NCERT Solutions Class 10 Science Chapter 1 — Chemical Reactions and Equations

NCERT Class 10 Science — Chemical Reactions and Equations, Chapter 1 Exercises. All 20 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The exercise divides into three groups. Questions 1 to 3 are multiple-choice checks on reading a reaction correctly. Questions 4 to 8 and 11 to 15 are the heart of the chapter — turning words into balanced equations and naming the type of reaction. Questions 16 to 20 are about oxidation and reduction and their two everyday consequences, corrosion and rancidity. Three ideas carry all of it:

• A balanced equation has the same number of atoms of every element on both sides, because mass is conserved in a chemical reaction. Balance by adjusting coefficients only, never subscripts — changing a subscript changes the substance.
• The five types of reaction: combination (two or more become one), decomposition (one becomes two or more), displacement (one element pushes another out of its compound), double displacement (two compounds exchange ions), and oxidation–reduction.
• Oxidation is gain of oxygen or loss of hydrogen; reduction is loss of oxygen or gain of hydrogen. In this chapter the oxygen half is what is tested.

Key insight. Balance an equation in a fixed order and it stops being trial and error: leave hydrogen and oxygen until last, and balance polyatomic groups such as $\text{SO}_4$ or $\text{NO}_3$ as whole units rather than atom by atom. Question 6(a) balances in one step that way — two $\text{NO}_3$ groups are needed on the left because $\text{Ca(NO}_3)_2$ carries two on the right — and the same habit makes Question 5(c), the ugliest on the page, straightforward.

Question 1

Which of the statements about the reaction below are incorrect?

$$2\text{PbO(s)} + \text{C(s)} \longrightarrow 2\text{Pb(s)} + \text{CO}_2\text{(g)}$$

(a) Lead is getting reduced. (b) Carbon dioxide is getting oxidised. (c) Carbon is getting oxidised. (d) Lead oxide is getting reduced.

Solution. Track the oxygen, which is what this chapter’s definitions are built on. On the left, lead is combined with oxygen as $\text{PbO}$; on the right it is free metal. So lead oxide has lost oxygen — lead oxide is being reduced. Carbon starts free and ends as $\text{CO}_2$, so carbon has gained oxygen — carbon is being oxidised.

Now test each statement.

(a) “Lead is getting reduced” — incorrect. It is lead oxide that is reduced. Lead is the product of that reduction; it is not itself losing oxygen, because it has none to lose.

(b) “Carbon dioxide is getting oxidised” — incorrect. Carbon dioxide is what carbon becomes after oxidation. Nothing is happening to $\text{CO}_2$ in this reaction; it is a product.

(c) “Carbon is getting oxidised” — correct, as shown above.

(d) “Lead oxide is getting reduced” — correct, as shown above.

The incorrect statements are therefore (a) and (b).

Question 2

$$\text{Fe}_2\text{O}_3 + 2\text{Al} \longrightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$$

The above reaction is an example of a

Solution. Count what goes in and what comes out. Two substances react and two are formed, so it is neither combination (which would give one product) nor decomposition (which would need one reactant).

That leaves the two displacement types. In a double displacement, two compounds exchange partners. Here only one reactant is a compound; the other, aluminium, is a free element. A free element taking the place of another element in its compound is a displacement reaction — aluminium, being the more reactive metal, displaces iron from iron(III) oxide.

This particular reaction is the thermite reaction, used to weld railway tracks, and it releases enough heat to leave the iron molten.

Question 3

What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.

Solution. A metal above hydrogen in the reactivity series displaces hydrogen from a dilute acid, giving a salt and hydrogen gas. Iron is such a metal, so a reaction certainly takes place — which rules out (c).

$$\text{Fe(s)} + 2\text{HCl(aq)} \longrightarrow \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

The salt is iron(II) chloride and the gas is hydrogen, which is (a). Option (b) is wrong because chlorine stays in the salt rather than being released, and there is no source of the hydroxide it names. Option (d) describes neutralisation of an acid by a base, which is not what a metal does — a metal releases hydrogen rather than forming water.

Question 4

What is a balanced chemical equation? Why should chemical equations be balanced?

Solution. A chemical equation is balanced when the number of atoms of each element is the same on the reactant side as on the product side. The equation in Question 1 is balanced: two lead, two oxygen and one carbon on each side.

The reason is the law of conservation of mass: in a chemical reaction, matter is neither created nor destroyed. Atoms are only rearranged — bonds break and new ones form — so every atom present at the start must still be present at the end. An unbalanced equation claims atoms have appeared from nowhere or vanished, which cannot happen.

