`n_{Na}={4,6}/23=0,2\ (mol)`
`2Na+2H_2O\to 2NaOH+H_2`
Theo phương trình:
`n_{NaOH}=n_{Na}=0,2\ (mol)`
Coi $V_{dd\ NaOH}=V_{H_2O}=200\ ml=0,2\ l$
$\to C_{M_{dd\ NaOH}}=\dfrac{0,2}{0,2}=1\ (M)$
Theo phương trình:
`n_{H_2}=1/2n_{Na}=1/2 .0,2=0,1\ (mol)`
$\to V_{khí}=0,1.22,4=2,24\ (l)$