NCERT Exemplar Solutions Class 10 Science Chemical Reactions and Equations — Short and Long Answer Questions Q19–44

NCERT Exemplar Problems, Class 10 Science — Chemical Reactions and Equations, Short Answer Questions 19 to 38 and Long Answer Questions 39 to 44. All 26 questions solved.

This is where the chapter stops being recognition and starts being production. In the multiple-choice run you were handed an equation and asked what type it was; here you are given a sentence of English and have to produce the equation yourself, balanced, with state symbols, and then classify it. Almost every mark lost in this chapter is lost in one of three places, and they are worth naming before you start:

• Getting the formula wrong. An equation built on $\mathrm{MgCl}$ instead of $\mathrm{MgCl_2}$ cannot be balanced, and no amount of fiddling with coefficients will rescue it. Fix the formulae from the valencies first, then balance.
• Changing a subscript while balancing. Coefficients go in front; subscripts are part of the substance and are never touched. Writing $\mathrm{H_2O_2}$ where you need $2\mathrm{H_2O}$ changes what the reaction produces.
• Naming a type without checking the shape. Count reactants and products before you commit. One product means combination, one reactant means decomposition, a free element on the left means displacement, two compounds swapping partners means double displacement.

Key insight. Balancing is not guesswork if you take the elements in the right order: leave hydrogen and oxygen until last, and balance whichever element appears in the fewest formulae first. In almost every equation on this page that single rule turns a trial-and-error problem into two or three forced steps.

A note on two answers in the printed key. Questions 33 and 37 are both marked “write a balanced chemical equation”, and the key’s own answers to them are not balanced. Both are corrected below, and in each case the key contradicts itself elsewhere, so this is a typesetting slip rather than a disagreement about chemistry. The corrections are explained where they occur.

Short Answer Questions

Question 19

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

(a) Nitrogen gas is treated with hydrogen gas in the presence of a catalyst at 773 K to form ammonia gas. (b) Sodium hydroxide solution is treated with acetic acid to form sodium acetate and water. (c) Ethanol is warmed with ethanoic acid to form ethyl acetate in the presence of concentrated $\mathrm{H_2SO_4}$. (d) Ethene is burnt in the presence of oxygen to form carbon dioxide, water and releases heat and light.

Solution. Take each sentence and extract three things in order: what goes in, what comes out, and what sits over the arrow.

(a) The reactants are $\mathrm{N_2}$ and $\mathrm{H_2}$, both diatomic, and the product is $\mathrm{NH_3}$. Nitrogen balances with a coefficient of 2 on the ammonia, which then needs 6 hydrogen atoms, so 3 molecules of $\mathrm{H_2}$:

$$\mathrm{N_2}(g) + 3\mathrm{H_2}(g) \xrightarrow[\text{773 K}]{\text{Catalyst}} 2\mathrm{NH_3}(g)$$

Two reactants, one product — a combination reaction. (This is the Haber process, and the catalyst and temperature belong over the arrow because they are conditions, not reactants.)

(b) An acid and a base giving a salt and water:

$$\mathrm{NaOH}(aq) + \mathrm{CH_3COOH}(aq) \rightarrow \mathrm{CH_3COONa}(aq) + \mathrm{H_2O}(l)$$

It balances as written. Sodium and hydrogen have exchanged partners, so it is a double displacement reaction, and because the reactants are an acid and a base it is specifically a neutralisation reaction.

(c) Esterification. The $\mathrm{-OH}$ of the alcohol and the $\mathrm{-OH}$ of the acid combine to give water, leaving the ester:

$$\mathrm{C_2H_5OH}(l) + \mathrm{CH_3COOH}(l) \xrightarrow{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_3COOC_2H_5}(l) + \mathrm{H_2O}(l)$$

A double displacement reaction, known by its own name, esterification. The concentrated sulphuric acid is a dehydrating agent and catalyst, so it goes over the arrow.

(d) Ethene is $\mathrm{C_2H_4}$. Burning any hydrocarbon gives carbon dioxide and water; balance carbon first, then hydrogen, then oxygen:

$$\mathrm{C_2H_4}(g) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 2\mathrm{H_2O}(g) + \text{Heat} + \text{Light}$$

Two carbons give $2\mathrm{CO_2}$; four hydrogens give $2\mathrm{H_2O}$; that needs $4 + 2 = 6$ oxygen atoms, so $3\mathrm{O_2}$. It is a combustion reaction and therefore also a redox reaction.

Question 20

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

(a) Thermit reaction, iron (III) oxide reacts with aluminium and gives molten iron and aluminium oxide. (b) Magnesium ribbon is burnt in an atmosphere of nitrogen gas to form solid magnesium nitride. (c) Chlorine gas is passed in an aqueous potassium iodide solution to form potassium chloride solution and solid iodine. (d) Ethanol is burnt in air to form carbon dioxide, water and releases heat.

Solution.

(a) Iron(III) oxide is $\mathrm{Fe_2O_3}$; aluminium oxide is $\mathrm{Al_2O_3}$. The two oxides carry the same number of oxygen atoms, so the oxides balance one-for-one and only the metals need coefficients:

$$\mathrm{Fe_2O_3}(s) + 2\mathrm{Al}(s) \rightarrow \mathrm{Al_2O_3}(s) + 2\mathrm{Fe}(l) + \text{Heat}$$

Aluminium has taken the oxygen from the iron, so it is a displacement reaction — and, because oxygen has moved from one element to another, a redox reaction. The heat released is enough to melt the iron, which is why the equation writes $\mathrm{Fe}(l)$ and why the reaction is used to weld railway tracks.

(b) Magnesium is $\mathrm{Mg^{2+}}$ and the nitride ion is $\mathrm{N^{3-}}$, so magnesium nitride is $\mathrm{Mg_3N_2}$:

$$3\mathrm{Mg}(s) + \mathrm{N_2}(g) \rightarrow \mathrm{Mg_3N_2}(s)$$

One product — a combination reaction.

