Question

Answer 1

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
          0,15                  0,15        0,15 
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

Answer 2

\(a,n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

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

           0,15-------------->0,15---->0,15

\(V_{H_2}=0,15.24,79=3,7185\left(l\right)\\ b,m_{muối}=0,15.127=19,05\left(g\right)\)