Sửa 20ml thành 200ml
Giải :
Ta có:
\(\left\{{}\begin{matrix}n_{Fe}=\frac{2,8}{56}=0,05\left(mol\right)\\n_S=\frac{0,8}{32}=0,025\left(mol\right)\end{matrix}\right.\)
\(PTHH:Fe+S\rightarrow FeS\)
Lập tỉ lệ : \(\frac{0,05}{1}>\frac{0,025}{1}\) nên Fe dư
\(n_{Fe}=0,05-0,025=0,025\left(mol\right)\Rightarrow n_{FeS}=0,025\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
\(n_{HCl}=0,025.2+0,025.2=0,1\left(mol\right)\)
\(\Rightarrow CM_{HCl}=\frac{0,1}{0,2}=0,5M\)
\(\left\{{}\begin{matrix}n_{H2}=n_{Fe}=0,025\left(mol\right)\\n_{H2S}=n_{FeS}=0,025\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{H2}=0,025.2=0,05\left(g\right)\\m_{H2S}=0,025.34=0,85\left(g\right)\end{matrix}\right.\)