\(a) 2Na + 2H_2O \to 2NaOH + H_2\\ Na_2O + H_2O \to 2NaOH\\ n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ n_{Na} = 2n_{H_2} = 0,3(mol) \Rightarrow m_{Na} = 0,3.23 = 6,9(gam)\\ b) n_{Na_2O} = \dfrac{19,3-6,9}{62} = 0,2(mol)\\ n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,7(mol)\\ m_{dd} = 19,3 + 181 - 0,15.2 = 200(gam)\\ C\%_{NaOH} = \dfrac{0,7.40}{200}.100\% = 14\%\)