(c) Chlorine is more reactive than iodine, so it takes the potassium and turns the iodide into free iodine:

$$\mathrm{Cl_2}(g) + 2\mathrm{KI}(aq) \rightarrow 2\mathrm{KCl}(aq) + \mathrm{I_2}(s)$$

A displacement reaction.

(d) Ethanol is $\mathrm{C_2H_5OH}$. Two carbons give $2\mathrm{CO_2}$; six hydrogens give $3\mathrm{H_2O}$; that is $4 + 3 = 7$ oxygen atoms on the right, and the ethanol itself supplies one, so six more are needed — $3\mathrm{O_2}$:

$$\mathrm{C_2H_5OH}(l) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(l) + \text{Heat}$$

A combustion and therefore a redox reaction.

Question 21

Complete the missing components/variables given as $x$ and $y$ in the following reactions

(a) $\mathrm{Pb(NO_3)_2}(aq) + 2\mathrm{KI}(aq) \rightarrow \mathrm{PbI_2}\,(x) + 2\mathrm{KNO_3}\,(y)$ (b) $\mathrm{Cu}(s) + 2\mathrm{AgNO_3}(aq) \rightarrow \mathrm{Cu(NO_3)_2}(aq) + x(s)$ (c) $\mathrm{Zn}(s) + \mathrm{H_2SO_4}(aq) \rightarrow \mathrm{ZnSO_4}(x) + \mathrm{H_2}(y)$ (d) $\mathrm{CaCO_3}(s) \xrightarrow{\;x\;} \mathrm{CaO}(s) + \mathrm{CO_2}(g)$

Solution. Three of these are missing state symbols and one is missing a condition, so read what kind of blank each is before filling it.

(a) Lead iodide is the famous yellow precipitate, so it is a solid: $x = (s)$. Potassium nitrate is soluble and stays dissolved: $y = (aq)$.

(b) Here the blank is a substance, not a state. Copper displaces silver from silver nitrate, so the missing product is silver metal — and the coefficient must match the two silver atoms on the left: $x = 2\mathrm{Ag}$.

(c) Zinc sulphate is soluble, so $x = (aq)$; hydrogen leaves as a gas, so $y = (g)$.

(d) The blank sits above the arrow, which is where conditions go. Limestone decomposing to quicklime and carbon dioxide is the thermal decomposition used in lime kilns, so $x = \text{Heat}$.

Question 22

Which among the following changes are exothermic or endothermic in nature?

(a) Decomposition of ferrous sulphate (b) Dilution of sulphuric acid (c) Dissolution of sodium hydroxide in water (d) Dissolution of ammonium chloride in water

Solution. The test is which way heat flows, and there is a useful shortcut: a change that has to be driven absorbs heat, and a change that happens by itself as soon as the substances meet usually releases it.

(a) Ferrous sulphate does not decompose on its own — it has to be heated strongly, and it stops when you stop heating. That continuous demand for energy is what endothermic means. Endothermic.

(b) Diluting sulphuric acid warms the mixture noticeably; pouring water onto concentrated acid can make it spit. Exothermic.

(c) Sodium hydroxide pellets make water distinctly hot as they dissolve. Exothermic.

(d) Ammonium chloride makes water distinctly cold as it dissolves — it is the salt used in instant cold packs. Endothermic.

Notice that (c) and (d) are both simply dissolving a salt, and they go opposite ways. Dissolution is not inherently exothermic or endothermic: it depends on whether the energy released when the ions become hydrated exceeds the energy needed to break the crystal apart.

Question 23

Identify the reducing agent in the following reactions

(a) $4\mathrm{NH_3} + 5\mathrm{O_2} \rightarrow 4\mathrm{NO} + 6\mathrm{H_2O}$ (b) $\mathrm{H_2O} + \mathrm{F_2} \rightarrow \mathrm{HF} + \mathrm{HOF}$ (c) $\mathrm{Fe_2O_3} + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO_2}$ (d) $2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}$

Solution. The reducing agent is the substance that causes reduction — and it does so by being oxidised itself. At this level, “oxidised” means it gained oxygen or lost hydrogen. So the rule to apply is: find the reactant that gained oxygen or lost hydrogen; that one is the reducing agent.

(a) Ammonia loses its hydrogen and gains oxygen to become $\mathrm{NO}$. It is oxidised, so $\mathrm{NH_3}$ is the reducing agent.

(b) Fluorine goes from $\mathrm{F_2}$ to $\mathrm{HF}$, gaining hydrogen — so fluorine is reduced. Whatever supplied that hydrogen is the reducing agent, and the only other reactant is water: $\mathrm{H_2O}$.

(c) Carbon monoxide gains oxygen to become carbon dioxide. CO is the reducing agent — which is exactly its industrial job in the blast furnace, where it strips oxygen from iron ore.

(d) Hydrogen gains oxygen to become water. $\mathrm{H_2}$ is the reducing agent.

Question 24

Identify the oxidising agent (oxidant) in the following reactions

(a) $\mathrm{Pb_3O_4} + 8\mathrm{HCl} \rightarrow 3\mathrm{PbCl_2} + \mathrm{Cl_2} + 4\mathrm{H_2O}$ (b) $2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}$ (c) $\mathrm{CuSO_4} + \mathrm{Zn} \rightarrow \mathrm{Cu} + \mathrm{ZnSO_4}$ (d) $\mathrm{V_2O_5} + 5\mathrm{Ca} \rightarrow 2\mathrm{V} + 5\mathrm{CaO}$ (e) $3\mathrm{Fe} + 4\mathrm{H_2O} \rightarrow \mathrm{Fe_3O_4} + 4\mathrm{H_2}$ (f) $\mathrm{CuO} + \mathrm{H_2} \rightarrow \mathrm{Cu} + \mathrm{H_2O}$

Solution. This is the mirror image of Question 23. The oxidising agent is the reactant that is itself reduced — it lost oxygen or gained hydrogen.

