$n_{OH^-} = 2n_{H_2} = 0,195.2 = 0,39(mol)$
$n_{Fe^{3+}} = n_{FeCl_3} = 0,1(mol)$
$n_{H^+} = n_{HCl} = 0,15(mol)$
$H^+ + OH^- \to H_2O$
$Fe^{3+} + 3OH^- \to Fe(OH)_3$
$n_{OH^-} = n_{H^+} + 3n_{Fe(OH)_3} \Rightarrow n_{Fe(OH)_3} = 0,08(mol)$
$m_{Fe(OH)_3} = 0,08.107 = 8,56(gam)$