ScienceCBSE Class 10 · Chemical Reactions and Equations

Balancing Chemical Equations: Hit-and-Trial Method

Updated 14 October 2026 · Fact-checked

A balanced chemical equation has the same number of atoms of each element on both sides of the arrow. It must balance because mass is neither created nor destroyed in a reaction. You balance it by hit-and-trial, changing only the numbers in front of formulae, never the formulae themselves, then add state symbols.

Equation 2H2 + O2 → 2H2O with molecule models and two seesaws showing 4 H atoms and 2 O atoms balanced on each side. — Count the atoms on each side: 4 hydrogen and 2 oxygen both before and after the arrow, so the equation is balanced.
Picture: Kvr.lohith · CC BY-SA 4.0 · Wikimedia Commons

Understand Balancing Chemical Equations

A chemical equation is a short way of writing a reaction using formulae. When magnesium burns, the first thing you might write is $M g + O_{2} \to M g O$ . Count the atoms: there are two oxygen atoms on the left but only one on the right. An equation like this, with correct formulae but unequal atom counts, is called a skeletal equation.

Why does it matter? From Class 9 you know the law of conservation of mass: in a chemical reaction mass is not created or destroyed. Atoms only rearrange; they do not appear from nowhere or vanish. So every atom present in the reactants must still be there in the products. An equation that shows this is a balanced chemical equation: for every element, the number of atoms on the left-hand side (reactants) equals the number on the right-hand side (products).

The method in NCERT is called hit-and-trial. You place whole-number coefficients (the big numbers in front of formulae) and keep adjusting until every element matches, using the smallest whole numbers possible. The one rule you can never break: do not touch the formulae. To get four oxygen atoms you may write $4 H_{2} O$ , but never $H_{2} O_{4}$ or $(H_{2} O)_{4}$ , because those would be different substances.

A practical order helps. Start with the formula that has the most atoms, and inside it pick the element with the most atoms. Balance that element, then the next one that is still unequal, and finish with an element that appears on its own (such as $F e$ or $H_{2}$ ), because changing its coefficient disturbs nothing else. Finally count every element again.

A balanced equation becomes more useful when you add state symbols: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water. For example, $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$ tells you the water is used as steam. Conditions such as heat, sunlight, a catalyst or a pressure are written above or below the arrow.

In the board exam this topic rarely stands alone. You will be asked to turn a word statement into a balanced equation, balance an equation and then name the reaction type, or spot which of four equations is balanced in an MCQ.

Balancing Chemical Equations: formulas and key facts

Condition for a balanced equation: $atoms of each element on LHS = atoms of that element on RHS$; Follows from the law of conservation of mass: mass of reactants = mass of products.
Counting atoms in a term: $atoms = coefficient \times subscript$; In $3 C a (O H)_{2}$ there are 3 Ca, 6 O and 6 H atoms: a subscript outside a bracket multiplies everything inside it.
Worked model from NCERT: $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$; Iron with steam: balance O first (from $F e_{3} O_{4}$ ), then H, then Fe.
State symbols: $(s) solid, (l) liquid, (g) gas, (a q) aqueous solution$; Write (aq) only for a substance dissolved in water; a precipitate is (s).

How to solve Balancing Chemical Equations questions

Use this routine for every 'write the balanced equation' question, whether it starts from words or from a skeletal equation.

1Write correct formulae for every reactant and product (from the word statement if needed). Get the formulae right first; balancing cannot fix a wrong formula.
2Mentally box each formula: from now on nothing inside a box may change.
3Make a quick table of atoms of each element on the left and on the right.
4Start with the formula that has the most atoms, and the element in it with the most atoms. Put a coefficient to equalise that element.
5Move to the next element that is still unequal. Treat a polyatomic ion that stays intact on both sides (such as $S O_{4}$ or $N O_{3}$ ) as one unit to save time.
6Balance elements that appear alone (like $F e$ , $C u$ , $O_{2}$ , $H_{2}$ ) last, since their coefficients affect nothing else.
7If you are left with a half (for example $\frac{7}{2} O_{2}$ ), multiply every coefficient by 2 to get whole numbers.
8Recount every element on both sides, then add state symbols and any condition over the arrow if the question asks.

