Chemical Reactions and Equations – Case-based Questions with Answers

CBSE Class 10 Chemistry — Chapter 1: Chemical Reactions & Equations | 15 Case-Based Questions

Content Bank — Topics covered in cases:

Balancing equations, conservation of mass, stoichiometry
Types of reactions: combination, decomposition, displacement, double displacement
Redox basics, catalysts, energetics, precipitation, gas tests
Corrosion, rancidity, electrolysis and laboratory techniques

How to use: Read the scenario, attempt sub-questions, then click Show Answer to reveal model answers and explanations. Use for classroom discussion or self-study.

Case 1

Scenario: A student mixes blue copper(II) sulfate solution with iron filings. After some time the blue colour fades and a reddish-brown deposit forms on iron.

(a) Write the balanced chemical equation for the reaction.
(b) Identify the type of reaction and name the substance oxidised and reduced.

(a) Balanced equation: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s).

(b) This is a single displacement (displacement) reaction (a redox reaction). Iron is oxidised (Fe → Fe²⁺ + 2 e⁻) and Cu²⁺ is reduced to Cu(s) (Cu²⁺ + 2 e⁻ → Cu).

Note: The blue CuSO₄ fades as Cu²⁺ concentration decreases; FeSO₄ solution is pale green.

Case 2

Scenario: On heating limestone in a crucible, gas is evolved which turns limewater milky. The solid residue is used in building material.

(a) Write the chemical equation for the thermal decomposition.
(b) Identify the gas evolved and the industrial name for the residue.

(a) Balanced equation: CaCO₃(s) → CaO(s) + CO₂(g) (on heating).

(b) The gas evolved is carbon dioxide (CO₂) (confirmed by limewater turning milky due to CaCO₃). The solid residue CaO is quicklime, used to make cement and mortar (slaked to Ca(OH)₂ as needed).

Case 3

Scenario: Students perform the decomposition of hydrogen peroxide with and without MnO₂. With MnO₂ oxygen is produced rapidly.

(a) Write the decomposition equation for H₂O₂.
(b) Explain the role of MnO₂ and classify it.

(a) Decomposition: 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g).

(b) MnO₂ acts as a catalyst — it provides an alternative pathway with lower activation energy and is not consumed. It speeds up the reaction; MnO₂ remains chemically unchanged at the end of reaction.

Case 4

Scenario: A mixture of aqueous solutions of silver nitrate and potassium chloride produced a white solid which settled on standing.

(a) Provide the ionic equation and the net ionic equation.
(b) Name the type of reaction and the precipitate formed.

(a) Molecular equation: AgNO₃(aq) + KCl(aq) → AgCl(s)↓ + KNO₃(aq).
Ionic equation: Ag⁺(aq) + NO₃⁻(aq) + K⁺(aq) + Cl⁻(aq) → AgCl(s)↓ + K⁺(aq) + NO₃⁻(aq).
Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)↓.

(b) This is a double displacement (precipitation) reaction. The precipitate formed is silver chloride (AgCl), a white insoluble solid.

Case 5

Scenario: During electrolysis of molten sodium chloride in the lab model, two products are observed: a gas at the anode and a liquid metal at the cathode.

(a) Identify the anode and cathode products and write half-equations.
(b) Why is molten NaCl used instead of solid?

(a) At cathode (reduction): Na⁺ + e⁻ → Na(l). At anode (oxidation): 2 Cl⁻ → Cl₂(g) + 2 e⁻. Net: 2 NaCl(l) → 2 Na(l) + Cl₂(g).

(b) Molten NaCl is required because ions must be free to move. In a solid lattice ions are fixed and cannot conduct electricity; melting provides a mobile ionic medium for electrolysis.

Case 6

Scenario: A student burns magnesium ribbon in air producing a white ash which dissolves in dilute acid to give colourless solution.

(a) Write the equation for combustion and for dissolution in acid.
(b) Identify the white ash.

(a) Combustion: 2 Mg(s) + O₂(g) → 2 MgO(s) (white ash). Dissolution in dilute HCl: MgO(s) + 2 HCl(aq) → MgCl₂(aq) + H₂O(l).

(b) The white ash is magnesium oxide (MgO), a basic oxide that dissolves in acid to give magnesium chloride which is colourless in solution.

Case 7

Scenario: A farmer adds sodium carbonate to hard water and observes a white solid forming and settling out.

(a) Explain the chemistry behind the observation and write the equation.
(b) Which type of hardness is removed and why?

(a) Temporary (carbonate) hardness due to Ca²⁺ and Mg²⁺ salts is removed by Na₂CO₃ forming insoluble carbonates: Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)↓ (white precipitate). Similarly for Mg²⁺.

(b) Temporary hardness is removed because carbonate ions precipitate calcium and magnesium as insoluble carbonates; permanent (non-carbonate) hardness due to sulfates or chlorides is not removed by this method.

