CBSE Class 10 Science HOTs Chemical Reactions and Equations Set 06

This is Set 06 of higher-order thinking (HOTS) questions for CBSE Class 10 Science Chapter 1, Chemical Reactions and Equations, with step-by-step answers and a Teacher's Note for each question. The questions ask students to identify unknown metals, salts and gases from clues, to reason about heat changes and reactivity, and to balance and classify equations that need careful thinking.

Very Short Answer Type HOTS Questions

Question 1. What type of chemical reactions are represented by the following reactions? (i) Cl₂ + 2KI → 2KCl + I₂ (ii) PQ + RS → PS + RQ (iii) A₂O₃ + 2B → B₂O₃ + 2A
Answer:
(i) Chlorine is more reactive than iodine and displaces it from KI, so it is a displacement reaction.
(ii) Two compounds exchange their parts (P and R swap), so it is a double displacement reaction.
(iii) Element B displaces element A from its oxide, so it is a displacement reaction.
In simple words: Look at whether one element or two compounds change partners.

Teacher's Note:
1. An element replacing an element is displacement.
2. Two compounds exchanging parts is double displacement.

Question 2. A substance X, which is an oxide of a group 2 element, is used intensively in the cement industry. This element is also present in bones. On treatment with water it forms a solution which turns red litmus blue. Identify X and also write the chemical reaction involved.
Answer:
Step 1: The group 2 element present in bones and cement is calcium, so its oxide X is calcium oxide (quicklime), CaO.
Step 2: It reacts vigorously with water to form calcium hydroxide, which is a base and turns red litmus blue.
CaO(s) + H₂O(l) → Ca(OH)₂(aq) + heat
In simple words: X is quicklime; with water it makes slaked lime, a base.

Teacher's Note:
1. The reaction is a combination reaction and is exothermic.
2. Slaked lime solution is called lime water.

Question 3. Identify the reducing agent in each of the following reactions: (a) 4NH₃ + 5O₂ → 4NO + 6H₂O (b) H₂O + F₂ → HF + HOF (c) Fe₂O₃ + 3CO → 2Fe + 3CO₂ (d) 2H₂ + O₂ → 2H₂O
Answer:
A reducing agent is the substance that gets oxidised (gains oxygen or loses hydrogen).
(a) NH₃ gains oxygen to form NO, so NH₃ is the reducing agent.
(b) F₂ is reduced to HF, and H₂O is oxidised (it gains oxygen in HOF), so H₂O is the reducing agent.
(c) CO gains oxygen to form CO₂, so CO is the reducing agent.
(d) H₂ gains oxygen to form H₂O, so H₂ is the reducing agent.
In simple words: The reducing agent is the one that is itself oxidised.

Teacher's Note:
1. Reducing agent: oxidised.
2. Oxidising agent: reduced.

Question 4. Write balanced equations for the reactions between each of the following oxides and water: (a) Li₂O (b) CaO (c) SO₃
Answer:
(a) Li₂O + H₂O → 2LiOH (a metal oxide gives a base)
(b) CaO + H₂O → Ca(OH)₂ (slaked lime)
(c) SO₃ + H₂O → H₂SO₄ (a non-metal oxide gives an acid)
In simple words: Metal oxides give bases with water; non-metal oxides give acids.

Teacher's Note:
1. Balance the metal atoms first.
2. Each reaction is a combination reaction.

Question 5. Represent the following chemical reaction as a balanced equation: liquid rubbing alcohol (C₃H₈O) reacts with oxygen to form carbon dioxide gas and water.
Answer:
Step 1: Skeleton: C₃H₈O(l) + O₂(g) → CO₂(g) + H₂O(l)
Step 2: Balance carbon by taking 3CO₂ and hydrogen by taking 4H₂O: C₃H₈O + O₂ → 3CO₂ + 4H₂O
Step 3: Oxygen on the right = 6 + 4 = 10. C₃H₈O already has 1, so O₂ must supply 9 atoms, i.e. 9/2 molecules.
Step 4: Multiply by 2 to remove the fraction.
2C₃H₈O(l) + 9O₂(g) → 6CO₂(g) + 8H₂O(l)
In simple words: Balance C, then H, then O, and double everything to clear the fraction.

Teacher's Note:
1. Balance oxygen last because it occurs in several places.
2. Check: C 6, H 16, O 20 on both sides.

