Câu hỏi của ...........................

\(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)PTHH: Fe + H2SO4 ---> FeSO4 + H2 0,5---------------->0,5------->0,5\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,5.152=76\left(g\right)\\V_{H_2}=0,5.22,4=11,2\left(l\right)\end{matrix}\right.\)Câu trả lời: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)PTHH: Fe + H2SO4 ---> FeSO4 + H2 0,5---------------->0,5------->0,5\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,5.152=76\left(g\right)\\V_{H_2}=0,5.22,4=11,2\left(l\right)\end{matrix}\right.\)