\(n_Z=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
\(\overline{M}_Z=\dfrac{34+2}{2}=18\left(g/mol\right)\)
H2S 34 16
18
H2 2 16
=> \(\dfrac{n_{H_2S}}{n_{H_2}}=\dfrac{16}{16}=1\Rightarrow n_{H_2S}=n_{H_2}=\dfrac{0,15}{2}=0,075\left(mol\right)\)
\(Fe_{dư}+2HCl\rightarrow FeCl_2+H_2\)
0,075<--------------------- 0,075
=> \(n_{Fe.dư}=0,075\left(mol\right)\)
Bảo toàn S: \(n_{FeS}=n_{H_2S}=0,075\left(mol\right)\)
=> \(n_{Fe.bđ}=0,075+0,075=0,15\left(mol\right)\)
=> \(a=m_{Fe}=0,15.56=8,4\left(g\right)\)
\(n_{S.dư}=n_{S.bđ}-0,075=\dfrac{1,28}{32}=0,04\left(mol\right)\\ \Rightarrow n_{S.bđ}=0,115\left(mol\right)\Rightarrow b=m_S=0,115.32=3,68\left(g\right)\)
b) \(H\%_{Fe,S}=\dfrac{0,115.56.100\%}{0,15.56}=76,67\%\)