Short Answer Type HOTS Questions

Question 6. Why does the oxidation of iron to rust cause a problem, while the oxidation of aluminium to aluminium oxide does not?
Answer:
Iron reacts with oxygen and moisture to form rust (hydrated iron(III) oxide). Rust is porous and flaky, so it peels off and exposes fresh iron, and rusting continues until the iron is destroyed.
Aluminium forms a thin, tough and non-porous layer of aluminium oxide (Al₂O₃) on its surface. This layer sticks to the metal and protects it from further oxidation.
In simple words: Rust flakes off and lets rusting go on; aluminium oxide sticks on and protects the metal.

Teacher's Note:
1. This is why aluminium is used for utensils and foils.
2. Painting or galvanising protects iron.

Question 7. Why does sodium metal react violently with water?
Answer:
Sodium is a very reactive metal and displaces hydrogen from water: 2Na + 2H₂O → 2NaOH + H₂.
The reaction is highly exothermic. The heat produced is enough to ignite the hydrogen gas formed, so the reaction is violent and may even catch fire.
In simple words: Sodium is very reactive and the reaction gives out a lot of heat.

Teacher's Note:
1. Sodium is stored under kerosene to keep it away from air and moisture.
2. Never touch sodium with wet hands.

Question 8. Why are antioxidants added to foods containing fats and oils?
Answer:
Fats and oils get oxidised by air and turn rancid, which changes their taste and smell.
Antioxidants are reducing agents. They get oxidised in place of the fats and oils, so rancidity is delayed and the food stays fresh for longer.
In simple words: Antioxidants protect food by taking up the oxygen first.

Teacher's Note:
1. Packing in nitrogen also prevents rancidity.
2. Refrigeration slows down oxidation.

Question 9. Why does a magnesium ribbon burn in air with a dazzling flame and form white ash?
Answer:
Magnesium is a reactive metal and combines with oxygen of the air: 2Mg + O₂ → 2MgO.
The reaction is highly exothermic and releases a large amount of heat and light, which gives the dazzling white flame. The white ash is magnesium oxide.
In simple words: Mg burns with a bright flame and leaves white MgO.

Teacher's Note:
1. Clean the ribbon with sandpaper before burning.
2. Do not look at the flame directly.

Question 10. Why does the colour of the solution change and a black precipitate form when hydrogen sulphide gas is passed through an aqueous solution of copper sulphate? Write the equation and name the type of reaction.
Answer:
Hydrogen sulphide reacts with copper sulphate and the blue solution fades because colourless sulphuric acid is formed. Black insoluble copper sulphide settles as a precipitate.
H₂S(g) + CuSO₄(aq) → CuS(s)↓ + H₂SO₄(aq)
The ions are exchanged and a precipitate forms, so it is a double displacement (precipitation) reaction.
In simple words: A black solid forms and the blue colour fades.

Teacher's Note:
1. The black precipitate is copper sulphide.
2. Blue Cu²⁺ ions are removed from the solution.

Question 11. Anhydrous copper sulphate is dissolved in one beaker and hydrated copper sulphate (CuSO₄·5H₂O) in another beaker. What heat changes do you expect in the two beakers?
Answer:
Anhydrous copper sulphate combines strongly with water as it dissolves and releases heat, so the solution in the first beaker becomes warm (exothermic).
Hydrated copper sulphate already holds water of crystallisation, so no hydration heat is released; its dissolution absorbs a little heat from the surroundings and the second beaker becomes slightly cooler (endothermic).
In simple words: The anhydrous salt warms the water; the hydrated salt slightly cools it.

Teacher's Note:
1. Dissolving the anhydrous salt releases the energy of hydration.
2. Anhydrous CuSO₄ is white and turns blue on taking up water.

Question 12. When CaO is added to water in a beaker, a rise in temperature is observed. However, when Ba(OH)₂ is mixed with NH₄Cl, a fall in temperature is observed. Why?
Answer:
The reaction of CaO with water is exothermic and releases heat: CaO + H₂O → Ca(OH)₂ + heat, so the temperature rises.
The reaction of Ba(OH)₂ with NH₄Cl is endothermic. It absorbs heat from the surroundings: Ba(OH)₂ + 2NH₄Cl → BaCl₂ + 2NH₃ + 2H₂O, so the temperature falls.
In simple words: One reaction gives out heat; the other takes it in.

