ideal stoichiometric interpretation. The coefficients do not directly tell you the mass ratios without accounting for relative molecular masses.

Take 2Mg + O₂ → 2MgO. Using relative atomic masses Mg 24 and O 16 for an introductory calculation, two moles of magnesium have mass 48 g, one mole of oxygen molecules has mass 32 g, and two moles of magnesium oxide have mass 80 g. Conservation appears again: 48 + 32 = 80.

If 12 g of magnesium reacts completely with sufficient oxygen, it is 0.5 mol Mg. The 2:2 ratio gives 0.5 mol MgO, with mass approximately 20 g using MgO relative mass 40. The oxygen that joins contributes about 8 g, explaining why the mass of the solid product can be greater than the starting magnesium alone.

A newly balanced equation is the foundation for these calculations, not their conclusion. The separate G3 limiting-reactant workshop extends coefficient ratios to decide which reactant runs out first. This volume owns the earlier, narrower task of producing and verifying the equation correctly.

16. When a correct atom count can still be a wrong answer

A student can balance a plausible-looking equation that represents the wrong chemistry. If the question describes magnesium oxide, a formula for a different oxide will not become acceptable just because the new atom totals match. Correct formulae must come before balancing. This is the first place to check when a reaction answer seems puzzling.

A second error is omitting a stated reactant or product. If acid reacts with a carbonate, the expected products normally include a salt, water and carbon dioxide in the relevant context. An equation that balances only a proposed salt may be missing part of the chemical event, even if some atom counts happen to be forced into agreement.

A third error is confusing observed mass in an open apparatus with total mass conserved. If a gas escapes, the contents of the container no longer form the full system. Ask which material crossed the boundary before claiming a loss of matter. Similarly, a reaction may use oxygen from the surrounding air, making the solid product heavier than the starting metal.

A fourth error is treating coefficients as exact masses. “Two moles of hydrogen” and “two grams of hydrogen” are different statements. The balanced equation gives mole proportions; mass calculations require molar masses. The proper units should be visible in every numerical answer.

17. A complete worked case: magnesium combustion

Suppose a class burns a small piece of magnesium ribbon in oxygen. First identify the reagents: magnesium and oxygen. The main intended product is magnesium oxide. The initial symbolic equation is Mg + O₂ → MgO. Because there are two oxygen atoms on the reactant side and only one in the product formula, write 2MgO, then supply the matching 2Mg.

The balanced equation with states is 2Mg(s) + O₂(g) → 2MgO(s). Audit: magnesium 2:2, oxygen 2:2. The state symbols identify magnesium and magnesium oxide as solids and oxygen as a gas. In a real laboratory, a bright white light is associated with burning magnesium, so safe handling and suitable eye protection matter; this passage is conceptual learning, not an invitation to repeat a combustion experiment at home.

If a reaction uses 0.20 mol Mg and oxygen is sufficient, the 2:2 coefficient ratio means 0.20 mol MgO can form in the ideal calculation. Using Ar(Mg) = 24 and Ar(O) = 16, the mass of MgO is 0.20 × 40 = 8.0 g. Starting magnesium mass is 0.20 × 24 = 4.8 g; the rest of the product mass came from oxygen.

If a measured mass differs from the ideal result, do not rewrite the equation to fit it. Consider whether product escaped, the metal reacted completely, the sample contained impurities, or the weighing method captured the whole product. A balanced reaction gives a model under specified assumptions; a practical evaluation asks how well the evidence matches that model.

18. Read the question before you start changing numbers

Some questions ask you to “write a balanced equation”, which calls for correct formulae and coefficients. Others ask you to “state the observation”, “deduce the salt formula”, “explain the mass change” or “calculate moles”. Even when they concern the same reaction, the required answers differ.

If a problem asks for state symbols, add them correctly and check against the specified conditions. If it asks for an ionic equation, identify spectators and verify charge as well as atoms. If it asks for reacting masses, begin with a balanced equation and then use molar masses. Do not insert every piece of chemical knowledge into every short response.

When faced with an unfamiliar reaction, work from the named substances and the principles in the syllabus. If necessary, write a word equation first, then the formulae, then coefficients. Guessing a product merely to complete an attractive-looking line can waste more time than identifying what the question actually supplied.

A short, honest answer with the correct equation is better than a confident paragraph attached to the wrong species. Chemical precision comes from managing distinct jobs in the right order.

19. Twenty-eight original equation-writing questions

These original exercises progress from simple balance checks to acid reactions, states, ionic equations and basic mole-ratio interpretation. Write your answer before looking at the model solution. A few questions deliberately contain an already-balanced equation; noticing that no changes are required is part of the skill.

If an exercise names an unfamiliar substance, do not invent a formula: use the formula given or consult the appropriate syllabus material. These practice cases are teaching examples, not a claim that a particular equation will appear in the 2027 examination paper.

