$n_{Fe}=1,12/56=0,02mol$
$n_{HCl}=1,095/36,5=0,03mol$
a/ $Fe + 2HCl\to FeCl_2+H_2$
$\text{Theo pt : 1 mol 2 mol}$
$\text{Theo đbài :0,02mol 0,03mol}$
$\text{⇒Sau pư Fe dư 0,005mol}$
$⇒m_{Fe dư}=0,005.56=0,28g$
b/
Theo pt :
$n_{H_2}=1/2.n_{HCl}=1/2.0,03=0,015mol$
$⇒V_{H_2}=0,015.22,4=0,336l$
c/
$n_{H_2}=0,015.80\%=0,012mol$
$⇒V_{H_2}=0,012.22,4=0,2688l$