Question

If one of the industrial processes used for the manufacture of sodium hydroxide, a gas \[{\rm{X}}\] is formed by products. The gas \[{\rm{X}}\] reacts with lime water to give a component \[{\rm{Y}}\] which is used as a bleaching agent in a chemical industry. Identify \[{\rm{X}}\] and \[{\rm{Y}}\] giving the chemical equation of the reaction is involved.

Answer 1

Hint: As we know that, the if any salt is dissolved in water it dissociates into ions. Strong acid and strong base give a salt which completely dissociates into ions. That ions again can be hydrolysed.

Complete step by step solution
The industrial process to manufacture sodium hydroxide is known as chloral-alkali process. In this process, we use sodium chloride initially. The sodium chloride is a salt which dissociates into sodium ion and a chloride ion when voltage is applied as.
\[{\rm{NaCl + }}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{N}}{{\rm{a}}^ + }{\rm{(aq)}} + {\rm{C}}{{\rm{l}}^ - }{\rm{(aq)}}\]
Now, this sodium ion reacts with water to give sodium hydroxide \[{\rm{NaOH}}\] and hydrogen gas \[{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]. But as we have read in our given statement that the gas when reacts with lime water it gives bleaching agent (\[{\rm{CaOC}}{{\rm{l}}_2}\]), so this hydrogen gas is not the gas in our answer as.
\[{\rm{N}}{{\rm{a}}^{\rm{ + }}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{NaOH}} + {{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]
Now, we are left with chloride ion, it takes electron from sodium atom and converted into chlorine gas as-
\[{\rm{2C}}{{\rm{l}}^ - }{\rm{ + 2}}{{\rm{e}}^ - } \to {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\]
So, if we write the whole equation as
\[{\rm{NaCl}} + {\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{NaOH}} + {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(X)(g)}} + {{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]
Now, we proceed with the reaction of chlorine gas with lime water (\[{\rm{Ca(OH}}{{\rm{)}}_2}\]) to give bleaching agent as-
\[{\rm{C}}{{\rm{l}}_2}{\rm{(g)}} + {\rm{Ca(OH}}{{\rm{)}}_2} \to {\rm{CaOC}}{{\rm{l}}_{\rm{2}}}{\rm{(Y)}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}\]
Hence, the compound \[{\rm{X}}\] and \[{\rm{Y}}\] are chlorine gas (\[{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\]) and bleaching powder (\[{\rm{CaOC}}{{\rm{l}}_2}\]) respectively.

Note:
The sodium chloride salt is electrolysed by anode-cathode reaction. At anode, chlorine gas is formed by reduction and at cathode, hydrogen gas is released.