Question

Answer 1

\(a,PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow n_{FeSO_4}=n_{Fe}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,2\cdot152=30,4\left(g\right)\\m_{Fe}=0,2\cdot56=11,2\left(g\right)\end{matrix}\right.\)

Answer 2

a. \(PTHH:Fe+H_2SO_4--->FeSO_4+H_2\)

b. Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{FeSO_4}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)

\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)