Nếu số mol C${\mathrm{O}}_2$ bằng 0,005 mol

$$\begin{gathered} \mathrm{Vì} {\mathrm{n}}_{{\mathrm{CO}}_2}<{\mathrm{n}}_{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}=>\hspace{0.17em}\mathrm{chỉ} \mathrm{xảy} \mathrm{ra} \mathrm{phản} \mathrm{ứng} \mathrm{tạo} \mathrm{kết} \mathrm{tủa} \\ \mathrm{PTHH}:\hspace{0.17em}{\mathrm{CO}}_2+\hspace{0.17em}\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\to {\mathrm{CaCO}}_3+\hspace{0.17em}{\mathrm{H}}_2\mathrm{O} \\ 0,005 0,005 \mathrm{mol} \\ {\mathrm{m}}_{{\mathrm{CaCO}}_3}=0,005.100=0,5 \mathrm{g} \\ \mathrm{Nếu} \mathrm{số} \mathrm{mol} {\mathrm{CO}}_2 \mathrm{bằng} 0,024 \mathrm{mol} \\ \mathrm{Vì} {\mathrm{n}}_{{\mathrm{CO}}_2}>{\mathrm{n}}_{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}=>\mathrm{xảy} \mathrm{ra} \mathrm{phản} \mathrm{ứng} \mathrm{tạo} \mathrm{kết} \mathrm{tủa} \\ \mathrm{PTHH}:\hspace{0.17em}{\mathrm{CO}}_2+\hspace{0.17em}\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\to {\mathrm{CaCO}}_3+\hspace{0.17em}{\mathrm{H}}_2\mathrm{O} \\ 0,02 0,02 0,02 \mathrm{mol} \\ {\mathrm{CO}}_2+\hspace{0.17em}{\mathrm{CaCO}}_3+\hspace{0.17em}{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2^{} \\ 0,004 0,004 \mathrm{mol} \\ => {\mathrm{n}}_{{\mathrm{CaCO}}_3}=0,02-0,004=0,016 \mathrm{mol} \\ {\mathrm{m}}_{{\mathrm{CaCO}}_{3 }}=0,016.100=1,6 \mathrm{g} \\ \mathrm{Vậy} 0,5 \mathrm{g}<{\mathrm{m}}_{{\mathrm{CaCO}}_3}< 1,6 \mathrm{g} \end{gathered}$$