\(\begin{array}{l}
Dat\,halogen\,la\,X\\
Mg+X_2\xrightarrow{t^o}MgX_2\\
Theo\,PT:\,n_{X_2}=n_{Mg}=\frac{6,48}{24}=0,27(mol)\\
\to M_{X_2}=\frac{43,2}{0,27}=160(g/mol)\\
\to M_X=80(g/mol)\\
\to X:\,Brom\,(Br)\\
BTKL:\,m_{MgBr_2}=6,48+43,2=49,68(g)
\end{array}\)