CaCO_3 xrightarrow[excess H_2O]{in CO_2} A | vedclass

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(D) When $CaCO_3$ is treated with $CO_2$ and excess $H_2O$, it forms calcium bicarbonate, which is soluble in water: $CaCO_3(s) + CO_2(g) + H_2O(l) \r

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$Ca(HCO_3)_2, CaCO_3$

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(D) When $CaCO_3$ is treated with $CO_2$ and excess $H_2O$, it forms calcium bicarbonate, which is soluble in water: $CaCO_3(s) + CO_2(g) + H_2O(l) \rightarrow Ca(HCO_3)_2(aq)$. Thus, $A = Ca(HCO_3)_2$. When $Ca(HCO_3)_2$ is heated, it decomposes back into calcium carbonate, water, and carbon dioxide: $Ca(HCO_3)_2(aq) \xrightarrow{\Delta} CaCO_3(s) + H_2O(l) + CO_2(g)$. Thus, $B = CaCO_3$. Therefore, the correct option is $D$.

$CaCO_3$ $\xrightarrow[excess \ H_2O]{in \ CO_2} A$ $\xrightarrow{\Delta} B$
Identify $A$ and $B$.

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$CaCO_3$ $\xrightarrow[excess \ H_2O]{in \ CO_2} A$ $\xrightarrow{\Delta} B$ Identify $A$ and $B$.