Question

A white solid Y, on heating gives off a gas which turns lime water milky; the residue is yellow when hot; white when cold .The solid Y is probably:
(A) $ZnC{{O}_{3}}$
(B) $PbC{{O}_{3}}$
(C) $ZnS{{O}_{4}}$
(D) $Zn{{\left( N{{O}_{3}} \right)}_{2}}$

Answer 1

Hint: The metal carbonates decompose when heated in order to form the oxide of metal and carbon dioxide gas. The gas which turns the lime water milky is carbon dioxide. The unknown compound Y is also referred to as calamine or smithsonite.

Complete step by step answer:
- We are asked to find a solid which on heating gives off a gas which can turn the lime water milky. The residue of the solid after heating is yellow in color when it is hot and white when cold.
- The unknown solid Y is zinc carbonate which has the molecular formula $ZnC{{O}_{3}}$.When zinc carbonate is heated carbon dioxide gas and zinc oxide are formed. The reaction is represented below
\[ZnC{{O}_{3}}\xrightarrow{heat}ZnO+C{{O}_{2}}\]

-. Since a single compound is broken down to two components, this is a decomposition reaction. The gas which turns the lime water milky is carbon dioxide. It reacts with the calcium oxide (CaO) from lime water to form calcium carbonate and the reaction is shown below
\[CaO+C{{O}_{2}}\to CaC{{O}_{3}}\]

-The zinc oxide ($ZnO$) is the residue formed and it’s yellow in colour when it’s hot and white in colour when cold. The reason for the yellow colour of the zinc oxide is the defects in the solid structure which keep on increasing with temperature.
 From the above discussions it’s clear that the white solid Y is zinc carbonate.
So, the correct answer is “Option A”.

Note: It should be noted that the compound lead carbonate is also white in colour when it’s solid and yellow in colour after heating. But it does not become white in colour when cooled. Hence keep in mind that the only compound whose colour is white when it’s solid and yellow in colour after heating and white after cooling.