NCERT Solutions Class 10 Science Chapter 1 In-text Questions — Chemical Reactions and Equations

NCERT Class 10 Science — Chemical Reactions and Equations, the three in-text QUESTIONS boxes on pages 6, 10 and 13. All 8 questions solved.

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

The three boxes follow the chapter’s three stages: writing and balancing equations, then the types of reaction, then oxidation and reduction in everyday life. Two skills cover all eight questions:

• Turning a sentence into a balanced equation — write the correct formulae first, balance by changing only the coefficients, and add state symbols (s), (l), (g) and (aq) when they are asked for.
• Reading a reaction for what has gained or lost oxygen — that is all “oxidised” and “reduced” mean in this chapter.

Key insight. A balanced equation is a statement about atoms, not about words. Once the formulae are written down correctly, balancing is bookkeeping — and the bookkeeping is easiest if polyatomic groups like $\text{SO}_4$ and $\text{NO}_3$ are counted as single units. Question 2(ii) of the first box looks the worst on the page and is the clearest example: three sulphate groups on the left fix almost everything else in one step.

Questions, page 6. The box that closes Section 1.1, on writing and balancing chemical equations.

Page 6, Question 1

Why should a magnesium ribbon be cleaned before burning in air?

Solution. Magnesium is a reactive metal, so a ribbon left in the air does not stay pure magnesium. Its surface reacts slowly with the oxygen around it and becomes coated with a dull white layer of magnesium oxide.

That layer is the problem. It is already an oxide, so it will not burn, and it keeps the air away from the metal underneath — so the ribbon is hard to light and does not burn with the brilliant white flame it should.

Rubbing the ribbon with sandpaper scrapes the oxide coating off, exposing clean metal, which then burns readily:

$$2\text{Mg(s)} + \text{O}_2\text{(g)} \xrightarrow{\ \text{burn}\ } 2\text{MgO(s)}$$

Page 6, Question 2

Write the balanced equation for the following chemical reactions.

(i) Hydrogen + Chlorine → Hydrogen chloride

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

(iii) Sodium + Water → Sodium hydroxide + Hydrogen

Solution. Write the formulae first, remembering that hydrogen and chlorine are diatomic gases.

(i) $\text{H}_2 + \text{Cl}_2 \longrightarrow \text{HCl}$. Each molecule of $\text{H}_2$ and $\text{Cl}_2$ supplies two atoms, so two molecules of $\text{HCl}$ are formed.

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

(ii) $\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \longrightarrow \text{BaSO}_4 + \text{AlCl}_3$. Count $\text{SO}_4$ as one unit: aluminium sulphate carries three, so three $\text{BaSO}_4$ are formed, which needs three $\text{BaCl}_2$. Those supply six chlorine atoms, which make two $\text{AlCl}_3$ — and the two aluminium atoms match.

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

(iii) $\text{Na} + \text{H}_2\text{O} \longrightarrow \text{NaOH} + \text{H}_2$. Hydrogen leaves as $\text{H}_2$, so take two water molecules; they give two $\text{NaOH}$ and one $\text{H}_2$, so two sodium atoms are needed.

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

Page 6, Question 3

Write a balanced chemical equation with state symbols for the following reactions.

(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

Solution. This question adds one thing to the last: the state symbols. The wording gives them away — “in water” means (aq), “insoluble” means the substance comes out as a solid, (s), and water itself is a liquid, (l).

(i) Both reactants are solutions, so both are (aq). Barium sulphate is insoluble, so it is (s) — a precipitate. Sodium chloride stays dissolved, so (aq).

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

Counting: one barium, one sulphate, two sodium and two chlorine on each side. Balanced.

(ii) Both reactants are solutions again; the salt formed stays in solution and the water is a liquid.

$$\text{NaOH(aq)} + \text{HCl(aq)} \longrightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$$

One of each atom on both sides — this one is balanced as written. It is a neutralisation reaction.

Questions, page 10. The box that follows Section 1.2.2, on decomposition reactions.

Page 10, Question 1

A solution of a substance ‘X’ is used for whitewashing.

(i) Name the substance ‘X’ and write its formula.

(ii) Write the reaction of the substance ‘X’ named in (i) above with water.

Solution. (i) The substance used for whitewashing is calcium oxide, also called quick lime, formula CaO. It is not applied as it is: it is first added to water.

(ii) Calcium oxide reacts vigorously with water, giving out a great deal of heat, to form calcium hydroxide — slaked lime:

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

It is that slaked lime solution which is put on the walls. Once on the wall it slowly reacts with carbon dioxide from the air to form a thin, hard layer of calcium carbonate, and it is this that gives whitewash its shiny white finish:

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

The first reaction is a combination reaction, and so is the second.