There is a practical reason too. Only a balanced equation gives the correct ratio in which the substances react, so only a balanced equation can be used to work out how much of a reactant is needed or how much product is formed.

Question 5

Translate the following statements into chemical equations and then balance them.

Solution. Write the formulae first and balance afterwards; trying to do both at once is where mistakes come from. Remember that hydrogen, nitrogen, oxygen and chlorine exist as diatomic molecules.

(a) Hydrogen gas combines with nitrogen to form ammonia.

Unbalanced: $\text{H}_2 + \text{N}_2 \longrightarrow \text{NH}_3$. Nitrogen needs a coefficient of 2 on the right to use both atoms of $\text{N}_2$; that gives 6 hydrogen atoms on the right, so 3 molecules of $\text{H}_2$ on the left.

$$3\text{H}_2 + \text{N}_2 \longrightarrow 2\text{NH}_3$$

(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

Unbalanced: $\text{H}_2\text{S} + \text{O}_2 \longrightarrow \text{H}_2\text{O} + \text{SO}_2$. Take 2 molecules of $\text{H}_2\text{S}$, which fixes 2 sulphur and so 2 $\text{SO}_2$, and 2 $\text{H}_2\text{O}$. Oxygen last: $2 + 4 = 6$ atoms on the right, so 3 $\text{O}_2$ on the left.

$$2\text{H}_2\text{S} + 3\text{O}_2 \longrightarrow 2\text{H}_2\text{O} + 2\text{SO}_2$$

(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.

Unbalanced: $\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \longrightarrow \text{AlCl}_3 + \text{BaSO}_4$. Treat $\text{SO}_4$ as one unit: there are 3 of them on the left, so 3 $\text{BaSO}_4$ on the right, which needs 3 $\text{BaCl}_2$ on the left. That gives 6 chlorine, so 2 $\text{AlCl}_3$ — and the 2 aluminium match.

$$3\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \longrightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4$$

(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Unbalanced: $\text{K} + \text{H}_2\text{O} \longrightarrow \text{KOH} + \text{H}_2$. Hydrogen is released as $\text{H}_2$, so take 2 water molecules, giving 2 $\text{KOH}$ and one $\text{H}_2$, and therefore 2 potassium.

$$2\text{K} + 2\text{H}_2\text{O} \longrightarrow 2\text{KOH} + \text{H}_2$$

Question 6

Balance the following chemical equations.

Solution. All four are easiest if the polyatomic groups — $\text{NO}_3$, $\text{SO}_4$ — are balanced as whole units.

(a) $\text{HNO}_3 + \text{Ca(OH)}_2 \longrightarrow \text{Ca(NO}_3)_2 + \text{H}_2\text{O}$. The right-hand side carries two $\text{NO}_3$ groups, so two $\text{HNO}_3$ are needed. That gives 4 hydrogen and, with the 2 from the hydroxide, 4 hydrogen on the right as 2 water molecules.

$$2\text{HNO}_3 + \text{Ca(OH)}_2 \longrightarrow \text{Ca(NO}_3)_2 + 2\text{H}_2\text{O}$$

(b) $\text{NaOH} + \text{H}_2\text{SO}_4 \longrightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$. Two sodium on the right means 2 $\text{NaOH}$; the hydrogen then comes to 4 on the left, so 2 water.

$$2\text{NaOH} + \text{H}_2\text{SO}_4 \longrightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

(c) $\text{NaCl} + \text{AgNO}_3 \longrightarrow \text{AgCl} + \text{NaNO}_3$. Counting: one Na, one Cl, one Ag and one $\text{NO}_3$ on each side. It is already balanced.

$$\text{NaCl} + \text{AgNO}_3 \longrightarrow \text{AgCl} + \text{NaNO}_3$$

(d) $\text{BaCl}_2 + \text{H}_2\text{SO}_4 \longrightarrow \text{BaSO}_4 + \text{HCl}$. Two chlorine on the left need 2 $\text{HCl}$ on the right, and the 2 hydrogen then match.

$$\text{BaCl}_2 + \text{H}_2\text{SO}_4 \longrightarrow \text{BaSO}_4 + 2\text{HCl}$$

Question 7

Write the balanced chemical equations for the following reactions.

Solution.

