\(n_{CO_2}=0,12\left(mol\right);n_{Ca\left(OH\right)_2}=0,1\left(mol\right)\Rightarrow n_{OH^-}=0,2\left(mol\right)\\ Tacó:\dfrac{n_{OH^-}}{n_{CO_2}}=1,67\\ \Rightarrow Tạo2muốiCaCO_3vàCa\left(HCO_3\right)_2\\ Đặt:\left\{{}\begin{matrix}n_{CaCO_3}=x\left(mol\right)\\n_{Ca\left(HCO_3\right)_2}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+2y=0,12\left(BTNT\left(C\right)\right)\\x+y=0,1\left(BTNT\left(Ca\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,08\\y=0,02\end{matrix}\right.\\ \Rightarrow m_{CaCO_3}=0,08.100=8\left(g\right)\)