a) 2Na + 2H2O $\to$ 2NaOH + H2
b)
Theo PTHH :
n NaOH = n Na = 3,9/23 = 39/230(mol)
=> CM NaOH = \(\dfrac{\dfrac{39}{230}}{0,15} = 1,13M\)
\(n_{Na}=\dfrac{3.9}{23}=\dfrac{39}{230}\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(\dfrac{39}{230}..........\dfrac{39}{230}\)
\(C_{M_{NaOH}}=\dfrac{\dfrac{39}{230}}{0.15}=1.13\left(M\right)\)