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Mastersheet Solutions: Chemical Reactions and Equations
Student Name: Class: 10 CBSE Subject: Science (Chemistry)
Topic 1 Solutions: Chemical Equations and Characteristics
1.
Ans: A chemical reaction is a process in which one or more substances (reactants) react to form new substances (products) with entirely different chemical properties. The four observations that help determine whether a chemical reaction has occurred are:
  1. Change in state
  2. Change in color
  3. Evolution of a gas
  4. Change in temperature (exothermic/endothermic) or formation of a precipitate.
2.
Ans: A chemical change is permanent because new chemical bonds are formed, resulting in new substances with new chemical identities, which cannot be reversed by simple physical methods (e.g., burning wood to ash). A physical change is temporary because it only affects the physical state, shape, or appearance without altering the chemical composition, and can easily be reversed (e.g., melting of ice to water: $\text{H}_2\text{O}(s) \rightleftharpoons \text{H}_2\text{O}(l)$).
3.
Ans: Black copper oxide reacts with dilute hydrochloric acid to form blue-green copper chloride and water.
Balanced equation: $\text{CuO}(s) + 2\text{HCl}(aq) \rightarrow \text{CuCl}_2(aq) + \text{H}_2\text{O}(l)$
Observation: The black copper oxide powder dissolves and the solution turns a beautiful blue-green color due to the formation of copper(II) chloride ($\text{CuCl}_2$).
4.
Ans: The reaction is a highly exothermic Combination Reaction.
Balanced equation: $\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{Heat}$
Calcium oxide (quicklime) combines vigorously with water to form calcium hydroxide (slaked lime), releasing a massive amount of heat energy.
5.
Ans: Heating white lead nitrate powder undergoes thermal decomposition:
Balanced equation: $2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$
Observations:
  1. Brown fumes of nitrogen dioxide ($\text{NO}_2$) gas are evolved.
  2. The white crystalline powder decomposes to form a yellow residue of lead oxide ($\text{PbO}$) which is reddish-brown when hot.
  3. A crackling sound is heard during heating.
6.
Ans: Hydrogen gas ($\text{H}_2$) is evolved.
Balanced equation: $\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)$
Gas Test: Bring a burning matchstick or candle near the mouth of the gas collection tube; the gas burns with a characteristic pop sound, confirming it is hydrogen.
7.
Ans: Nitrogen is an inert, non-reactive gas. Food items containing fats and oils react easily with atmospheric oxygen, leading to oxidation, which causes rancidity (unpleasant smell and bad taste). Flushing packets with nitrogen displaces oxygen, preventing oxidation. Antioxidants are substances (like BHA and BHT) that retard oxidation by reacting with oxygen themselves, preserving food quality.
8.
Ans: Characteristics of the reactions:
  1. HCl + Sodium Carbonate: Rapid effervescence due to the evolution of colorless, odorless carbon dioxide ($\text{CO}_2$) gas: $\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$.
  2. KI + Lead Nitrate: Formation of a bright yellow precipitate of lead iodide ($\text{PbI}_2$): $\text{Pb(NO}_3)_2(aq) + 2\text{KI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)$.
9.
Ans: Quicklime ($\text{CaO}$) undergoes a highly exothermic chemical combination with water to produce slaked lime ($\text{Ca(OH)}_2$).
Balanced equation: $\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{Heat}$
It is highly exothermic because a large amount of heat is liberated, causing the temperature of the reaction mixture to rise significantly.
10.
Ans: Invert a dry gas jar or test tube over a burning candle. Condensation of water droplets is seen on the cooler walls, showing hydrogen combustion. When a small amount of lime water is added to the jar and shaken, it turns milky, showing the presence of carbon dioxide ($\text{CO}_2$) gas.
Lime water reaction: $\text{Ca(OH)}_2(aq) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)$ (white precipitate).
Topic 2 Solutions: Balancing Chemical Equations
11.
