\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2..............0,4.............0,2...............0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{MgCl_2}=95.0,2=19\left(g\right)\\ c,a=C_{MddHCl}=\dfrac{0,4}{0,2}=2\left(M\right)\)