91Ó°ÊÓ

An ore is the compound of the metal from which it can be extracted commercially.

Aluminium is a common element in the Earth's crust, but it's also found in a variety of minerals and ores. Bauxite is the mineral utilised in the commercial manufacturing of aluminium. Because ${\mathrm{Al}}_2{\mathrm{O}}_3$is soluble in the combination, but ${\mathrm{Fe}}_2{\mathrm{O}}_3$and ${\mathrm{TiO}}_2$are not, the ore is treated with hot $\mathrm{NaOH}$after mining to remove any potential iron or titanium oxide residue. The Bаyer process is the name for this. Silicium oxide,${\mathrm{SiO}}_2$, is also soluble in the combination, in addition to aluminium.

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3+2\mathrm{NaOH}+3{\mathrm{H}}_2\mathrm{O}→{2\mathrm{Al}\left(\mathrm{OH}\right)}_4^{-}+{\mathrm{Na}}^{+} \\ {\mathrm{SiO}}_2+2\mathrm{NaOH}+2{\mathrm{H}}_2\mathrm{O}→{\mathrm{Na}}_2\mathrm{Si}(\mathrm{OH}{)}_6 \end{gathered}$$

As, ${\mathrm{Na}}_2\mathrm{Si}(\mathrm{OH}{)}_6$precipitates as aluminosilicate when heated further, and can be filtered out of the mixture along with ${\mathrm{TiO}}_2$and ${\mathrm{Fe}}_2{\mathrm{O}}_3$.

Aluminium precipitates as ${\mathrm{Al}}_2{\mathrm{O}}_3$when the filtrate is acidified. To lower the melting temperature of the mixture from $2073°\mathrm{C}$and $950°\mathrm{C}$the oxide is melted in cryolite with ammonium fluoride added. The Hall-Heroult technique is then used to electrolyze the mixture in a graphite-lined furnace.

$2{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{sol}\right)+3\mathrm{C}\left(\mathrm{s}\right)→4\mathrm{Al}\left(\mathrm{s}\right)+3\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)$

Nitrogen, often known as${\mathrm{N}}_2$makes up around$70\%$

of the air we breathe. The most frequent method for obtaining nitrogen is air liquefaction followed by fraction distillation.

Nitrogen can also be derived from a variety of substances, such as:

$$\begin{gathered} 2{\mathrm{NaN}}_3\left(\mathrm{s}\right)→2\mathrm{Na}\left(\mathrm{s}\right)+3{\mathrm{N}}_2(\mathrm{g}) \\ {\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)→{\mathrm{N}}_2\left(\mathrm{g}\right)+2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_3+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \end{gathered}$$

Chlorine is created and used in greater quantities than any other halogen. It can be found in halite or $\mathrm{NaCl}$, which is generally known as culinary salt and is made from sea water. ${\mathrm{Cl}}_2$can be made by electrolysis from molten or from its aqueous solution, making it significantly easier and less expensive to create. On the anode, the half-reaction of chloride ion oxidation is:

$2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)→{\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}$

Water reduction on the cathode:

$2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}→{\mathrm{H}}_2\left(\mathrm{g}\right)+\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{aq}\right)$

Overall:

${\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)→{\mathrm{Cl}}_2\left(\mathrm{g}\right)+\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{aq}\right)$

Calcium is made by reducing the oxide, $\mathrm{CaO}$, with aluminium as a reducing agent.

$6\mathrm{CaO}+4\mathrm{Al}→2{\mathrm{Al}}_2{\mathrm{O}}_3+6\mathrm{Ca}$

At high temperatures $\left(1200°\mathrm{C}\right)$and low pressure, the reaction is possible.

Sodium can be recovered from halite in the same way as chlorine can, but the method is a little more complicated and expensive. When an aqueous solution of $\mathrm{NaCl}$is electrolyzed, the half-cell potential of chloride ions $(\mathrm{E}=2.71\mathrm{V})$causes water reduction $(\mathrm{E}=-0.42\mathrm{V})$on the cathode. As a result, pure melted $\mathrm{NaCl}$must be electrolyzed to obtain sodium on the cathode. After that, the reaction is as follows:

$$\begin{gathered} \mathrm{R}:2{\mathrm{Na}}^{+}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}→2\mathrm{Na}\left(\mathrm{s}\right) \\ \mathrm{O}:2{\mathrm{Cl}}^{-}\left(\mathrm{l}\right)→{\mathrm{Cl}}_2\left(\mathrm{g}\right)+2\mathrm{e}- \end{gathered}$$

Overall:

$2\mathrm{NaCl}\left(\mathrm{l}\right)→2\mathrm{Na}\left(\mathrm{s}\right)+\mathrm{C}{\mathrm{l}}_2\left(\mathrm{g}\right)$

Since pure$\mathrm{NaCl}$has a melting temperature of$801°\mathrm{C}$, it's typical to add ${\mathrm{CaCl}}_2$to reduce heating costs (the mixture's melting point is$580°\mathrm{C}$).