\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
a) PT: 2Na + 2H2O \(\rightarrow\) 2NaOH + H2
--------0,1----------------0,1--------0,05 (mol)
b) mNaOH = 0,1.40 = 4(g)
c) đktc: \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
d) 200 ml = 0,2 l
\(C_M=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)