Teacher's Note:
1. Touch the outside of the test tube to feel the change.
2. Heat absorbed means the surroundings feel cold.

Long Answer Type HOTS Questions

Question 13. In an experiment, 20 g of zinc (A) was reacted with 20 g of iodine (B) in a 250 mL beaker containing 50 mL of water. The reaction proceeded with evolution of heat and at the end the reaction mixture did not give any colour on adding starch solution, but some zinc was found settling at the bottom. Zn(s) + I₂(s) → ZnI₂(s) + heat. (i) Is the reaction endothermic or exothermic? (ii) Out of the two reactants, which one is left unreacted and why? (iii) How much of (A) and (B) should be taken so that no reactant is left at the end?
Answer:
(i) Heat is evolved, so the reaction is exothermic.
(ii) Moles of Zn = 20 ÷ 65.4 ≈ 0.31 mol and moles of I₂ = 20 ÷ 254 ≈ 0.079 mol. Since Zn and I₂ react in the ratio 1 : 1, iodine is used up completely (no blue-black colour with starch) and zinc is left unreacted.
(iii) Equal moles are needed: 65.4 g of Zn require 254 g of I₂. So the masses should be in the ratio 65.4 : 254 (about 1 : 3.9), for example 20 g of Zn and about 77.7 g of I₂.
In simple words: Iodine is the limiting reactant; zinc is in excess.

Teacher's Note:
1. Starch turns blue-black with iodine, so no colour means no iodine is left.
2. Use moles to compare amounts, not masses.

Teacher's Note:
1. Write the reactivity order before filling the table.
2. The same metal and its own oxide never react.

Question 15. When metal P is treated with a dilute acid Q, a gas G is evolved which burns readily with a little explosion. (i) Name any two metals which can behave like metal P. (ii) Name any two acids which can behave like acid Q. (iii) Name the gas G. (iv) Is the gas G lighter or heavier than air? (v) Is the reaction between metal P and dilute acid Q exothermic or endothermic?
Answer:
(i) Zinc and iron (or magnesium).
(ii) Dilute hydrochloric acid and dilute sulphuric acid.
(iii) The gas G is hydrogen; it burns with a pop sound.
(iv) Hydrogen is lighter than air.
(v) The reaction is exothermic, for example Zn + H₂SO₄ → ZnSO₄ + H₂ + heat.
In simple words: A metal and a dilute acid give hydrogen gas and heat.

Teacher's Note:
1. The pop sound test identifies hydrogen.
2. Metals below hydrogen in the reactivity series do not react with dilute acids.

Question 16. A metal X forms a salt XSO₄. The salt forms a clear solution in water which reacts with sodium hydroxide solution to form a blue precipitate Y. Metal X is used in making electric wires and alloys like brass. (i) What could metal X be? (ii) Write the name, formula and colour of salt XSO₄. (iii) What is the blue precipitate Y? (iv) Write the chemical equation for the reaction of XSO₄ with sodium hydroxide, with state symbols.
Answer:
(i) Metal X is copper (used in electric wires and brass).
(ii) XSO₄ is copper sulphate, CuSO₄, which is blue in colour.
(iii) The blue precipitate Y is copper hydroxide, Cu(OH)₂.
(iv) CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s)↓ + Na₂SO₄(aq)
In simple words: Copper sulphate with NaOH gives a blue solid, copper hydroxide.

Teacher's Note:
1. This is a double displacement (precipitation) reaction.
2. Blue colour of the solution fades as Cu²⁺ is used up.

Question 17. A silvery-white metal X taken in the form of a ribbon, when ignited, burns in air with a dazzling white flame to form a white powder Y. When water is added to powder Y, it dissolves partially to form another substance Z. (i) What could metal X be? (ii) What is powder Y? (iii) With which substance does metal X combine to form powder Y? (iv) What is substance Z? Name one domestic use of Z. (v) Write the balanced chemical equation for the burning of X in air.
Answer:
(i) X is magnesium.
(ii) Y is magnesium oxide, MgO.
(iii) X combines with oxygen of the air.
(iv) Z is magnesium hydroxide, Mg(OH)₂, used as an antacid to relieve indigestion (milk of magnesia).
(v) 2Mg + O₂ → 2MgO
In simple words: Mg burns to MgO, which with water gives Mg(OH)₂.

