n Fe =0,1(mol)
n S =0,025(mol)
PTHH: Fe + S -to-> FeS
(mol)__0,025_0,025__0,025
=> Fe dư 0,1 - 0,025=0,075(mol)
hh sau pứ: Fe dư và FeS
PTHH: Fe + 2HCl -> FeCl2 + H2
(mol)__0,075_0,15____________0,075__
PTHH:FeS+2HCl -> FeCl2+ H2S
(mol)__0,025_0,05___________0,025
V khí = (0,075+0,025).22,4= 2,24(l)
%V H2 = (0,075.22,4)/2,24.100%=75%
=> %V H2S = 100-75=25%
n HCl pứ = 0,15+0,05=0,2(mol)
n NaOH =0,25(mol)
PTHH: NaOH + HCl -> NaCl + H2O
(mol)___0,25____0,25_________________
n HCl dư = 0,25(mol)
∑n HCl = 0,25+0,2=0,45(mol)
CM HCl = 0,45/0,5=0,9(M)