\(n_{Mg}=\dfrac{8,4}{24}=0,35\left(mol\right)\)
Pt: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,35mol\(\rightarrow\)0,7mol\(\rightarrow\)0,35mol\(\rightarrow\)0,35mol
\(C\%_{HCl}=\dfrac{0,7.36,5}{146}.100=17,5\%\)
\(m_{dd\left(spu\right)}=m_{Mg}+m_{HCl}-m_{H_2}\)
= 8,4 + 146 - 0,35.2 = 153,7 (g)
\(C\%_{MgCl_2}=\dfrac{0,35.95}{153,7}.100=21,63\%\)