Chemical Reactions and Equations

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.

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.

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.

Rusting requires iron to be in contact with both oxygen and water (moisture). A coating of paint or oil forms a protective barrier on the iron surface and prevents air and moisture from reaching the metal. Therefore, option B is correct. Heating and drying may remove moisture temporarily, but unlike paint or oil, it does not provide a lasting protective coating.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The reddish-brown coating is copper, and the light-green solution is due to iron sulphate. This is not a combination reaction because the reactants do not combine to form a single product.

Oils and fats in oily food react with oxygen in air and undergo oxidation. This oxidation produces an unpleasant smell and taste; the process is called rancidity. Heat increases the reaction rate, so oily food becomes rancid faster in warm, open air. Option B is incorrect because warmth does not make fats unreactive; it can increase the rate of oxidation.

In this reaction, the two reactants, magnesium and oxygen, combine to form only one product, \(\mathrm{MgO}\). Therefore, it is a combination reaction. In a displacement reaction, an element replaces another element present in a compound, for example \(\mathrm{Zn}+\mathrm{CuSO}_4\rightarrow\mathrm{ZnSO}_4+\mathrm{Cu}\). No compound containing a replaceable element is present here.

Zinc is more reactive than copper, so it displaces copper from copper sulphate: \(\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}\). As colourless zinc sulphate forms, the blue colour fades, and displaced copper deposits as a reddish-brown coating on zinc. In contrast, this reaction does not produce a gas or a white precipitate.

The oxidation number of Sn in \(\mathrm{SnCl_2}\) increases from \(+2\) to \(+4\), so \(\mathrm{SnCl_2}\) is oxidised. It donates electrons and reduces \(\mathrm{Fe^{3+}}\) in \(\mathrm{FeCl_3}\) to \(\mathrm{Fe^{2+}}\). Therefore, \(\mathrm{SnCl_2}\) is the reducing agent, whereas \(\mathrm{FeCl_3}\) is the oxidising agent.

Iron is more reactive than copper. Therefore, iron displaces copper from copper sulphate and forms iron sulphate. The green solution is iron sulphate, \(\mathrm{FeSO_4}\), and the reddish-brown coating is copper metal. The reaction is \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). It is a displacement reaction because a more reactive metal replaces a less reactive metal from its compound. In a double displacement reaction, ions of two compounds exchange partners, which does not occur here.

Iron is more reactive than copper, so it displaces copper from copper sulfate solution. In this reaction, iron normally forms \(Fe^{2+}\) ions; therefore, the salt formed is ferrous sulfate, \(FeSO_4\): \(Fe + CuSO_4 \rightarrow FeSO_4 + Cu\). Option A correctly refers to the relative reactivity, but \(Fe_2(SO_4)_3\) contains iron in the \(Fe^{3+}\) state and is not the product of this usual displacement reaction.

In \(Fe_2O_3\), iron has oxidation state \(+3\), which becomes \(0\) in the product \(Fe\). A decrease in oxidation state indicates reduction; therefore, iron(III) oxide is reduced. In contrast, aluminium changes from \(0\) to \(+3\), so it is oxidised.