a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b) \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)
Theo phương trình hóa học: \(n_{Fe}=n_{H_2}=0,15mol\rightarrow m_{Fe}=0,15.56=8,4g\)
c) Theo pt: \(n_{HCl}=2n_{Fe}=2.0,15=0,3mol,V_{HCl}=50ml=0,05l\)
\(C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,3}{0,05}=6M\)
