nHCl= 0.05*2=0.1 mol
Fe + 2HCl --> FeCl2 + H2
_____0.1_____0.05____0.05
mFeCl2= 0.05*127=6.35g
VH2= 0.05*22.4=1.12l
VHCl = 50 ml = 0.05 l
nHCl = CM . V = 2 . 0.05 = 0.1 mol
a). Fe + 2HCl ➝ FeCl2 + H2
0.2 ←0.1 ➝ 0.05 : 0.05 mol
b). mFeCl2 = n . M = 0.05 . 127 = 6.35 mol
c). VH2 = n . 22.4 = 0.05 . 22.4 = 1.12 l