Ans: The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. Therefore, the total mass of the reactants must equal the total mass of the products. To satisfy this law, the number of atoms of each element on the reactant side (left) must equal the number of atoms of that same element on the product side (right).
12.
Ans: Balancing step-by-step:
1. Unbalanced: $\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$
2. Balance Oxygen: 4 O on right, so multiply $\text{H}_2\text{O}$ by 4: $\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$
3. Balance Iron: 3 Fe on right, so multiply $\text{Fe}$ by 3: $3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$
4. Balance Hydrogen: 8 H on left ($4 \times 2$), so multiply $\text{H}_2$ by 4: $3\text{Fe}(s) + 4\text{H}_2\text{O}(g) \rightarrow \text{Fe}_3\text{O}_4(s) + 4\text{H}_2(g)$ (Balanced).
13.
Ans: Balanced chemical equations:
  1. $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$
  2. $2\text{H}_2\text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{SO}_2(g) + 2\text{H}_2\text{O}(g)$
14.
Ans: Balanced equation: $2\text{HNO}_3(aq) + \text{Ca(OH)}_2(aq) \rightarrow \text{Ca(NO}_3)_2(aq) + 2\text{H}_2\text{O}(l)$.
15.
Ans: Balanced equation: $2\text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$.
16.
Ans: The equation is already balanced: $\text{NaCl}(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$.
17.
Ans: Balanced equation: $\text{BaCl}_2(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{HCl}(aq)$.
18.
Ans: Balanced equation: $\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq)$.
19.
Ans: Balanced equation: $\text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) + \text{Energy}$.
20.
Ans: Balanced equation: $2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$.
21.
Ans: Balanced equation: $2\text{FeSO}_4(s) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)$.
22.
Ans: Balanced equation: $2\text{H}_2\text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{SO}_2(g) + 2\text{H}_2\text{O}(l)$.
23.
Ans: Balanced equation: $2\text{Al}(s) + 3\text{CuCl}_2(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{Cu}(s)$.
24.
Ans: State symbols ($(s)$, $(l)$, $(g)$, $(aq)$) show the physical state of the reactants and products. In thermodynamics, they are vital because the enthalpy of a reaction depends on state. For example, in the formation of water:
$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$ ($\Delta H = -286 \text{ kJ/mol}$)
$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g)$ ($\Delta H = -242 \text{ kJ/mol}$)
Omitting the state symbols would hide the $44 \text{ kJ}$ difference representing the latent heat of vaporization of water.
25.
Ans: Balanced equation: $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)$.
Topic 3 Solutions: Types of Chemical Reactions
26.
Ans: A combination reaction is a reaction where two or more reactants combine to form a single product.
  1. Between two elements: Burning of magnesium in oxygen to form magnesium oxide: $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$.
  2. Between an element and a compound: Oxidation of carbon monoxide to carbon dioxide: $2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g)$.
27.
Ans: In a combination reaction, two or more substances combine to form a single product, whereas in a decomposition reaction, a single complex reactant splits into two or more simpler products.
Combination equation: $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$
Decomposition equation: $2\text{H}_2\text{O}(l) \xrightarrow{\text{Electricity}} 2\text{H}_2(g) + \text{O}_2(g)$.
28.
Ans: When a decomposition reaction is carried out by heating, it is called thermal decomposition.
Balanced equation: $\text{CaCO}_3(s) \xrightarrow{\Delta} \text{CaO}(s) + \text{CO}_2(g)$
Industrial application: Calcium oxide ($\text{CaO}$ or quicklime) obtained is used on a large scale in the manufacture of cement and glass.
29.
Ans: Electrolysis of Water activity:
Equation: $2\text{H}_2\text{O}(l) \xrightarrow{\text{Electricity}} 2\text{H}_2(g) + \text{O}_2(g)$
Reason for double volume: As seen from the balanced equation, water molecules decompose to yield hydrogen and oxygen gases in a 2:1 ratio by volume. Thus, the volume of hydrogen collected at the cathode is double the volume of oxygen collected at the anode.
30.
