$PTHH :$
$Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O$
$a.n_{Fe}=\dfrac{79}{56}≈1,41mol$
$\text{Theo pt :}$
$n_{Fe_2O_3}=1/2.n_{Fe}=1/2.1,41=0,705mol$
$⇒m_{Fe_2O_3}=0,705.160=112,8g$
$\text{b.Theo pt :}$
$n_{H_2O}=3/2.n_{Fe}=3/2.1,41=2,115mol$
$⇒m_{H_2O}=2,115.18=38,07g$
$\text{c.Theo pt :}$
$n_{H_2}=3/2.n_{Fe}=3/2.1,41=2,115mol$
$⇒V_{H_2}=2,115.22,4=47,376l$