Ta có: \(n_{CO_2}=0,3\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,15\left(mol\right)\)
\(n_{NaOH}=0,1\left(mol\right)\)
\(\Rightarrow n_{OH^-}=2n_{Ca\left(OH\right)_2}+n_{NaOH}=0,4\left(mol\right)\)
\(\dfrac{n_{OH^-}}{n_{CO_2}}=1,3\)=> Thu được 2 muối sau phản ứng
\(n_{CaCO_3}=n_{OH^-}-n_{CO_2}=0,4-0,3=0,1\left(mol\right)\)
\(\Rightarrow m=0,1.100=10\left(g\right)\)