Looking for classes? Ksquare Career Institute, Bengaluru →
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
A
$CaCO_3$
B
$Na_2CO_3$
C
$K_2CO_3$
D
$CaSO_4\cdot2H_2O$
Detailed Solution
$Na_2CO_3$ and $K_2CO_3$ are stable to heat and do not decompose, and gypsum only loses water, so A must be $CaCO_3$.
Heating A: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2\uparrow$ (colourless gas, residue CaO)
Residue in water: $CaO + H_2O \rightarrow Ca(OH)_2$, so B is $Ca(OH)_2$
Excess $CO_2$ through B: $Ca(OH)_2 + 2CO_2 \rightarrow Ca(HCO_3)_2$, so C is $Ca(HCO_3)_2$
Gentle heating of C: $Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 + CO_2 + H_2O$, which gives back A
Hence the compound A is $CaCO_3$.
Heating A: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2\uparrow$ (colourless gas, residue CaO)
Residue in water: $CaO + H_2O \rightarrow Ca(OH)_2$, so B is $Ca(OH)_2$
Excess $CO_2$ through B: $Ca(OH)_2 + 2CO_2 \rightarrow Ca(HCO_3)_2$, so C is $Ca(HCO_3)_2$
Gentle heating of C: $Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 + CO_2 + H_2O$, which gives back A
Hence the compound A is $CaCO_3$.