20. Worked solutions: twenty-nine original reaction cases

Balance and audit the reaction yourself before reading the answer. In the explanation, focus on the first element or condition that would expose a mistake, not just the final row of coefficients.

Question 1. Balance Mg + O₂ → MgO.

Worked answer. 2Mg + O₂ → 2MgO.

Why it is correct. Left and right each contain two Mg and two O atoms. Preserve the MgO formula.

Question 2. Balance H₂ + O₂ → H₂O.

Worked answer. 2H₂ + O₂ → 2H₂O.

Why it is correct. The 2 before water gives two oxygen atoms and four hydrogen atoms on the right.

Question 3. Balance Fe + O₂ → Fe₂O₃.

Worked answer. 4Fe + 3O₂ → 2Fe₂O₃.

Why it is correct. Oxygen totals are six on each side; iron totals are four.

Question 4. Balance Al + O₂ → Al₂O₃.

Worked answer. 4Al + 3O₂ → 2Al₂O₃.

Why it is correct. Use six oxygen atoms as a common total for O₂ and Al₂O₃.

Question 5. Is C + O₂ → CO₂ already balanced?

Worked answer. Yes, the coefficients are 1:1:1.

Why it is correct. No adjustment is required when every element already matches.

Question 6. Balance C + O₂ → CO.

Worked answer. 2C + O₂ → 2CO.

Why it is correct. Carbon monoxide and dioxide are different products; do not change CO into CO₂.

Question 7. Balance H₂ + Cl₂ → HCl.

Worked answer. H₂ + Cl₂ → 2HCl.

Why it is correct. Two H and two Cl atoms occur on either side.

Question 8. What does the 3 mean in 3H₂O?

Worked answer. It represents three units or moles of water in the equation’s stoichiometric context.

Why it is correct. The coefficient multiplies the entire formula; total H atoms = 6 and O atoms = 3.

Question 9. Count atoms in 2Ca(OH)₂.

Worked answer. There are two Ca, four O and four H atoms in the formula-unit count.

Why it is correct. Multiply the entire parenthesised OH group and then by the external coefficient.

Question 10. Balance Ca(OH)₂ + HCl → CaCl₂ + H₂O.

Worked answer. Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O.

Why it is correct. The two hydroxides require two acid hydrogen ions and form two waters.

Question 11. Balance HCl + NaOH → NaCl + H₂O.

Worked answer. It is already balanced with all coefficients 1.

Why it is correct. There is one Na, one Cl, two H and one O on both sides.

Question 12. Balance H₂SO₄ + NaOH → Na₂SO₄ + H₂O.

Worked answer. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

Why it is correct. Sodium needs two units; water then balances the hydrogens and oxygens.

Question 13. Balance Mg + HCl → MgCl₂ + H₂.

Worked answer. Mg + 2HCl → MgCl₂ + H₂.

Why it is correct. Two chloride ions in magnesium chloride require two HCl units.

Question 14. Balance Zn + H₂SO₄ → ZnSO₄ + H₂.

Worked answer. This equation is already balanced 1:1:1:1.

Why it is correct. The sulfate group remains intact and the hydrogen atoms form H₂.

Question 15. Balance Na₂CO₃ + HCl → NaCl + H₂O + CO₂.

Worked answer. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.

Why it is correct. Count sodium, chlorine, hydrogen, carbon and oxygen independently.

Question 16. Is CaCO₃ → CaO + CO₂ balanced?

Worked answer. Yes, coefficients 1:1:1.

Why it is correct. One Ca, one C and three O occur on each side.

Question 17. Balance CuO + H₂ → Cu + H₂O.

Worked answer. This equation is balanced 1:1:1:1.

Why it is correct. There is one Cu, one O and two H on both sides.

Question 18. Balance Fe₂O₃ + CO → Fe + CO₂.

Worked answer. Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Why it is correct. Both sides have two Fe, three C and six O.

Question 19. What state does (aq) represent?

Worked answer. A substance dissolved in water, forming an aqueous solution.

Why it is correct. It is not a count of atoms or a guarantee that every salt is soluble.

Question 20. Write the balanced equation with states for magnesium in hydrochloric acid.

Worked answer. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).

Why it is correct. The solid metal reacts with aqueous acid to give aqueous salt and hydrogen gas.

Question 21. Write a precipitation equation for silver nitrate and sodium chloride solutions.

Worked answer. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).

Why it is correct. Silver chloride precipitates as a solid while sodium nitrate remains aqueous.

Question 22. Write the net ionic equation for silver chloride precipitation.

Worked answer. Ag⁺(aq) + Cl⁻(aq) → AgCl(s).

Why it is correct. Sodium and nitrate are spectator ions; net electric charge is zero on both sides.

