PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
Ta có: \(n_{CO_2}=\frac{1,81}{22,4}=\frac{181}{2240}\left(mol\right)\) \(\Rightarrow n_{CaCO_3}=\frac{181}{2240}\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=\frac{181}{2240}\cdot100=\frac{905}{112}\left(g\right)\) \(\Rightarrow\%m_{CaCO_3}=\frac{\frac{905}{112}}{10}\cdot100\approx80,8\%\)