$2K + 2H_2O \to 2KOH + H_2$
$n_{KOH} = n_K = \dfrac{3,9}{39} = 0,1(mol)$
$n_{H_2} = \dfrac{1}{2}n_K = 0,05(mol)$
$m_{dd\ sau\ pư} = 3,9 + 101,8 - 0,05.2 = 105,6(gam)$
$C\%_{KOH} = \dfrac{0,1.56}{105,6}.100\% = 5,303\%$
\(n_K=0,1\left(mol\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
0,1 ----- 0,1 ------ 0,1 ----- 0,05 (mol)
\(m_{dd}\left(saupứ\right)=3,9+101,8-0,05.2=105,6\left(g\right)\)
\(\Rightarrow C\%\left(KOH\right)=\dfrac{0,1.56}{105,6}.100\%=5,3\%\)