$\text{a) PTHH: $Fe + 2HCl → FeCl_{2} + H_{2}↑$}$
$\text{b) -Theo bài ra, ta có: }$
$\text{$n_{Fe} = \frac{11,2}{56}$ = 0,2 (mol) }$
$\text{-Theo PTHH, ta có: }$
$\text{$n_{H_{2}} = n_{Fe}$ = 0,2 mol }$
$\text{⇒ $V_{H_{2}đktc}$ = 0,2.22,4 = 4,48 (l) }$