Chapter 1 · Chemical Reactions and Equations

1.1 Chemical Reactions and Balanced Chemical Equations

Spot a chemical change, write it as an equation, and balance it so that every atom on the left turns up on the right: $\mathrm{2Mg + O_{2} \rightarrow 2MgO}$.

Leave a glass of milk on the kitchen shelf on a hot day and by evening it has turned sour. An iron gate left in the monsoon rain grows a flaky brown coat. Food on your plate is broken down in your stomach, and the LPG flame under the pressure cooker turns gas and air into something invisible that rises from the burner. In each case the substances at the end are different substances from the ones at the start — they have new properties that the starting materials did not have. These are chemical changes, and chemistry is the science of describing and explaining them.

To talk about chemical changes precisely, chemists use a short, universal language: the chemical equation. A good equation tells you, in one line, which substances react, what they form, in what states, under what conditions, and — once it is balanced — in what proportions. In this section you will learn to recognise a chemical change, write it first in words and then in formulas, and balance it so that it obeys the most basic rule in chemistry: atoms are never created or destroyed.

Recognising a Chemical Change

When you melt ice or dissolve sugar in tea, the substance is still the same substance: you can freeze the water again or evaporate the tea to get the sugar back. These are physical changes. In a chemical change, the atoms of the starting substances are rearranged to form new substances.

📘 Definition 1.1 — Chemical reaction, reactants and products
A chemical reaction is a process in which one or more substances are changed into one or more new substances with different properties. The substances we start with are the reactants; the new substances formed are the products. In a reaction the atoms of the reactants are regrouped by breaking and forming bonds; no atom is created, destroyed or turned into an atom of a different element.

You cannot see atoms regrouping, so how do you know a reaction has happened? Look for one or more of these signs:

🫧 Evolution of a gasBubbles (effervescence) appear, e.g. zinc in dilute sulphuric acid gives hydrogen.
🎨 Change in colourA new colour appears or an old one fades, e.g. a blue solution turning green.
🌡️ Change in temperatureThe mixture becomes warm or cold; some reactions give out light as well.
⬇️ Formation of a precipitateAn insoluble solid appears when two clear solutions are mixed.
🧊 Change in stateFor example, a solid ribbon burns to a powder, or a gas forms a liquid or solid.

Three classic laboratory activities show these signs clearly.

✏️ Example 1 — Physical or chemical?
Decide whether each change is physical or chemical, and give the evidence: (a) butter melting in a hot pan; (b) milk setting into curd; (c) common salt dissolving in water; (d) a candle burning; (e) a cut apple turning brown.
Test Ask: is a new substance with different properties formed, or is the same substance just in a different form or place?
(a), (c) Physical. Melted butter sets again when cooled; the salt can be recovered by evaporating the water. No new substance is formed.
(b) Chemical. Curd has a sour taste and a thick texture that milk does not have, and it cannot be turned back into milk. (Bacteria convert milk sugar into lactic acid.)
(d) Both. Some wax only melts (physical), but the wax vapour burns: heat and light are given out and new gases (carbon dioxide and water vapour) are formed — a chemical change.
(e) Chemical. A new brown substance forms on the cut surface when it reacts with oxygen in the air — a change in colour.
⚠️ A sign is a clue, not a proof

Bubbles also appear when water boils, and a colour can change when you simply dilute a solution. The real test is always whether a new substance has been formed. In an exam, name the sign you observe and the new substance responsible: “a yellow precipitate of lead iodide is formed”, not just “it turned yellow”.

From Words to Formulas: Chemical Equations

The quickest way to describe the burning of magnesium is a word equation, with the reactants on the left, the products on the right, and an arrow meaning “react to give”: $$\text{magnesium} + \text{oxygen} \rightarrow \text{magnesium oxide}$$ Writing the formula of each substance in place of its name gives a chemical equation: $$\mathrm{Mg + O_{2} \rightarrow MgO}$$ Remember from Class 9 that oxygen gas exists as molecules $\mathrm{O_{2}}$, and magnesium oxide has the formula $\mathrm{MgO}$ because the valencies of magnesium and oxygen are both 2.

📘 Definition 1.2 — Chemical equation and skeletal equation
A chemical equation represents a chemical reaction using the symbols and formulas of the reactants (written on the left, joined by $+$) and products (on the right), separated by an arrow $\rightarrow$ pointing towards the products. An equation in which the formulas are correct but the number of atoms of some element is not the same on both sides is a skeletal equation (an unbalanced equation).

Count the atoms in $\mathrm{Mg + O_{2} \rightarrow MgO}$: on the left there are 2 oxygen atoms, on the right only 1. One oxygen atom seems to have vanished — and atoms do not vanish. The equation is only a skeleton; it still has to be balanced. The reason it must be balanced is one of the founding laws of chemistry.

📐 Law 1.1 — Law of conservation of mass
In a chemical reaction mass can neither be created nor destroyed. Provided nothing enters or leaves the system (a closed container, with every gas counted), the total mass of the products equals the total mass of the reactants. The reason is that the atoms are only rearranged: the same atoms, in the same numbers, are present before and after the reaction.

The law was established in the 1770s–1780s by the French chemist Antoine Lavoisier, who heated metals in sealed glass vessels and found that the total mass did not change, even though the metal turned into a powdery “calx” (an oxide). Earlier experimenters, working in open vessels, had been puzzled that metals seemed to gain weight on heating: the extra mass was oxygen taken from the air.