(a) Calcium hydroxide + carbon dioxide → calcium carbonate + water. One of each already balances — this is the limewater test, and the calcium carbonate is what turns it milky.

$$\text{Ca(OH)}_2 + \text{CO}_2 \longrightarrow \text{CaCO}_3 + \text{H}_2\text{O}$$

(b) Zinc + silver nitrate → zinc nitrate + silver. Zinc nitrate carries two $\text{NO}_3$ groups, so 2 $\text{AgNO}_3$ are needed, releasing 2 silver.

$$\text{Zn} + 2\text{AgNO}_3 \longrightarrow \text{Zn(NO}_3)_2 + 2\text{Ag}$$

(c) Aluminium + copper chloride → aluminium chloride + copper. Aluminium chloride needs 3 chlorine per formula unit and copper chloride supplies 2, so take 2 $\text{AlCl}_3$ (6 chlorine) against 3 $\text{CuCl}_2$ (6 chlorine), giving 2 aluminium and 3 copper.

$$2\text{Al} + 3\text{CuCl}_2 \longrightarrow 2\text{AlCl}_3 + 3\text{Cu}$$

(d) Barium chloride + potassium sulphate → barium sulphate + potassium chloride. Two potassium and two chlorine give 2 $\text{KCl}$.

$$\text{BaCl}_2 + \text{K}_2\text{SO}_4 \longrightarrow \text{BaSO}_4 + 2\text{KCl}$$

Question 8

Write the balanced chemical equation for the following and identify the type of reaction in each case.

Solution. Balance first, then classify by counting reactants and products and asking what has swapped places.

(a) Potassium bromide(aq) + barium iodide(aq) → potassium iodide(aq) + barium bromide(s).

$$2\text{KBr(aq)} + \text{BaI}_2\text{(aq)} \longrightarrow 2\text{KI(aq)} + \text{BaBr}_2\text{(s)}$$

Two compounds exchange their ions — potassium takes the iodide, barium the bromide — so this is a double displacement reaction. The solid product also makes it a precipitation reaction.

(b) Zinc carbonate(s) → zinc oxide(s) + carbon dioxide(g).

$$\text{ZnCO}_3\text{(s)} \longrightarrow \text{ZnO(s)} + \text{CO}_2\text{(g)}$$

One substance breaks up into two: a decomposition reaction.

(c) Hydrogen(g) + chlorine(g) → hydrogen chloride(g).

$$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \longrightarrow 2\text{HCl(g)}$$

Two substances join to give one: a combination reaction.

(d) Magnesium(s) + hydrochloric acid(aq) → magnesium chloride(aq) + hydrogen(g).

$$\text{Mg(s)} + 2\text{HCl(aq)} \longrightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

A free element, magnesium, pushes hydrogen out of its compound: a displacement reaction.

Question 9

What does one mean by exothermic and endothermic reactions? Give examples.

Solution. The distinction is about which way energy flows during the reaction.

An exothermic reaction releases energy — usually as heat, sometimes as light — along with the products, so the surroundings get warmer. Burning is the obvious case, and so is respiration, and so is quicklime slaking:

$$\text{CaO(s)} + \text{H}_2\text{O(l)} \longrightarrow \text{Ca(OH)}_2\text{(aq)} + \text{heat}$$

$$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \longrightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)} + \text{heat}$$

An endothermic reaction absorbs energy from the surroundings, which therefore get cooler, and it usually needs that energy supplied continuously to keep going. Thermal decomposition is the standard example:

$$\text{CaCO}_3\text{(s)} \xrightarrow{\ \text{heat}\ } \text{CaO(s)} + \text{CO}_2\text{(g)}$$

Question 10

Why is respiration considered an exothermic reaction? Explain.

Solution. Every activity of the body — moving, thinking, keeping warm — needs energy, and that energy has to come from somewhere. It comes from food.

Carbohydrates in food are broken down into glucose, which then combines with the oxygen brought in by breathing:

$$\text{C}_6\text{H}_{12}\text{O}_6\text{(aq)} + 6\text{O}_2\text{(g)} \longrightarrow 6\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} + \text{energy}$$

Energy appears on the product side: it is given out by the reaction, not taken in. That is exactly what makes a reaction exothermic. It is also why the body stays warm, and why you feel warmer after exertion — the reaction is running faster.

Question 11

Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.

Solution. Compare what each does to the number of substances.

In a combination reaction two or more substances join to form a single product, and energy is usually given out:

$$\text{CaO(s)} + \text{H}_2\text{O(l)} \longrightarrow \text{Ca(OH)}_2\text{(aq)}$$

In a decomposition reaction a single substance breaks up into two or more products, and energy must usually be supplied:

$$\text{CaCO}_3\text{(s)} \xrightarrow{\ \text{heat}\ } \text{CaO(s)} + \text{CO}_2\text{(g)}$$

So the two are opposites in both respects: many becoming one against one becoming many, and energy released against energy absorbed. The pair above shows it neatly — the calcium oxide made by decomposing limestone is the very substance that combines with water in the first equation.