Page 10, Question 2

Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.

Solution. Activity 1.7 is the electrolysis of water: a current is passed through acidified water and the gases given off at the two carbon electrodes are collected in separate test tubes.

Water decomposes as

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

The equation settles it. For every two molecules of hydrogen released, only one molecule of oxygen is released, because a water molecule contains twice as many hydrogen atoms as oxygen atoms. Equal numbers of molecules of any gas occupy equal volumes, so twice as many molecules means twice the volume.

The gas collected in double amount is therefore hydrogen, and it is collected at the cathode, the electrode joined to the negative terminal — as Fig. 1.6 shows. The oxygen appears at the anode in half that volume.

Questions, page 13. The last box, after Section 1.3 on the effects of oxidation in everyday life.

Page 13, Question 1

Why does the colour of copper sulphate solution change when an iron nail is dipped in it?

Solution. Copper sulphate solution is blue because of the copper ions dissolved in it. Iron is more reactive than copper, so when an iron nail is put in, the iron displaces copper from the solution:

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

The copper leaves the solution and is deposited on the nail as a reddish-brown coating; the iron takes its place in solution as iron(II) sulphate, which is pale green. So the blue fades and the solution turns light green — and the nail, which went in grey, comes out brown.

This is a displacement reaction, and it is the standard demonstration that a more reactive metal displaces a less reactive one from its salt solution.

Page 13, Question 2

Give an example of a double displacement reaction other than the one given in Activity 1.10.

Solution. In a double displacement reaction two compounds exchange their ions. Activity 1.10 is the sodium sulphate and barium chloride reaction, so any other pair of solutions that swap partners will do — and the clearest examples are those that give a precipitate, because the exchange is then visible.

Lead nitrate and potassium iodide solutions, both colourless, give a bright yellow precipitate of lead iodide:

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

The lead has taken the iodide and the potassium the nitrate — two exchanges at once, which is what makes it double displacement.

Neutralisation is another example, with no precipitate:

$$\text{NaOH(aq)} + \text{HCl(aq)} \longrightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$$

Page 13, Question 3

Identify the substances that are oxidised and the substances that are reduced in the following reactions.

(i) $4\text{Na(s)} + \text{O}_2\text{(g)} \longrightarrow 2\text{Na}_2\text{O(s)}$

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

Solution. Follow the oxygen in each case: the substance that gains oxygen is oxidised, and the one that loses it is reduced.

(i) Sodium starts as the free metal, with no oxygen, and ends combined in $\text{Na}_2\text{O}$. It has gained oxygen, so sodium is oxidised. The oxygen itself has combined with the metal, so oxygen is reduced.

(ii) Copper(II) oxide starts with oxygen and ends as free copper, so it has lost oxygen: copper oxide is reduced. Hydrogen starts free and ends as water, so it has gained oxygen: hydrogen is oxidised.

Notice that in both reactions the two happen together — something is oxidised precisely because something else is reduced. That is why they are called redox reactions.

Common mistakes

• Balancing by changing a subscript. Turning $\text{HCl}$ into $\text{H}_2\text{Cl}_2$ to balance the first equation makes a substance that does not exist. Only the number in front of a formula may be altered.
• Forgetting the state symbols in Question 3 of the first box. The question asks for them specifically, and the wording supplies each one: “in water” is (aq), “insoluble” is (s), water is (l).
• Answering the whitewashing question with slaked lime alone. The substance used is calcium oxide, and the question then asks for its reaction with water. Naming $\text{Ca(OH)}_2$ as X leaves the second part with nothing to answer.
• Saying the doubled gas is oxygen. A water molecule has two hydrogen atoms to one oxygen, so hydrogen comes off in double the volume — and it is collected at the cathode.
• Saying in Question 1 of the last box that “the nail rusts”. Rusting is a reaction with air and moisture. What happens here is displacement: iron pushes copper out of the solution, which is why copper is deposited on the nail.
• Naming only the oxidised substance in Question 3. Oxidation and reduction always occur together; each part has two answers, not one.

Practise next

• The end-of-chapter Exercises for this chapter, where Questions 5 to 8 balance harder equations and ask for the type of each reaction, and Questions 16 to 20 develop oxidation into corrosion and rancidity.
• Chapter 2, Acids, Bases and Salts, whose in-text questions take the neutralisation reaction met here and build the whole chapter on it.