Ans: A chemical decomposition reaction carried out in the presence of light energy is called photolytic decomposition.
Examples:
  1. $2\text{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\text{Ag}(s) + \text{Cl}_2(g)$ (white silver chloride turns grey).
  2. $2\text{AgBr}(s) \xrightarrow{\text{Sunlight}} 2\text{Ag}(s) + \text{Br}_2(g)$ (pale yellow silver bromide turns grey).
31.
Ans: Silver chloride decomposes slowly when exposed to light, yielding metallic grey silver and chlorine gas. To prevent this photolytic decomposition by blocking sunlight, it is stored in dark-colored glass bottles.
Equation: $2\text{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\text{Ag}(s) + \text{Cl}_2(g)$.
32.
Ans: A displacement reaction is a reaction in which a highly reactive element displaces a less reactive element from its salt solution.
Equation: $\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$
Fading of color: Iron is more reactive than copper. It displaces copper from the blue copper sulphate ($\text{CuSO}_4$) solution, forming light green iron sulphate ($\text{FeSO}_4$) and depositing reddish-brown copper metal on the iron nail.
33.
Ans: Reactivity order: Copper < Iron < Zinc.
Supporting displacement equations:
  1. $\text{Zn}(s) + \text{FeSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Fe}(s)$ (Zinc displaces Iron).
  2. $\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$ (Iron displaces Copper).
  3. Zinc also easily displaces Copper: $\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$, while Copper cannot displace either.
34.
Ans: Heating brown copper powder in a china dish causes it to react with oxygen in the air, forming a layer of black copper(II) oxide.
Balanced equation: $2\text{Cu}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{CuO}(s)$
Observation: The brown surface of copper powder turns completely black.
35.
Ans: Copper reacts slowly with moist carbon dioxide, oxygen, and water in the atmosphere, forming a basic green coating of basic copper carbonate ($\text{CuCO}_3 \cdot \text{Cu(OH)}_2$).
Equation: $2\text{Cu}(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g) + \text{O}_2(g) \rightarrow \text{CuCO}_3 \cdot \text{Cu(OH)}_2(s)$ (Basic Copper Carbonate).
36.
Ans: Exothermic reactions release heat energy to the surroundings, raising the temperature of the mixture. Endothermic reactions absorb heat energy from the surroundings, lowering the temperature.
Exothermic: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) + \text{Heat}$
Endothermic: $\text{N}_2(g) + \text{O}_2(g) \xrightarrow{\text{High Temp}} 2\text{NO}(g)$ or $\text{CaCO}_3(s) \xrightarrow{\Delta} \text{CaO}(s) + \text{CO}_2(g)$.
37.
Ans: During digestion, food is broken down into simple glucose ($\text{C}_6\text{H}_{12}\text{O}_6$). In cellular respiration, glucose combines with oxygen in the cells of our body, undergoing slow oxidation to release energy in the form of ATP and heat.
Equation: $\text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) + \text{Energy (Heat)}$. Since energy is liberated, it is exothermic.
38.
Ans: Microorganisms (bacteria and fungi) break down complex organic molecules in vegetable matter into simpler substances. This decomposition process releases heat energy due to biological respiration inside the compost pile, making it an exothermic reaction. The rich compost formed serves as a highly nutritious organic manure for crops.
39.
Ans: When burning magnesium ribbon is placed in a nitrogen atmosphere, it reacts to form a green-yellow solid called magnesium nitride.
Balanced equation: $3\text{Mg}(s) + \text{N}_2(g) \xrightarrow{\Delta} \text{Mg}_3\text{N}_2(s)$ (Magnesium Nitride).
40.
Ans: A neutralization reaction is a reaction between an acid and a base to form salt and water. It is classified as a type of Double Displacement Reaction.
Balanced equation: $\text{NaOH}(aq) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$.
41.
Ans: Heating sodium hydrogen carbonate (baking soda) undergoes thermal decomposition to yield sodium carbonate, carbon dioxide, and water vapor.