(a) Chloride ion becomes free chlorine, so chlorine is oxidised and hydrochloric acid is the reducing agent. The other reactant, $\mathrm{Pb_3O_4}$, loses oxygen and is the oxidant.

(b) Oxygen is added to magnesium, so $\mathrm{O_2}$ is the oxidant. Elemental oxygen is the oxidising agent in every combustion.

(c) Copper goes from a compound to the free metal — it loses its sulphate and is reduced — so $\mathrm{CuSO_4}$ is the oxidant, and zinc is the reducing agent.

(d) Vanadium goes from $\mathrm{V_2O_5}$ to free vanadium, losing oxygen: $\mathrm{V_2O_5}$ is the oxidant.

(e) Water loses oxygen to become hydrogen, so $\mathrm{H_2O}$ is the oxidant. This is the same reaction as Question 3 of the multiple-choice set, which is worth noting: water is the oxidising agent there too.

(f) Copper oxide loses oxygen to become copper: CuO is the oxidant.

In every pair, the agent and its own fate are opposite — the oxidising agent gets reduced. Checking that the other reactant was oxidised is a free way to confirm each answer.

Question 25

Write the balanced chemical equations for the following reactions

(a) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogencarbonate. (b) Sodium hydrogencarbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide. (c) Copper sulphate on treatment with potassium iodide precipitates cuprous iodide $(\mathrm{Cu_2I_2})$, liberates iodine gas and also forms potassium sulphate.

Solution.

(a) The phrase in equal molar concentrations is the whole difficulty. With excess acid, sodium carbonate goes all the way to carbon dioxide and water; with one mole of acid per mole of carbonate, only one of the two sodium atoms is replaced, and the reaction stops at the hydrogencarbonate:

$$\mathrm{Na_2CO_3} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{NaHCO_3}$$

It balances as written — count sodium (2 each side), chlorine, hydrogen, carbon and oxygen in turn.

(b) Now the second sodium goes, and the unstable carbonic acid that would form breaks up immediately into water and carbon dioxide:

$$\mathrm{NaHCO_3} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} + \mathrm{CO_2}$$

Reactions (a) and (b) together are the two stages of the single reaction $\mathrm{Na_2CO_3} + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H_2O} + \mathrm{CO_2}$, which is worth writing out to see that the parts add up.

(c) This is the hardest balance in the exercise, because copper is reduced from $\mathrm{Cu^{2+}}$ to $\mathrm{Cu^{+}}$ while iodide is oxidised to iodine. Balance copper first: $\mathrm{Cu_2I_2}$ needs two coppers, so $2\mathrm{CuSO_4}$. Two sulphates need $2\mathrm{K_2SO_4}$, which needs 4 potassium, so $4\mathrm{KI}$. That supplies 4 iodine atoms; two of them are in $\mathrm{Cu_2I_2}$, leaving two for one $\mathrm{I_2}$:

$$2\mathrm{CuSO_4} + 4\mathrm{KI} \rightarrow \mathrm{Cu_2I_2} + 2\mathrm{K_2SO_4} + \mathrm{I_2}$$

Question 26

A solution of potassium chloride when mixed with silver nitrate solution, an insoluble white substance is formed. Write the chemical reaction involved and also mention the type of the chemical reaction?

Solution. Two soluble salts are mixed and an insoluble solid appears, so the partners must have exchanged. The white solid is silver chloride, which is the standard insoluble silver salt:

$$\mathrm{KCl}(aq) + \mathrm{AgNO_3}(aq) \rightarrow \mathrm{AgCl}(s)\downarrow + \mathrm{KNO_3}(aq)$$

It is already balanced. Potassium has taken the nitrate and silver has taken the chloride, so it is a double displacement reaction, and because one product is insoluble and separates out it is also a precipitation reaction.

The white precipitate is the same silver chloride that darkens in sunlight in Question 34, which is a useful thread to notice: the exemplar keeps returning to a small number of substances from different directions.

Question 27

Ferrous sulphate decomposes with the evolution of a gas having a characteristic odour of burning sulphur. Write the chemical reaction involved and identify the type of reaction.

Solution. The clue in the question is the smell: burning sulphur is sulphur dioxide. So $\mathrm{SO_2}$ must be among the products, and since green ferrous sulphate crystals leave a reddish-brown solid behind, that residue is ferric oxide.

Balancing needs care. Two formula units of $\mathrm{FeSO_4}$ supply the two iron atoms for $\mathrm{Fe_2O_3}$ and two sulphur atoms; the eight oxygen atoms then split as three in $\mathrm{Fe_2O_3}$, two in $\mathrm{SO_2}$ and three in $\mathrm{SO_3}$ — which is why both oxides of sulphur appear:

$$2\mathrm{FeSO_4}(s) \xrightarrow{\text{Heat}} \mathrm{Fe_2O_3}(s) + \mathrm{SO_2}(g) + \mathrm{SO_3}(g)$$

One reactant breaking into three products, driven by heat: a thermal decomposition reaction.

Iron has gone from $+2$ to $+3$ here, so it is a redox reaction as well — but the type the question is after is decomposition.

Question 28

Why do fire flies glow at night?

Solution. Fireflies carry a protein in their abdomen which, in the presence of an enzyme, undergoes aerial oxidation — that is, it is oxidised by the oxygen of the air. The chemical energy released by that oxidation is given out as visible light rather than as heat, so the insect glows.

The point of the question for this chapter is the classification. This is a chemical reaction, not a physical effect: a substance is consumed and a different one is produced. It is an oxidation, and it is exothermic in the sense that it releases energy — but the energy leaves as light. Reactions that release energy as light are called chemiluminescent, and when the reaction happens inside a living organism it is bioluminescence.

