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I) The reaction in chamber Q is the dissolution of aluminium oxide from bauxite in sodium hydroxide solution.

Chemistry

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Answer

a(iv)

236.37 g\text{236.37 g}

Step-by-step solution

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Part a(i)

The reaction in chamber Q is the dissolution of aluminium oxide from bauxite in sodium hydroxide solution.

Al2O3(s)+2NaOH(aq)+3H2O(l)→2NaAl(OH)4(aq)\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \to 2\text{NaAl(OH)}_4\text{(aq)}

Part a(ii)

Residue P is Iron(III) oxide (Fe2O3\text{Fe}_2\text{O}_3).

Part a(iii)

R: Precipitation (or Hydrolysis)
C: Electrolysis

Part a(iv)

236.37 g\text{236.37 g}

  1. Convert time from hours to seconds.

    t=8 hours×60minuteshour×60secondsminute=28800 st = 8 \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} \times 60 \frac{\text{seconds}}{\text{minute}} = 28800 \text{ s}

  2. Calculate the total charge (Q) passed.

    Q=I×t=88.0 A×28800 s=2534400 CQ = I \times t = 88.0 \text{ A} \times 28800 \text{ s} = 2534400 \text{ C}

  3. Determine the moles of electrons.

    Moles of electrons=QFaraday constant=2534400 C96500 C/mol=26.2632 mol\text{Moles of electrons} = \frac{Q}{\text{Faraday constant}} = \frac{2534400 \text{ C}}{96500 \text{ C/mol}} = 26.2632 \text{ mol}

  4. Use the stoichiometry of the electrolysis reaction (Al3++3e−→Al\text{Al}^{3+} + 3\text{e}^- \to \text{Al}) to find moles of aluminium.

    Moles of Al=Moles of electrons3=26.2632 mol3=8.7544 mol\text{Moles of Al} = \frac{\text{Moles of electrons}}{3} = \frac{26.2632 \text{ mol}}{3} = 8.7544 \text{ mol}

  5. Calculate the mass of aluminium produced.

    Mass of Al=Moles of Al×Molar mass of Al=8.7544 mol×27 g/mol=236.3688 g\text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} = 8.7544 \text{ mol} \times 27 \text{ g/mol} = 236.3688 \text{ g}

    The mass of aluminium produced is 236.37 g\boxed{\text{236.37 g}}.

Part a(v)

Process B is the conversion of aluminium hydroxide to aluminium oxide. This is accomplished by

  • Heating (or calcination)
  • Dehydration

Part a(vi)

The equation for the conversion of aluminium hydroxide into aluminium oxide is

2Al(OH)3(s)→heatAl2O3(s)+3H2O(g)2\text{Al(OH)}_3\text{(s)} \xrightarrow{\text{heat}} \text{Al}_2\text{O}_3\text{(s)} + 3\text{H}_2\text{O(g)}

Part b

i) Aluminium is an amphoteric metal, meaning it reacts with both acids and strong bases. Strongly alkaline washing liquids will react with the aluminium metal to form soluble sodium aluminate, which corrodes the utensil.

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