\(a,2Na+2H_2O\rightarrow2NaOH+H_2\\ 2K+2H_2O\rightarrow2KOH+H_2\\ b,n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right);n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\\ n_{H_2}=\dfrac{1}{2}.\left(0,2+0,1\right)=0,15\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,Dung.dịch.sau.phản.ứng.có.KOH.và.NaOH.đều.là.kiềm.\\ \Rightarrow Quỳ.tím.hoá.xanh\)
\(n_{Na}=\dfrac{m}{M}=\dfrac{4,6}{23}=0,2mol\)
\(n_K=\dfrac{m}{M}=\dfrac{3,9}{39}=0,1mol\)
\(Na+2H_2O\rightarrow Na\left(OH\right)_2+H_2\)
\(K+2H_2O\rightarrow K\left(OH\right)_2+H_2\)
\(V_{H_2}=n_{H_2}.22,4=\left(0,2+0,1\right).22,4=6,72l\)
Dung dịch sau phản ứng làm quỳ tím chuyển sang màu xanh