Contrast this with a glowing hot iron rod, which emits light because it is hot and undergoes no chemical change at all. The firefly stays cool.

Question 29

Grapes hanging on the plant do not ferment but after being plucked from the plant can be fermented. Under what conditions do these grapes ferment? Is it a chemical or a physical change?

Solution. While the grape is attached to the vine it is living tissue, and its own immune system keeps the microorganisms that cause fermentation from growing in it. Pluck it and that defence stops working, so yeasts and bacteria can multiply in the sugary juice.

The condition fermentation needs is the absence of air — anaerobic conditions. The yeast breaks the sugar down into ethanol and carbon dioxide:

$$\mathrm{C_6H_{12}O_6} \xrightarrow{\text{Yeast}} 2\mathrm{C_2H_5OH} + 2\mathrm{CO_2}$$

It is a chemical change. Glucose disappears and two substances that were not there before — ethanol and carbon dioxide — take its place, and nothing will turn the wine back into grape juice.

Question 30

Which among the following are physical or chemical changes?

(a) Evaporation of petrol (b) Burning of Liquefied Petroleum Gas (LPG) (c) Heating of an iron rod to red hot. (d) Curdling of milk (e) Sublimation of solid ammonium chloride

Solution. The question to ask of each is whether a new substance exists afterwards.

(a) Evaporating petrol turns liquid petrol into petrol vapour. Same molecules, different state. Physical.

(b) Burning LPG destroys the hydrocarbon molecules and makes carbon dioxide and water. Chemical.

(c) A red-hot iron rod is still iron; it glows because it is hot, and it returns to normal on cooling. Physical. (If it were heated in air long enough to form a black oxide scale, that would be chemical — but the change described here is only the heating.)

(d) Curdling is bacteria converting the lactose of milk into lactic acid, which makes the protein coagulate. Milk cannot be recovered from curd. Chemical.

(e) Solid ammonium chloride heated in a test tube vanishes and re-forms as a white solid higher up the tube. NCERT classes this as physical, and that is the answer to give.

It is worth knowing that (e) is subtler than it looks. What actually happens is that $\mathrm{NH_4Cl}$ dissociates on heating into ammonia and hydrogen chloride gases, which recombine on the cooler glass:

$$\mathrm{NH_4Cl}(s) \rightleftharpoons \mathrm{NH_3}(g) + \mathrm{HCl}(g)$$

So chemical bonds really are broken and remade. It is classed as physical at this level because the substance you start with and the substance you end with are the same, and the process is completely reversible — but the honest description is that it is a reversible chemical dissociation wearing the appearance of sublimation.

Question 31

During the reaction of some metals with dilute hydrochloric acid, following observations were made.

(a) Silver metal does not show any change (b) The temperature of the reaction mixture rises when aluminium (Al) is added. (c) The reaction of sodium metal is found to be highly explosive (d) Some bubbles of a gas are seen when lead (Pb) is reacted with the acid.

Explain these observations giving suitable reasons.

Solution. All four observations are explained by one idea: a metal displaces hydrogen from an acid only if it lies above hydrogen in the reactivity series, and the further above it lies, the more violently it does so.

(a) Silver lies below hydrogen. It cannot displace hydrogen from hydrochloric acid, so nothing happens — no bubbles, no warming, no change in the metal.

(b) Aluminium lies well above hydrogen and reacts readily:

$$2\mathrm{Al}(s) + 6\mathrm{HCl}(aq) \rightarrow 2\mathrm{AlCl_3}(aq) + 3\mathrm{H_2}(g)$$

The reaction releases heat, which is why the mixture warms up. All metal–acid reactions are exothermic.

(c) Sodium is near the top of the reactivity series, so the same reaction runs enormously faster and releases its heat all at once. The heat is sufficient to ignite the hydrogen being produced, and the result is an explosion. This is why sodium is never deliberately added to an acid in a school laboratory.

(d) Lead is above hydrogen, but only just — it sits immediately above it — so it does react, but slowly and gently, and the only visible sign is a slow stream of hydrogen bubbles:

$$\mathrm{Pb}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{PbCl_2}(aq) + \mathrm{H_2}(g)$$

Read in order, the four observations are a demonstration of the reactivity series itself: sodium violent, aluminium vigorous, lead sluggish, silver inert.

Question 32

A substance X, which is an oxide of a group 2 element, is used intensively in the cement industry. This element is present in bones also. On treatment with water it forms a solution which turns red litmus blue. Identify X and also write the chemical reactions involved.

Solution. Work through the clues in the order given and they converge on one element.

Group 2 narrows it to the alkaline earth metals. Used intensively in the cement industry and present in bones both point at calcium — calcium oxide is the principal ingredient of cement, and bone mineral is a calcium phosphate. So X is calcium oxide, $\mathrm{CaO}$, known as quick lime.

The fourth clue confirms it. Calcium oxide reacts with water in the reaction called slaking of lime:

$$\mathrm{CaO}(s) + \mathrm{H_2O}(l) \rightarrow \mathrm{Ca(OH)_2}(aq) + \text{Heat}$$

Calcium hydroxide is a base, and a basic solution turns red litmus blue — exactly as described. The solution is lime water, and it gives the familiar test for carbon dioxide:

$$\mathrm{Ca(OH)_2}(aq) + \mathrm{CO_2}(g) \rightarrow \mathrm{CaCO_3}(s)\downarrow + \mathrm{H_2O}(l)$$

Question 33

Write a balanced chemical equation for each of the following reactions and also classify them.

(a) Lead acetate solution is treated with dilute hydrochloric acid to form lead chloride and acetic acid solution. (b) A piece of sodium metal is added to absolute ethanol to form sodium ethoxide and hydrogen gas. (c) Iron (III) oxide on heating with carbon monoxide gas reacts to form solid iron and liberates carbon dioxide gas. (d) Hydrogen sulphide gas reacts with oxygen gas to form solid sulphur and liquid water.

