a) 2Na+2H2O--->2NaOH+H2
b) n\(_{Na}=\frac{9,2}{23}=0,4\left(mol\right)\)
Theo pthh
n\(_{NaOH}=n_{Na}=0,4\left(mol\right)\)
m\(_{NaOH}=0,4.40=16\left(g\right)\)
c)n\(_{H2}=\frac{1}{2}n_{Na}=0,2\left(mol\right)\)
V\(_{H2}=0,2.22,4=4,48\left(l\right)\)