Question

A solid compound ${}^{\prime }{X}^{\prime }$ on heating gives $C{O}_2$ gas and a residue. The residue mixed with water form ${}^{\prime }{Y}^{\prime }$. On passing $C{O}_2$ it turn milky and in excess of $C{O}_2$ through ${}^{\prime }{Y}^{\prime }$ in the water, a clear solution ${}^{\prime }{Z}^{\prime }$ is obtained. On boiling ${}^{\prime }{Z}^{\prime }$ compound ${}^{\prime }{X}^{\prime }$ is reformed. The compound ${}^{\prime }{X}^{\prime }$ is :

• A. $K_{2}C{O}_3$
• B. $Ca(OH{)}_2$
• C. $CaC{O}_3$
• D. None of the above

Answer 1

The correct option is C $CaC{O}_3$

The given compound ${}^{\prime }{X}^{\prime }$ is $CaC{O}_3$
$\begin{array}{ccccccccccc}CaC{O}_3& \stackrel{\Delta }{\to }& CaO& +& C{O}_2\uparrow & ;& CaO& +& H_{2}O& \to & Ca(OH{)}_2\\ \left(X\right)& & \text{(residue)}& & & & & & & & \left(Y\right)\\ Ca(OH{)}_2& +& C{O}_2& +& H_{2}O& \to & Ca(HC{O}_3{)}_2& & & & \\ & & & & & & Z& & & & \\ Ca(HC{O}_3{)}_2& \stackrel{\Delta }{\to }& CaC{O}_3& +& C{O}_2\uparrow & +& H_{2}O& & & & \\ Z& & & & & & & & & & \end{array}$