a,
$Fe+2HCl\to FeCl_2+H_2$
$n_{Fe}=\dfrac{2,8}{56}=0,05 mol$
$\Rightarrow n_{H_2}=0,05 mol$
$V_{H_2}=0,05.22,4=1,12l$
b,
$n_{HCl}=0,05.2=0,1 mol$
$\Rightarrow V_{HCl}=\dfrac{0,1}{2}=0,05l=50ml$
c,
$n_{FeCl_2}=0,05 mol$
$\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,05}{0,05}=1M$