`Fe + H_2 SO_4 -> FeSO_4 + H_2`
`0,3` `0,3` `(mol)`
`n_[Fe]=[16,8]/56=0,3(mol)`
`=>V_[H_2]=0,3.22,4=6,72(l)`
`->B`
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(V_{H_2}=22,4\times0,3=6,72\left(l\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{H_2}=n_{Fe}=0,3\left(mol\right)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\\ \Rightarrow ChọnB\)