The sBlock Elements - Result Question 41
####47. The compound $A$ on heating gives a colourless gas and a residue that is dissolved in water to obtain $B$. Excess of $CO_2$ is bubbled through aqueous solution of $B, C$ is formed which is recovered in the solid form. Solid $C$ on gentle heating gives back $A$. The compound is [2010]
(a) $CaSO_4 \cdot 2 H_2 O$
(b) $CaCO_3$
(c) $Na_2 CO_3$
(d) $K_2 CO_3$
Show Answer
Solution:
- (b) $CaCO_3$ (s) $\xrightarrow{\Delta} CO_2(g) \uparrow+CaO$ (s)
(A) colourless residue
$CaO(s)+H_2 O \longrightarrow Ca(OH)_2$
$Ca(OH)_2+2 CO_2+H_2 O \longrightarrow Ca(HCO_3)_2$
$\underset{(C)}{Ca(HCO_3) _{2(s)}} \xrightarrow[(A)]{\Delta} CaCO _{3(s)}+CO _{2(g)}+H_2 O$
48
(b)
The given properties coincide with $CaCO_3$
