The Sblock Elements Ques 41
41. 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) $K_2 CO_3$
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Answer:
Correct Answer: 41.(b)
Solution:
(b) $ \underset{(A)}{CaCO_3(s)} \xrightarrow{\Delta} \underset{\text{colourless}}{Co_2(g)^\uparrow} + \underset{\text{residue}}{CaO(s)} $
$ CaO(s) + H_2 O \longrightarrow \underset{(B)}{Ca(OH)_2} $
$ Ca(OH)_2 + 2CO_2 + H_2 O \longrightarrow \underset{(C)}{Ca(HCO_3)_2} $
$\underset{(C)}{Ca(HCO_3) _{2(s)}} \xrightarrow{\Delta} \underset{(A)}{CaCO _{3(s)}}+CO _{2(g)}+H_2 O$
BETA
