a) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
b) Ta có: \(n_{Fe}=\frac{0,56}{56}=0,01\left(mol\right)=n_{FeSO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,01\cdot152=1,52\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{H_2SO_4}=n_{Fe}=0,01mol\)
\(\Rightarrow m_{H_2SO_4}=0,01\cdot98=0,98\left(g\right)\) \(\Rightarrow m_{ddH_2SO_4}=\frac{0,98}{19,6\%}=5\left(g\right)\)