a) PTHH: $Fe+2HCl→FeCl_2+H_2$
b) Ta có: $\dfrac{m_{HCl}}{182,5}×100$% $=$ $5$%
⇒ $m_{HCl}=9,125(g)$
$n_{HCl}=\dfrac{9,125}{36,5}=0,25(mol)$
$n_{Fe}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}×0,25=0,125(mol)$
→ $m_{Fe}=0,125×56=7(g)$
$n_{H_2}=n_{Fe}=0,125(mol)$
→ $V_{H_2}=0,125×22,4=2,8(l)$
c) $m_{\text{dd sau p/ứ}}=7+182,5-(0,125×2)=189,25$
→ $C$%$_{FeCl_2}=\dfrac{0,125×127}{189,25}×100$% $=$ $8,4$%