Chemical Reactions and Equations

In a chemical equation, a downward arrow \(\downarrow\) is used for an insoluble solid product that separates from a solution and settles down; this is called a precipitate. An upward arrow \(\uparrow\) is used for a gas evolved during the reaction. Option B reverses the meanings of these two symbols.

In chemical equations, \((aq)\) means the substance is dissolved in water (aqueous solution). \((l)\) means the substance is present as a liquid (pure liquid state).

Hydrogen has oxidation number 0 in \(\mathrm{H_2}\), which becomes +1 in \(\mathrm{H_2O}\); therefore, hydrogen is oxidised. Copper has oxidation number +2 in \(\mathrm{CuO}\) and 0 in \(\mathrm{Cu}\); therefore, copper(II) oxide is reduced. Option B states the opposite changes.

In option A, zinc is more reactive than copper and displaces copper from \(\mathrm{CuSO_4}\). Zinc is oxidised because its oxidation number changes from \(0\) to \(+2\), whereas copper is reduced because its oxidation number changes from \(+2\) to \(0\). Therefore, this is both a displacement and a redox reaction. Option B is a double-displacement (precipitation) reaction, not a displacement reaction.

The oxidation number of hydrogen in \(H_2\) increases from 0 to +1 in \(H_2O\), so \(H_2\) is oxidised. At the same time, copper in \(CuO\) is reduced from +2 to 0 in \(Cu\). A substance that is itself oxidised while causing reduction of another substance is the reducing agent; therefore, \(H_2\) is the reducing agent. In contrast, \(CuO\) is the oxidising agent.

Oxidation can be identified by an increase in oxidation number or addition of oxygen, while reduction involves a decrease in oxidation number or removal of oxygen. In H₂, hydrogen has oxidation number 0; in H₂O it becomes +1, so hydrogen is oxidised. Copper in CuO is +2 and becomes elemental copper with oxidation number 0, so CuO is reduced. Since both occur together, this is a redox reaction and option A is correct.

According to the law of conservation of mass, mass is neither created nor destroyed during a reaction in a closed container. Since no gas or other substance leaves the container, the total mass of the products equals the total mass of the reactants, which is 20 g. A value such as 10 g would be possible only if some material escaped from the container.

According to the law of conservation of mass, the total mass remains constant during a chemical reaction. In an open container, a gas produced may escape, so its mass will not be included in the measurement. Therefore, the reaction should be performed in a closed container and the total mass should be measured before and after the reaction. Measuring only the remaining solid is incorrect because the gas is also a product of the reaction.

The product side contains \(2\mathrm{MgO}\). Each formula unit of \(\mathrm{MgO}\) has 1 oxygen atom, and the coefficient 2 means there are 2 such units. Thus, the total number of oxygen atoms is \(2\times1=2\). Do not confuse this with the subscript 2 in \(\mathrm{O}_2\), which shows the number of atoms in one oxygen molecule on the reactant side.

Subscripts specify the fixed number of atoms of each element in a compound. For example, changing the subscript 2 in \(\mathrm{H_2O}\) to form \(\mathrm{H_2O_2}\) changes water into hydrogen peroxide, a different substance. Therefore, only coefficients are changed to balance an equation, because they change the amount of a substance without changing its identity.

The reaction is \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2 + heat}\). Calcium oxide and water combine to form a single product, calcium hydroxide; therefore, it is a combination reaction. Since heat is released, it is exothermic. In contrast, in a decomposition reaction, one compound breaks down into two or more substances.

Calcium carbonate undergoes thermal decomposition: \(\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}\). The \(\mathrm{CO_2}\) produced reacts with lime water, \(\mathrm{Ca(OH)_2}\), to form an insoluble white precipitate of \(\mathrm{CaCO_3}\), making the solution milky. Hydrogen, nitrogen, and oxygen do not produce this milkiness.

On heating, calcium carbonate undergoes thermal decomposition: \(\mathrm{CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)}\). Carbon dioxide is released as a gas, while calcium oxide remains as the solid product. Calcium hydroxide is formed when calcium oxide reacts with water, not in this decomposition reaction.