Teacher's Note:
1. MgO + H₂O → Mg(OH)₂.
2. The solution is basic.

Question 18. When the solution of substance A is added to a solution of potassium iodide, a yellow solid separates out. (i) What do you think substance A would be? (ii) Name the yellow substance. (iii) Write the characteristic of chemical reactions illustrated by this example. (iv) Write a balanced chemical equation with the physical states.
Answer:
(i) A is lead nitrate, Pb(NO₃)₂.
(ii) The yellow solid is lead iodide, PbI₂.
(iii) Formation of a precipitate (a double displacement reaction).
(iv) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s)↓ + 2KNO₃(aq)
In simple words: Lead nitrate and potassium iodide give yellow lead iodide.

Teacher's Note:
1. The yellow precipitate is the clue to lead iodide.
2. Both reactants are colourless solutions.

Case Based HOTS Questions

Question 19. When hydrogen gas is passed into a chamber containing a gas X, both gases combine to form gas Y. White fumes of ammonium chloride are formed when gas Y combines with gas Z. Write the suitable chemical equations for the process and find the gases X, Y and Z.
Answer:
Step 1: Hydrogen combines with nitrogen to form ammonia, so X is nitrogen (N₂) and Y is ammonia (NH₃): N₂ + 3H₂ → 2NH₃
Step 2: Ammonia gives white fumes of ammonium chloride with hydrogen chloride gas, so Z is hydrogen chloride (HCl): NH₃ + HCl → NH₄Cl
In simple words: N₂ and H₂ give NH₃; NH₃ and HCl give white NH₄Cl fumes.

Teacher's Note:
1. Both reactions are combination reactions.
2. White fumes of NH₄Cl are the test for ammonia.

Question 20. When ferrous sulphate having green colour is heated strongly it breaks down to form a brown coloured compound along with two gases X and Y. These gases form sulphurous acid and sulphuric acid respectively when passed into water. Write the chemical equations and find the molar masses of X and Y.
Answer:
Step 1: 2FeSO₄(s) →^heat Fe₂O₃(s) + SO₂(g) + SO₃(g). The brown compound is ferric oxide.
Step 2: SO₂ + H₂O → H₂SO₃ (sulphurous acid), so X = SO₂. SO₃ + H₂O → H₂SO₄ (sulphuric acid), so Y = SO₃.
Step 3: Molar mass of SO₂ = 32 + 2 × 16 = 64 g/mol. Molar mass of SO₃ = 32 + 3 × 16 = 80 g/mol.
In simple words: X is SO₂ (64 g/mol) and Y is SO₃ (80 g/mol).

Teacher's Note:
1. It is a thermal decomposition reaction.
2. Use atomic masses S = 32 and O = 16.

Question 21. A metal salt MX, when decomposed by passing electric current through its molten state, forms metal M and gas X₂. Metal M is stored in kerosene due to its highly reactive nature, whereas X₂ is used to disinfect drinking water. (i) What is the colour of salt MX? (ii) What could the metal M and gas X₂ be? (iii) What type of chemical reaction takes place? Write the chemical equation.
Answer:
(i) The salt is sodium chloride, NaCl, which is white.
(ii) Metal M is sodium (stored in kerosene) and gas X₂ is chlorine, Cl₂ (used to disinfect water).
(iii) It is an electrolytic decomposition reaction: 2NaCl(l) →^electricity 2Na(l) + Cl₂(g)
In simple words: Electricity splits molten NaCl into sodium and chlorine.

Teacher's Note:
1. Electrolysis is decomposition by electricity.
2. Chlorine is also a bleaching agent.

Question 22. A strip of metal X is dipped in a blue coloured solution of YSO₄. After some time a layer of metal Y from the salt solution is formed on the surface of X. Metal X is used in galvanisation whereas metal Y is used in making electric wires. (i) What are metals X and Y? (ii) Write the name and formula of the salt YSO₄. (iii) What type of reaction takes place? Write the equation.
Answer:
(i) X is zinc (used in galvanisation) and Y is copper (used in electric wires).
(ii) YSO₄ is copper sulphate, CuSO₄.
(iii) It is a displacement reaction: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Zinc is more reactive, so it displaces copper and the blue colour fades.
In simple words: Zinc pushes copper out of copper sulphate solution.

Teacher's Note:
1. Blue solution turns colourless as ZnSO₄ forms.
2. Copper settles on the zinc strip.