\(n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Fe}=0,6(mol)\\ \Rightarrow m_{HCl}=0,6.36,5=21,9(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{21,9}{7,3\%}=300(g)\\ b,n_{H_2}=n_{Fe}=0,3(mol)\\ \Rightarrow V_{H_2(đktc)}=0,3.22,4=6,72(l)\)