a) $n_{Fe_2O_3} = \dfrac{40}{160} = 0,25(mol)$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$
$n_{HCl} = 6n_{Fe_2O_3} = 0,25.6 = 1,5(mol)$
$m_{dd\ HCl} = \dfrac{1,5.36,5}{7,3\%} = 750(gam)$
b) $n_{FeCl_3} = 2n_{Fe_2O_3} = 0,5(mol)$
$m_{FeCl_3} = 0,5.162,5 = 81,25(gam)$