a)
$n_{HCl} = \dfrac{100.7,3\%}{36,5} = 0,2(mol)$
$F e+ 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$m_{Fe} = 0,1.56 = 5,6(gam)$
b) $n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
c) $m_{dd\ sau\ pư} = 5,6 + 100 - 0,1.2 = 105,4(gam)$
$C\%_{FeCl_2} = \dfrac{0,1.127}{105,4}.100\% = 12,05\%$