Chemical Reactions and Equations

According to the law of conservation of mass, atoms are neither created nor destroyed in a chemical reaction. Therefore, in a balanced chemical equation, the number of atoms of every element is the same on the reactant and product sides, so total mass is conserved. Option B is incorrect because products may be gases, liquids, or solids.

The reaction is \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2}\). Two reactants combine to form a single product, calcium hydroxide, so it is a combination reaction. The container becoming hot indicates that heat is released; therefore, it is also exothermic. In contrast, a decomposition reaction involves one compound breaking down into two or more products.

Heating calcium carbonate produces carbon dioxide:
\[\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}\]
The carbon dioxide reacts with calcium hydroxide in lime water to form an insoluble precipitate of calcium carbonate:
\[\mathrm{CO_2 + Ca(OH)_2 \rightarrow CaCO_3\downarrow + H_2O}\]
This insoluble calcium carbonate precipitate makes lime water appear milky. In contrast, passing excess carbon dioxide can form soluble calcium bicarbonate, causing the milkiness to decrease.

Thermal decomposition requires an input of energy to break the bonds in calcium carbonate. On heating, calcium carbonate decomposes as CaCO₃ → CaO + CO₂, so heat supplies the activation energy and is written over the reaction arrow as a condition, not as a product. Hence option B is correct. The products are calcium oxide and carbon dioxide, not heat or a precipitate formed by heat.

Lead nitrate undergoes thermal decomposition on heating:
\[
2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}
\]
\(\mathrm{NO_2}\) is the brown gas, and \(\mathrm{PbO}\) is the solid residue. Although \(\mathrm{O_2}\) is also formed, it is colourless, so it cannot be the brown gas.

Lead nitrate undergoes thermal decomposition: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). \(\mathrm{NO_2}\) is a brown gas, so the brown fumes confirm the formation of nitrogen dioxide. Although \(\mathrm{O_2}\) is also produced, it is colourless and therefore does not account for the brown colour.

In sunlight, silver chloride decomposes: \(2\mathrm{AgCl}(s) \xrightarrow{\text{sunlight}} 2\mathrm{Ag}(s) + \mathrm{Cl}_2(g)\). This is a photolytic decomposition reaction. The silver metal formed is grey, so white silver chloride gradually appears grey. Chlorine gas is also produced, but it does not cause the grey colour.

Silver bromide is photosensitive. Light supplies energy that can cause it to decompose, producing silver and bromine; this property is related to photochemical reactions and was used in photographic materials. A dark or amber bottle absorbs or blocks much of the incident light, reducing decomposition during storage. Therefore A is correct; the bottle is not intended to promote burning, dissolving, or mass gain from air.

In option A, calcium oxide and water combine to form a single product, calcium hydroxide, and heat is released during the process. Therefore, it is an exothermic combination reaction. In contrast, option B shows one compound breaking into two products on heating, so it is a thermal decomposition reaction.

Pure water contains very few ions, so it conducts electricity poorly and electrolysis proceeds very slowly. Adding dilute sulphuric acid supplies \(\mathrm{H^+}\) and other ions, increasing the electrical conductivity of the solution so that current can pass. The acid is added to improve conduction, not to increase the amount of oxygen produced.

The balanced equation for electrolysis of water is 2H₂O → 2H₂ + O₂. It shows that hydrogen and oxygen are produced in a volume ratio of 2:1 under the same conditions. Therefore, for 5 mL of oxygen, hydrogen volume = 2 × 5 = 10 mL. Thus option B is correct; the other options do not follow the stoichiometric gas ratio.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution. The reaction is \(\mathrm{Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s)}\). The released copper deposits as a reddish-brown coating on the nail, while the iron sulphate formed makes the solution pale green. Rusting in option B is a different process and does not produce a copper coating.

The blue colour of copper sulphate solution is due to hydrated copper(II) ions, \(\mathrm{Cu^{2+}}\). Since iron is more reactive, it displaces copper(II) ions from the solution: \(\mathrm{Fe(s)+CuSO_4(aq)\rightarrow FeSO_4(aq)+Cu(s)}\). Thus, the concentration of \(\mathrm{Cu^{2+}}\) decreases and the blue colour fades. In contrast, \(\mathrm{Fe^{2+}}\) ions are formed during the reaction; they do not decrease.

In a displacement reaction, a more reactive metal removes a less reactive metal from its salt solution. Since A displaces B, A is more reactive than B, written A > B. Since B displaces C, B > C. Combining both relationships gives A > B > C, so option A is correct. Equal reactivity would not produce these displacements.

In a displacement reaction, a more reactive metal displaces a less reactive metal from its salt solution. Zinc displaces copper from copper sulphate, so zinc is more reactive than copper. Copper cannot displace zinc from zinc sulphate because copper is the less reactive metal.

In the double-displacement reaction between barium chloride and sodium sulphate, insoluble white barium sulphate is formed:
\(\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)}\).
Therefore, sodium chloride remains in solution. Option B can also form a white precipitate, but it is calcium carbonate, not barium sulphate.

A precipitation reaction occurs when ions present in two aqueous solutions combine to form a product that is insoluble in water. The product separates as a solid precipitate, often making the mixture cloudy or visibly solid. Therefore option A is correct. Gas evolution indicates another type of reaction, while melting and freezing are physical changes.

The reaction is \(\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)}\). The \(\mathrm{Ba^{2+}}\) and \(\mathrm{Na^+}\) ions exchange anionic partners, so it is a double-displacement reaction. The product \(\mathrm{BaSO_4}\) is an insoluble solid that separates as a precipitate, so it is also a precipitation reaction. Unlike option B, no oxidation or reduction occurs in this reaction.

Copper oxide contains oxygen bonded to copper. In the reaction CuO + H₂ → Cu + H₂O, oxygen is removed from copper oxide and combines with hydrogen to form water. Removal of oxygen is defined as reduction, so option A is correct. Addition of oxygen would indicate oxidation, while precipitation and evaporation do not describe this chemical change.

Option B is correct. In the balanced equation, \(2KClO_3\) contains 2 K atoms, 2 Cl atoms, and 6 O atoms. The products \(2KCl\) and \(3O_2\) also contain 2 K atoms, 2 Cl atoms, and 6 O atoms. \(MnO_2\) is not a reactant; it is a catalyst that increases the reaction rate and is written above the arrow. In option A, oxygen atoms are not balanced, and the reaction is a decomposition reaction, not a displacement reaction.