\(n_{khi}=\frac{1,792}{22,4}=0,08\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:x\left(mol\right)\\n_{FeS}:y\left(mol\right)\end{matrix}\right.\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
x_____________x_________
\(FeS+H_2SO_4\rightarrow FeSO_4+H_2S\)
y__________________y_______
\(n_{H2SO4}=n_{khi}=0,08\left(mol\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}56x+88y=6,4\\x+y=0,08\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,06\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{H2}=\%n=\frac{0,02}{0,08}.100\%=25\%\\\%V_{H2S}=\%n=100\%-25\%=75\%\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:x\left(mol\right)\\n_{FeS}:y\left(mol\right)\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x____2x_____x________x
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
y_______2y_____y________y
Giải hệ PT:
\(\left\{{}\begin{matrix}56x+88y=6,4\\x+y=\frac{1,792}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,06\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%n_{Fe}=\frac{0,02}{0,02+0,06}.100\%=25\%\\\%n_{FeS}=100\%-25\%=75\text{%}\end{matrix}\right.\)
\(n_{HCl}=2n_{H2}+2n_{H2S}=0,16\)
\(m_{dd\left(HCl\right)}=\frac{0,16.36,5}{20\%}=29,2\left(g\right)\)
\(m_{dd\left(sau\right)}=29,2+6,4-0,2.2-0,06.24=33,52\left(g\right)\)
\(C\%_{FeCl2}=\frac{\left(0,02+0,06\right).127}{33,52}.100\%=30,31\%\)
\(M_B=\frac{2.0,02+0,06.34}{0,08}=26\)
\(d_{B/kk}=\frac{26}{29}\)
