a) Fe + 2HCl → FeCl2 + H2
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
b) Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15\times56=8,4\left(g\right)\)
c) Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)