Cu ko td với HCl
\(n_{HCl}=2\cdot0,3=0,6\left(mol\right);n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \text{Vì }\dfrac{n_{Fe}}{1}< \dfrac{n_{HCl}}{2}\text{ nên HCl dư}\\ \Rightarrow n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,2\cdot22,4=4,48\left(l\right)\)