a/ PTHH: $Fe+2HCl\to FeCl_2+H_2$
b/ $n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{5,6}{56}=0,1(mol)$
Theo pt: $n_{H_2}=0,1(mol)$
$\to V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24(l)$
c/ Theo pt: $n_{HCl}=0,2(mol)$
$\to m_{HCl}=n_{HCl}.M_{HCl}=0,2.36,5=7,3(g)$
d/ Theo pt: $n_{FeCl_2}=0,1(mol)$
$\to m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,1.127=12,7(g)$