$1/$
$a/$
$Fe+2HCl→FeCl2+H2$
$b/$
$n_{Fe}=8,4/56=0,15mol$
$⇒nFeCl2=nFe=0,15mol$
$⇒mFeCl2=0,15.127=19,05g$
$c/$
$nH2=nFe=0,15mol$
$⇒V_{H2}=0,15.22,4=3,36l$
$2/$
$a/$
$mCu=80.80\%=64g⇒nCu=64/64=1mol$
$mO=80.20\%=16g⇒nO=16/16=1mol$
$⇒CTHH$ của hợp chất là CuO
$b/$
$CuO+2HCl→CuCl2+H2$
$ n_{CuO}=16/80=0,2mol$
$⇒nHCl=2.nCuO=2.0,2=0,4mol$
$⇒mHCl=0,4.36,5=14,6g.$