$n_{Fe_2O_3}=24/160=0,15mol$
$n_{H_2O}=\dfrac{1,8.10²³}{6,02.10²³}=0,3mol$
$a./$
$Fe2O3+3H2→2Fe+3H2O$
$\text{Theo pt :}$
$n_{H_2}=n_{H2O}=0,3mol$
$⇒V_{H_2}=0,3.22,4=6,72l$
$b/$
$n_{Fe_2O_3 dư}=0,15-0,1=0,05mol $
$⇒m_{rắn}=0,2.56+0,05.160=19,2g$
$c/$
$\%mFe2O3 =0,1/0,15=66,67\%$