Solution.

(a) Lead acetate is $\mathrm{Pb(CH_3COO)_2}$ — lead is divalent, so it carries two acetate groups. Those two acetate groups leave as two molecules of acetic acid, and the two chlorides that replace them come from two molecules of $\mathrm{HCl}$:

$$\mathrm{Pb(CH_3COO)_2} + 2\mathrm{HCl} \rightarrow \mathrm{PbCl_2} + 2\mathrm{CH_3COOH}$$

A double displacement reaction.

The printed key writes this with only one $\mathrm{CH_3COOH}$, which is not balanced. The key’s own coefficient of 2 on the hydrochloric acid settles the matter: two chlorides arrive, so two acetates must leave, and each of them picks up one of the two hydrogens. Carbon and hydrogen do not balance otherwise. Since the question explicitly asks for a balanced equation, the coefficient belongs there.

(b) Sodium displaces the hydrogen of the $\mathrm{-OH}$ group. Each ethanol molecule gives up one hydrogen atom and hydrogen gas is diatomic, so two molecules of ethanol are needed:

$$2\mathrm{Na} + 2\mathrm{C_2H_5OH} \rightarrow 2\mathrm{C_2H_5ONa} + \mathrm{H_2}$$

A displacement reaction.

(c) Carbon monoxide takes the oxygen from the ore. Three oxygens must move, so three molecules of CO:

$$\mathrm{Fe_2O_3} + 3\mathrm{CO} \xrightarrow{\text{Heat}} 2\mathrm{Fe} + 3\mathrm{CO_2}$$

A redox reaction — iron is reduced and carbon monoxide oxidised. This is the central reaction of the blast furnace.

(d) Two hydrogen sulphide molecules supply the four hydrogens for two waters, and the single $\mathrm{O_2}$ supplies both oxygen atoms:

$$2\mathrm{H_2S} + \mathrm{O_2} \rightarrow 2\mathrm{S} + 2\mathrm{H_2O}$$

A redox reaction — sulphur is oxidised from $-2$ to $0$, oxygen reduced from $0$ to $-2$.

Question 34

Why do we store silver chloride in dark coloured bottles?

Solution. Silver chloride is decomposed by light. Left in a clear bottle on a sunlit bench it slowly breaks down into silver and chlorine:

$$2\mathrm{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\mathrm{Ag}(s) + \mathrm{Cl_2}(g)$$

The finely divided silver left behind is grey, so the white solid darkens and the reagent is no longer pure silver chloride. Storing it in a dark or amber bottle keeps light out and prevents this photochemical decomposition.

The same reasoning applies to silver bromide, and it is the reason the reaction was the basis of black-and-white photography: the light that falls on the film is precisely what decomposes the silver salt and darkens it.

Question 35

Balance the following chemical equations and identify the type of chemical reaction.

(a) $\mathrm{Mg}(s) + \mathrm{Cl_2}(g) \rightarrow \mathrm{MgCl_2}(s)$ (b) $\mathrm{HgO}(s) \xrightarrow{\text{Heat}} \mathrm{Hg}(l) + \mathrm{O_2}(g)$ (c) $\mathrm{Na}(s) + \mathrm{S}(s) \xrightarrow{\text{Fuse}} \mathrm{Na_2S}(s)$ (d) $\mathrm{TiCl_4}(l) + \mathrm{Mg}(s) \rightarrow \mathrm{Ti}(s) + \mathrm{MgCl_2}(s)$ (e) $\mathrm{CaO}(s) + \mathrm{SiO_2}(s) \rightarrow \mathrm{CaSiO_3}(s)$ (f) $\mathrm{H_2O_2}(l) \xrightarrow{\text{UV}} \mathrm{H_2O}(l) + \mathrm{O_2}(g)$

Solution. Two of these are already balanced, which is part of the test — the instruction to balance does not guarantee there is anything to do.

(a) One magnesium, one chlorine molecule, one $\mathrm{MgCl_2}$. Already balanced. One product from two reactants: combination.

(b) Oxygen is the problem: one atom on the left, two on the right. Doubling the mercuric oxide fixes it and then the mercury needs doubling too:

$$2\mathrm{HgO}(s) \xrightarrow{\text{Heat}} 2\mathrm{Hg}(l) + \mathrm{O_2}(g)$$

One reactant, two products: decomposition.

(c) $\mathrm{Na_2S}$ needs two sodium atoms:

$$2\mathrm{Na}(s) + \mathrm{S}(s) \xrightarrow{\text{Fuse}} \mathrm{Na_2S}(s)$$

Combination.

(d) Four chlorine atoms on the left have to appear as $\mathrm{MgCl_2}$ on the right, so two of them — and that needs two magnesium atoms:

$$\mathrm{TiCl_4}(l) + 2\mathrm{Mg}(s) \rightarrow \mathrm{Ti}(s) + 2\mathrm{MgCl_2}(s)$$

Magnesium has displaced titanium: displacement. (This is the Kroll process, the industrial route to titanium metal.)

(e) Calcium 1, silicon 1, oxygen $1 + 2 = 3$ on each side. Already balanced. Combination.

(f) Two hydrogen peroxide molecules give two waters and one oxygen molecule:

$$2\mathrm{H_2O_2}(l) \xrightarrow{\text{UV}} 2\mathrm{H_2O}(l) + \mathrm{O_2}(g)$$

Decomposition — a photochemical one, since ultraviolet light drives it, which is why hydrogen peroxide is also sold in dark bottles.

Question 36

A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by emission of light. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y.

(a) Write the chemical formulae of X and Y. (b) Write a balanced chemical equation, when X is dissolved in water.

Solution.

(a) Burning magnesium in oxygen gives magnesium oxide, the white ash left on the bench:

$$2\mathrm{Mg}(s) + \mathrm{O_2}(g) \rightarrow 2\mathrm{MgO}(s)$$

so X is $\mathrm{MgO}$.

