Chemical Reactions and Equations

The reaction is \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2 + heat}\). Two reactants combine to form one product, calcium hydroxide, so it is a combination reaction. Since heat is released, it is also exothermic. In a decomposition reaction, one compound breaks down into simpler substances, which does not occur here.

This is a redox reaction, in which oxidation and reduction occur together. Hydrogen changes from oxidation number 0 in H₂ to +1 in H₂O, so it is oxidised by gaining oxygen or losing electrons conceptually. Copper in CuO changes from +2 to 0 in Cu, so CuO is reduced. Thus option A is correct; B reverses the changes, while C and D incorrectly assign the same process to both substances.

Lead nitrate undergoes thermal decomposition: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). The solid residue is lead oxide, \(\mathrm{PbO}\), and the brown gas is nitrogen dioxide, \(\mathrm{NO_2}\). Oxygen is also produced, but it is colourless, unlike the brown \(\mathrm{NO_2}\) gas.

Thermal decomposition of lead nitrate produces lead(II) oxide, nitrogen dioxide, and oxygen: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). \(\mathrm{NO_2}\) is a brown gas, whereas \(\mathrm{O_2}\) is colourless. Therefore, the visible brown fumes are nitrogen dioxide, not oxygen. Oxygen is confirmed by its ability to relight a glowing splint.

Silver chloride is photosensitive and decomposes in sunlight: \(2\mathrm{AgCl}(s) \xrightarrow{\text{sunlight}} 2\mathrm{Ag}(s) + \mathrm{Cl}_2(g)\). The silver metal formed is grey, so the silver chloride appears grey. Chlorine is produced as a gas, not as a solid.

Silver bromide decomposes in the presence of light to form silver and bromine: \(2\mathrm{AgBr}\xrightarrow{\text{light}}2\mathrm{Ag}+\mathrm{Br}_2\). A dark-coloured bottle reduces the light reaching the substance and thus helps prevent this photochemical decomposition. This is not related to dissolving it in water or increasing its hardness.

The balanced equation for electrolysis of water is \(2H_2O \rightarrow 2H_2 + O_2\). At the same temperature and pressure, gas volumes are proportional to their mole ratios. Therefore, the volume ratio of hydrogen to oxygen is \(2:1\). For 7 mL of oxygen, the volume of hydrogen is \(2 \times 7 = 14\) mL. The value 21 mL would result from incorrectly using a \(3:1\) ratio.

The oxidation number of zinc changes from \(0\) to \(+2\), so it loses two electrons and is oxidised: \(\mathrm{Zn \rightarrow Zn^{2+}+2e^-}\). The oxidation number of \(\mathrm{Cu^{2+}}\) changes from \(+2\) to \(0\); it gains electrons and is reduced: \(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\). Therefore, option A is correct. Option B reverses both changes.

Electrolysis requires the solution to conduct electric current. Pure or distilled water contains very few ions, so its electrical conductivity is too low for efficient electrolysis. Adding a small amount of a suitable acid increases the concentration of mobile ions, allowing current to pass and the water to decompose into hydrogen and oxygen. Therefore option B is correct; the other choices are not problems of electrical conduction.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution:
\[\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\]
The displaced copper forms a reddish-brown coating on the iron nail. Iron sulphate is formed in the solution; it does not deposit on the nail.

The oxidation state of Al increases from 0 to +3, so Al is oxidised and loses electrons. These electrons reduce Fe³⁺ in \(Fe_2O_3\) to Fe. Therefore, Al is the reducing agent. In contrast, \(Fe_2O_3\) accepts electrons and acts as the oxidising agent.

In a displacement reaction, a more reactive metal displaces a less reactive metal from an aqueous solution of its salt. Since zinc displaces copper, zinc is more reactive than copper. In contrast, copper cannot displace zinc because copper is less reactive.

Barium chloride and sodium sulphate undergo a double-displacement reaction in aqueous solution:
\(\mathrm{BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)}\).
Barium sulphate is the white insoluble precipitate, while sodium chloride remains in solution. In option C, sodium chloride is formed, but the precipitate is barium carbonate rather than barium sulphate.

The reaction is:
\[\mathrm{BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)\downarrow+2NaCl(aq)}\]
In solution, \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) ions combine to form insoluble \(\mathrm{BaSO_4}\), which separates as a solid precipitate. In contrast, \(\mathrm{NaCl}\) remains soluble in water.

A double displacement reaction generally occurs between two ionic compounds. The positive ions, or cations, exchange partners with the negative ions, or anions, of the other compound. For example, in AB + CD, the new combinations are AD and CB, provided the resulting substances are chemically possible, such as a precipitate, gas, or water. Therefore option B gives the governing principle; the other choices describe unrelated physical or chemical processes.

During thermal decomposition, \(\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}\). Since 100 g of \(\mathrm{CaCO_3}\) produces 56 g of \(\mathrm{CaO}\), 12.0 g should produce \(12.0 \times 56/100 = 6.72\) g of solid, consistent with 6.7 g. The \(\mathrm{CO_2}\) gas escapes from the open crucible, so weighing only the solid gives a lower mass. In a closed system, the combined mass of the solid and gas remains conserved.

The reaction is \(\mathrm{CuO + H_2 \rightarrow Cu + H_2O}\). Oxygen is removed from copper(II) oxide to form water, and \(\mathrm{CuO}\) changes into copper. Removal of oxygen is reduction; therefore, copper(II) oxide is reduced. In contrast, hydrogen gains oxygen to form water, so hydrogen is oxidised.

The reaction is \(\mathrm{CuO + H_2 \rightarrow Cu + H_2O}\). Hydrogen gains oxygen from copper(II) oxide and forms water. Gain of oxygen is oxidation, so hydrogen is oxidised. In contrast, copper(II) oxide loses oxygen to form copper and is therefore reduced.

In option C, the ions of \(\mathrm{AgNO_3}\) and \(\mathrm{NaCl}\) exchange partners, producing insoluble \(\mathrm{AgCl}\) as a precipitate. Hence, it is both a double displacement and a precipitation reaction. In contrast, in option A, zinc displaces hydrogen, so it is a single displacement reaction.

The correct pair is \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\). These ions form insoluble, white barium sulfate in water: \(\mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)}\). \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) remain as spectator ions in solution and do not form the white precipitate.