(a) The following reaction scheme is an illustration of the contact process. Study the scheme and answer the questions that follow. (i) Name X and Y (ii) Wr...

(a) The Contact process. The scheme reads: X(gas) + Y(solid) \(\xrightarrow{I}\) Sulphur(IV) oxide; Oxygen + Sulphur(IV) oxide \(\xrightarrow{II}\) Sulphur(VI) oxide \(\xrightarrow{III}\) Oleum \(\xrightarrow{IV}\) Concentrated H₂SO₄.

(i) X and Y. The first step burns a solid in a gas to give SO₂, so X is oxygen (air) and Y is sulphur.

(ii) Balanced equations.

Process I: \[ S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)} \]

Process II: \[ 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \]

Process III (SO₃ absorbed in concentrated H₂SO₄ to form oleum): \[ SO_{3(g)} + H_2SO_{4(l)} \rightarrow H_2S_2O_{7(l)} \]

Process IV (oleum diluted with water): \[ H_2S_2O_{7(l)} + H_2O_{(l)} \rightarrow 2H_2SO_{4(l)} \]

(iii) Catalyst in process II: vanadium(V) oxide, V₂O₅ (platinum can also be used).

(iv) Le Chatelier's principle. Process II is exothermic in the forward direction. By Le Chatelier's principle, raising the temperature shifts the equilibrium position in the endothermic (backward) direction so as to absorb the added heat. This reduces the yield of SO₃, so a high temperature does not favour the forward reaction; a moderate temperature (about 450 C) is used instead.

(v) Two uses of SO₂:

• As a bleaching agent (for wood pulp, straw, silk).
• As a food preservative and fumigant/disinfectant (it also serves as the starting gas for making H₂SO₄).

(b) Volume of unused oxygen. From \(2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}\), H₂ reacts with O₂ in the ratio 2 : 1 by volume.

\[ \text{O}_2\ \text{needed} = \frac{1}{2} \times 40 = 20\ \text{cm}^3 \]

\[ \text{Unused O}_2 = 30 - 20 = \mathbf{10\ cm^3} \]

(c) Heating calcium trioxocarbonate(IV).

\[ CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)} \]

Moles of CaCO₃: molar mass \(= 40.0 + 12.0 + (3\times16.0) = 100\ g\,mol^{-1}\).

\[ n = \frac{1.0}{100} = 0.01\ \text{mol} \]

Mass of residue (CaO): molar mass of CaO \(= 40.0 + 16.0 = 56\ g\,mol^{-1}\).

\[ m_{CaO} = 0.01 \times 56 = \mathbf{0.56\ g} \]

Volume of CO₂ at s.t.p.: \(n_{CO_2} = 0.01\ mol\).

\[ V = 0.01 \times 22.4 = 0.224\ dm^3 = \mathbf{224\ cm^3} \]

Volume at 15 C and 760 mmHg: pressure is unchanged (760 mmHg = s.t.p. pressure), so use \(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\) with \(T_1 = 273\ K,\ T_2 = 273 + 15 = 288\ K\).

\[ V_2 = 0.224 \times \frac{288}{273} = \mathbf{0.236\ dm^3\ (\approx 236\ cm^3)} \]