The striking part of the question is that the ribbon keeps burning in nitrogen. Nitrogen is normally so unreactive that it is used to prevent oxidation — Question 14 of the multiple-choice set relies on that. Magnesium is reactive enough to break the nitrogen triple bond anyway, and forms magnesium nitride with the nitride ion $\mathrm{N^{3-}}$:

$$3\mathrm{Mg}(s) + \mathrm{N_2}(g) \rightarrow \mathrm{Mg_3N_2}(s)$$

so Y is $\mathrm{Mg_3N_2}$.

(b) Magnesium oxide is a basic oxide, and a basic oxide plus water gives the hydroxide:

$$\mathrm{MgO}(s) + \mathrm{H_2O}(l) \rightarrow \mathrm{Mg(OH)_2}(aq)$$

It balances as written, and it is a combination reaction. Magnesium hydroxide is the “milk of magnesia” used as an antacid.

Question 37

Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Explain why?

Solution. The reactivity series decides it. Zinc lies above hydrogen, so zinc atoms give up electrons more readily than hydrogen does and can push hydrogen out of its compound:

$$\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(g)$$

Copper lies below hydrogen. It holds its electrons more tightly than hydrogen does, so it cannot displace hydrogen from the acid, and nothing happens:

$$\mathrm{Cu}(s) + \mathrm{HCl}(aq) \rightarrow \text{No reaction}$$

The general rule worth carrying away is that a metal displaces hydrogen from a dilute acid if and only if it stands above hydrogen in the reactivity series. It is the same rule that explains all four observations in Question 31.

The printed key gives this equation as $\mathrm{Zn} + \mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}$, which is not balanced — zinc chloride carries two chlorine atoms, so two molecules of acid are needed. The same answer key writes it correctly as $\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(g)$ five answers later, at Question 42(b), which is how we can be sure it is a typesetting slip and not a claim about the chemistry.

Question 38

A silver article generally turns black when kept in the open for a few days. The article when rubbed with toothpaste again starts shining.

(a) Why do silver articles turn black when kept in the open for a few days? Name the phenomenon involved. (b) Name the black substance formed and give its chemical formula.

Solution.

(a) Air contains traces of hydrogen sulphide, and silver reacts with it. When a metal is slowly eaten into by the moisture, acids and gases around it, the process is called corrosion. Silver corroding in this way is usually called tarnishing, and it is a chemical reaction — a new compound forms on the surface.

(b) The black coating is silver sulphide, $\mathrm{Ag_2S}$:

$$2\mathrm{Ag}(s) + \mathrm{H_2S}(g) \rightarrow \mathrm{Ag_2S}(s) + \mathrm{H_2}(g)$$

Toothpaste restores the shine because it is a mild abrasive: rubbing physically removes the thin sulphide layer and exposes clean metal underneath. Notice that the two halves of this question are different kinds of change — the tarnishing is chemical, the cleaning is physical.

It is worth contrasting this with rusting. Iron rusting is also corrosion, but the rust flakes away and lets the attack continue into the metal; silver sulphide forms a thin, adherent layer, which is why a tarnished spoon is disfigured rather than destroyed.

Long Answer Questions

Question 39

On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed

(a) Write a balanced chemical equation of the reaction. (b) Identify the brown gas X evolved. (c) Identify the type of reaction. (d) What could be the pH range of aqueous solution of the gas X?

Solution.

(b) Start here, because the equation cannot be written until X is known. The only brown gas that a nitrate can give on heating is nitrogen dioxide, $\mathrm{NO_2}$ — and its brown colour is the standard identification.

(a) Copper(II) nitrate is $\mathrm{Cu(NO_3)_2}$. Balance nitrogen before oxygen: two nitrate groups per formula unit give two $\mathrm{NO_2}$, and doubling the whole equation is what lets the leftover oxygen come out as a whole $\mathrm{O_2}$ molecule:

$$2\mathrm{Cu(NO_3)_2}(s) \xrightarrow{\text{Heat}} 2\mathrm{CuO}(s) + 4\mathrm{NO_2}(g) + \mathrm{O_2}(g)$$

Check the oxygen: $2 \times 6 = 12$ on the left; $2 + 8 + 2 = 12$ on the right.

(c) One reactant giving three products, driven by heat: a thermal decomposition reaction.

(d) Nitrogen dioxide is an oxide of a non-metal, and non-metal oxides are acidic. Dissolved in water it gives nitric and nitrous acids:

$$2\mathrm{NO_2}(g) + \mathrm{H_2O}(l) \rightarrow \mathrm{HNO_3}(aq) + \mathrm{HNO_2}(aq)$$

So the solution is acidic and its pH is less than 7 — in practice somewhere in the range 1 to 6 depending on concentration. This is the chemistry behind the nitric acid component of acid rain.

Question 40

Give the characteristic tests for the following gases

(a) $\mathrm{CO_2}$   (b) $\mathrm{SO_2}$   (c) $\mathrm{O_2}$   (d) $\mathrm{H_2}$

Solution. A good gas test gives an unmistakable change that no other common gas produces, and each of these four has one.

(a) Carbon dioxide — lime water turns milky. Pass the gas through lime water, a solution of calcium hydroxide. Insoluble calcium carbonate forms as a fine white suspension:

$$\mathrm{Ca(OH)_2}(aq) + \mathrm{CO_2}(g) \rightarrow \mathrm{CaCO_3}(s)\downarrow + \mathrm{H_2O}(l)$$

Passing excess carbon dioxide turns the milkiness clear again, as the carbonate dissolves to the soluble hydrogencarbonate — a useful confirmation.

