a) PTHH: $Fe+2HCl→FeCl_2+H_2↑$
b) $n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$
$n_{H_2}=n_{Fe}=0,1(mol)$
⇒ $V_{H_2}=0,1×22,4=2,24(l)$
c) Ta có: $m_{\text{dd sau p/ứ}}=5,6+100-(0,1×2)=105,4(g)$
Mặt khác ta lại có: $n_{FeCl_2}=n_{Fe}=0,1(mol)$
⇒ $C$%$_{FeCl_2}=\dfrac{0,1×127}{105,4}×100$% $=$ $12,05$%