✏️ Example 2 — Using the law of conservation of mass
(a) $2.4\ \mathrm{g}$ of magnesium burns completely in $1.6\ \mathrm{g}$ of oxygen. What mass of magnesium oxide is formed?
(b) A student heats $10.0\ \mathrm{g}$ of calcium carbonate (marble chips) strongly in a sealed tube. It decomposes into calcium oxide and carbon dioxide; $5.6\ \mathrm{g}$ of calcium oxide is formed. What mass of carbon dioxide is formed?
(c) The student repeats (b) in an open test tube and finds that the solid now weighs only $5.6\ \mathrm{g}$. Does this break the law?
(a) Total mass of reactants $=2.4\ \mathrm{g}+1.6\ \mathrm{g}=4.0\ \mathrm{g}$. All of it ends up in the only product, so the mass of magnesium oxide is $4.0\ \mathrm{g}$.
(b) $$\text{calcium carbonate} \rightarrow \text{calcium oxide} + \text{carbon dioxide}.$$ Mass of carbon dioxide $=10.0\ \mathrm{g}-5.6\ \mathrm{g}=4.4\ \mathrm{g}$.
(c) No. The $4.4\ \mathrm{g}$ of carbon dioxide is a gas and escaped into the air, so it was not weighed. The law applies to the whole system; in an open container gases can leave (or, as with burning magnesium, oxygen can join from the air).
Check Product masses $5.6+4.4=10.0\ \mathrm{g}$, equal to the mass of reactant. ✓
⚠️ A spelling note
This course uses the spellings sulphur, sulphuric acid, sulphate, as in the Class 10 syllabus and the board exams. The international (IUPAC) spellings are sulfur, sulfuric acid, sulfate; you will meet them in Class 11–12 and in many other books. Both name exactly the same substances.

Balancing a Chemical Equation

To balance an equation we put a whole number, the coefficient, in front of a formula. A coefficient multiplies everything in the formula after it: $\mathrm{2H_{2}O}$ means two water molecules, containing $2\times2=4$ hydrogen atoms and $2\times1=2$ oxygen atoms. For burning magnesium: $$\mathrm{2Mg + O_{2} \rightarrow 2MgO}\qquad\text{Mg: } 2=2,\quad\text{O: } 2=2.$$

📘 Definition 1.3 — Balanced chemical equation
A balanced chemical equation is one in which the number of atoms of each element is the same on the reactant side and on the product side. It is written with the smallest possible whole-number coefficients (a coefficient of 1 is not written).
🧭 Problem-solving strategy — balancing by hit and trial
  1. Write the skeletal equation with correct formulas. Imagine each formula in a box: you may put numbers in front of the box but never change what is inside (the subscripts).
  2. Make an atom table: list each element with the number of its atoms on the left and on the right.
  3. Start with the formula that has the most atoms, and within it the element with the most atoms. Balance that element by choosing a coefficient.
  4. Balance the remaining elements one at a time, updating the table after every change. Elements that occur as free elements (like $\mathrm{Fe}$, $\mathrm{H_{2}}$, $\mathrm{O_{2}}$) are usually easiest to fix last. Treat a polyatomic ion such as $\mathrm{NO_{3}}$ or $\mathrm{SO_{4}}$ that appears unchanged on both sides as a single unit.
  5. Check every element once more, and make sure the coefficients are the smallest whole numbers.
  6. Add state symbols and any conditions (next subsection).
✏️ Example 3 — Iron and steam
When steam is passed over red-hot iron, iron(II,III) oxide $\mathrm{Fe_{3}O_{4}}$ and hydrogen are formed. Balance the equation $\mathrm{Fe + H_{2}O \rightarrow Fe_{3}O_{4} + H_{2}}$.
Atom table
ElementLeft (reactants)Right (products)
Fe13
H22
O14
Largest formula $\mathrm{Fe_{3}O_{4}}$ has the most atoms (7), and oxygen is its element with the most atoms (4). To get 4 oxygen atoms on the left, write $\mathrm{4H_{2}O}$: $$\mathrm{Fe + 4H_{2}O \rightarrow Fe_{3}O_{4} + H_{2}}\qquad\text{O: } 4=4.$$
Hydrogen Now there are $4\times2=8$ H atoms on the left and 2 on the right. Write $\mathrm{4H_{2}}$: $$\mathrm{Fe + 4H_{2}O \rightarrow Fe_{3}O_{4} + 4H_{2}}\qquad\text{H: } 8=8.$$
Iron 1 Fe on the left, 3 on the right: write $\mathrm{3Fe}$. $$\mathrm{3Fe + 4H_{2}O \rightarrow Fe_{3}O_{4} + 4H_{2}}$$
Check Fe: $3=3$; H: $8=8$; O: $4=4$. ✓ With state symbols (steam is a gas): $\mathrm{3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{4}(s) + 4H_{2}(g)}$.
Interactive 1 — Equation Balancer: Count Every Atom

Each row of balls is one element: atoms on the left of the arrow (reactants) and on the right (products). Use the − / + buttons to change the coefficients — you cannot touch the subscripts. A row turns green when both sides match. Press ▶ Play to watch the hit-and-trial method solve the equation step by step.

atoms on the left—
atoms on the right—
elements balanced—
smallest whole numbers?—
⚠️ Common mistakes when balancing

1. Changing a subscript. To fix $\mathrm{H_{2} + O_{2} \rightarrow H_{2}O}$ you might be tempted to write $\mathrm{H_{2} + O_{2} \rightarrow H_{2}O_{2}}$. The atoms now balance, but $\mathrm{H_{2}O_{2}}$ is hydrogen peroxide, a different substance! Changing a subscript changes the substance. The correct equation is $\mathrm{2H_{2} + O_{2} \rightarrow 2H_{2}O}$.

