`n_(Fe)=\frac{8,4}{56}=0,15(mol)`
`Fe+2HCl->FeCl_2+H_2`
`0,15` `0,3` `0,15` `0,15`
`a,`
`V_(H_2)=0,15.22,4=3,36(l)`
`b,`
`m_(HCl)=0,3.36,5=10,95(g)`
`m_(dd HCl)=\frac{10,95}{10,95%}=100(g)`
`c,`
`m_(dd)=100+8,4-0,3=108,1(g)`
`m_(FeCl_2)=0,15.127=19,05(g)`
`C%_(FeCl_2)=\frac{19,05}{108,1}.100=17,62%`