Chemical Equations And Balancing

Light the kitchen stove and LPG bursts into a clean blue flame — within a second, new substances are born that were not there before. Chemists everywhere record such changes in one compact, universal line of shorthand called a chemical equation. Writing that line correctly — and balancing it so that not a single atom goes missing — is the first and most scoring skill of Class 10 chemistry.

Definition: A chemical equation is the symbolic representation of a chemical reaction using chemical formulae: the reactants are written on the left and the products on the right, separated by an arrow (→) pointing towards the products.

Definition: A balanced chemical equation contains an equal number of atoms of each element on both sides of the arrow, as demanded by the Law of Conservation of Mass — mass can neither be created nor destroyed in a chemical reaction.

From word equation to skeletal equation

Hold a magnesium ribbon in a flame with a pair of tongs — after cleaning it with sandpaper, which removes the unreactive coating of magnesium oxide/carbonate so that the fresh metal beneath can react. It burns with a dazzling white flame and leaves behind a white powder, magnesium oxide. The change can first be recorded as a word equation:

Magnesium + Oxygen → Magnesium oxide

The substances that react — magnesium and oxygen — are the reactants; the new substance formed — magnesium oxide — is the product. The plus sign joins substances on the same side, and the arrow, read as "gives" or "forms", separates reactants from products and shows the direction of the change.

Word equations are long, and they change from language to language. Chemistry solves this by replacing names with chemical formulae:

Mg + O2 → MgO

An equation written like this, before any atom-checking, is called a skeletal equation. Note that oxygen is written as O2, never as a lone O — hydrogen, oxygen, nitrogen and chlorine all exist as diatomic molecules (H2, O2, N2, Cl2). A formula equation is compact, understood worldwide, and shows exactly which elements take part in the reaction.

The law that forces us to balance

Count the atoms in Mg + O2 → MgO. Left side: 1 Mg and 2 O. Right side: 1 Mg but only 1 O. One oxygen atom has quietly vanished — and that is forbidden. The Law of Conservation of Mass (established by Antoine Lavoisier) states that mass is neither created nor destroyed in a chemical reaction: atoms are only rearranged into new combinations. Two consequences follow directly:

• the number of atoms of each element must be equal on the two sides, and
• the total mass of the reactants must equal the total mass of the products.

An equation that passes the atom count is a balanced chemical equation; the raw skeletal form is unbalanced. Balancing the magnesium equation gives 2Mg + O2 → 2MgO. Verify with masses (Mg = 24 u, O = 16 u): reactants 2 × 24 + 32 = 80 u; products 2 × 40 = 80 u. The chemical account book tallies to the last atom.

Balancing by hit and trial — the five-step drill

NCERT balances equations by the hit-and-trial method: try the smallest whole-number coefficients until every element's count matches. Watch the full method on the burning of hydrogen.

Step 1 — Write the skeletal equation. H2 + O2 → H2O

Step 2 — Count each element on both sides. Left: H = 2, O = 2. Right: H = 2, O = 1. Hydrogen already matches; oxygen does not. (NCERT draws a box around each formula while counting — the box is a reminder that nothing inside it may be altered.)

Step 3 — Start with the compound having the most atoms and fix its unbalanced element. Here that compound is H2O. Make oxygen equal by placing the coefficient 2 before it: H2 + O2 → 2H2O. Remember, a coefficient multiplies the entire formula — 2H2O means 4 H atoms and 2 O atoms.

Step 4 — Rebalance whatever got disturbed; keep lone elements for the end. Hydrogen is now 2 on the left but 4 on the right, so place 2 before H2: 2H2 + O2 → 2H2O. Free elements such as O2, H2 or a lone metal are balanced last, because changing their coefficient disturbs no other element.

Step 5 — Verify, reduce, and polish. H: 4 = 4 ✓; O: 2 = 2 ✓. The coefficients 2 : 1 : 2 are already the smallest whole numbers. With state symbols: 2H2(g) + O2(g) → 2H2O(l).

Golden rule: balance only with coefficients written in front of formulae. Never touch a subscript. Writing H2O2 in place of H2O would "balance" the oxygen instantly — but H2O2 is hydrogen peroxide, a completely different substance, so the equation would then describe a different (and wrong) reaction.

Making the equation more informative — states and conditions

A board answer earns full marks when the balanced equation also communicates the physical states of the substances and the conditions of the reaction.

