\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
_____0,3<------------------------0,15
=> mNa = 0,3.23 = 6,9 (g)
=> \(\left\{{}\begin{matrix}\%Na=\dfrac{6,9}{21}.100\%=32,857\%\\\%K_2O=100\%-32,857\%=67,143\%\end{matrix}\right.\)