Đặt hóa trị \(R\) là \(n\)
\(\begin{array}{l}
336ml=0,336l\\
n_{H_2}=\frac{0,336}{22,4}=0,015(mol)\\
2R+2nH_2O\to 2R(OH)_n+nH_2\\
Theo\,PT:\,n_R.n=2n_{H_2}=0,03\\
\to \frac{1,17n}{M_R}=0,03\\
\to M_R=39n\\
\to M_R=39;n=1\\
\to R:\,kali\,(K)
\end{array}\)