Question

How is sodium hydroxide manufactured in industries? Name the process. In this process a gas ${\rm{X}}$ is formed as a by-product. This gas reacts with lime water to give a compound ${\rm{Y}}$, which is used as a bleaching agent in the chemical industry. Identify ${\rm{X}}$ and ${\rm{Y}}$ and write the chemical equation of the reactions involved.

Answer 1

We have different industrial manufacturing processes for different compounds including ${\rm{NaOH}}$. The process becomes more useful if the by-products can also be of further use.

Complete step by step solution
We can define sodium hydroxide as a strong base being a hydroxide of an alkali metal. The chemical formula is ${\rm{NaOH}}$. It has varied applications including being used in the manufacturing of soap, detergents, paper and many other different chemicals; it is also used in petroleum refining, in laboratories, or in the purification of aluminum ore, bauxite and many more. This has led to manufacturing of ${\rm{NaOH}}$ on industrial scale.

The industrial manufacturing process for ${\rm{NaOH}}$ involves electrolysis of brine which is basically common salt dissolved in water. The process is known as chlor-alkali process. We will get a better understanding about the suitability of this name after going through the process briefly.

Brine is basically $NaCl\left( {aq} \right)$ and its electrolysis leads to its decomposition for which the chemical reaction can be written as follows:
$2NaCl\left( {aq} \right) + 2{H_2}O\left( l \right) \to 2NaOH\left( {aq} \right) + C{l_2}\left( g \right) + {H_2}\left( g \right)$

As it is evident that during manufacturing of ${\rm{NaOH}}$ that it is an alkali, $C{l_2}$ gas is also produced as a by-product giving the name chlor-alkali to the process. So, we can infer that $X$ is $C{l_2}$.

Now, as it is given that $C{l_2}$ reacts with lime water for which we have chemical formula $Ca{\left( {OH} \right)_2}$ and the reaction between the two can be shown by the following chemical equation:
$C{l_2}\left( g \right) + Ca{\left( {OH} \right)_2}\left( {aq} \right) \to CaOC{l_2}\left( {aq} \right) + {H_2}O\left( l \right)$

We know that $CaOC{l_2}$ is called bleaching powder for its bleaching properties. So, we can say that $Y$ is $CaOC{l_2}$.

Note:
Here, we have one more by-product, ${H_2}\left( g \right)$ but we have to frame our answer by considering the further related reactions as well which are given by $C{l_2}$ not ${H_2}$.