(b) Sulphur dioxide — decolourises acidified potassium permanganate. Sulphur dioxide is a strong reducing agent, so it reduces the purple permanganate ion to the colourless manganese(II) ion:

$$2\mathrm{KMnO_4} + 2\mathrm{H_2O} + 5\mathrm{SO_2} \rightarrow \mathrm{K_2SO_4} + 2\mathrm{MnSO_4} + 2\mathrm{H_2SO_4}$$

Equivalently it turns orange acidified potassium dichromate green. Sulphur dioxide also has a sharp choking smell of burning sulphur, which is how it is recognised in Question 27.

(c) Oxygen — relights a glowing splint. Oxygen supports combustion, so a splint that has been blown out and is merely glowing bursts back into flame in it, and a burning candle burns more brightly. Oxygen itself does not burn.

(d) Hydrogen — burns with a pop sound. Bring a burning splint to the mouth of the tube and the hydrogen ignites with a squeaky pop, forming water:

$$2\mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow 2\mathrm{H_2O}(l)$$

The distinction between (c) and (d) is the one to hold on to: hydrogen burns and oxygen supports burning, so the splint goes into the tube glowing for oxygen and alight for hydrogen.

Question 41

What happens when a piece of

(a) zinc metal is added to copper sulphate solution? (b) aluminium metal is added to dilute hydrochloric acid? (c) silver metal is added to copper sulphate solution?

Also, write the balanced chemical equation if the reaction occurs

Solution. Every part is decided by the reactivity series, and the answer to (c) is that nothing happens — which is a complete answer, not a gap.

(a) Zinc is more reactive than copper, so it displaces copper from its sulphate:

$$\mathrm{Zn}(s) + \mathrm{CuSO_4}(aq) \rightarrow \mathrm{ZnSO_4}(aq) + \mathrm{Cu}(s)$$

There are two visible changes, and both are worth stating: the blue colour of the copper sulphate solution fades, because zinc sulphate is colourless, and a reddish-brown deposit of copper appears on the zinc. A displacement reaction.

(b) Aluminium is above hydrogen, so it displaces hydrogen from the acid and the gas is seen as brisk bubbling:

$$2\mathrm{Al}(s) + 6\mathrm{HCl}(aq) \rightarrow 2\mathrm{AlCl_3}(aq) + 3\mathrm{H_2}(g)$$

A displacement reaction. The mixture also warms up, as Question 31(b) observes.

(c) Silver is less reactive than copper, so it cannot displace copper from a copper salt. The blue solution stays blue and the silver stays bright:

$$\mathrm{Ag}(s) + \mathrm{CuSO_4}(aq) \rightarrow \text{No reaction}$$

Comparing (a) and (c) is the point of the question. The same salt solution, two different metals, and only the one that sits above copper does anything.

Question 42

What happens when zinc granules are treated with dilute solution of $\mathrm{H_2SO_4}$, $\mathrm{HCl}$, $\mathrm{HNO_3}$, $\mathrm{NaCl}$ and $\mathrm{NaOH}$, also write the chemical equations if reaction occurs.

Solution. Five reagents, and they divide into three different outcomes. Taking them in the order given hides the pattern, so it is worth sorting them.

(a) Dilute sulphuric acid — zinc is above hydrogen, so hydrogen is displaced:

$$\mathrm{Zn}(s) + \mathrm{H_2SO_4}(aq) \rightarrow \mathrm{ZnSO_4}(aq) + \mathrm{H_2}(g)$$

(b) Dilute hydrochloric acid — the same reaction with a different acid:

$$\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(g)$$

(c) Dilute nitric acid — this one is the exception, and it is the reason the question exists. Nitric acid is a strong oxidising agent, so any hydrogen produced is immediately oxidised to water and never escapes as a gas. What comes off instead is an oxide of nitrogen — with dilute acid, dinitrogen oxide:

$$4\mathrm{Zn}(s) + 10\mathrm{HNO_3}(aq) \rightarrow 4\mathrm{Zn(NO_3)_2}(aq) + 5\mathrm{H_2O}(l) + \mathrm{N_2O}(g)$$

So zinc does react, but no hydrogen gas is evolved. The general rule is that nitric acid does not give hydrogen with metals, with very dilute acid and magnesium or manganese as the only exceptions.

(d) Sodium chloride solution — sodium is far more reactive than zinc, so zinc cannot displace it. No reaction.

(e) Sodium hydroxide solution — zinc is amphoteric, meaning it reacts with alkalis as well as acids, dissolving to give a zincate and hydrogen:

$$\mathrm{Zn}(s) + 2\mathrm{NaOH}(aq) \rightarrow \mathrm{Na_2ZnO_2}(aq) + \mathrm{H_2}(g)$$

Read together, the five cases say something worth remembering: zinc gives hydrogen with ordinary dilute acids and with alkali, but not with nitric acid — and not with a salt of a more reactive metal.

Question 43

On adding a drop of barium chloride solution to an aqueous solution of sodium sulphite, white precipitate is obtained.

(a) Write a balanced chemical equation of the reaction involved. (b) What other name can be given to this precipitation reaction? (c) On adding dilute hydrochloric acid to the reaction mixture, white precipitate disappears. Why?

Solution.

(a) Sodium sulphite is $\mathrm{Na_2SO_3}$ — note the ending, which distinguishes it from sulphate, $\mathrm{Na_2SO_4}$, and is the commonest slip in this question. Barium takes the sulphite and sodium takes the chloride:

$$\mathrm{Na_2SO_3}(aq) + \mathrm{BaCl_2}(aq) \rightarrow \mathrm{BaSO_3}(s)\downarrow + 2\mathrm{NaCl}(aq)$$

The white precipitate is barium sulphite.

(b) Two compounds have exchanged partners, so it is a double displacement reaction. Precipitation reactions are always double displacements in which one product is insoluble; “precipitation” describes what you see, “double displacement” describes what happened.