2. Writing a coefficient in the middle. $\mathrm{H_{2}2O}$ or $\mathrm{Fe_{3}4O}$ mean nothing. Coefficients go only in front of a complete formula.

3. Forgetting brackets. In $\mathrm{Pb(NO_{3})_{2}}$ the subscript 2 multiplies the whole bracket: 1 Pb, 2 N and 6 O atoms.

4. Not reducing. $\mathrm{4H_{2} + 2O_{2} \rightarrow 4H_{2}O}$ is balanced, but every coefficient can be halved; write $\mathrm{2H_{2} + O_{2} \rightarrow 2H_{2}O}$.

✏️ Example 4 — When the numbers do not divide: aluminium and oxygen
A freshly cut aluminium surface quickly forms a thin protective coat of aluminium oxide. Balance $\mathrm{Al + O_{2} \rightarrow Al_{2}O_{3}}$.
Oxygen first $\mathrm{Al_{2}O_{3}}$ is the largest formula. Oxygen atoms come in twos on the left ($\mathrm{O_{2}}$) and in threes on the right ($\mathrm{O_{3}}$ inside $\mathrm{Al_{2}O_{3}}$). The smallest number that is a multiple of both 2 and 3 is their LCM, 6.
Make 6 on each side $\mathrm{3O_{2}}$ gives $3\times2=6$ O, and $\mathrm{2Al_{2}O_{3}}$ gives $2\times3=6$ O: $$\mathrm{Al + 3O_{2} \rightarrow 2Al_{2}O_{3}}$$
Aluminium Right side: $2\times2=4$ Al. Write $\mathrm{4Al}$: $$\mathrm{4Al(s) + 3O_{2}(g) \rightarrow 2Al_{2}O_{3}(s)}$$
Check Al: $4=4$; O: $6=6$. ✓ The coefficients 4, 3, 2 have no common factor, so they are the smallest whole numbers.
✏️ Example 5 — Keeping a polyatomic ion together
Balance the equation for the yellow-precipitate reaction, $\mathrm{Pb(NO_{3})_{2} + KI \rightarrow PbI_{2} + KNO_{3}}$, and add state symbols.
Spot the unit The nitrate group $\mathrm{NO_{3}}$ appears unchanged on both sides, so count it as one unit: 2 on the left (inside $\mathrm{Pb(NO_{3})_{2}}$), 1 on the right.
Nitrate Write $\mathrm{2KNO_{3}}$: $\mathrm{Pb(NO_{3})_{2} + KI \rightarrow PbI_{2} + 2KNO_{3}}$. Now K is 1 on the left and 2 on the right.
Potassium and iodine Write $\mathrm{2KI}$: K $2=2$ and I $2=2$. $$\mathrm{Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)}$$
Check Pb $1=1$; N $2=2$; O $6=6$; K $2=2$; I $2=2$. ✓ The state symbol (s) on $\mathrm{PbI_{2}}$ records the yellow precipitate; the other three substances stay dissolved.

State Symbols and Reaction Conditions

A balanced equation becomes much more informative when we add the physical state of each substance and the conditions needed for the reaction.

(s) solid$\mathrm{Mg(s)}$, $\mathrm{PbI_{2}(s)}$ — including precipitates
(l) liquid$\mathrm{H_{2}O(l)}$ — a pure liquid
(g) gas$\mathrm{H_{2}(g)}$, $\mathrm{CO_{2}(g)}$; water as steam is $\mathrm{H_{2}O(g)}$
(aq) aqueousdissolved in water: $\mathrm{H_{2}SO_{4}(aq)}$, $\mathrm{KI(aq)}$

For zinc and dilute sulphuric acid: $$\mathrm{Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)}$$ Conditions are written above or below the arrow: heat (sometimes shown as $\Delta$), sunlight, an electric current, a temperature, a pressure, or a catalyst — a substance that speeds up a reaction without itself being used up. For example, carbon monoxide and hydrogen combine to make methanol (an industrial fuel and solvent) at a pressure of about 340 atmospheres: $$\mathrm{CO(g) + 2H_{2}(g) \xrightarrow{340\ atm} CH_{3}OH(l)}$$

📘 Definition 1.4 — State symbols and precipitate
State symbols are the labels $\mathrm{(s)}$, $\mathrm{(l)}$, $\mathrm{(g)}$ and $\mathrm{(aq)}$ written immediately after a formula to show whether the substance is a solid, a liquid, a gas, or dissolved in water. A precipitate is an insoluble solid that separates out of a solution during a reaction; it is given the state symbol $\mathrm{(s)}$.