Lead nitrate undergoes thermal decomposition: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). Lead oxide, \(\mathrm{PbO}\), is the solid residue, and nitrogen dioxide, \(\mathrm{NO_2}\), is the brown gas. Although oxygen is also produced, it is colourless, so option A does not identify the brown gas correctly.

Silver bromide is photosensitive. In the presence of light, it can decompose: \(2\mathrm{AgBr}\xrightarrow{\text{light}}2\mathrm{Ag}+\mathrm{Br}_2\). A dark-coloured bottle reduces exposure to light and therefore prevents photolytic decomposition. It does not increase the solubility of silver bromide in water.

The balanced equation for electrolysis of water is \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). At the same temperature and pressure, gas volumes are proportional to their mole ratio. Therefore, \(H_2:O_2 = 2:1\). Hence, for 20 mL of hydrogen, the oxygen volume is \(20/2 = 10\) mL. Choosing 20 mL would be incorrect because equal volumes of the two gases are not produced.

Calcium carbonate undergoes thermal decomposition: CaCO₃ → CaO + CO₂. In the open test tube, carbon dioxide escapes, so the measured solid loses the mass of the gas and appears lighter. For example, 100 g of calcium carbonate can produce 56 g of calcium oxide and 44 g of carbon dioxide; the total remains 100 g when all products are included. Calcium atoms are not destroyed, and CO₂ has mass. Therefore option A correctly explains the observation.

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 deposits as a reddish-brown coating on the nail, while the iron sulphate formed makes the solution green. Unlike option A, copper cannot displace iron because copper is less reactive.

The blue colour of aqueous copper sulphate is associated mainly with hydrated Cu²⁺ ions. When an iron strip is placed in the solution, iron is more reactive and displaces copper: Fe + CuSO₄ → FeSO₄ + Cu. As Cu²⁺ ions are removed from the solution and copper metal is deposited, the intensity of the blue colour decreases; Fe²⁺ may also give the solution a pale green colour. Iron ions do not disappear, and sodium or barium ions are not involved. Therefore option C is correct.

Hydrogen removes oxygen from \(\mathrm{CuO}\), converting it to copper \(\mathrm{Cu}\); therefore, \(\mathrm{H_2}\) is the reducing agent. Hydrogen is itself oxidised to \(\mathrm{H_2O}\). In contrast, \(\mathrm{CuO}\) is reduced, so it is the oxidising agent rather than the reducing agent.

In option A, zinc, \(\mathrm{Zn}\), displaces copper from copper sulfate, so the reaction is a displacement reaction. The oxidation state of Zn changes from \(0\) to \(+2\), showing oxidation, while that of \(\mathrm{Cu^{2+}}\) changes from \(+2\) to \(0\), showing reduction. Thus, it is a redox displacement reaction. In contrast, option C also involves oxidation and reduction, but it is a photochemical decomposition reaction, not a displacement reaction.

CuO loses oxygen, so it is reduced; the oxidation number of copper changes from \\( +2 \\) to \\(0\\). H₂ gains oxygen to form H₂O, so it is oxidised; the oxidation number of hydrogen changes from \\(0\\) to \\( +1 \\). Therefore, this is a redox reaction, not merely a displacement reaction.

In \(H_2O_2\), oxygen has an oxidation number of \(-1\). Some oxygen becomes \(-2\) in water, \(H_2O\), so it is reduced, while some becomes \(0\) in \(O_2\), so it is oxidised. Thus, oxygen undergoes both changes in the same reaction. In contrast, during the decomposition of \(KClO_3\), chlorine is reduced and oxygen is oxidised; the changes occur in different elements.

From the balanced equation \(2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu}\), the mole ratio \(\text{Al} : \text{Cu} = 2 : 3\). So for \(1\) mol Al, \(\text{Cu} = 1 \times \frac{3}{2} = \frac{3}{2} = 1.5\) mol. Hence, \(1\) mol Al does not produce \(1\) mol Cu; it produces \(\frac{3}{2}\) mol Cu.

In option B, zinc is more reactive than copper and displaces copper from \(\mathrm{CuSO_4}\). Heat is released during this reaction, so it is an exothermic displacement reaction. In contrast, option D involves exchange of ions and is a double-displacement reaction.