Equations — Concepts, Formulas & Examples

Chemical equations are the language of reactions. Balancing them is a core skill. CBSE Class 9 and 10 introduce the rules; NEET assumes mastery and tests application in stoichiometry.

Core Concepts

What a chemical equation shows

Reactants on the left, products on the right, arrow between. Physical states in brackets — (s) solid, (l) liquid, (g) gas, (aq) aqueous. Numbers before formulas are coefficients.

A balanced equation tells us three things simultaneously:

Qualitative: What reacts and what forms
Quantitative (molecular): The ratio of molecules/formula units involved
Quantitative (molar): The ratio of moles (and therefore grams, via molar mass)

For example: $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$

This means: 2 molecules of H $_2$ react with 1 molecule of O $_2$ to form 2 molecules of H $_2$ O. Or equivalently: 2 moles of H $_2$ + 1 mole of O $_2$ → 2 moles of H $_2$ O. In mass: 4 g H $_2$ + 32 g O $_2$ → 36 g H $_2$ O (mass conserved: 36 = 36).

Law of conservation of mass

Mass is neither created nor destroyed in a chemical reaction. So atoms of each element on the left must equal atoms on the right. This is the principle behind balancing.

Antoine Lavoisier established this in the 1780s by carefully weighing reactants and products in sealed containers. Every balanced equation is a direct application of this law.

Balancing by inspection

Write the skeletal equation. 2. Count atoms of each element on both sides. 3. Adjust coefficients to balance. 4. Leave hydrogen and oxygen for last, usually. 5. Verify every element.

Systematic approach:

\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3

Fe appears as 1 on the left, 2 on the right. Put coefficient 2 before Fe:

2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3

O: 2 on left, 3 on right. LCM of 2 and 3 is 6. Need 3 O $_2$ on left (6 O atoms) and 2 Fe $_2$ O $_3$ on right (6 O atoms):

4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3

Fe: 4 = 4. O: 6 = 6. Balanced.

Types of chemical reactions

Combination (A + B → AB), decomposition (AB → A + B), displacement (A + BC → AC + B), double displacement (AB + CD → AD + CB), redox, neutralisation, combustion.

Type	Pattern	Example
Combination	A + B → AB	$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$
Decomposition	AB → A + B	$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$
Single displacement	A + BC → AC + B	$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$
Double displacement	AB + CD → AD + CB	$\text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} \downarrow + \text{NaNO}_3$
Combustion	Fuel + O $_2$ → CO $_2$ + H $_2$ O	$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$
Neutralisation	Acid + Base → Salt + Water	$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Balancing redox equations

Use oxidation number method or half-reaction method. Balance atoms other than H and O first, then O using water, then H using H $^+$ (or OH $^-$ in basic medium), then electrons for charge balance, then combine half-reactions.

Half-reaction method (acidic medium):

Example: $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$

Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$ Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$

Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Multiply oxidation by 5: $5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$ Add: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$

Stoichiometry from balanced equations

Once balanced, the coefficients give the mole ratio. This ratio is the key to all stoichiometric calculations — mass-to-mass, volume-to-volume, and limiting reagent problems.

\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{\text{Volume at STP}}{22.4\;\text{L}} = \frac{\text{Number of particles}}{6.022 \times 10^{23}}

The balanced equation gives the mole ratio. From there, convert to any other quantity.

Limiting reagent: In a reaction with two or more reactants, the one that runs out first limits the amount of product. To find it: calculate how many moles of product each reactant can produce. The one that produces less is the limiting reagent.

Information not given by a balanced equation

A balanced equation does NOT tell us:

The rate of the reaction (fast or slow)
Whether the reaction is spontaneous or not ( $\Delta G$ )
The mechanism (which bonds break and form, in what order)
The conditions needed (temperature, pressure, catalyst)
Whether the reaction goes to completion or reaches equilibrium

Worked Examples

$\text{H}_2\text{O}_2 \to \text{H}_2\text{O} + \text{O}_2$ . Left has 2H 2O. Right has 2H 3O. Multiply H $_2$ O $_2$ by 2, H $_2$ O by 2 → $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$ . Now 4H 4O both sides. Done.

The coefficient 2 in 2H $_2$ means two molecules of hydrogen gas. In mass terms, 2 × 2 = 4 grams of H $_2$ . Stoichiometry problems depend on these ratios.

Given: 10 g of H $_2$ and 64 g of O $_2$ react to form water.

$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Moles of H $_2$ = 10/2 = 5 mol. Moles of O $_2$ = 64/32 = 2 mol.

From the equation: 2 mol H $_2$ needs 1 mol O $_2$ . So 5 mol H $_2$ needs 2.5 mol O $_2$ .

We only have 2 mol O $_2$ — so O $_2$ is the limiting reagent.

Product: 2 mol O $_2$ × (2 mol H $_2$ O / 1 mol O $_2$ ) = 4 mol H $_2$ O = 72 g.