State symbols, written in brackets after each formula: (s) solid, (l) liquid, (g) gas, and (aq) aqueous solution — a substance dissolved in water.

• Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
• 2H2(g) + O2(g) → 2H2O(l)
• 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) — the water here is steam, so it is H2O(g), not H2O(l).

Reaction conditions — temperature, pressure, catalyst, light — are written above or below the arrow:

• CO(g) + 2H2(g) --(340 atm)--> CH3OH(l)
• 6CO2 + 12H2O --(sunlight, chlorophyll)--> C6H12O6 + 6O2 + 6H2O (photosynthesis)
• Heating is commonly shown by the Greek letter delta (Δ) written over the arrow.

Many examiners also accept an upward arrow (↑) after a gas that escapes and a downward arrow (↓) after a precipitate that settles, in place of (g) and (s).

Five traps that cost marks

1. Changing a subscript to balance. Illegal — it changes the substance itself (turning H2O into H2O2 converts water into peroxide).
2. Forgetting diatomic elements. Reactant oxygen is O2, hydrogen is H2, nitrogen is N2, chlorine is Cl2 — never a lone atom.
3. Counting polyatomic ions atom by atom. When a group such as SO4 or NO3 remains intact on both sides, count the whole group as one unit — faster and far safer.
4. Leaving fractions or non-minimal coefficients. 4H2 + 2O2 → 4H2O is atom-balanced but must be reduced to 2H2 + O2 → 2H2O; a fraction such as 7/2 must be cleared by multiplying the whole equation by 2.
5. Reading too much into the equation. A balanced equation is only an atom account — it does not tell you the speed of the reaction or the conditions required unless these are written on the arrow.

Work through the examples below in order. They begin with straight conversions and finish with the twists CBSE papers love — polyatomic ions, odd oxygen counts, subscript traps and mass audits.

Example 1 — Word equation to chemical equation

Q: Write the balanced chemical equation, with state symbols, for: zinc + dilute sulphuric acid → zinc sulphate + hydrogen.
Given: Formulae — Zn, H2SO4, ZnSO4, H2.
Solve: Skeletal equation: Zn + H2SO4 → ZnSO4 + H2. Count both sides: Zn 1 = 1; H 2 = 2; S 1 = 1; O 4 = 4. Every element already matches — some skeletal equations are born balanced, but you must still verify by counting.
Answer: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g). Quick check: acid and salt are in solution (aq), hydrogen escapes as a gas (g). ✓

Example 2 — Burning magnesium

Q: Balance Mg + O2 → MgO and add the condition and state symbols.
Solve: O: left 2, right 1 → place 2 before MgO: Mg + O2 → 2MgO. Mg: left 1, right 2 → place 2 before Mg: 2Mg + O2 → 2MgO. Verify: Mg 2 = 2, O 2 = 2.
Answer: 2Mg(s) + O2(g) --heat--> 2MgO(s). Sanity check with masses: 48 + 32 = 80 u on both sides. ✓

Example 3 — Combustion of methane (count-table method)

Q: Balance the equation for the burning of methane (natural gas): CH4 + O2 → CO2 + H2O, and add state symbols.
Solve: Count first — C: 1 = 1 ✓; H: 4 vs 2 ✗; O: 2 vs 3 ✗. Balance H by placing 2 before H2O: CH4 + O2 → CO2 + 2H2O (H: 4 = 4). Recount O on the right: 2 + 2 = 4. Balance O last (it stands alone on the left) with 2 before O2: CH4 + 2O2 → CO2 + 2H2O. Verify: C 1 = 1, H 4 = 4, O 4 = 4.
Answer: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). In a flame the water leaves as vapour, hence H2O(g). ✓

Example 4 — NCERT favourite: iron and steam

Q: Balance Fe + H2O → Fe3O4 + H2 (iron reacting with steam) and add state symbols.
Solve: The biggest compound is Fe3O4 — start with its oxygen. O: left 1, right 4 → 4H2O: Fe + 4H2O → Fe3O4 + H2. H: left 8, right 2 → 4H2: Fe + 4H2O → Fe3O4 + 4H2. Fe: left 1, right 3 → 3Fe. Verify: Fe 3 = 3, H 8 = 8, O 4 = 4.
Answer: 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g). Trap alert: the water is steam, so its state symbol is (g), not (l). ✓

