The reaction involved in converting calcium carbonate to calcium ions using carbon dioxide is:
\[
\text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Ca}^{2+} + 2\text{HCO}_3^{-}
\]
From the reaction, 1 mole of CaCO\(_3\) requires exactly 1 mole of CO\(_2\).
Given:
1 mole/L of CaCO\(_3\) in 1 L solution \(\Rightarrow\) 1 mole of CaCO\(_3\).
Thus, required CO\(_2\) = 1 mole.
Molecular weight of CO\(_2\) = 44 g/mole.
Hence, amount of CO\(_2\) needed = \(1 \times 44 = 44\) grams.
Final Answer: 44 grams