\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

C: 3 on left → need 3CO $_2$ . H: 8 on left → need 4H $_2$ O. O on right: 3(2) + 4(1) = 10 O atoms → need 5O $_2$ .

\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}

Verify: C: 3=3, H: 8=8, O: 10=10. Balanced.

Solved Problems (Exam Style)

Problem 1 (CBSE Class 10): Balance: $\text{Fe}_2\text{O}_3 + \text{Al} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}$

Fe: 2 on left, 1 on right → put 2 before Fe: $\text{Fe}_2\text{O}_3 + \text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$ Al: 1 on left, 2 on right → put 2 before Al: $\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$ O: 3 = 3. Balanced. This is the thermite reaction — used to weld railway tracks.

Problem 2 (JEE Main): How many grams of CO $_2$ are produced when 44 g of propane burns completely?

$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$

Moles of C $_3$ H $_8$ = 44/44 = 1 mol. From equation: 1 mol C $_3$ H $_8$ → 3 mol CO $_2$ . Mass of CO $_2$ = 3 × 44 = 132 g.

Common Mistakes

Changing subscripts while balancing. Only coefficients may change; subscripts are fixed by the formula.

Forgetting that polyatomic ions (like SO $_4$ ) can be balanced as a unit.

Writing fractional coefficients in the final equation. Multiply through to get whole numbers.

Forgetting to include physical states. CBSE boards deduct marks if you write just “NaCl” instead of “NaCl(aq)” or “NaCl(s)”.

Not identifying the limiting reagent before calculating product mass. If you use the excess reagent, your answer will be wrong.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

CBSE Class 10 boards ask balancing equations in every paper — expect 2-3 marks. JEE Main tests stoichiometry (limiting reagent, percent yield) heavily — 1-2 questions per paper. NEET expects you to balance equations mentally while solving problems.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Always verify by counting atoms on both sides at the end. One extra check catches most errors.

Practice Questions

Q1. Balance: $\text{KMnO}_4 + \text{HCl} \rightarrow \text{KCl} + \text{MnCl}_2 + \text{H}_2\text{O} + \text{Cl}_2$

$2\text{KMnO}_4 + 16\text{HCl} \rightarrow 2\text{KCl} + 2\text{MnCl}_2 + 8\text{H}_2\text{O} + 5\text{Cl}_2$

Verify: K: 2=2, Mn: 2=2, O: 8=8, H: 16=16, Cl: 16 = 2+4+10 = 16. Balanced.

Q2. What mass of oxygen is needed to burn 16 g of methane completely?

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ . Moles of CH $_4$ = 16/16 = 1 mol. From the equation: 1 mol CH $_4$ needs 2 mol O $_2$ . Mass of O $_2$ = 2 × 32 = 64 g.

Q3. Identify the type of reaction: $\text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2$

This is a decomposition reaction — one compound breaks down into two simpler substances when heated. Specifically, it is thermal decomposition. CaCO $_3$ (limestone) decomposes to CaO (quicklime) and CO $_2$ at about 840°C.

Q4. In the reaction $2\text{Al} + 3\text{CuSO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{Cu}$ , which element is displaced?

Copper is displaced by aluminium. Al is more reactive than Cu (higher in the reactivity series), so Al displaces Cu from CuSO $_4$ solution. This is a single displacement reaction. You can observe it as a reddish copper deposit forming on aluminium foil.

Q5. 5 g of calcium carbonate reacts with excess HCl. What volume of CO $_2$ is produced at STP?

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

Moles of CaCO $_3$ = 5/100 = 0.05 mol. From equation: 1 mol CaCO $_3$ → 1 mol CO $_2$ . So 0.05 mol CO $_2$ is produced. Volume at STP = 0.05 × 22.4 = 1.12 L.

FAQs

Can a chemical equation have fractional coefficients? Intermediate steps can use fractions (e.g., $\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$ ), and this is valid in thermodynamics where we define $\Delta H$ per mole of a specific substance. However, the final balanced equation in stoichiometry should use whole numbers — multiply through to clear fractions.

What is the difference between a skeletal equation and a balanced equation? A skeletal equation shows the correct formulas of reactants and products but has no coefficients adjusted. A balanced equation has coefficients that ensure equal numbers of each type of atom on both sides.

Why do we need to balance equations? Because atoms are neither created nor destroyed in chemical reactions (law of conservation of mass). An unbalanced equation violates this fundamental law and gives incorrect stoichiometric predictions — you would calculate wrong amounts of reactants and products.

How do you balance equations with polyatomic ions? If the same polyatomic ion appears on both sides of the equation (e.g., SO $_4^{2-}$ ), balance it as a single unit rather than balancing S and O separately. This is faster and less error-prone.

Balancing equations is like arithmetic — it must be right every time. Build the habit of checking and mistakes disappear.