Example 5 — Tricky: polyatomic ions as single units

Q: Balance BaCl2 + Al2(SO4)3 → AlCl3 + BaSO4, the reaction in which solutions of barium chloride and aluminium sulphate give a white precipitate of barium sulphate.
Solve: The sulphate group SO4 stays intact on both sides — count it as one unit. SO4: left 3, right 1 → 3BaSO4. Ba: right 3, left 1 → 3BaCl2. Al: left 2, right 1 → 2AlCl3. Cl: left 3 × 2 = 6, right 2 × 3 = 6 — it balances automatically, a good sign the earlier steps were correct.
Answer: 3BaCl2(aq) + Al2(SO4)3(aq) → 2AlCl3(aq) + 3BaSO4(s). Verify: Ba 3 = 3, Cl 6 = 6, Al 2 = 2, SO4 3 = 3. ✓

Example 6 — Tricky: the odd-oxygen fraction trick

Q: Balance the combustion of ethane: C2H6 + O2 → CO2 + H2O.
Solve: C: 2 → 2CO2. H: 6 → 3H2O. Oxygen needed on the right = 2 × 2 + 3 × 1 = 7 — an odd number, but O2 supplies oxygen only in pairs. Temporarily allow a fraction: C2H6 + 7/2 O2 → 2CO2 + 3H2O. Fractions are not allowed in a final equation, so multiply every coefficient by 2.
Answer: 2C2H6 + 7O2 → 4CO2 + 6H2O. Verify: C 4 = 4, H 12 = 12, O 14 = 14. ✓

Example 7 — Tricky: why changing a subscript is illegal

Q: To balance H2 + O2 → H2O, student A writes H2 + O2 → H2O2 and student B writes 2H2 + O2 → 2H2O. Both versions have equal atoms on the two sides. Who is correct, and why?
Solve: H2O2 is hydrogen peroxide — a different compound with different properties, not water. By changing the subscript, student A changed the identity of the product, so the equation no longer describes the burning of hydrogen at all. Subscripts are fixed by a compound's composition; balancing may adjust only the coefficients placed in front of formulae.
Answer: Student B is correct: 2H2(g) + O2(g) → 2H2O(l). Rule of thumb: coefficients — yes; subscripts — never.

Example 8 — Tricky: mass audit using conservation of mass

Q: 3.0 g of magnesium ribbon is burnt completely in exactly 2.0 g of oxygen. What mass of magnesium oxide forms? Name the law used, and show that the balanced equation supports the numbers. (Mg = 24 u, O = 16 u)
Formula: Total mass of reactants = total mass of products.
Solve: By conservation of mass, mass of MgO = 3.0 + 2.0 = 5.0 g. Cross-check with 2Mg + O2 → 2MgO: the mass ratio is 48 : 32 : 80, which simplifies to 3 : 2 : 5 — so 3 g of Mg needs exactly 2 g of O2 and yields 5 g of MgO. The data sits perfectly on the balanced ratio.
Answer: 5.0 g of MgO, by the Law of Conservation of Mass. Sanity check: 48 + 32 = 80 u on each side of the balanced equation. ✓

Cover the answer, recall, then check. 8 cards on writing and balancing chemical equations.

Q1. What is a chemical equation?
A1. A symbolic representation of a chemical reaction using formulae of reactants and products, with an arrow pointing from reactants to products.

Q2. Why must every chemical equation be balanced?
A2. Because of the law of conservation of mass — atoms are neither created nor destroyed, so the number of atoms of each element must be equal on both sides.

Q3. What is the difference between a word equation and a balanced equation?
A3. A word equation names reactants and products; a balanced equation uses formulae and equal atom counts on both sides.

Q4. What do (s), (l), (g) and (aq) indicate?
A4. The physical states: solid, liquid, gas and aqueous (dissolved in water).

Q5. How do you show a gas evolved or a precipitate formed?
A5. An upward arrow (↑) for a gas released and a downward arrow (↓) for an insoluble precipitate.

Q6. Balance: Fe + H2O → Fe3O4 + H2
A6. 3Fe + 4H2O → Fe3O4 + 4H2 (heated iron reacting with steam).

Q7. What is the hit-and-trial method of balancing?
A7. Adjusting the coefficients (numbers before formulae), never the subscripts, until each element has equal atoms on both sides.

Q8. How are reaction conditions such as heat or a catalyst shown?
A8. They are written above or below the arrow (e.g. Δ for heat, or the catalyst/temperature/pressure).