\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,1---------------------->0,1
=> VH2 =0,1.22,4 = 2,24(l)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ Theo.PTHH:n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)