\(n_{Fe}=0,1\left(mol\right)\) \(n_{HCl}=0,1\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
bđ:_____0,1___0,1_____________
pứ:_____0,05__0,1____0,05____0,05_
spứ:____0,05___0_____0,05____0,05_
Chất dư là Fe: \(n_{Fe\left(du\right)}=0,05\left(mol\right)\rightarrow m_{Fe\left(du\right)}=2,8\left(g\right)\)