a) \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\left(1\right)\)
\(n\left(Fe\right)=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\left(1\right)\Rightarrow n\left(H_2\right)=n\left(FeCl_2\right)=0,1\left(mol\right)\)
\(V\left(H_2\right)=0,1.22,4=2,24\left(lít\right)\)
b) \(m\left(FeCl_2\right)=0,1.\left(56+2.35,5\right)=12,7\left(g\right)\)
\(m_{dd}\left(sau.PƯ\right)=m_{Fe}+m_{dd}\left(HCl\right)-m_{H_2}=5,6+100-0,1.2=105,4\left(g\right)\)
\(C\%\left(FeCl_2\right)=\dfrac{12,7}{105,4}.100\%=12,05\left(\%\right)\)