Question

A metal carbonate X on treatment with a mineral acid liberates a gas which when passed through an aqueous solution of a substance Y gives back X. The substance Y in reaction with the gas obtained at the anode during electrolysis of Brine gives a compound Z which can decolourise coloured fabrics. The compounds X, Y and Z respectively are:
(A) $CaC{{O}_{3}}\text{,}Ca{{\left( OH \right)}_{2}}\text{,}CaOC{{l}_{2}}$
(B) $Ca{{\left( OH \right)}_{2}}$,$\text{CaO}$,$\text{CaOCl}$
(C) $CaC{{O}_{3}}$,$CaOC{{l}_{2}}\text{,}$$Ca{{\left( OH \right)}_{2}}$
(D) $Ca{{\left( OH \right)}_{2}}$,$CaC{{O}_{3}}$,$CaOC{{l}_{2}}$

Answer 1

Hint: Calcium is an alkaline earth metal, which is the most abundant metallic element in the human body. We can see that calcium doesn’t occur naturally in the free state, but its compounds are widely distributed.

Complete step by step solution:
- Let’s solve the answer step by step:
\[CaC{{O}_{3}}\to C{{O}_{2}}+CaO\]
\[CaO+{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}\]
- A metal carbonate X that is $CaC{{O}_{3}}$ on treatment with a mineral acid liberates a gas that is $C{{O}_{2}}$ which when passed through an aqueous solution of a substance Y that is $Ca{{\left( OH \right)}_{2}}$ gives back X that is$CaC{{O}_{3}}$.
$CaC{{O}_{3}}$is called calcium carbonate, CaO is called calcium oxide.
\[Ca{{\left( OH \right)}_{2}}+C{{l}_{2}}\to CaOC{{l}_{2}}\]
- The substance Y that is $Ca{{\left( OH \right)}_{2}}$ on reaction with the gas Chlorine $C{{l}_{2}}$ obtained at the anode during electrolysis of Brine gives a compound Z that is $CaOC{{l}_{2}}$ which can decolourise coloured fabrics.
$Ca{{\left( OH \right)}_{2}}$is called calcium hydroxide or slaked lime.

Hence, we can see that the correct option is (A) that is the compounds X, Y and Z respectively are $CaC{{O}_{3}}\text{,}Ca{{\left( OH \right)}_{2}}\text{,}CaOC{{l}_{2}}$.

Note: We should not get confused in between quick lime and slaked lime. The quick lime is having molecular formula CaO and slaked lime is having molecular formula $Ca{{\left( OH \right)}_{2}}$.