\(n_{NaOH\left(bđ\right)}=0,5.0,5=0,25\left(mol\right)\\ n_{NaOH\left(thêm\right)}=0,5.1,5-0,25=0,5\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{Na\left(thêm\right)}=n_{NaOH\left(thêm\right)}=0,5\left(mol\right)\\ m_{Na\left(thêm\right)}=0,5.23=11,5\left(g\right)\)