a)
n_{HCl}={200.10,95\%}/{36,5}=0,6(mol)
Fe+2HCl->FeCl_2+H_2
Theo PT: n_{Fe}=n_{H_2}=0,6(mol)
->m_{\text{đinh giảm}}=m_{Fe}-m_{H_2}=0,6.56-0,6.2=32,4(g)
b)
Theo PT: n_{FeCl_2}=n_{Fe}=0,6(mol)
m_{dd\ spu}=200+32,4=232,4(g)
->C\%_{FeCl_2}={0,6.127}/{232,4}.100\%\approx 32,79\%