Balancing Chemical Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Balancing a combustion equation

Suppose your family switches to a natural gas (methane, $C H_{4}$ ) connection for cooking, and your younger cousin asks why the flame needs plenty of air. You decide to show her the balanced equation for methane burning completely in oxygen to give carbon dioxide and water.

Write the balanced equation with state symbols for complete burning of methane. How many molecules of oxygen are needed for each molecule of methane, and why would less air be a problem?

Try it first, then show the solutionSolution

Skeletal equation: $C H_{4} + O_{2} \to C O_{2} + H_{2} O$ .
Carbon: 1 on each side, already equal.
Hydrogen: 4 on the left, 2 on the right, so write $2 H_{2} O$ , giving the still unbalanced $C H_{4} + O_{2} \to C O_{2} + 2 H_{2} O$ .
Oxygen: right side now has 2 (in $C O_{2}$ ) + 2 (in $2 H_{2} O$ ) = 4 atoms, so write $2 O_{2}$ on the left.
Check: C 1 = 1, H 4 = 4, O 4 = 4. Add states: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$ .
The ratio 1 : 2 means each methane molecule needs two oxygen molecules; with too little air there is not enough oxygen for every carbon atom to become $C O_{2}$ , so burning is incomplete.

Answer: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$ . Two oxygen molecules per methane molecule; less air leaves the burning incomplete.

Memory hook: One gas molecule, two oxygens: the balanced equation tells the stove how much air it needs.

Tricks to remember Balancing Chemical Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Big first, loners last
Balance the biggest formula first and single elements ( $F e$ , $C u$ , $H_{2}$ , $O_{2}$ ) last.
Use it when: Gives the order of steps in any hit-and-trial question.
Coefficients, not subscripts
You may change the number in front (how many molecules), never the small number inside (what the molecule is).
Use it when: Avoids the most costly error: writing $H_{2} O_{2}$ or $M g O_{2}$ to force a balance.
Ions travel in packs
If $S O_{4}$ , $N O_{3}$ , $C O_{3}$ or $O H$ appears unchanged on both sides, count it as one block.
Use it when: Fast balancing of double displacement and neutralisation equations.

Balancing Chemical Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Word statement to balanced equation · 2 marks · Very short answerBoard level

Translate into a balanced chemical equation with state symbols: sodium metal reacts with water to give sodium hydroxide solution and hydrogen gas.

Try it first, then show the solutionSolution

Skeletal equation: $N a + H_{2} O \to N a O H + H_{2}$ (½ mark).
Hydrogen: 2 on the left, 1 + 2 = 3 on the right. Put 2 before $N a O H$ and 2 before $H_{2} O$ : left H = 4, right H = 2 + 2 = 4.
Sodium: now 2 on the right, so write $2 N a$ . Oxygen: 2 = 2 (1 mark).
With states: $2 N a (s) + 2 H_{2} O (l) \to 2 N a O H (a q) + H_{2} (g)$ (½ mark).

Answer: $2 N a (s) + 2 H_{2} O (l) \to 2 N a O H (a q) + H_{2} (g)$

Example 2: Balancing with a bracketed ion · 3 marks · Short answerHarder board level

Barium chloride reacts with aluminium sulphate to give a precipitate of barium sulphate and a solution of aluminium chloride. Write the balanced equation with state symbols and name the type of reaction.

Try it first, then show the solutionSolution

Formulae: $B a C l_{2}$ , $A l_{2} (S O_{4})_{3}$ , $B a S O_{4}$ , $A l C l_{3}$ (½ mark).
Treat $S O_{4}$ as a block: 3 on the left, so write $3 B a S O_{4}$ . That needs 3 Ba, so write $3 B a C l_{2}$ (1 mark).
Chlorine: 6 on the left, so write $2 A l C l_{3}$ ; aluminium then 2 = 2 (½ mark).
Equation: $3 B a C l_{2} (a q) + A l_{2} (S O_{4})_{3} (a q) \to 3 B a S O_{4} (s) + 2 A l C l_{3} (a q)$ (½ mark).
Ions are exchanged and an insoluble solid forms: a double displacement (precipitation) reaction (½ mark).