Balanced equation: $2\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)$.
42.
Ans: The shiny brown element 'X' is Copper (Cu). The black colored compound formed is Copper(II) oxide (CuO).
Balanced equation: $2\text{Cu}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{CuO}(s)$.
43.
Ans: Iron reacts with oxygen and moisture in the air to form hydrated iron(III) oxide (rust). Applying paint creates a physical barrier that prevents iron from coming into direct contact with air and moisture, thereby preventing rusting.
44.
Ans: Two commonly used antioxidants are:
  1. BHA (Butylated Hydroxyanisole)
  2. BHT (Butylated Hydroxytoluene).
45.
Ans: Electrolytic decomposition of molten sodium chloride yields sodium metal and chlorine gas.
Balanced equation: $2\text{NaCl}(l) \xrightarrow{\text{Electricity}} 2\text{Na}(s) + \text{Cl}_2(g)$.
Topic 4 Solutions: Redox, Double Displacement & Precipitation
46.
Ans:
  • Oxidation: The addition of oxygen to a substance or the removal of hydrogen from a substance.
  • Reduction: The removal of oxygen from a substance or the addition of hydrogen to a substance.
47.
Ans:
  • Oxidation: The process involving loss of electrons by an atom or ion (e.g., $\text{Na} \rightarrow \text{Na}^+ + e^-$).
  • Reduction: The process involving gain of electrons by an atom or ion (e.g., $\text{Cl} + e^- \rightarrow \text{Cl}^-$).
48.
Ans: For the reaction: $\text{CuO}(s) + \text{H}_2(g) \xrightarrow{\Delta} \text{Cu}(s) + \text{H}_2\text{O}(l)$:
  1. Substance Oxidized: $\text{H}_2$ (gains oxygen to become $\text{H}_2\text{O}$).
  2. Substance Reduced: \text{CuO} (loses oxygen to become \text{Cu} ).
  3. Oxidizing Agent: \text{CuO} (provides oxygen for oxidation).
  4. Reducing Agent: $\text{H}_2$ (removes oxygen from \text{CuO} ).
49.
Ans: A redox reaction is a reaction in which both oxidation and reduction occur simultaneously. In the reaction:
$\text{ZnO}(s) + \text{C}(s) \rightarrow \text{Zn}(s) + \text{CO}(g)$
  • Reduction step: $\text{ZnO}$ is reduced to $\text{Zn}$ by losing oxygen ($\text{ZnO} \rightarrow \text{Zn}$).
  • Oxidation step: $\text{C}$ is oxidized to $\text{CO}$ by gaining oxygen ($\text{C} \rightarrow \text{CO}$).
Since both steps happen in the same reaction, it is a redox process.
50.
Ans: An oxidizing agent is a substance that provides oxygen (or removes hydrogen) in a chemical reaction. In the reaction:
$\text{H}_2\text{S}(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) + \text{S}(s)$
Oxidizing Agent: Chlorine ($\text{Cl}_2$) because it removes hydrogen from hydrogen sulphide ($\text{H}_2\text{S}$) to form hydrochloric acid ($\text{HCl}$), getting reduced in the process.
51.
Ans: In the reaction: $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)$:
Reducing Agent: Carbon Monoxide ($\text{CO}$) because it removes oxygen from iron(III) oxide ($\text{Fe}_2\text{O}_3$), converting it into iron metal while itself being oxidized to carbon dioxide ($\text{CO}_2$).
52.
Ans: Observations: On mixing the two solutions, a rapid precipitation occurs, and a white solid precipitate separates out immediately.
  • Precipitate formed: Barium Sulphate ($\text{BaSO}_4$) (insoluble in water).
  • Balanced Equation: $\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq)$.
53.
Ans: No, all double displacement reactions are not precipitation reactions. A double displacement reaction can also result in the formation of a soluble salt and water, such as in neutralization reactions.
Neutralization double displacement equation:
$\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$ (no precipitate is formed as sodium chloride is highly soluble).
