a) \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{H_2}=0,4\left(mol\right)\)
\(V_{H_2}=0,4.22,4=8,96\left(lít\right)\)
b) \(n_{FeCl_2}=0,4\left(mol\right)\)
\(m_{FeCl_2}=n.M=0,4.127=50,8\left(g\right)\)