Question

Answer 1

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{20}{56}=\dfrac{5}{14}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=\dfrac{5}{7}\left(mol\right)\\n_{FeCl_2}=\dfrac{5}{14}\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=\dfrac{5}{14}\cdot22,4=8\left(l\right)\\m_{FeCl_2}=\dfrac{5}{14}\cdot127\approx45,36\left(g\right)\\C\%_{HCl}=\dfrac{\dfrac{5}{7}\cdot36,5}{200}\cdot100\%\approx13,04\%\end{matrix}\right.\)