$pthh:Na_2O+H_2O \rightarrow NaOH$
Theo bài ra $m_{NaOH}=100.\dfrac{8}{100}=8(g)$
$\rightarrow m_{H_2O}=100-8=92(g)$
$\rightarrow n_{H_2O}=\dfrac{92}{18} \approx 5,1(mol)$
$n_{Na_2O}=\dfrac{9,3}{62}=0,15(mol)$
Vì theo pthh $n_{H_2O}=n_{Na_2O}$ mà trên thực tế $n_{H_2O}>n_{Na_2O}$ nên $H_2O$ dư. Tính theo $Na_2O$
Theo pthh: $n_{NaOH}=n_{Na_2O}=0,15(mol)$
$\rightarrow m_{NaOH}=0,15.40+8=14(g)$
$\rightarrow C$%$NaOH=\dfrac{14}{100+9,3}.100$%$ \approx
12,81$%