[tex]n_{Na}=\frac{2,3}{23}=0,1(mol)[/tex]
Phương trình phản ứng:
[tex]2Na+2H_2O\rightarrow 2NaOH+H_2[/tex]
[tex]n_{Na}=n_{NaOH}=n_{H_2O}=2n_{H_2}=0,1(mol)\rightarrow n_{H_2}=0,05(mol)[/tex]
[tex]m_{NaOH}=0,1.40=4(g)[/tex]
[tex]m_{dd}=m_{Na}+m_{H_2O}-m_{H_2}=2,3+47,8-0,05.2=50(g)[/tex]
C%ddNaOH=[tex]\frac{4}{50}.100=8[/tex]%