Answer: $3 B a C l_{2} (a q) + A l_{2} (S O_{4})_{3} (a q) \to 3 B a S O_{4} (s) + 2 A l C l_{3} (a q)$ ; double displacement (precipitation) reaction.

Example 3: Thermal decomposition with three products · 3 marks · Short answerHarder board level

Balance: $P b (N O_{3})_{2} \to P b O + N O_{2} + O_{2}$ (heated). State what you would observe and why the equation must be balanced before it is used.

Try it first, then show the solutionSolution

Nitrogen: 2 on the left, 1 on the right, so write $2 N O_{2}$ . Oxygen now: left 6, right 1 + 4 + 2 = 7, odd, so double the lead nitrate.
Try $2 P b (N O_{3})_{2}$ : Pb 2, N 4, O 12. Right: $2 P b O$ (Pb 2, O 2) + $4 N O_{2}$ (N 4, O 8) + $O_{2}$ (O 2) = O 12 (1½ marks).
Balanced: $2 P b (N O_{3})_{2} (s) heat 2 P b O (s) + 4 N O_{2} (g) + O_{2} (g)$ .
Observation: brown fumes of nitrogen dioxide are given off (½ mark).
Mass is conserved in a reaction, so atoms of each element must be equal on both sides; only a balanced equation shows this and gives the correct ratio of substances (1 mark).

Answer: $2 P b (N O_{3})_{2} (s) heat 2 P b O (s) + 4 N O_{2} (g) + O_{2} (g)$ ; brown fumes of $N O_{2}$ ; balancing obeys the law of conservation of mass.

Example 4: Halves and doubling · 3 marks · Short answerHarder board level

Ethane ( $C_{2} H_{6}$ ) burns completely in oxygen to give carbon dioxide and water. Write the balanced equation using the smallest whole-number coefficients. Show how you remove a fraction.

Try it first, then show the solutionSolution

Skeletal: $C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$ .
Carbon: write $2 C O_{2}$ . Hydrogen: 6 on the left, so write $3 H_{2} O$ (1 mark).
Oxygen on the right: 4 + 3 = 7 atoms, which needs $\frac{7}{2} O_{2}$ , not a whole number (½ mark).
Multiply everything by 2: $2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ (1 mark).
Check: C 4 = 4, H 12 = 12, O 14 = 8 + 6 = 14 (½ mark).

Answer: $2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$

Example 5: Spot the student's error · 3 marks · Short answerHOTS / competency

A student balanced the burning of magnesium as $M g + O_{2} \to M g O_{2}$ and said 'Now both sides have one Mg and two O, so it is balanced.' (a) What is wrong? (b) Write the correct balanced equation. (c) Using atomic masses Mg = 24 u and O = 16 u, show that mass is conserved in your equation.

Try it first, then show the solutionSolution

(a) The student changed the formula of the product. Magnesium oxide is $M g O$ ; $M g O_{2}$ is a different (wrong) formula, so the equation now describes a reaction that does not happen (1 mark).
(b) Keep $M g O$ and use coefficients: oxygen needs $2 M g O$ , then magnesium needs $2 M g$ : $2 M g + O_{2} \to 2 M g O$ (1 mark).
(c) Left: $2 \times 24 + 2 \times 16 = 48 + 32 = 80$ u. Right: $2 \times (24 + 16) = 2 \times 40 = 80$ u. Equal masses (1 mark).

Answer: (a) A subscript was changed; formulae must not change. (b) $2 M g + O_{2} \to 2 M g O$ . (c) 80 u on each side.

Where students lose marks in Balancing Chemical Equations

The slips examiners see most often, and what to write instead.

