a) PTHH: \(2Fe+3Cl_2\rightarrow2FeCl_3\)
b) \(n_{Fe}=\dfrac{11,2}{56}=0,2\) (mol)
Theo PTHH \(\Rightarrow n_{Cl_2}=\dfrac{3}{2}.n_{Fe}=\dfrac{3}{2}.0,2=0,3\) (mol)
\(\Rightarrow m_{Cl_2}=0,3.71=21,3\left(g\right)\)
c) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PTHH thì \(n_{HCl}=2.n_{Fe}=2.0,2=0,4\) (mol)
\(\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200ml\)