\(\begin{array}{l}
Dat\,n_{CaCO_3}=x(mol);n_{MgCO_3}=y(mol)\\
\to 100x+84y=28,4(1)\\
CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\
MgCO_3+2HCl\to MgCl_2+CO_2+H_2O\\
Theo\,PT:\,x+y=n_{CO_2}=\frac{6,72}{22,4}=0,3(2)\\
(1)(2)\to x=0,2;y=0,1\\
\to\begin{cases}m_{CaCO_3}=0,2.100=20(g)\\m_{MgCO_3}=0,1.84=8,4(g)\end{cases}
\end{array}\)