Question 23. Write the net ionic equation for simple aqueous neutralisation.

Worked answer. H⁺(aq) + OH⁻(aq) → H₂O(l).

Why it is correct. Hydrogen and oxygen counts balance and overall charge is zero.

Question 24. Is Mg²⁺ + OH⁻ → Mg(OH)₂ correct as written?

Worked answer. No. Write Mg²⁺ + 2OH⁻ → Mg(OH)₂.

Why it is correct. Two hydroxide ions supply two groups and balance the +2 charge.

Question 25. In 2H₂ + O₂ → 2H₂O, what is the H₂:O₂ mole ratio?

Worked answer. 2:1.

Why it is correct. Coefficients represent amount-of-substance ratios, not equal gram masses.

Question 26. How many moles of water can 3 mol of H₂ ideally produce with excess O₂?

Worked answer. 3 mol of water, using the 2:2 mole ratio.

Why it is correct. The products depend also on having sufficient oxygen and appropriate reaction conditions.

Question 27. How much magnesium oxide forms ideally from 12 g Mg using Ar(Mg)=24 and Ar(O)=16, with enough oxygen?

Worked answer. Mg amount = 12/24 = 0.5 mol. MgO amount = 0.5 mol. Mass = 0.5 × 40 = 20 g.

Why it is correct. The oxide’s extra 8 g comes from oxygen; mass is conserved across the full system.

Question 28. A solid loses mass in an open acid-carbonate experiment. Does matter disappear?

Worked answer. No. Carbon dioxide may have escaped the measured container.

Why it is correct. The measured boundary excludes some products; conservation still applies to the complete reaction.

Question 29. A learner writes 2Mg + O₂ → 2MgO₂. Is it balanced and chemically suitable?

Worked answer. No. Oxygen count is two on the left and four on the right, and MgO₂ is not the intended magnesium oxide formula.

Why it is correct. Do not alter subscripts. The balanced magnesium oxide equation is 2Mg + O₂ → 2MgO.

21. How to audit errors without rewriting the whole chapter

Different wrong equations reveal different needs. A learner who changes H₂O to H₂O₂ needs to practise formula identity. A learner who forgets to multiply the oxygen in 3H₂O needs coefficient multiplication. A learner who balances atoms but not total charge in an ionic equation needs a charge check. Each error has a specific small repair.

Make a compact inventory with two columns labelled left and right. Write each element once and compare its total on the two sides. If the error remains, recheck whether a coefficient applies to a multi-atom group or whether parentheses were read correctly. Avoid randomly adding numbers until the totals appear to match.

For a mole-ratio mistake, underline the coefficients and separately write the molar masses. A ratio of formula units is not a ratio of gram masses. For a state-symbol mistake, identify whether the species is solid, liquid, gas or dissolved under the actual conditions. The appropriate test is not always a longer balancing worksheet.

After repairing the error, attempt a different equation with the same structure. A correction has truly transferred when you can handle a new pair of reactants without relying on having memorised the previous coefficients.

22. A practical seven-day equation route

On day one, practise reading formulae and counting atoms in substances such as H₂O, Ca(OH)₂, MgCl₂ and Al₂O₃. On day two, balance simple synthesis reactions, checking each element after every coefficient. On day three, practise diatomic elements and decomposition reactions.

On day four, write acid-metal, acid-alkali and acid-carbonate equations with correct products, checking the salt formula before balancing. On day five, attach state symbols to familiar reactions and explain what the states mean. On day six, practise simple ionic equations and verify net charge.

On day seven, complete a mixed timed set of the original questions and reflect on the first point where any wrong answer diverged: product prediction, formula, coefficient, state, charge or calculation. A parent or tutor can help by asking which element has unequal totals and whether a correction would change the substance’s identity.

The learner does not need every equation in the world to be ready. Consistent conservation checks, a reliable set of core chemical formulae, and calm interpretation of the requested task provide a strong platform for unfamiliar examples.

23. What a correct equation lets you understand next

The best reason to balance chemical equations is not to satisfy a visual rule. It is to preserve the chemical identities and atom counts while revealing how substances participate in a reaction. Once those relationships are correct, you can interpret masses, moles, gas volumes and experimental changes on a much stronger foundation.

Continue through the Science Learning Hub and the G3 SEC Learner Guide. For earlier foundations, the PSLE Learning Guide covers conservation, fair comparisons and basic substances in learner-friendly language. For reading symbols across subjects, Vol 0137 shows why the same ordinary word can carry a technical meaning in English, Mathematics or Science.

A useful last check takes less than a minute: formulae unchanged, atoms balanced, ionic charge balanced where relevant, states meaningful, coefficients reduced to simplest whole-number ratios. With that small routine, the equation begins to explain the reaction instead of merely looking complete.