54.
Ans: When magnesium ribbon burns in air, it reacts with oxygen to form magnesium oxide.
Reaction: $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$
  • Oxidizing Agent: Oxygen ($\text{O}_2$) because it gains electrons and oxidizes magnesium.
  • Reducing Agent: Magnesium ($\text{Mg}$) because it loses electrons and reduces oxygen.
55.
Ans: In the reaction: $\text{MnO}_2(s) + 4\text{HCl}(aq) \rightarrow \text{MnCl}_2(aq) + 2\text{H}_2\text{O}(l) + \text{Cl}_2(g)$:
  • Substance Oxidized: Hydrochloric acid ($\text{HCl}$) (loses hydrogen to form $\text{Cl}_2$ gas).
  • Oxidizing Agent: Manganese dioxide ($\text{MnO}_2$) (loses oxygen to form $\text{MnCl}_2$, getting reduced in the process).
56.
Ans: Corrosion is the slow, progressive destruction of metals by the action of air, moisture, or chemical acids in their environment. For rusting of iron, both Oxygen (Air) and Water (Moisture) are essential conditions.
Formula of Rust: Hydrated iron(III) oxide: $\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$.
57.
Ans: Three methods to prevent corrosion of iron:
  1. Painting: Coat the metal surface with paint to block air and water contact.
  2. Galvanization: Coat iron with a thin protective layer of highly reactive zinc metal ($\text{Zn}$). Even if the zinc coating is scratched, it oxidizes preferentially, protecting the underlying iron.
  3. Alloying: Combine iron with carbon, chromium, and nickel to form Stainless Steel, which is highly resistant to rusting.
58.
Ans: Rancidity is the chemical oxidation of unsaturated fats and oils present in food items, resulting in highly unpleasant odors, foul tastes, and stale quality.
Factors accelerating rancidity:
  1. Exposure to direct sunlight (photolysis).
  2. High ambient temperatures.
  3. Presence of oxygen/air in packaging.
  4. High moisture levels.
59.
Ans: For the reaction: $4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)$:
  • Substance Oxidized: Sodium ($\text{Na}$) because it gains oxygen to form $\text{Na}_2\text{O}$ (electronically, it loses electrons: $\text{Na} \rightarrow \text{Na}^+ + e^-$).
  • Substance Reduced: Oxygen ($\text{O}_2$) because it is gained by sodium (electronically, it gains electrons: $\text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-}$).
60.
Ans: For the reaction: $2\text{H}_2\text{S}(g) + \text{SO}_2(g) \rightarrow 3\text{S}(s) + 2\text{H}_2\text{O}(l)$:
  • Oxidizing Agent: Sulphur dioxide ($\text{SO}_2$) (it is reduced by losing oxygen to form sulphur).
  • Reducing Agent: Hydrogen sulphide ($\text{H}_2\text{S}$) (it is oxidized by losing hydrogen to form sulphur).
Topic 5 Solutions: Case Studies & Integrated Puzzles
61.
Ans: Based on Case Study 1 (Electrolysis of Water):
  1. Pure water is a poor conductor of electricity (insulator). Dilute sulphuric acid is added to increase the electrical conductivity of water by releasing free hydronium and sulphate ions for current flow.
  2. Hydrogen gas ($\text{H}_2$) is evolved at the Cathode (negatively charged electrode) as $\text{H}^+$ ions gain electrons.
  3. The volume ratio of gases collected at the anode ($\text{O}_2$) and cathode ($\text{H}_2$) is 1:2 (Oxygen:Hydrogen).
62.
Ans: Based on Case Study 2 (Industrial Metal Extraction):
  1. $\text{Fe}_2\text{O}_3$ (Iron(III) oxide) is reduced as it loses oxygen to become iron metal.
  2. CO (Carbon monoxide) is the reducing agent as it removes oxygen from $\text{Fe}_2\text{O}_3$.
  3. Carbon monoxide ($\text{CO}$) is a gas, which allows it to mix and react uniformly with solid iron oxide ore particles inside the blast furnace, achieving a much higher reduction efficiency than solid carbon.
