\(a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{Mg}=0,15\left(mol\right)\\ \Rightarrow m=m_{Mg}=0,15\cdot24=3,6\left(g\right)\\ c,n_{HCl}=2n_{H_2}=0,3\left(mol\right)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,1}=3M\)