$m_{H_2O} = D.V = 1.4,459 = 4,459(gam)$
Gọi $n_X = a(mol)$
$2X + 2H_2O \to 2XOH + H_2$
Theo PTHH : $n_{H_2} = \dfrac{1}{2}n_X = 0,5a(mol)$
$n_{XOH} = n_X = a(mol)$
Sau phản ứng, $m_{dd} = 0,897 + 4,459 - 0,5a.2 = 5,356 - a(gam)$
Suy ra :
$C\%_{XOH} = \dfrac{a(X + 17)}{5,356 - a}.100\% = 29,34\%$
mà $X.a = 0,897$ nên $a = 0,039 ; X = 23(Natri)$