Chemical Reactions and Equations

Calcium oxide, \(\mathrm{CaO}\) (quicklime), reacts with water to form calcium hydroxide, \(\mathrm{Ca(OH)_2}\) (slaked lime):
\[\mathrm{CaO + H_2O \rightarrow Ca(OH)_2}\]
This is an exothermic reaction. Calcium carbonate requires \(\mathrm{CO_2}\) for its formation, so it is not produced by adding water alone.

Calcium oxide reacts with water as CaO + H₂O → Ca(OH)₂. Two reactants form one product, calcium hydroxide, so the reaction is a combination reaction. Considerable heat is released during the reaction, which gives it the additional classification exothermic. It is not decomposition, displacement or photochemical breakdown. Thus option A states both required reasons and is correct.

Calcium carbonate undergoes thermal decomposition: \(\mathrm{CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)}\). Therefore, calcium oxide is the solid product and carbon dioxide is the gaseous product. Calcium hydroxide would require water, which is not a reactant in this reaction.

Thermal decomposition of calcium carbonate produces carbon dioxide: \(\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}\). When this gas is passed through lime water, \(\mathrm{Ca(OH)_2}\), it forms an insoluble precipitate of calcium carbonate, making the lime water milky: \(\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O}\). Oxygen does not form a calcium carbonate precipitate under these conditions.

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 the brown gas, while \(\mathrm{O_2}\) is colourless. Option B incorrectly names nitrogen gas; nitrogen is released in this reaction as \(\mathrm{NO_2}\), not as \(\mathrm{N_2}\).

On heating, lead nitrate undergoes thermal decomposition: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). The yellow solid residue is lead(II) oxide, \(\mathrm{PbO}\). In contrast, lead(II) chloride contains chloride ions, which cannot be formed by merely heating lead nitrate.

On heating, lead nitrate decomposes according to 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂. Nitrogen dioxide, NO₂, is a reddish-brown gas, so its brown fumes provide a visible clue during the reaction. Oxygen is colourless, as are hydrogen and carbon dioxide under ordinary conditions. Therefore the gas identified by the brown fumes is nitrogen dioxide, making option B correct.

In the presence of light, silver chloride decomposes to form metallic silver and chlorine gas: \(2AgCl \xrightarrow{sunlight} 2Ag + Cl_2\). The metallic silver formed causes the grey colour. Since light causes the decomposition, it is photolytic decomposition. In contrast, thermal decomposition is caused by heat, while electrolytic decomposition is caused by electric current.

In the presence of light, silver chloride undergoes photodecomposition: \(2\mathrm{AgCl}\xrightarrow{\text{light}}2\mathrm{Ag}+\mathrm{Cl_2}\). The metallic silver formed is grey, so AgCl appears grey. Chlorine is also produced, but it does not cause the grey colour.

Many silver salts, such as silver chloride and silver bromide, are photosensitive. Light can supply energy for their decomposition, producing changed substances; for example, silver chloride darkens as silver forms. A dark or amber bottle absorbs or blocks much of the incident light and slows this photochemical reaction. It does not increase solubility, cause combustion, or remove mass. Hence option A is correct.

The electrolysis equation is 2H₂O → 2H₂ + O₂. It shows that hydrogen and oxygen are produced in a volume ratio of 2:1, provided temperature and pressure are the same. Therefore, for 15 mL of oxygen, hydrogen volume = 2 × 15 = 30 mL. Option A is half the oxygen volume, while B ignores the ratio and D uses an incorrect factor of three.

In hydrogen peroxide, oxygen has oxidation number −1. During decomposition, some oxygen atoms become −2 in H₂O, so they are reduced, while other oxygen atoms become 0 in O₂, so they are oxidised. The same compound therefore undergoes simultaneous oxidation and reduction; this is disproportionation. The other choices are displacement, combination, and double-displacement reactions, not disproportionation.

During electrolysis of water, hydrogen is produced at the cathode and oxygen at the anode. Their volume ratio is hydrogen : oxygen = 2 : 1, so oxygen is the gas formed in the lower volume, at the anode. The cathode produces the larger volume of hydrogen. Exam tip: Remember: H₂ at the cathode and O₂ at the anode during electrolysis of water.

Pure water contains very few ions and is therefore a poor conductor of electricity. Adding a suitable acid, such as dilute sulphuric acid, increases the concentration of mobile ions in the solution. These ions carry charge between the electrodes, allowing electrolysis to proceed more readily. The acid is added to improve conductivity, not to increase water mass or give colour or hardness to the gases.

Decomposition means breaking one compound into simpler substances, but the energy supplied can differ. In electrolytic decomposition, electrical energy drives the reaction, as in the electrolysis of water. In photolytic decomposition, light energy causes the compound to split, as in the decomposition of silver chloride in sunlight. Thus the source of energy is the distinguishing basis; precipitate colour, metal lustre, and gas mass are not defining criteria.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution: Fe + CuSO₄ → FeSO₄ + Cu. The deposited copper appears reddish brown, while aqueous iron(II) sulphate is pale green, causing the solution to become greenish. Sodium chloride, barium sulphate, and magnesium oxide contain elements not involved in this reaction, so they cannot explain the observation.

The blue colour of aqueous copper sulphate is mainly due to hydrated copper(II) ions, \(\mathrm{Cu^{2+}}\). When these ions are converted into copper metal during a reaction, their concentration in the solution decreases and the blue colour fades. Iron(II) ions may form and change the solution’s appearance, but the direct reason for loss of blue colour is the decrease in \(\mathrm{Cu^{2+}}\) ions. Exam tip: Link the colour of a solution with the metal ion responsible for it.

In a displacement reaction, a solid metal can displace the metal present in a salt solution only if it is more reactive. Since metal A is less reactive than metal B, A cannot remove B from B’s salt solution. Therefore, no displacement reaction occurs. Option B would be possible only if A were more reactive than B. Exam tip: A metal higher in the reactivity series displaces a metal lower in the series from its salt solution.

Carbon dioxide normally suppresses fire by reducing contact between fuel and oxygen, but burning magnesium is an important exception. At high temperature magnesium removes oxygen from carbon dioxide: 2Mg + CO₂ → 2MgO + C. The reaction supplies magnesium with oxygen and produces carbon, so the ribbon can continue burning. CO₂ is not combustible, and it does not simply decompose into oxygen.

For example, in
\(BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)\), the ions exchange partners, so it is a double displacement reaction. The \(BaSO_4\) formed is an insoluble white solid in water; therefore, its formation is called precipitation. A combination reaction usually forms one product from the reactants, so option A is not suitable. Exam tip: When mixing aqueous solutions produces an insoluble solid, identify that solid as a precipitate.