In many school-book descriptions you will also see an upward arrow $\uparrow$ after a gas that escapes and a downward arrow $\downarrow$ after a precipitate, e.g. $\mathrm{Zn + H_{2}SO_{4} \rightarrow ZnSO_{4} + H_{2}\uparrow}$. In this course balanced equations use state symbols; recognise the arrows, but do not mix the two conventions in one equation.

✏️ Example 6 — A complete equation from a description
Write a balanced equation, with state symbols and conditions, for each process.
(a) Hydrogen and oxygen gases combine to form liquid water. (ISRO's cryogenic upper-stage engines burn liquid hydrogen in liquid oxygen; the product is water.)
(b) In photosynthesis, green leaves use carbon dioxide and water in the presence of sunlight and chlorophyll to make glucose $\mathrm{C_{6}H_{12}O_{6}}$ and oxygen.
(a) Skeleton $\mathrm{H_{2} + O_{2} \rightarrow H_{2}O}$. O: 2 vs 1, so write $\mathrm{2H_{2}O}$; then H: 2 vs 4, so write $\mathrm{2H_{2}}$. $$\mathrm{2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)}$$ Check: H $4=4$, O $2=2$. ✓
(b) Skeleton $\mathrm{CO_{2} + H_{2}O \rightarrow C_{6}H_{12}O_{6} + O_{2}}$. Glucose is the largest formula. Carbon: write $\mathrm{6CO_{2}}$. Hydrogen: 12 in glucose, so write $\mathrm{6H_{2}O}$.
(b) Oxygen Left: $6\times2+6\times1=18$ O. Right: 6 in glucose, so $\mathrm{O_{2}}$ must supply 12 atoms: $\mathrm{6O_{2}}$. $$\mathrm{6CO_{2}(g) + 6H_{2}O(l) \xrightarrow[\text{chlorophyll}]{\text{sunlight}} C_{6}H_{12}O_{6}(aq) + 6O_{2}(g)}$$
Check C $6=6$; H $12=12$; O $18=6+12$. ✓ (You may also see a version with $\mathrm{12H_{2}O}$ on the left and $\mathrm{6H_{2}O}$ on the right, because the oxygen released actually comes from water; cancelling $\mathrm{6H_{2}O}$ from both sides gives the equation above.)
Interactive 2 — Atoms Regroup, Mass Stays: Open vs Sealed Vessel

Watch the reactant molecules break apart and their atoms regroup into products; every atom you see at the start is still there at the end. The digital balance shows the mass (in atomic mass units, u) of everything inside the vessel. Switch between a sealed and an open vessel: when a gas can enter from the air or escape into it, the reading changes — but the total mass of all the atoms never does.

balance reading at start (u)—
balance reading now (u)—
change in reading (u)—
total mass of all atoms (u)—
⚠️ What a balanced equation does not tell you

A balanced equation tells you the reactants and products, their formulas and states, and the ratio in which molecules react. It does not tell you whether the reaction will actually happen when you mix the substances, how fast it goes, whether heat is given out or taken in (unless you add that information), or what the molecules look like. You can write a perfectly balanced equation for copper dissolving in water to give hydrogen — it simply does not happen. Chemistry needs experiments as well as equations.