Mistake: Changing subscripts to force a balance
It looks quicker than finding coefficients.
Wrong working
$H_{2} + O_{2} \to H_{2} O_{2}$
Right working
$2 H_{2} + O_{2} \to 2 H_{2} O$
Do this: Box every formula before you start; only the number in front may change.
Usually costs: The full mark for the equation, since the product is now a different substance.
Mistake: Forgetting that a bracket subscript multiplies everything inside
Students count $C a (O H)_{2}$ as one O and one H.
Wrong working
$H N O_{3} + C a (O H)_{2} \to C a (N O_{3})_{2} + H_{2} O$ called balanced.
Right working
$2 H N O_{3} + C a (O H)_{2} \to C a (N O_{3})_{2} + 2 H_{2} O$
Do this: Expand brackets in your atom table: $C a (O H)_{2}$ = 1 Ca, 2 O, 2 H.
Usually costs: ½ to 1 mark per equation.
Mistake: Leaving fractions or non-simplest coefficients
Students stop at $\frac{7}{2} O_{2}$ , or double twice by mistake.
Wrong working
$C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$
Right working
$2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$
Do this: Clear any fraction by multiplying through, then check that the coefficients have no common factor.
Usually costs: ½ mark in most marking schemes.
Mistake: Wrong or missing state symbols when asked
A precipitate is marked (aq), or water used as steam is marked (l).
Wrong working
$B a S O_{4} (a q)$ for the white precipitate.
Right working
$B a S O_{4} (s)$ for the precipitate; $H_{2} O (g)$ when water reacts as steam.
Do this: Ask: is it dissolved in water (aq), a solid formed in the tube (s), a gas given off (g) or a pure liquid (l)?
Usually costs: ½ mark when the question says 'with state symbols'.

Exam tips

If the question says 'write the balanced equation', a correct skeletal equation earns little; always finish the balancing and recount.
Write formulae carefully from valencies before balancing: $A l C l_{3}$ , $A l_{2} (S O_{4})_{3}$ , $C a (N O_{3})_{2}$ , $N a_{2} S O_{4}$ .
Add state symbols and conditions (heat, sunlight, electricity, catalyst) whenever the question asks for them or a reaction type depends on them.
For 'balance and identify the type' questions, give the type in a separate line so the examiner can award that mark easily.
In an MCQ asking which equation is balanced, count the element with the most atoms first; it rules out options fastest.

