A factory extracts aluminium from bauxite by electrolysis of aluminium oxide, Al2O3. Use the data in the table to answer the questions. Quantity Value relat...

This is a quantitative question on the electrolysis of aluminium oxide, testing relative formula mass, reacting masses, percentage yield, and electrode half-equations.

(a) The relative formula mass of \( \mathrm{Al_2O_3} \):

\[ M_r = (2 \times 27) + (3 \times 16) = 54 + 48 = 102 \]

so \( M_r = 102 \) [1].

(b) Work in moles (tonnes are consistent throughout, so we can treat them like grams here):

\[ \text{moles of } Al_2O_3 = \frac{510}{102} = 5 \]

Each formula unit contains 2 aluminium atoms, so [1] for the moles and [1] for using the 1:2 ratio:

\[ \text{mass of Al} = 5 \times 2 \times 27 = 270 \text{ tonnes} \]

The maximum mass of aluminium is 270 tonnes [1].

(c) Using \( 2Al_2O_3 \rightarrow 4Al + 3O_2 \), the mass ratio of oxide to oxygen is \( (2 \times 102) : (3 \times 32) = 204 : 96 \):

\[ \text{mass of } O_2 = 510 \times \frac{96}{204} = 240 \text{ tonnes} \]

giving 240 tonnes of oxygen [1] for the method, [1] for the answer.

(d) The bauxite is only 45% aluminium oxide, so:

\[ \text{mass of bauxite} = \frac{510}{0.45} = 1133 \text{ tonnes (to the nearest tonne)} \]

[1] for the working, [1] for the answer.

(e) Percentage yield compares the actual mass obtained (243 tonnes) with the maximum from part (b) (270 tonnes):

\[ \text{percentage yield} = \frac{243}{270} \times 100 = 90\% \]

[1] for the expression, [1] for 90%.

(f) At the electrodes:

\[ \text{cathode: } Al^{3+} + 3e^- \rightarrow Al \] \[ \text{anode: } 2O^{2-} \rightarrow O_2 + 4e^- \]

One mark each [2]. Metal ions are reduced (gain electrons) at the negative cathode; oxide ions are oxidised (lose electrons) at the positive anode.

(g) Any two reasons the real yield falls short, one mark each up to [2]: not all the oxide reacts / the reaction does not go to completion; some aluminium is lost during handling; impurities are present; some product is left in the cell.

(h) The extraction uses a large amount of energy because aluminium is above carbon in the reactivity series and must be extracted by electrolysis, which needs a large electrical input, and the cell must be kept molten at high temperature (any one) [1].