$1)PTHH:Na_2O+H_2O\to 2NaOH$
$n_{Na_2O}=\dfrac{9,3}{62}=0,15(mol)$
Theo PT: $n_{NaOH}=2n_{Na_2O}=0,3(mol)$
$\Rightarrow C\%_{NaOH}=\dfrac{0,3.40}{9,3+90,7}.100\%=12\%$
$2)PTHH:2NaOH+FeSO_4\to Fe(OH)_2\downarrow+Na_2SO_4$
$n_{FeSO_4}=\dfrac{200.16\%}{152}\approx 0,21(mol)$
Lập tỉ lệ: $\dfrac{n_{NaOH}}{2}<n_{FeSO_4}\Rightarrow FeSO_4$ dư
$\Rightarrow n_{FeSO_4(dư)}=0,21-0,3:2=0,06(mol)$
Theo PT: $n_{Na_2SO_4}=n_{Fe(OH)_2}=0,15(mol)$
$\Rightarrow m_{Fe(OH)_2}=0,15.90=13,5(g)$
$m_{dd_{sau}}=9,3+90,7+200-13,5=286,5(g)$
$\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,15.142}{286,5}.100\%\approx 7,43\%$
$\Rightarrow C\%_{FeSO_4}=\dfrac{0,06.152}{286,5}.100\%\approx 3,18\%$
$3)PTHH:Fe(OH)_2\xrightarrow{t^o}FeO+H_2O$
$FeO+2HCl\to FeCl_2+H_2O$
Theo PT: $n_{HCl}=2n_{Fe(OH)_2}=0,3(mol)$
$\Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{1,5}=0,2(l)=200(ml)$
