$n_{Fe}$ `=` `(5.6)/56``= 0,1(mol)`
`a)`
`Fe + 2HCl → ` $FeCl_{2}$ `+` $H_{2}$
`0,1(mol)`-------------------→`0,1(mol)`
`b)`
$V_{H_{2}}$ `=` `0,1.22,4``=2,24(l)`
`c)`
$m_{HCl}$ `=` `0,2.36,5=7,3(g)`
$mdd_{HCl}$ `=` `( 7,3.100%)/(18,25%)` `= 40(g)`
@cuthilien