Exercises

I
Chemical changes and the law of conservation of mass.
  • 1 Milk is turned into paneer by adding lemon juice to hot milk. Is this a physical or a chemical change? Give one piece of evidence.
  • 2 Classify as physical or chemical: (a) a wet towel drying; (b) a matchstick burning; (c) an iron nail rusting; (d) glass breaking.
  • 3 Name the sign of a chemical change observed in each case: (a) a blue solution turns green when an iron nail is dipped into it; (b) baking soda fizzes with lemon juice; (c) a mixture of quicklime and water gets very hot.
  • 4 Why is magnesium ribbon rubbed with sandpaper before it is burnt?
  • 5 Describe what you would observe when potassium iodide solution is added to lead nitrate solution, and name the product responsible.
  • 6 $12\ \mathrm{g}$ of carbon burns completely in $32\ \mathrm{g}$ of oxygen to form carbon dioxide only. What mass of carbon dioxide is formed?
  • 7 $5.0\ \mathrm{g}$ of calcium carbonate is heated in a sealed tube and gives $2.8\ \mathrm{g}$ of calcium oxide and carbon dioxide. Find the mass of carbon dioxide. If the tube were open, what would the residue weigh, and why?
  • 8 $4.8\ \mathrm{g}$ of magnesium burns to give $8.0\ \mathrm{g}$ of magnesium oxide. What mass of oxygen combined with the magnesium? What is the ratio of the mass of magnesium to the mass of oxygen, in its simplest form?
  • 9 A student places a flask containing zinc and dilute sulphuric acid on an electronic balance, with no stopper, and sees the reading fall by $0.10\ \mathrm{g}$. Explain the observation. Does it disprove the law of conservation of mass?
  • 10 Applied. A sealed plastic bottle containing vinegar, with a small packet of baking soda taped inside the cap, weighs $152.0\ \mathrm{g}$. The bottle is tipped so they mix and fizz. What does the balance read when the fizzing stops? The cap is then opened and the reading becomes $151.0\ \mathrm{g}$. What mass of gas escaped?
1 Chemical. Paneer is a new solid with different properties; the milk proteins clump together (curdle) when acid is added and cannot be turned back into milk.
2 (a) physical (water only evaporates); (b) chemical (heat, light and new gases and ash); (c) chemical (a new brown substance, rust); (d) physical (still glass, only smaller pieces).
3 (a) change in colour; (b) evolution of a gas (carbon dioxide); (c) change in temperature (heat is given out).
4 Magnesium slowly reacts with air to form a dull layer of magnesium oxide (and carbonate) on its surface. This layer stops the metal from catching fire easily, so it is rubbed off to expose fresh magnesium.
5 A bright yellow precipitate forms at once in the colourless solutions. It is lead iodide, $\mathrm{PbI_{2}}$: $$\mathrm{Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)}.$$
6 Mass of products = mass of reactants $=12+32=44\ \mathrm{g}$ of carbon dioxide.
7 Carbon dioxide $=5.0-2.8=2.2\ \mathrm{g}$. In an open tube the residue would still be $2.8\ \mathrm{g}$ of calcium oxide, because the $2.2\ \mathrm{g}$ of carbon dioxide gas escapes into the air.
8 Oxygen $=8.0-4.8=3.2\ \mathrm{g}$. Ratio $\text{Mg}:\text{O}=4.8:3.2=3:2$.
9 Hydrogen gas formed in the reaction escapes from the open flask, so it is no longer weighed. The law is not broken: if the flask were sealed (or the hydrogen collected and weighed), the total mass would be unchanged.
10 While sealed the reading stays $152.0\ \mathrm{g}$ — the carbon dioxide formed is trapped inside. After opening, $152.0-151.0=1.0\ \mathrm{g}$ of carbon dioxide escaped.
II
Balance each skeletal equation. Show your atom table for at least three of them.
  • 1 $\mathrm{H_{2} + Cl_{2} \rightarrow HCl}$
  • 2 $\mathrm{Na + O_{2} \rightarrow Na_{2}O}$
  • 3 $\mathrm{Fe + Cl_{2} \rightarrow FeCl_{3}}$
  • 4 $\mathrm{SO_{2} + O_{2} \rightarrow SO_{3}}$
  • 5 $\mathrm{KClO_{3} \rightarrow KCl + O_{2}}$
  • 6 $\mathrm{CaCO_{3} + HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}}$
  • 7 $\mathrm{NH_{3} + O_{2} \rightarrow NO + H_{2}O}$
  • 8 $\mathrm{Al + H_{2}SO_{4} \rightarrow Al_{2}(SO_{4})_{3} + H_{2}}$
  • 9 $\mathrm{C_{2}H_{5}OH + O_{2} \rightarrow CO_{2} + H_{2}O}$
  • 10 $\mathrm{Fe_{2}O_{3} + CO \rightarrow Fe + CO_{2}}$
  • 11 $\mathrm{P_{4} + O_{2} \rightarrow P_{4}O_{10}}$
  • 12 $\mathrm{MnO_{2} + HCl \rightarrow MnCl_{2} + Cl_{2} + H_{2}O}$
  • 13 $\mathrm{C_{6}H_{12}O_{6} + O_{2} \rightarrow CO_{2} + H_{2}O}$ (respiration)
  • 14 $\mathrm{Cu + HNO_{3} \rightarrow Cu(NO_{3})_{2} + NO_{2} + H_{2}O}$ (challenging)
Check: type your balanced equation and count the atoms on each side
Write formulas with ordinary digits, coefficients in front, and -> or = for the arrow, e.g. Pb(NO3)2 + 2KI -> PbI2 + 2KNO3.
1 Two H and two Cl on the left, so two HCl: $\mathrm{H_{2} + Cl_{2} \rightarrow 2HCl}$.
2 O: write $\mathrm{2Na_{2}O}$ (2 O); then Na: $2\times2=4$, write $\mathrm{4Na}$: $\mathrm{4Na + O_{2} \rightarrow 2Na_{2}O}$.
3 Cl comes in twos (left) and threes (right); LCM 6: $\mathrm{3Cl_{2}}$ and $\mathrm{2FeCl_{3}}$, then $\mathrm{2Fe}$: $\mathrm{2Fe + 3Cl_{2} \rightarrow 2FeCl_{3}}$.
4 S is balanced; O is $2+2=4$ vs 3. Try $\mathrm{2SO_{3}}$ (6 O, 2 S), then $\mathrm{2SO_{2}}$ (4 O) leaves 2 O for $\mathrm{O_{2}}$: $\mathrm{2SO_{2} + O_{2} \rightarrow 2SO_{3}}$.
5 O: 3 vs 2, LCM 6: $\mathrm{2KClO_{3}}$ and $\mathrm{3O_{2}}$; then K and Cl: $\mathrm{2KCl}$. $\mathrm{2KClO_{3} \rightarrow 2KCl + 3O_{2}}$.
6 Ca, C and O start balanced; Cl is 1 vs 2, so $\mathrm{2HCl}$, which also balances H (2 = 2): $\mathrm{CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}}$.
7 H: 3 vs 2, LCM 6: $\mathrm{2NH_{3}}$ and $\mathrm{3H_{2}O}$; N: $\mathrm{2NO}$; O on the right $=2+3=5$, needing $\tfrac52\mathrm{O_{2}}$. Double everything to clear the fraction: $\mathrm{4NH_{3} + 5O_{2} \rightarrow 4NO + 6H_{2}O}$. Check: N 4, H 12, O 10 on each side.
