nCO2 = 6,72: 22,4 = 0,3 (mol)
Ta có: \(k = \frac{{{n_{C{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \frac{{0,3}}{{0,25}} = 1,2\)
=> 1< k < 2 => Tạo 2 muối
CO2 + Ca(OH)2 → CaCO3↓ + H2O
x x x (mol)
2CO2 + Ca(OH)2 → Ca(HCO3)2
y y y (mol)
\(\left\{ \begin{gathered}
{n_{Ca{{(OH)}_2}}} = x + y \hfill \\
{n_{C{O_2}}} = x + 2y \hfill \\
\end{gathered} \right. = > \left\{ \begin{gathered}
x = 0,2 \hfill \\
y = 0,05 \hfill \\
\end{gathered} \right.\)
=> mCaCO3 = 0,2.100 = 20 (gam).
Đáp án C