a/ PTHH: $Fe+2HCl\xrightarrow{t^o} FeCl_2+H_2$
b/ $n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{8,4}{56}=0,15(mol)$
Theo pt ta có tỉ lệ: 1:2:1:1
$→n_{H_2}=0,15(mol)$
$→V_{H_2}=n_{H_2}.22,4=3,36(l)$
c/ $n_{Fe}=0,15(mol)→n_{FeCl_2}=0,15(mol)$
$→m_{FeCl_2}=M_{FeCl_2}.n_{FeCl_2}=0,15.127=19,05(g)$