\(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
de: 0,4 \(\rightarrow0,4\) \(\rightarrow0,2\) (mol)
\(V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(m_{NaOH}=40.0,4=16\left(g\right)\)
\(m_{ddA}=m_{Na}+m_{H_2O}-m_{H_2}\)
= 9,2 + 110,4 - 0,2.2 = 119,2 (g)
\(C\%_{ddA}=\dfrac{m_{NaOH}}{m_{ddA}}.100\%\approx13,42\%\)