(c) Barium sulphite is the salt of a weak acid, sulphurous acid $\mathrm{H_2SO_3}$. Adding a strong acid displaces the weak one, and sulphurous acid is unstable and breaks up at once into water and sulphur dioxide:

$$\mathrm{BaSO_3}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{BaCl_2}(aq) + \mathrm{H_2O}(l) + \mathrm{SO_2}(g)$$

The precipitate disappears because the barium ends up as barium chloride, which is soluble — and the escaping sulphur dioxide can be recognised by its smell of burning sulphur.

This is a useful diagnostic in its own right. Barium sulphate would not dissolve in hydrochloric acid, because sulphuric acid is strong and cannot be displaced by hydrochloric acid. So the fact that this precipitate dissolves proves it was the sulphite.

Question 44

You are provided with two containers made up of copper and aluminium. You are also provided with solutions of dilute $\mathrm{HCl}$, dilute $\mathrm{HNO_3}$, $\mathrm{ZnCl_2}$ and $\mathrm{H_2O}$. In which of the above containers these solutions can be kept?

Solution. A solution can be kept in a container only if it does not attack the metal, so there are eight questions here and each is decided by comparing the reactivity of the container metal with what the solution can do to it.

In the copper container. Copper is below hydrogen in the reactivity series, and that settles three of the four cases at once.

• Dilute HCl — can be kept. Copper cannot displace hydrogen from an acid, so ordinary dilute acids leave it alone.
• Dilute $\mathrm{HNO_3}$ — cannot be kept. Nitric acid attacks copper not as an acid but as an oxidising agent, and the reactivity series does not protect against that. The copper dissolves, giving a blue solution of copper nitrate.
• $\mathrm{ZnCl_2}$ — can be kept. Zinc is more reactive than copper, so copper cannot displace zinc from its salt. No reaction.
• $\mathrm{H_2O}$ — can be kept. Copper does not react with water, hot or cold.

In the aluminium container. Aluminium is a reactive metal well above hydrogen, and the answers mostly reverse.

• Dilute HCl — cannot be kept. Aluminium displaces hydrogen briskly: $2\mathrm{Al} + 6\mathrm{HCl} \rightarrow 2\mathrm{AlCl_3} + 3\mathrm{H_2}$.
• Dilute $\mathrm{HNO_3}$ — can be kept. This is the surprise, and it is the best part of the question. Nitric acid oxidises the aluminium surface to a thin, tough, impervious layer of $\mathrm{Al_2O_3}$, which then seals the metal beneath it and stops any further attack. The metal is said to be passivated. Note that the very property that destroys the copper container protects the aluminium one.
• $\mathrm{ZnCl_2}$ — cannot be kept. Aluminium is more reactive than zinc, so it displaces zinc from the solution: $2\mathrm{Al} + 3\mathrm{ZnCl_2} \rightarrow 2\mathrm{AlCl_3} + 3\mathrm{Zn}$.
• $\mathrm{H_2O}$ — can be kept. Aluminium does not react with cold or hot water; it is attacked only by steam, $2\mathrm{Al}(s) + 3\mathrm{H_2O}(g) \rightarrow \mathrm{Al_2O_3}(s) + 3\mathrm{H_2}(g)$, and the same oxide layer slows even that.

Common mistakes

• Changing a subscript to balance an equation. Balancing is done with coefficients in front of formulae only. Turning $\mathrm{H_2O}$ into $\mathrm{H_2O_2}$ to find an extra oxygen, as happens in Questions 19(d) and 35(f), produces a different substance and a different reaction.
• Writing lead acetate with one acetate group in Question 33(a). Lead is divalent, so the formula is $\mathrm{Pb(CH_3COO)_2}$ and two molecules of acetic acid come out. The printed key drops that coefficient; the balance proves it is needed.
• Forgetting the 2 on HCl in Question 37. $\mathrm{ZnCl_2}$ needs two chlorine atoms, so $\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}$. The key prints this one unbalanced too, and then writes it correctly at Question 42(b) — so check every equation against the atom count rather than against your memory of having seen it.
• Confusing sulphite with sulphate in Question 43. $\mathrm{Na_2SO_3}$ is the sulphite; $\mathrm{Na_2SO_4}$ is the sulphate. Part (c) only works for the sulphite, because barium sulphate does not dissolve in hydrochloric acid — the difference of one oxygen atom changes the whole answer.
• Expecting hydrogen from nitric acid in Question 42. Nitric acid is an oxidising acid and oxidises the hydrogen to water as fast as it forms, so an oxide of nitrogen comes off instead. Writing $\mathrm{Zn} + 2\mathrm{HNO_3} \rightarrow \mathrm{Zn(NO_3)_2} + \mathrm{H_2}$ is the single most common error in the long-answer band.
• Missing the reversal in Question 44. Nitric acid destroys the copper container and protects the aluminium one. Oxidising power is not the same thing as position in the reactivity series, and a question that mixes the two is testing exactly that distinction.
• Ignoring “in equal molar concentrations” in Question 25(a). With excess acid the products are sodium chloride, water and carbon dioxide; with one mole of acid the reaction stops at sodium hydrogencarbonate. The stated proportion is part of the question.
• Giving the reactant instead of the agent in Questions 23 and 24. The reducing agent is whatever got oxidised, and vice versa. Confirming that the other reactant went the opposite way takes three seconds and catches the slip every time.

Practise next

• Exemplar Multiple Choice Questions — the same reaction types in recognition form, which is the faster way to drill the classification this page depends on.
• NCERT textbook Chapter 1 — Chemical Reactions and Equations for the end-of-chapter exercise, where the balancing is set as straightforward practice rather than buried in a word problem.
• Textbook in-text questions for corrosion and rancidity, the background to Questions 38 and to the oil-storage question in the multiple-choice set.
• Textbook activities for the experiments behind Questions 32, 41 and 43 — the lime-water test, the zinc-in-copper-sulphate displacement and the barium precipitate are all done there.