Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(PTHH:Fe+H_2SO_4--->FeSO_4+H_2\uparrow\)
Theo PT: \(n_{FeSO_4}=n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
\(V_{H_2}=0,2.22,4=4,48\left(lít\right)\)