a) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15\cdot56=8,4\left(g\right)\)
c) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(V_{H_2SO_4}=50ml=0,05\left(l\right)\)
\(\Rightarrow C_{M,H_2SO_4}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_4}}=\dfrac{0,15}{0,05}=3M\)