Explain the leaching of:
$(i)$ Alumina from Bauxite (Bayer's Process)
$(ii)$ Gold and Silver (Cyanide Process)

(N/A) Principle: It works on the difference in the solubilities of ore and impurities in a suitable solvent.
$(i)$ Leaching of Alumina from Bauxite (Bayer's process): Bauxite is the principal ore of aluminium. It contains impurities of ferric oxide $(Fe_{2}O_{3})$,silica $(SiO_{2})$,and titanium oxide $(TiO_{2})$. The powdered ore is digested with concentrated $NaOH$ solution at $473-523 \ K$ temperature and $35-36 \ bar$ pressure. The alumina dissolves as sodium aluminate,while $Fe_{2}O_{3}$ and $TiO_{2}$ remain as solid residue. $SiO_{2}$ dissolves as sodium silicate.
$Al_{2}O_{3(s)} + 2NaOH_{(aq)} + 3H_{2}O_{(l)} \xrightarrow[473-523 \ K]{35 \ bar} 2Na[Al(OH)_{4}]_{(aq)}$
$SiO_{2(s)} + 2NaOH_{(aq)} \rightarrow Na_{2}SiO_{3(aq)} + H_{2}O_{(l)}$
The sodium aluminate solution is neutralized by passing $CO_{2}$ gas,causing hydrated $Al_{2}O_{3}$ to precipitate. $A$ small amount of freshly prepared hydrated $Al_{2}O_{3}$ is added to induce precipitation,a process called seeding.
$2Na[Al(OH)_{4}]_{(aq)} + 2CO_{2(g)} \rightarrow Al_{2}O_{3} \cdot xH_{2}O_{(s)} + 2NaHCO_{3(aq)}$
Sodium silicate remains in the solution,and the hydrated alumina is filtered and heated to obtain pure $Al_{2}O_{3}$.
$Al_{2}O_{3} \cdot xH_{2}O_{(s)} \xrightarrow{1470 \ K} Al_{2}O_{3(s)} + xH_{2}O_{(g)}$
$(ii)$ Cyanide Process (Leaching of Gold and Silver): In the metallurgy of gold and silver,the metal is leached with a dilute solution of $KCN$ or $NaCN$ in the presence of air (which supplies $O_{2}$). The metal is later recovered by a displacement reaction.
$4M_{(s)} + 8CN^{-}_{(aq)} + 2H_{2}O_{(l)} + O_{2(g)} \rightarrow 4[M(CN)_{2}]^{-}_{(aq)} + 4OH^{-}_{(aq)}$
$(M = Ag \text{ or } Au)$