Bauxite purification

Crude bauxite is essentially hydrated aluminium oxide, Al₂O₃·2H₂O, mined together with iron(III) oxide (Fe₂O₃) as the major impurity, plus smaller amounts of silica (sand) and titanium oxide. None of these impurities can simply be picked out or washed away — they are chemically bound into the same lump of ore. The purification therefore has to exploit a difference in chemical behaviour, not a difference in size, density or colour. That difference is amphoterism. Aluminium oxide reacts with both acids and strong alkalis, so it dissolves when the crushed ore is boiled with hot, concentrated sodium hydroxide:

where: Al₂O₃ is the wanted oxide, NaOH is the hot concentrated alkali used to attack it, NaAlO₂ is soluble sodium aluminate, and H₂O is released as the by-product. Iron(III) oxide is a simple basic oxide — it has no acidic character at all — so it does not react with NaOH and stays behind as an insoluble solid. Because one oxide has dissolved and the other has not, a single filtration now does all the separating: the Fe₂O₃ (with the silica and titanium oxide) is left on the filter paper as residue, while the aluminium leaves the process dissolved, as sodium aluminate, in the filtrate. The filtrate is then diluted or seeded so that pure aluminium hydroxide, Al(OH)₃, precipitates back out of solution — the reverse of the dissolving step, now purified of everything the residue carried away. Finally this hydroxide precipitate is filtered off and heated strongly (calcined), driving off water and leaving pure alumina behind:

where: Al(OH)₃ is the precipitated hydroxide, $\Delta$ denotes strong heating, Al₂O₃ is the pure alumina finally obtained, and H₂O escapes as steam. This last step is the same type of reaction as the calcination of bauxite ore itself — heat drives off a volatile substance, leaving the oxide behind — so it connects naturally to the general metallurgy idea of "roasting/calcining to the oxide" met earlier in the unit. The pure Al₂O₃ produced here is what is dissolved in molten cryolite and electrolysed in the next stage of aluminium extraction; nothing about that electrolysis step is covered here, only the purification that must happen first.

Question: A sample of 780 g of pure aluminium hydroxide precipitate, obtained after filtering the sodium aluminate solution, is heated strongly until decomposition is complete. Calculate the mass of pure aluminium oxide obtained.

Step 1. Write the balanced equation and its molar masses: $2\text{Al(OH)}_3 \xrightarrow{\Delta} \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O}$, with $156\ \text{g}$ Al(OH)₃ giving $102\ \text{g}$ Al₂O₃, as found above.

Step 2. Set up the proportion for the given mass: $780\ \text{g}$ of Al(OH)₃ is exactly $780/156 = 5$ times the equation's base amount of $156\ \text{g}$.

Step 3. Scale the alumina mass by the same factor: mass of Al₂O₃ $= 5 \times 102\ \text{g} = 510\ \text{g}$.

Step 4 (check). The mass lost as water should be $780 - 510 = 270\ \text{g}$, which should equal $5 \times 3 \times 18\ \text{g} = 270\ \text{g}$ (three moles of H₂O, molar mass 18, scaled by 5) — it matches, confirming the arithmetic.

Answer: 510 g of pure aluminium oxide is obtained from 780 g of aluminium hydroxide.

Competitive stretch

The same 510 g of Al₂O₃ would have needed how much NaOH to dissolve it in the first place? From $\text{Al}_2\text{O}_3 + 2\text{NaOH} \rightarrow 2\text{NaAlO}_2 + \text{H}_2\text{O}$, one mole of Al₂O₃ (102 g) reacts with two moles of NaOH ($2 \times 40 = 80$ g). Since $510/102 = 5$ moles of Al₂O₃, the NaOH required is $5 \times 80 = 400\ \text{g}$ — a reminder that the two equations in this process chain to a single mass budget from ore to oxide.

• Thinking crude bauxite can be electrolysed directly. Impurities such as Fe₂O₃ must be removed first; only pure alumina is fed into the electrolytic cell.
• Treating the separation as purely physical. Filtration is only the last mechanical step; the real separation happens earlier, when Al₂O₃ dissolves selectively in NaOH and Fe₂O₃ does not.
• Confusing aluminium hydroxide and aluminium oxide. The precipitate from the sodium aluminate solution is Al(OH)₃, which still needs calcination to become Al₂O₃.
• Using the wrong mole ratio in stoichiometry. The calcination equation gives 2 mol Al(OH)₃ to 1 mol Al₂O₃, not 1:1 — dropping the coefficient of 2 is the commonest arithmetic slip.

Ask the tutor: want to see exactly how this purified alumina is then electrolysed into aluminium metal — ask the AI tutor to walk through the electrolysis step next.