8 Treat $\mathrm{SO_{4}}$ as a unit: 3 on the right, so $\mathrm{3H_{2}SO_{4}}$; then H: 6, so $\mathrm{3H_{2}}$; Al: $\mathrm{2Al}$. $\mathrm{2Al + 3H_{2}SO_{4} \rightarrow Al_{2}(SO_{4})_{3} + 3H_{2}}$.
9 C: $\mathrm{2CO_{2}}$; H (6 in ethanol): $\mathrm{3H_{2}O}$; O on the right $=4+3=7$, ethanol already has 1, so $\mathrm{O_{2}}$ supplies 6: $\mathrm{3O_{2}}$. $\mathrm{C_{2}H_{5}OH + 3O_{2} \rightarrow 2CO_{2} + 3H_{2}O}$.
10 Fe: $\mathrm{2Fe}$. Each CO picks up one O from $\mathrm{Fe_{2}O_{3}}$ to become $\mathrm{CO_{2}}$; there are 3 O to remove, so $\mathrm{3CO}$ and $\mathrm{3CO_{2}}$: $\mathrm{Fe_{2}O_{3} + 3CO \rightarrow 2Fe + 3CO_{2}}$. Check O: $3+3=6$ on each side.
11 P is balanced (4 = 4); O: 10 on the right, so $\mathrm{5O_{2}}$: $\mathrm{P_{4} + 5O_{2} \rightarrow P_{4}O_{10}}$.
12 O: 2 in $\mathrm{MnO_{2}}$, so $\mathrm{2H_{2}O}$; H: 4, so $\mathrm{4HCl}$; Cl: 4 on the left, $2+2$ on the right ✓. $\mathrm{MnO_{2} + 4HCl \rightarrow MnCl_{2} + Cl_{2} + 2H_{2}O}$.
13 C: $\mathrm{6CO_{2}}$; H: $\mathrm{6H_{2}O}$; O on the right $=12+6=18$, glucose has 6, so $\mathrm{6O_{2}}$. $\mathrm{C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O}$ — photosynthesis run backwards.
14 Cu is balanced (1 = 1). All the H of the acid ends up in water, so if we use $x$ molecules of $\mathrm{HNO_{3}}$ we get $\tfrac{x}{2}$ water molecules; $x$ must be even. Try $x=2$: water 1, but N would be 2 on the left and already 2 in $\mathrm{Cu(NO_{3})_{2}}$, leaving none for $\mathrm{NO_{2}}$ — no good. Try $x=4$: water 2, and N: $4=2+2$, so $\mathrm{2NO_{2}}$. O: left $4\times3=12$; right $6+4+2=12$ ✓. $\mathrm{Cu + 4HNO_{3} \rightarrow Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O}$.
III
Write a balanced chemical equation with state symbols (and conditions, where given) for each reaction.
  • 1 Zinc granules react with dilute sulphuric acid to give zinc sulphate solution and hydrogen gas.
  • 2 Calcium oxide (quicklime) reacts vigorously with water to form calcium hydroxide solution.
  • 3 Silver nitrate solution is added to sodium chloride solution; a white precipitate of silver chloride forms and sodium nitrate stays in solution.
  • 4 Copper powder is heated in air and turns black, forming copper(II) oxide.
  • 5 Sodium hydroxide solution reacts with hydrochloric acid to form sodium chloride solution and water.
  • 6 Methane (the main gas in CNG and biogas) burns in oxygen to form carbon dioxide and water vapour.
  • 7 An iron nail is placed in copper sulphate solution; iron sulphate solution and copper are formed.
  • 8 Water is split into hydrogen and oxygen gases by passing an electric current.
1 $\mathrm{Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)}$ (already balanced).
2 $\mathrm{CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(aq)}$ (heat is given out).
3 $$\mathrm{AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)}.$$
4 $\mathrm{2Cu(s) + O_{2}(g) \xrightarrow{\text{heat}} 2CuO(s)}$.
5 $\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_{2}O(l)}$.
6 $\mathrm{CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)}$. Check: C 1, H 4, O 4 on each side.
7 $\mathrm{Fe(s) + CuSO_{4}(aq) \rightarrow FeSO_{4}(aq) + Cu(s)}$ (the blue solution turns pale green and a brown coat of copper forms on the nail).
8 $\mathrm{2H_{2}O(l) \xrightarrow{\text{electric current}} 2H_{2}(g) + O_{2}(g)}$. The volume of hydrogen collected is twice that of oxygen, matching the coefficients 2 : 1.
IV
Reasoning, error-spotting and applications.
  • 1 A student balances $\mathrm{H_{2} + O_{2} \rightarrow H_{2}O}$ by writing $\mathrm{H_{2} + O_{2} \rightarrow H_{2}O_{2}}$. Explain what is wrong.
  • 2 Find and correct the error: $\mathrm{N_{2} + H_{2} \rightarrow 2NH_{3}}$ is “balanced because there are 2 N on each side”.
  • 3 How many atoms of each element are represented by $\mathrm{3Ca(OH)_{2}}$? How many atoms altogether?
  • 4 Give two pieces of information that a balanced chemical equation does not tell us.
  • 5 Applied. Domestic LPG is mostly butane, $\mathrm{C_{4}H_{10}}$. Balance its complete combustion, $\mathrm{C_{4}H_{10} + O_{2} \rightarrow CO_{2} + H_{2}O}$. How many oxygen molecules are needed for every butane molecule? Why can a blocked air vent on a gas stove be dangerous?
  • 6 Challenging. Balance $$\mathrm{KMnO_{4} + HCl \rightarrow KCl + MnCl_{2} + Cl_{2} + H_{2}O}.$$
1 The subscript was changed: $\mathrm{H_{2}O_{2}}$ is hydrogen peroxide, not water. Only coefficients may be changed: $\mathrm{2H_{2} + O_{2} \rightarrow 2H_{2}O}$.
2 Hydrogen is not balanced: 2 H on the left but $2\times3=6$ on the right. Correct: $\mathrm{N_{2} + 3H_{2} \rightarrow 2NH_{3}}$ (N 2 = 2, H 6 = 6).
3 Ca: 3; O: $3\times2=6$; H: $3\times2=6$; altogether $3+6+6=15$ atoms.
4 Any two of: whether the reaction actually takes place; how fast it is; whether heat is given out or absorbed; the concentrations of the solutions; what happens step by step at the molecular level.
5 C: $\mathrm{4CO_{2}}$; H: $\mathrm{5H_{2}O}$; O on the right $=8+5=13$, so $\tfrac{13}{2}\mathrm{O_{2}}$; doubling: $$\mathrm{2C_{4}H_{10}(g) + 13O_{2}(g) \rightarrow 8CO_{2}(g) + 10H_{2}O(g)}.$$ That is $13\div2=6.5$ oxygen molecules per butane molecule. With too little air the gas burns incompletely, forming soot and poisonous carbon monoxide instead of carbon dioxide.
6 O appears only in $\mathrm{KMnO_{4}}$ and $\mathrm{H_{2}O}$: with $\mathrm{2KMnO_{4}}$ (8 O) we need $\mathrm{8H_{2}O}$, hence $\mathrm{16HCl}$ (16 H). K and Mn: $\mathrm{2KCl}$, $\mathrm{2MnCl_{2}}$, using $2+4=6$ Cl; the remaining $16-6=10$ Cl form $\mathrm{5Cl_{2}}$. $$\mathrm{2KMnO_{4} + 16HCl \rightarrow 2KCl + 2MnCl_{2} + 5Cl_{2} + 8H_{2}O}$$ Check: K 2, Mn 2, O 8, H 16, Cl 16 on each side.
🎯 Exam corner BoardBoard · A–RBoard · Case-basedJEE Main / NEET

