\(Đặt.2.muối:ACO_3,B_2CO_3\\ n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ PTHH:ACO_3+2HCl\rightarrow ACl_2+CO_2+H_2O\\ B_2CO_3+2HCl\rightarrow2BCl+CO_2+H_2O\\ n_{CO^{2-}_3}=n_{muối.cacbonat}=n_{CO_2}=0,3\left(mol\right)\\ n_{Cl^-}=2.0,3=0,6\left(mol\right)\\ m_{muối.khan}=m_{muối.cacbonat}+\left(m_{Cl^-}-m_{CO^{2-}_3}\right)=10+\left(35,5.0,6-60.0,3\right)=13,3\left(g\right)\)