a)
Fe + 2HCl → FeCl2 + H2
x 2x x x
FeS + 2HCl → FeCl2 + H2S
x 2x x x
56x+88y = 6,4
x+y = 22,4:1,792
=> x=0,02
y=0,06
%nFe = 0,02:(0,02+0,06}.100\ = 25
%nFeS = 100%-25\% = 75%
b)
n_{HCl} = 2n_{H2}+2n_{H2S} = 0,16 mol
m_{ddHCl} = 0,16.36,5:20% = 29,2 g
m_{ddsau} = 29,2+6,4-0,2.2-0,06.24 = 33,52 g
C%_{FeCl2} = [(0,02+0,06).127.100]:33,52 = 30,31%
c)
M_B = (2.0,02+0,06.34):0,08 = 26 g/mol
d_{B/kk} = :26:29