Practice questions on Balancing Chemical Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following equations is balanced?
(a) $F e + H_{2} O \to F e_{3} O_{4} + H_{2}$
(b) $3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
(c) $3 F e + H_{2} O \to F e_{3} O_{4} + H_{2}$
(d) $F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
Hint
Count Fe, H and O on both sides for each option.
Answer
(b). Fe 3 = 3, H 8 = 8, O 4 = 4.
Q2 · 2 marks · Very short answerBoard level
Balance and name the type of reaction:
(a) $H N O_{3} + C a (O H)_{2} \to C a (N O_{3})_{2} + H_{2} O$
(b) $A l + C u C l_{2} \to A l C l_{3} + C u$
Hint
In (a) keep $N O_{3}$ as a block; in (b) chlorine sets the coefficients (2 and 3 have a lowest common multiple of 6).
Answer
(a) $2 H N O_{3} + C a (O H)_{2} \to C a (N O_{3})_{2} + 2 H_{2} O$ (neutralisation, a double displacement reaction). (b) $2 A l + 3 C u C l_{2} \to 2 A l C l_{3} + 3 C u$ (displacement reaction).
Q3 · 2 marks · Very short answerHarder board level
Hydrogen sulphide gas burns in air to give water and sulphur dioxide. Write the balanced equation and show your atom count.
Hint
Balance H and S first; oxygen ends up odd, so you will need to double.
Answer
$2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$ . Count: H 4 = 4, S 2 = 2, O 6 = 2 + 4 = 6.
Q4 · 3 marks · Short answerHarder board level
Aluminium reacts with dilute sulphuric acid to form aluminium sulphate and hydrogen gas. (a) Write the balanced equation with state symbols. (b) How many hydrogen molecules are released for every two aluminium atoms? (c) Why can we not write $A l S O_{4}$ to make balancing easier?
Hint
Aluminium sulphate is $A l_{2} (S O_{4})_{3}$ because aluminium forms $A l^{3 +}$ and sulphate is $S O_{4}^{2 -}$ .
Answer
(a) $2 A l (s) + 3 H_{2} S O_{4} (a q) \to A l_{2} (S O_{4})_{3} (a q) + 3 H_{2} (g)$ . (b) Three. (c) The formula is fixed by the charges of the ions ( $A l^{3 +}$ , $S O_{4}^{2 -}$ ); $A l S O_{4}$ would not be electrically neutral and is not the real compound. Only coefficients may change.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A chemical equation is balanced by changing the coefficients and not the subscripts of formulae.
Reason (R): Changing a subscript changes the identity of the substance.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think of $H_{2} O$ versus $H_{2} O_{2}$ .
Answer
(a). Changing a subscript turns, for example, water into hydrogen peroxide, a different substance; so only coefficients are adjusted.
Q6 · 3 marks · Short answerHOTS / competency
Ammonia burns in oxygen in the presence of a catalyst to form nitrogen monoxide ( $N O$ ) and water. Write the balanced equation using the smallest whole numbers. Then explain why the coefficients 8, 10, 8, 12 also give equal atom counts but are not accepted as the answer.
Hint
Balance H first ( $N H_{3}$ and $H_{2} O$ : lowest common multiple of 3 and 2 is 6), then N, then O.
Answer
$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$ . Check: N 4 = 4, H 12 = 12, O 10 = 4 + 6 = 10. The set 8, 10, 8, 12 is the same ratio doubled; it balances but the convention is the smallest whole-number coefficients, so it is not the simplest form.
Q7 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a student heats about 2 g of green ferrous sulphate crystals in a dry boiling tube. The crystals lose water, turn colour, and a smell of burning sulphur is noticed. The residue is a reddish-brown solid, ferric oxide ( $F e_{2} O_{3}$ ), and two gases, sulphur dioxide and sulphur trioxide, are given off.
(a) Write the skeletal equation for the decomposition of $F e S O_{4}$ . (1 mark)
(b) Balance it, adding state symbols and the condition. (1 mark)
(c) The student writes $F e S O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$ and claims it is balanced because 'there is one S on the left and two S on the right, which is close enough'. Explain the error and show the full atom count of your balanced equation. (2 marks)
Hint
Iron sets the coefficient: $F e_{2} O_{3}$ needs two Fe on the left.
Answer
(a) $F e S O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$ (skeletal). (b) $2 F e S O_{4} (s) heat F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$ . (c) Every element must match exactly; 'close enough' breaks the law of conservation of mass. Balanced count: Fe 2 = 2, S 2 = 1 + 1, O 8 = 3 + 2 + 3 = 8.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a skeletal chemical equation?Show answer
An equation with correct formulae in which the atoms of each element are not yet equal on both sides.
2.Which law makes balancing necessary?Show answer
The law of conservation of mass: mass is neither created nor destroyed in a chemical reaction.
3.Why is the method called hit-and-trial?Show answer
You try coefficients, check, and adjust until every element matches, using the smallest whole numbers.
4.What does (aq) mean?Show answer
Aqueous: the substance is present as a solution in water.
5.Balanced equation for iron and steam?Show answer
$3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$
6.Where are conditions such as heat or a catalyst written?Show answer
Above or below the arrow.

Balancing Chemical Equations: frequently asked questions

Why should a chemical equation be balanced?

Because atoms are neither created nor destroyed in a reaction, so the total mass of reactants equals the total mass of products. A balanced equation shows the same number of atoms of each element on both sides.

What is the easiest way to balance a chemical equation in Class 10?

Use the hit-and-trial method: start with the formula having the most atoms, balance one element at a time with coefficients, leave single elements for last, and recount everything at the end.

Can I change the subscript in a formula to balance an equation?

No. Changing a subscript changes the substance, for example water into hydrogen peroxide. Only the coefficients in front of formulae may change.

Are state symbols compulsory in the board exam?

Only when the question asks for them, but adding them correctly never loses marks and makes the equation more informative.

Is a balanced equation with coefficients 4, 6, 4, 2 wrong if 2, 3, 2, 1 also works?

It balances, but the accepted answer uses the smallest whole-number coefficients, so divide through by the common factor.