Gọi kim loại kiềm là R.
$2R+2H_2O\to 2ROH+H_2$
$n_R=\dfrac{3,9}{R} (mol)$
$\Rightarrow n_{ROH}=\dfrac{3,9}{R}; n_{H_2}=\dfrac{1,95}{R}(mol)$
$m_{H_2O}=136,2g$
$\Rightarrow m_{dd\text{spứ}}=3,9+136,2-2.\dfrac{1,95}{R}=140,1-\dfrac{3,9}{R} (g)$
$\Rightarrow \dfrac{3,9(R+17)}{R}=(140,1-\dfrac{3,9}{R}).4\%$
$\Leftrightarrow R=39(K)$