a) $Fe_{2}O_{3}$ + 6HCl -> 2Fe$Cl_{3}$ + 3$H_{2}O$
$n_{Fe_{2}O_{3}}$ = $\frac{16}{160}$ = 0,1mol
-> $n_{FeCl_{3}}$ = $2n_{Fe_{2}O_{3}}$ = 0.2 mol
$m_{FeCl_{3}}$ = 0,2.162,5= 32,5 g
b) $n_{HCl}$ = $6n_{Fe_{2}O_{3}}$= 0,6 mol
$C_{M_{HCl}}$ = $\frac{0,6}{0,2}$ = 3M
$C_{M_{FeCl_{3}}}$ = $\frac{0,2}{0,2}$ = 1M