Question 12

Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

Solution. A decomposition reaction needs energy from outside, and the three forms give the reaction its three names — thermal decomposition, photolysis and electrolysis.

Heat (thermal decomposition). Limestone decomposes on strong heating; this is the reaction used to make quicklime in a lime kiln.

$$\text{CaCO}_3\text{(s)} \xrightarrow{\ \text{heat}\ } \text{CaO(s)} + \text{CO}_2\text{(g)}$$

Light (photolysis). Silver chloride turns grey in sunlight as silver is set free — the reaction behind black-and-white photography.

$$2\text{AgCl(s)} \xrightarrow{\ \text{sunlight}\ } 2\text{Ag(s)} + \text{Cl}_2\text{(g)}$$

Electricity (electrolysis). Passing a current through acidified water splits it, giving twice as much hydrogen as oxygen by volume.

$$2\text{H}_2\text{O(l)} \xrightarrow{\ \text{electricity}\ } 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)}$$

Question 13

What is the difference between displacement and double displacement reactions? Write equations for these reactions.

Solution. The difference is in how many compounds are involved and what changes places.

In a displacement reaction, a free element takes the place of another element in its compound — one thing is displaced, in one direction. A more reactive metal displaces a less reactive one:

$$\text{Fe(s)} + \text{CuSO}_4\text{(aq)} \longrightarrow \text{FeSO}_4\text{(aq)} + \text{Cu(s)}$$

An iron nail left in blue copper sulphate solution comes out coated with reddish-brown copper, and the solution turns green.

In a double displacement reaction, two compounds exchange ions with each other — two things change places at once, and no free element is involved:

$$\text{Na}_2\text{SO}_4\text{(aq)} + \text{BaCl}_2\text{(aq)} \longrightarrow \text{BaSO}_4\text{(s)} + 2\text{NaCl(aq)}$$

The sodium takes the chloride and the barium the sulphate, and the insoluble barium sulphate comes down as a white precipitate.

Question 14

In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.

Solution. Copper is more reactive than silver, so it displaces silver from its salt solution — the copper goes into solution as copper(II) nitrate and the silver comes out as free metal.

Copper forms $\text{Cu}^{2+}$, so each copper atom releases two silver atoms, and two $\text{AgNO}_3$ are needed to supply the two nitrate groups:

$$\text{Cu(s)} + 2\text{AgNO}_3\text{(aq)} \longrightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{Ag(s)}$$

The colourless silver nitrate solution turns blue as copper(II) nitrate forms, and grey silver is deposited on the copper.

Question 15

What do you mean by a precipitation reaction? Explain by giving examples.

Solution. A precipitation reaction is one in which an insoluble solid — the precipitate — is formed when two solutions are mixed. It happens when the ions of the two solutions pair up in a new way and one of the new pairs will not dissolve.

Mixing sodium sulphate and barium chloride solutions, both clear and colourless, gives a white precipitate at once:

$$\text{Na}_2\text{SO}_4\text{(aq)} + \text{BaCl}_2\text{(aq)} \longrightarrow \text{BaSO}_4\text{(s)} + 2\text{NaCl(aq)}$$

The barium sulphate is insoluble and settles out; the sodium chloride stays in solution.

Mixing lead nitrate and potassium iodide gives a yellow precipitate of lead iodide in the same way:

$$\text{Pb(NO}_3)_2\text{(aq)} + 2\text{KI(aq)} \longrightarrow \text{PbI}_2\text{(s)} + 2\text{KNO}_3\text{(aq)}$$

Every precipitation reaction is also a double displacement reaction, since the two compounds have exchanged ions.

Question 16

Explain the following in terms of gain or loss of oxygen with two examples each.

(a) Oxidation (b) Reduction

Solution. In this chapter the two words are defined by what happens to oxygen, and nothing else is needed to answer.

