\(Fe+S\rightarrow FeS\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
0,2________0,4___0,2_______
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có :
\(n_{HCl}=\frac{146.10\%}{36,5}=0,4\left(mol\right)\)
\(d_{Z/H2}=13\Rightarrow M_Z=26\)
Áp dụng PP đường chéo :
\(\Rightarrow\frac{n_{H2S}}{n_{H2}}=3\)
\(\Rightarrow3n_{H2}=n_{H2S}\)
\(\Rightarrow n_{H2}=\frac{1}{15}\left(mol\right)\)
\(n_{Fe}=n_{FeS}+n_{Fe\left(\cdot dư\right)}=0,2+\frac{1}{15}=\frac{4}{15}\)
\(\Rightarrow n_S=n_{FeS}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=\frac{4}{15}.56=14,93\left(g\right)\)
\(\Rightarrow m_S=0,2.32=6,4\left(g\right)\)
Đề sửa lại rồi nha :
Gọi số mol H2S: x mol ; H2: y mol
\(Fe+S\rightarrow FeS\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
x____2x________x__________x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y____2y_______y________y
Ta có :
\(n_{HCl}=\frac{146.10\%}{36,5}=0,4\left(mol\right)\)
\(d_{Z/H2}=13\)
Áp dụng PPT đường chéo : (Như hình bên dưới nha )
\(\Rightarrow\frac{n_{H2S}}{n_{H2}}=3\)
\(\Rightarrow x-3y=0\left(1\right)\)
\(2x+2y=0,4\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,05\end{matrix}\right.\)
\(n_{Fe}=n_{FeS}+n_{Fe\cdot\left(dư\right)}=0,15+0,05=0,2\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(n_S=n_{FeS}=0,15\left(mol\right);m_S=0,15.32=4,8\left(g\right)\)
\(m_{Dd\left(spu\right)}=m_{dd\left(HCl\right)}+m_{Fe\left(Dư\right)}+m_{FeS}-m_{H2}-m_{H2S}\)
\(m_{dd\left(spu\right)}=146+0,05.56+0,15.88-0,05.2-0,15.34=156,8\)
\(\Rightarrow C\%_{FeCl2}=\frac{\left(0,15+0,05\right).127}{156,8}.100\%=16,2\%\)