$n_{H_2}=\frac{1,344}{22,4}=0,06 mol$
$2R+2H_2O\to 2ROH+H_2$
$\Rightarrow n_R=0,06.2=0,12 mol$
$M_R=\dfrac{4,68}{0,12}=39(K)$
Vậy R là kali (K)
$m_{H_2O}=27,44g$
$\Rightarrow m_{dd \text{spứ}}= 4,68+27,44-0,06.2=32g$
$n_{KOH}=0,12 mol$
$\Rightarrow C\%_{KOH}=\dfrac{0,12.56.100}{32}=21\%$