a) PTHH: 2Na + 2H2O \(\rightarrow\) 2NaOH + H2\(\uparrow\)
b) nNa = \(\frac{11,5}{23}=0,5\left(mol\right)\)
Theo PT: n\(H_2O\) = nNa = 0,5 (mol)
=> m\(H_2O\) = 0,5.18 = 9 (g)
c) Theo PT: n\(NaOH\) = nNa = 0,5 (mol)
=> m\(NaOH\) = 0,5.40 = 20 (g)