Em tham khảo nha :
\(\begin{array}{l}
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{CaC{O_3}}} = \dfrac{{12}}{{100}} = 0,12mol\\
{n_{C{O_2}}} > {n_{CaC{O_3}}}\\
\Rightarrow\text{Phản ứng tạo ra 2 muối } CaC{O_3},Ca{(HC{O_3})_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O(1)\\
Ca{(OH)_2} + 2C{O_2} \to Ca{(HC{O_3})_2}(2)\\
{n_{C{O_2}(1)}} = {n_{CaC{O_3}}} = 0,12mol\\
{n_{C{O_2}(2)}} = 0,3 - 0,12 = 0,18mol\\
{n_{Ca{{(HC{O_3})}_2}}} = \dfrac{{{n_{C{O_2}(2)}}}}{2} = 0,09mol\\
{n_{Ca{{(OH)}_2}}} = {n_{CaC{O_3}}} + {n_{Ca{{(HC{O_3})}_2}}} = 0,12 + 0,09 = 0,21mol\\
{C_{{M_{Ca{{(OH)}_2}}}}} = \dfrac{{0,21}}{{0,1}} = 2,1M
\end{array}\)