\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{NaOH}=n_{Na}=0,2\left(mol\right);n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,m_{ddNaOH}=4,6+95,6-0,1.2=100\left(g\right)\\ C\%_{ddNaOH}=\dfrac{0,2.40}{100}.100=8\%\)