\(n_{Fe}=\frac{2,8}{56}=0,05\left(mol\right)\)
\(2n_{Fe}=2n_{H2}\Rightarrow n_{H2}=0,05\left(mol\right)\)
\(V_{H2}=0,05.22,4=1,12\left(l\right)\)
Đổi 600ml = 0,6l
\(n_{HCl}=2n_{H2}=0,1\left(mol\right)\)
\(\Rightarrow CM_{HCl}=\frac{0,1}{0,6}=\frac{1}{6}M\)