E1. Board Which of the following correctly represents the burning of magnesium in air? (A) $\mathrm{Mg + O_{2} \rightarrow MgO_{2}}$   (B) $\mathrm{Mg + O \rightarrow MgO}$   (C) $\mathrm{2Mg(s) + O_{2}(g) \rightarrow 2MgO(s)}$   (D) $\mathrm{Mg_{2} + O_{2} \rightarrow 2MgO}$

Idea Formulas must be correct ($\mathrm{MgO}$, $\mathrm{O_{2}}$, and magnesium as $\mathrm{Mg}$) and atoms must balance. (A) has a wrong formula $\mathrm{MgO_{2}}$; (B) writes oxygen as single atoms; (D) invents $\mathrm{Mg_{2}}$.
Answer (C): Mg $2=2$, O $2=2$, correct formulas and state symbols.

E2. Board · A–R Assertion (A): A chemical equation must be balanced. Reason (R): In a chemical reaction atoms are neither created nor destroyed. Choose: (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.

Idea Both statements are true. Because atoms are only rearranged, the number of atoms of each element — and so the total mass — must be the same on both sides; that is exactly why we balance.
Answer (A).

E3. Board · Case-based In a school laboratory, Riya pours $3.31\ \mathrm{g}$ of lead nitrate dissolved in water into a beaker containing $3.32\ \mathrm{g}$ of potassium iodide dissolved in water. A bright yellow solid appears at once. After filtering and drying she collects $4.61\ \mathrm{g}$ of the yellow solid; the other product stays dissolved. (i) Name the yellow solid and the type of solid it is called. (ii) Write the balanced equation with state symbols. (iii) Use the law of conservation of mass to find the mass of the dissolved product. (iv) State two signs of a chemical change seen in this experiment.

(i) Lead iodide, $\mathrm{PbI_{2}}$; an insoluble solid formed from solutions is a precipitate.
(ii) $$\mathrm{Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)}.$$
(iii) Mass of reactants $=3.31+3.32=6.63\ \mathrm{g}$ (the water is unchanged, so it cancels). Mass of potassium nitrate $=6.63-4.61=2.02\ \mathrm{g}$.
Answer (iv) Formation of a precipitate and a change in colour (colourless solutions give a yellow solid).

E4. Board (3 marks) Translate into balanced chemical equations with state symbols: (a) Potassium metal reacts with water to give potassium hydroxide solution and hydrogen gas. (b) Ethane gas, $\mathrm{C_{2}H_{6}}$, burns in oxygen to give carbon dioxide and water vapour. (c) Carbon dioxide is bubbled through lime water (calcium hydroxide solution), which turns milky because insoluble calcium carbonate forms along with water.

