\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có:
\(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
\(\Rightarrow n_{H2}=\frac{1}{2}n_{Na}=0,1\left(mol\right)\)
\(\Rightarrow m_{H2}=0,1.2=0,2\left(g\right)\)
\(m_{dd\left(spu\right)}=m_{Na}+m_{H2O}-m_{H2}=4,6+100-0,2=104,4\left(g\right)\)