\(\begin{array}{l}
a)\\
Hh:\,Ba:x(mol);Mg:y(mol)\\
\to 137x+24y=18,5(1)\\
Bte:\,x+y=n_{H_2}=\frac{6,72}{22,4}=0,3(2)\\
(1)(2)\to x=0,1;y=0,2\\
m_{Ba}=0,1.137=13,7(g)\\
m_{Mg}=0,2.24=4,8(g)\\
b)\\
BTNT\,H:\,n_{H_2SO_4}=n_{H_2}=0,3(mol)\\
x=\frac{0,3.98}{500}.100\%=5,88\%\\
c)\\
BTNT\,Mg:\,n_{MgSO_4}=y=0,2(mol)\\
BTNT\,Ba:\,n_{BaSO_4}=x=0,1(mol)\\
\to C\%_{MgSO_4}=\frac{0,2.120}{18,5+500-0,1.233-0,3.2}.100\%\approx 4,85\%
\end{array}\)