\(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(m_{NaOH\left(bđ\right)}=\dfrac{150.10}{100}=15\left(g\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,4------------->0,4---->0,2
=> mNaOH(sau pư) = 0,4.40 + 15 = 31 (g)
mdd sau pư = 9,2 + 150 - 0,2.2 = 158,8 (g)
\(C\%_{dd.sau.pư}=\dfrac{31}{158,8}.100\%=19,52\%\)