\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ m_{HCl}=\dfrac{109,5\cdot10\%}{100\%}=10,95\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \text{Vì }\dfrac{n_{HCl}}{2}< \dfrac{n_{Mg}}{1}\text{ nên sau p/ứ }Mg\text{ dư}\\ \Rightarrow n_{Mg}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow n_{Mg\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\\ \Rightarrow m_{Mg\left(dư\right)}=0,05\cdot24=1,2\left(g\right)\)
\(b,n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,15\cdot22,4=3,36\left(l\right)\\ c,n_{MgCl_2}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)