Gọi \(n_{Na}=x\left(mol\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
x------------------>x---------->0,5x
=> \(m_{dd.sau.pứ}=23x+49,84-0,5x.2=22x+49,84\left(g\right)\)
\(n_{NaOH}=40x\left(g\right)\)
=> \(C\%_{NaOH}=\dfrac{40x}{22x+49,84}.100\%=20\%\)
=> x = 0,28 (mol)
=> mNa = 0,28.23 = 6,44 (g)