a) PTHH: Fe + 2HCl → $FeCl_{2}$ + $H_{2}$ ↑
1 mol 2 mol 1 mol 1 mol
0,15 mol 0,3 mol 0,15 mol 0,15 mol
b) n$H_{2}$ =3,36 : 22,4= 0,15 (mol)
$n_{Fe}$ = $\frac{0,15.1}{1}$ = 0,15 (mol)
$m_{Fe}$ =0,15.56=8,4 (g)
c) $n_{HCl}$ =$\frac{0,15.2}{1}$ =0,3 (mol)
$m_{HCl}$ =0,3.36,5=10,95 (g)
$m_{ddHCl}$ =$\frac{10,95}{73}$.100=150 (g)
d) n$FeCl_{2}$= $\frac{0,15.1}{1}$=0,15 (mol)
m$FeCl_{2}$=0,15.127=19,05 (g)
$m_{dd}$ =8,4+150-0,15.2=158,1 (g)
C%$FeCl_{2}$=$\frac{19.05}{158,1}$.100%=12,05%
