a)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
b)
Đổi \(200ml=0,2\left(l\right)\)
\(n_{HCl}=1\cdot0,2=0,2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo pt 1mol ..... 2mol....1mol....1mol
Theo đề 0,1mol....0,2mol....0,1mol....0,1mol
\(m_{FeCl_2}=0,1\cdot\left(56+35,5\cdot2\right)=12,7\left(g\right)\)
c) \(V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)