(a) H: $\mathrm{H_{2}O}$ has 2, $\mathrm{KOH}+\mathrm{H_{2}}$ has 3 — make it even with $\mathrm{2KOH}$, then $\mathrm{2H_{2}O}$, $\mathrm{2K}$: $\mathrm{2K(s) + 2H_{2}O(l) \rightarrow 2KOH(aq) + H_{2}(g)}$.
(b) C: $\mathrm{2CO_{2}}$; H: $\mathrm{3H_{2}O}$; O: $4+3=7$, so $\tfrac72\mathrm{O_{2}}$; double: $\mathrm{2C_{2}H_{6}(g) + 7O_{2}(g) \rightarrow 4CO_{2}(g) + 6H_{2}O(g)}$.
Answer (c) $$\mathrm{Ca(OH)_{2}(aq) + CO_{2}(g) \rightarrow CaCO_{3}(s) + H_{2}O(l)}$$ — already balanced (Ca 1, O 4, H 2, C 1 on each side).

E5. JEE Main / NEET Foundation. When the equation $a\,\mathrm{Al} + b\,\mathrm{HCl} \rightarrow c\,\mathrm{AlCl_{3}} + d\,\mathrm{H_{2}}$ is balanced with the smallest whole numbers, $a+b+c+d$ equals (A) 11   (B) 12   (C) 13   (D) 14

Idea Cl: 3 in $\mathrm{AlCl_{3}}$, so $b$ is a multiple of 3. H: $b$ must be even (hydrogen comes out as $\mathrm{H_{2}}$). Smallest $b=6$: then $c=2$, $a=2$, $d=3$.
Check $\mathrm{2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}}$: Al 2, H 6, Cl 6 on each side.
Answer $2+6+2+3=13$: (C).

E6. Board (2 marks) Why is it necessary to write state symbols in a chemical equation? Illustrate with the reaction of zinc with dilute sulphuric acid.

Idea State symbols make the equation more informative: they show which substances are solids, liquids, gases or solutions, so the reader knows, for example, which product bubbles off and which stays dissolved.
Answer $\mathrm{Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)}$ tells us that solid zinc reacts with an aqueous acid, zinc sulphate remains in solution, and hydrogen is given off as a gas.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much carbon dioxide does one domestic LPG cylinder put into the air — and how can a balanced equation tell us without measuring the gas?

Why it matters. Countries report their greenhouse-gas emissions to the United Nations (UNFCCC) mostly by calculation, not by measuring every chimney: the fuel burned is known, and balanced combustion equations plus the law of conservation of mass convert it into carbon dioxide. India's clean-cooking schemes have moved millions of families from wood and dung to LPG, and engineers compare fuels this way.

What to do.

  1. Take LPG as pure butane. Write and balance $\mathrm{C_{4}H_{10} + O_{2} \rightarrow CO_{2} + H_{2}O}$ (see Exercise IV.5).
  2. Using atomic masses $\mathrm{C}=12\ \mathrm{u}$, $\mathrm{H}=1\ \mathrm{u}$, $\mathrm{O}=16\ \mathrm{u}$, find the mass of 2 butane molecules and of 8 carbon dioxide molecules. Hence find the mass of carbon dioxide formed per gram of butane (a simple ratio).
  3. Check that the total mass of reactants equals the total mass of products in your equation.
  4. A standard Indian domestic cylinder holds $14.2\ \mathrm{kg}$ of LPG. Estimate the carbon dioxide released when it is used up, and for a family that uses 8 cylinders a year.
  5. Repeat step 2 for propane, $\mathrm{C_{3}H_{8}}$ (also present in LPG), and for methane, $\mathrm{CH_{4}}$ (piped natural gas in Indian cities, and the main cooking and heating gas in much of Europe and North America).

Tools: Calculator; optionally a spreadsheet to compare the three fuels.

Hand in: A half-page note: your balanced equations, the mass check, the carbon dioxide per cylinder and per year, and one reason why the real figure differs (e.g. LPG is a butane–propane mixture).

Going further. Burning a fuel gives energy as well as carbon dioxide. Look up the energy released per kilogram of LPG and of methane, and decide which fuel gives less carbon dioxide per unit of energy. Compare with burning wood, which also releases smoke particles indoors.

Instructor note: $\mathrm{2C_{4}H_{10} + 13O_{2} \rightarrow 8CO_{2} + 10H_{2}O}$; masses $116+416=352+180=532\ \mathrm{u}$. Carbon dioxide per gram of butane $=352/116\approx3.03\ \mathrm{g}$, so one cylinder gives about $43\ \mathrm{kg}$ and 8 cylinders about $345\ \mathrm{kg}$. Propane gives $132/44=3.0\ \mathrm{g}$ per gram and methane $44/16=2.75\ \mathrm{g}$ per gram; methane also releases slightly more energy per kilogram, so it emits the least carbon dioxide per unit of energy.

🧪 Self-check quiz

0/5 correct

Q1. Which observation, on its own, does not show that a chemical reaction has taken place?

Q2. The balanced form of $\mathrm{Fe + H_{2}O \rightarrow Fe_{3}O_{4} + H_{2}}$ has coefficients (in order)

Q3. We balance chemical equations because

Q4. Which state symbol is correct for sodium chloride dissolved in water?

Q5. How many atoms in total are represented by $\mathrm{2Ca(OH)_{2}}$?

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