\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\uparrow\)
Đặt nH2 = x mol; nH2S = y mol
Ta có: nkhí = x + y = 0,1 mol;
mkhí = 2x + 34y = 0,1.9.2 = 1,8 gam
Giải hệ ta có: x = 0,05 và y = 0,05
Suy ra nFe = 0,05. nFeS = 0,05 mol.
Vậy %nFe = 50%.
\(n_{\uparrow}=\dfrac{2,24}{22,4}=0,1mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{FeS}=y\left(mol\right)\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\uparrow\)
\(\Rightarrow x+y=0,1\left(1\right)\)
\(d_{\uparrow}\)/H2=9\(\Rightarrow\overline{M_{\uparrow}}=9\cdot2=18\)
Sơ đồ chéo:
Fe 56 70
18
FeS 88 38
\(\Rightarrow\dfrac{n_{Fe}}{n_{FeS}}=\dfrac{70}{38}=\dfrac{35}{19}=\dfrac{x}{y}\)(2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{108}\\y=\dfrac{19}{540}\end{matrix}\right.\)
\(\%Fe=\dfrac{\dfrac{7}{108}}{\dfrac{7}{108}+\dfrac{19}{540}}\cdot100\%=64,81\%\)
