a,
$n_{H_2}=\dfrac{0,48}{2}=0,24 (mol)$
$2M+2H_2O\to 2MOH+H_2$
$\Rightarrow n_M=0,24.2=0,48(mol)$
$M_M=\dfrac{11,04}{0,48}=23(Na)$
Vậy kim loại kiềm là natri.
b,
$m_{H_2O}=100g$
$\Rightarrow m_{dd}=11,04+100-0,48=110,56g$
$n_{NaOH}=n_{Na}=0,48 (mol)$
$\Rightarrow C\%_{NaOH}=\dfrac{0,48.40.100}{110,56}=17,37\%$