\(m_{H_2O}=87,6.1=87,6\left(g\right)\\ n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,2-------------------->0,4
\(m_{dd}=12,4+87,6=100\left(g\right)\\ m_{NaOH}=0,4.40=16\left(g\right)\\ C\%_{NaOH}=\dfrac{16}{100}.100\%=16\%\)