A set up is used to prepare hydrogen chloride gas and salt T. Identify Liquid M, Gas V, and Salt T. Write balanced chemical equations for the reactions. Name the process that formed salt T. Explain wh

here are the solutions to the questions:

33. The set up below was used to prepare hydrogen chloride gas and salt T.

    a) Identify the following

    • (i) Liquid M: $\boxed{\text{Concentrated Sulphuric (VI) acid}}$
    • (ii) Gas V: $\boxed{\text{Hydrogen chloride gas (HCl)}}$
    • (iii) Salt T: $\boxed{\text{Aluminium chloride (AlCl}_3\text{ or Al}_2\text{Cl}_6\text{)}}$

    b) Write balanced chemical equations for reactions that occur at:

    • (i) Flask I: $$\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \xrightarrow{\text{heat}} \text{NaHSO}_4\text{(s)} + \text{HCl(g)}$$ or at higher temperatures: $$2\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \xrightarrow{\text{heat}} \text{Na}_2\text{SO}_4\text{(s)} + 2\text{HCl(g)}$$
    • (ii) Combustion tube: $$2\text{Al(s)} + 6\text{HCl(g)} \xrightarrow{\text{heat}} 2\text{AlCl}_3\text{(s)} + 3\text{H}_2\text{(g)}$$

    c) Name the process that formed salt T as shown in the diagram.
    $\boxed{\text{Direct combination}}$ or $\boxed{\text{Synthesis}}$

    d) Sulphuric (VI) acid is used as a drying agent in this experiment. Explain why calcium oxide is unsuitable for the same purpose in this reaction.
    Sulphuric (VI) acid is suitable because it is an acidic drying agent and does not react with acidic hydrogen chloride gas. Calcium oxide is a basic oxide and would react with the acidic hydrogen chloride gas, forming calcium chloride and water, thus not effectively drying the gas.

    e) The water in the beaker was found to have a pH of 2.0 at the end of the experiment. Explain.
    Hydrogen chloride gas is highly soluble in water and ionizes completely to form hydrochloric acid. Hydrochloric acid is a strong acid, which causes the water to become acidic, resulting in a low pH of 2.0.

    f) Using dots(.) and cross diagram, draw the structure of hydrogen chloride gas

    $$\text{H} \overset{\times}{\cdot} \text{Cl} :$$

    (Where $\times$ represents an electron from Hydrogen and $\cdot$ represents an electron from Chlorine. Chlorine also has three lone pairs of electrons.)
    Alternatively:

    $$\text{H} \overset{\times}{\cdot} \underset{\cdot\cdot}{\overset{\cdot\cdot}{\text{Cl}}} \cdot\cdot$$

    g) In the space provided below, draw a well labelled diagram showing how you would dissolve hydrogen chloride gas in water.
    (Diagram should show an inverted funnel with its rim just touching the surface of the water in a beaker, with HCl gas being delivered into the funnel. This prevents suck-back due to the high solubility of HCl.)

    • Diagram features:
      • Gas jar/delivery tube supplying HCl gas.
      • Inverted funnel with its rim just touching the surface of the water.
      • Beaker containing water.
      • Labels for HCl gas, funnel, water.

    h) Explain why hydrogen chloride gas dissolved in methylbenzene does not react with calcium carbonate.
    Hydrogen chloride gas dissolves in methylbenzene to form a covalent solution where it exists as discrete HCl molecules. It does not ionize to produce $\text{H}^+$ ions. For a reaction with calcium carbonate to occur, $\text{H}^+$ ions are required.

    i) Using equation, state the observation made when a gas jar containing hydrogen chloride gas is opened near an open bottle of liquid ammonia.

    • Observation: $\boxed{\text{Dense white fumes are observed.}}$
    • Equation: $$\text{HCl(g)} + \text{NH}_3\text{(g)} \rightarrow \text{NH}_4\text{Cl(s)}$$

    j) Explain why the solution formed when hydrogen chloride gas dissolves in water conducts electricity yet the solution formed when hydrogen chloride gas dissolved in methylbenzene does not.
    When hydrogen chloride gas dissolves in water, it ionizes to form mobile $\text{H}^+$ (or $\text{H}_3\text{O}^+$) and $\text{Cl}^-$ ions, which are charge carriers, allowing the solution to conduct electricity. In contrast, when hydrogen chloride gas dissolves in methylbenzene, it remains as covalent molecules and does not ionize, meaning there are no free ions to conduct electricity.