Q22.5P - Chemistry: Molecular Nature Of Matter And Change

Question

What material is the source for commercial production of each of the following elements: 

(a) aluminium;

 (b) nitrogen;

 (c) chlorine; 

(d) calcium;

 (e) sodium?

Step-by-Step Solution

Answer

(a)  Bauxite for aluminium.

(b) Nitrogen-liquefaction of air and fractional distillation, or from $\left({\mathrm{NaN}}_3,{\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)$compounds.

(c)$\mathrm{NaCl}\left(\mathrm{aq}\right)$ electrolysis by chlorine.

(d) $\mathrm{CaO}$ for calcium.

(e) $\mathrm{NaCl}\left(\mathrm{l}\right)$ electrolysis for  sodium.

Step 1: Ore

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

Step 2: Subpart (a)

(a) Aluminium

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.

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

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.

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

Step 3: Subpart (b)

(b) Nitrogen

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:

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

Step 4: Subpart (c)

(c) Chlorine

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:

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

Water reduction on the cathode:

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

Overall:

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

Step 5: Subpart (d)

(d) Calcium

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

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

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

Step 6: Subpart (e)

(e) Sodium

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:

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

Overall:

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

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 $580°\mathrm{C}$).