\(\begin{array}{l}
1)\\
Dat\,hoa\,tri\,X\,la\,n\\
n_{H_2}=\frac{4,8-4,4}{2}=0,2(mol)\\
2X+2nHCl\to 2XCl_n+nH_2\\
Theo\,PT:\,n_X.n=2n_{H_2}=0,4(mol)\\
\to \frac{4,8n}{M_X}=0,4\to M_X=12n\\
\to n=2;M_X=24\to X:\,magie\,(Mg)\\
Mg+2HCl\to MgCl_2+H_2\\
MgCl_2+2NaOH\to Mg(OH)_2+2NaCl\\
Theo\,PT:\,n_{Mg(OH)_2}=n_{H_2}=0,2(mol)\\
\to m=m_{Mg(OH)_2}=0,2.58=11,6(g)
2)\\
a)\\
Dat\,n_{CaCO_3}=x(mol);n_{MgCO_3}=y(mol)\\
\to 100x+84y=28,4(1)\\
CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\
MgCO_3+2HCl\to MgCl_2+CO_2+H_2O\\
Theo\,PT:\,x+y=n_{CO_2}=\frac{6,72}{22,4}=0,3(2)\\
(1)(2)\to x=0,2;y=0,1\\
\to m_{CaCO_3}=0,2.100=20(g);m_{MgCO_3}=0,1.84=8,4(g)\\
b)\\
n_{HCl}=2n_{CO_2}=0,6(mol)\\
\to C\%_{HCl}=\frac{0,6.36,5}{500}.100\%=4,38\%
\end{array}\)