Fe+2HCl->FeCl2+H2
0,1----0,2-------0,1-----0,1
nFe=5,6\56=0,1 mol
=>VH2=0,1.22,4=2,24l
=>mdd=5,6+7,3-0,2=12,7g
=>C%=...
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) Ta có: \(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,1mol\) \(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
c) Ta có: \(m_{H_2}=0,1\cdot2=0,2\left(g\right)\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{Fe}:n_{HCl}=1:2\\n_{FeCl_2}:n_{Fe}=1:1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2mol\\n_{FeCl_2}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{HCl}-m_{H_2}=12,7\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\frac{12,7}{12,7}\cdot100=100\%\)