63.
Ans: Based on the milky gas description:
  1. The gas 'X' is Carbon dioxide ($\text{CO}_2$).
  2. Chemical equations:
    1. Turning milky: $\text{Ca(OH)}_2(aq) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)$ (white ppt).
    2. Milkiness disappearing: $\text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(\text{excess}) \rightarrow \text{Ca(HCO}_3)_2(aq)$.
  3. The soluble compound responsible is Calcium hydrogen carbonate ($\text{Ca(HCO}_3)_2$).
64.
Ans: Based on Case Study 3 (Magical Coin):
  1. Balanced chemical equation: $\text{Cu}(s) + 2\text{AgNO}_3(aq) \rightarrow \text{Cu(NO}_3)_2(aq) + 2\text{Ag}(s)$.
  2. It is a Single Displacement Reaction.
  3. Copper is more reactive than silver. It displaces silver ions from silver nitrate solution, forming soluble copper(II) nitrate ($\text{Cu(NO}_3)_2$), which imparts a characteristic blue color to the solution.
65.
Ans: Based on the reddish-brown metal puzzle:
  1. Element 'A' is Copper (Cu). Black compound 'B' is Copper(II) oxide (CuO). Blue solution 'C' is Copper(II) sulphate ($\text{CuSO}_4$).
  2. Chemical equations:
    1. $2\text{Cu}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{CuO}(s)$.
    2. $\text{CuO}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4(aq) + \text{H}_2\text{O}(l)$.
  3. The formation of 'B' from 'A' is a Combination Reaction (also classified as Oxidation).
66.
Ans: Based on Case Study 4 (Thermite Reaction):
  1. This reaction is called the Thermite Reaction.
  2. The reaction is highly exothermic. The massive amount of heat released raises the temperature beyond the melting point of iron ($1538^\circ\text{C}$), causing the displaced iron to be produced in liquid (molten) state.
  3. Substance Oxidized: Aluminium ($\text{Al}$). Oxidizing Agent: Iron(III) oxide ($\text{Fe}_2\text{O}_3$).
67.
Ans: Based on the stale chips complaint:
  1. The phenomenon is Rancidity (the oxidation of fats and oils present in potato chips by air).
  2. The packets should be flushed with Nitrogen gas ($\text{N}_2$).
  3. Nitrogen is inert and non-reactive. It displaces oxygen from the packet, preventing the chemical oxidation of oil and fats, thereby preserving crispness and flavor.
68.
Ans: Based on Case Study 5 (Lead Nitrate):
  1. The brown gas evolved is Nitrogen dioxide ($\text{NO}_2$).
  2. Balanced chemical equation: $2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$.
  3. Oxygen gas ($\text{O}_2$) is also formed. Test: Bring a glowing wooden splinter near the mouth of the tube; it will burst into flame, confirming the oxygen gas.
69.
Ans: Based on the exothermic hydration:
  1. Chemical equation: $\text{CuSO}_4(s) + 5\text{H}_2\text{O}(l) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s)$ (blue crystals).
  2. The container becomes warm because new chemical coordinate bonds are formed between anhydrous copper sulphate and water molecules, releasing hydration energy.
  3. It is an Exothermic Reaction because heat is liberated to the surroundings ($\Delta H < 0$).
70.
Ans: Based on Case Study 6 (Rusting Frames):
  1. Balanced chemical equation for rusting: $4\text{Fe}(s) + 3\text{O}_2(g) + 2x\text{H}_2\text{O}(l) \rightarrow 2\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}(s)$.
  2. Coastal areas (like Mumbai) have very high atmospheric humidity (water vapor) and presence of salt ions in the air, which increases the electrical conductivity of water droplets on iron, accelerating the electrochemical process of rusting compared to dry desert areas.
  3. Two methods: Galvanization (coating with zinc) or applying a protective layer of Anti-rust primer paint.
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