a) PTHH: $Fe+2HCl→FeCl_2+H_2$
b) $n_{HCl}=\dfrac{100}{1000}×2=0,2(mol)$
$n_{Fe}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}×0,2=0,1(mol)$
→ $m_{Fe}=0,1×56=5,6(g)$
$n_{H_2}=n_{Fe}=n_{FeCl_2}=0,1(mol)$
→ $V_{H_2}=0,1×22,4=2,24(l)$
c) $m_{ddHCl}=1,1×100=110(g)$
$m_{\text{dd sau p/ứ}}=5,6+110-(0,1×2)=115,4(g)$
$C$%$_{ddFeCl_2}=\dfrac{0,1×127}{115,4}×100$% $=$ $11,01$%