PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)=n_{H_2}=n_{FeSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,5\cdot22,4=11,2\left(l\right)\\m_{FeSO_4}=0,5\cdot152=76\left(g\right)\end{matrix}\right.\)