(a) Oxidation is the gain of oxygen by a substance during a reaction. Copper heated in air gains oxygen and its shiny brown surface turns black; methane gains oxygen when it burns.

$$2\text{Cu(s)} + \text{O}_2\text{(g)} \xrightarrow{\ \text{heat}\ } 2\text{CuO(s)}$$

$$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \longrightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$$

(b) Reduction is the loss of oxygen by a substance during a reaction. Copper(II) oxide loses its oxygen to hydrogen; zinc oxide loses its oxygen to carbon.

$$\text{CuO(s)} + \text{H}_2\text{(g)} \longrightarrow \text{Cu(s)} + \text{H}_2\text{O(g)}$$

$$\text{ZnO(s)} + \text{C(s)} \longrightarrow \text{Zn(s)} + \text{CO(g)}$$

Notice that the two always occur together: in the third equation the copper oxide is reduced and the hydrogen is oxidised, which is why such reactions are called redox reactions.

Question 17

A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.

Solution. Work from the description. A shiny brown element is a metal, and the metal that is brown rather than silvery is copper. Heated in air it gains oxygen — it is oxidised — and the black coating that forms is copper(II) oxide.

$$2\text{Cu(s)} + \text{O}_2\text{(g)} \xrightarrow{\ \text{heat}\ } 2\text{CuO(s)}$$

The chapter uses this pair the other way round as well: pass hydrogen over the black copper oxide and the brown colour returns as the oxide is reduced.

Question 18

Why do we apply paint on iron articles?

Solution. Iron corrodes — it rusts — and rusting needs two things present at the surface at once: air (oxygen) and moisture. Rust is not a protective layer like aluminium’s oxide; it is flaky, it falls away, and it exposes fresh metal underneath, so the damage keeps going until the article is eaten through.

A coat of paint seals the surface, so air and water cannot reach the iron. With one of the two necessary conditions removed, rusting cannot start. Repainting matters for the same reason — a scratch is enough to let corrosion begin at that spot.

Question 19

Oil and fat containing food items are flushed with nitrogen. Why?

Solution. Fats and oils left in contact with air are slowly oxidised, and the products of that oxidation smell and taste unpleasant. That spoiling is called rancidity.

Nitrogen is an unreactive gas. Flushing a sealed packet with nitrogen drives out the air and takes the oxygen with it, so the oxidation has nothing to work with and the food keeps far longer. This is why a packet of crisps is puffed up — the gas inside is nitrogen, not air.

Question 20

Explain the following terms with one example each.

(a) Corrosion (b) Rancidity

Solution.

(a) Corrosion is the slow eating away of a metal by its attack by the substances around it — air, moisture or acids — which turn the metal’s surface into a compound. Iron rusts to a reddish-brown coating of hydrated iron(III) oxide when exposed to damp air:

$$\text{iron} + \text{oxygen} + \text{water} \longrightarrow \text{rust (hydrated iron(III) oxide)}$$

Silver tarnishes black and copper turns green in the same way. Rusting matters most because iron’s corrosion product does not protect the metal beneath it.

(b) Rancidity is the spoiling of fats and oils in food by oxidation, which changes their smell and taste. Cooking oil left open for a long time, or fried food kept for days, develops a sharp unpleasant smell.

It is slowed by keeping the food in airtight containers, by refrigeration, by flushing packets with nitrogen, and by adding antioxidants.

Common mistakes

• Balancing by changing a subscript. Writing $\text{H}_2\text{O}_2$ instead of $\text{H}_2\text{O}$ to get another oxygen changes water into hydrogen peroxide — a different substance. Only the coefficients in front of a formula may be changed.
• Calling Question 2 a double displacement reaction. Double displacement needs two compounds to exchange ions. Here aluminium is a free element, so it is an ordinary displacement.
• Saying “lead is reduced” in Question 1. Lead has no oxygen to lose — it is lead oxide that is reduced. The distinction is the whole point of the question, and it is why (a) is one of the incorrect statements.
• Forgetting that hydrogen, oxygen, nitrogen and chlorine are diatomic in Question 5. Writing $\text{H} + \text{N} \longrightarrow \text{NH}_3$ balances easily and is wrong: the gases exist as $\text{H}_2$ and $\text{N}_2$.
• Balancing $\text{SO}_4$ and $\text{NO}_3$ atom by atom in Questions 5(c) and 6(a). Treated as whole units they balance in one step; split into sulphur and oxygen they take several and invite a slip.
• Explaining rancidity as “food going bad because of bacteria”. Rancidity is specifically the oxidation of fats and oils, which is why an unreactive gas like nitrogen prevents it.

Practise next

• The in-text questions of this chapter, which check the same balancing and the same reaction types in shorter form, eight questions across three boxes.
• Chapter 3, Metals and Non-metals, where the displacement reactions of Questions 2, 13 and 14 become the reactivity series and the extraction of metals.