Ta có: $n_{Fe}$ = $\frac{2,8}{56}$ =0,05(mol)
PTHH: Fe + 2HCl → Fe$Cl_{2}$ + $H_{2}$ O
$n_{pt}$: 1 2 1 1
$n_{đb}$: 0,05 →0,1 → 0,05 → 0,05
⇒$m_{HCl}$ =0,1.36,5=3,65(g)
$m_{ddHCl}$ = 50.1,18=59(g)
⇒C%$dd_{HCl}$ =$\frac{3,65}{59}$ .100%≈6,18%
Ta có: m$FeCl_{2}$ =0,05.127=6,35(g)
m$ddFeCl_{2}$ =2,8+59-(0,05.18)=60,9(g)
⇒C%$ddFeCl_{2}$ =$\frac{6,35}{60,9}$ .100%≈10,4269%
Chúc bạn học tốt!