Case 8

Scenario: A sealed container shows no mass change before and after a reaction between hydrogen and oxygen to form water vapour, but when cooled to liquid water mass remains the same.

(a) Which law is illustrated? Explain briefly.
(b) Give the balanced chemical equation.

(a) This illustrates the law of conservation of mass: in a closed system mass remains constant during a chemical reaction because atoms are neither created nor destroyed.

(b) Balanced equation: 2 H₂(g) + O₂(g) → 2 H₂O(l).

Case 9

Scenario: A student notes that AgCl formed in sunlight gradually darkens and silver is deposited.

(a) Write the photochemical change for AgCl.
(b) Explain how this principle was used historically.

(a) Photochemical decomposition: 2 AgCl(s) → 2 Ag(s) + Cl₂(g) (under sunlight; actual mechanism involves photoelectrons and partial reactions).

(b) This light sensitivity of silver halides was exploited in photography — silver halide coatings on film darken on exposure to light forming latent images later developed chemically to give visible photographs.

Case 10

Scenario: During a demonstration, iron nails are left in water and rust forms faster when salt was dissolved in the water.

(a) Explain why rusting accelerates in salty water.
(b) Suggest two preventive measures specifically useful for marine environments.

(a) Salts (e.g., NaCl) dissolve and increase the conductivity of water, facilitating electrochemical corrosion processes (electron and ion transfer). Chloride ions also disrupt protective oxide layers, accelerating rusting.

(b) Preventive measures: (1) Galvanisation (zinc coating) provides sacrificial protection; (2) Protective coatings and paints (epoxy or marine-grade paints) to exclude oxygen and moisture. Regular maintenance and cathodic protection (sacrificial anodes) are common in marine structures.

Case 11

Scenario: A laboratory reaction produces gas that relights a glowing splint. The teacher asks to identify the gas and write a balanced reaction which produces it from decomposition of a common solid.

(a) Identify the gas.
(b) Provide a decomposition equation that yields this gas (use a common lab reagent).

(a) The gas is oxygen (O₂), since it supports combustion and can relight a glowing splint.

(b) Thermal decomposition of potassium chlorate (often catalysed) yields O₂: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) (MnO₂ used as catalyst in demonstrations).

Case 12

Scenario: A student prepares CO₂ by reacting marble chips with dilute HCl and collects the gas to test for limewater. They measure mass before and after reaction in a closed, sealed setup and find no loss in mass.

(a) Write the balanced reaction.
(b) If the setup was perfectly sealed, why is mass unchanged even though gas forms?

(a) Reaction: CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l).

(b) In a sealed system mass remains constant because the produced CO₂ is still contained within the system—mass conservation holds. Mass only changes in an open system if gases escape.

Case 13

Scenario: A student notes that concentrated sulfuric acid can dehydrate sugar giving a black carbonaceous mass and heat.

(a) Suggest the type of reaction and write a qualitative equation.
(b) Explain why heat is produced.

(a) This is a dehydration (and exothermic) reaction where concentrated H₂SO₄ removes water from sugar (sucrose) producing carbon: C₁₂H₂₂O₁₁ (s) --(H₂SO₄)--> 12 C(s) + 11 H₂O(l) (qualitative representation).

(b) Heat is produced because the overall process is exothermic (strong acid interactions and formation of new bonds release energy) and concentrated H₂SO₄ is highly hygroscopic, extracting water vigorously.

Case 14

Scenario: Two solutions, A (barium chloride) and B (sodium sulfate), are mixed and a white solid forms immediately which is filtered and tested insoluble in water.

(a) Write molecular and net ionic equations.
(b) Explain how this reaction helps in qualitative inorganic analysis.

(a) Molecular: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s)↓ + 2 NaCl(aq).
Net ionic: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)↓.

(b) Formation of insoluble BaSO₄ (white) confirms presence of sulfate ions in qualitative tests; precipitation reactions and selective solubility are essential tools for identifying ions in solution.

Case 15

Scenario: A student mistakenly balances an equation by altering subscripts rather than coefficients, changing H₂O to H₂O₂ to match oxygen atoms. The teacher stops them and asks why this is incorrect.

(a) Explain why changing subscripts is unacceptable when balancing equations.
(b) Demonstrate the correct balancing approach for H₂ + O₂ → H₂O.

(a) Subscripts define the chemical identity and composition of a species. Changing them changes the substance (e.g., H₂O to H₂O₂ changes water into hydrogen peroxide). Balancing must use coefficients which adjust the numbers of molecules (moles) without altering formulas.

(b) Correct method uses coefficients: 2 H₂(g) + O₂(g) → 2 H₂O(l). Here coefficients balance H (4 on both sides) and O (2 on both sides) while keeping chemical identities intact.

Chemical Reactions and Equations – Case-based Questions with Answers

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