Oxide: $\rm R_2O_n$

Ta có $\rm pH = 12$

$\rm \to pOH = 14 - 12 = 2$

$\rm \to [OH^-] = 10^{-2} = 0,01M$

$\rm \to n_{OH^-} = 0,01 \times 1 =0,01$ mol

$\rm \to n_{R(OH)_n} = \dfrac{0,01}{n}(mol)$

$\rm R_2O_n + nH_2O \to 2R(OH)_n$

Theo PT

$\rm n_{R_2O_n} = \dfrac{1}{2}n_{R(OH)_n} = \dfrac{0,005}{n}(mol)$

$\rm \to 2M_R + 16n = \dfrac{0,31}{\dfrac{0,005}{n}} = 62n$

$\rm \to M_R = 23n$

Thỏa mãn $\rm n = 1 \to M_R = 23(Na)$

Vậy Oxide là $\rm Na_2O$

$\rm p_H=12 \to [H^+]=10^{-12}(M)$

 `=>`$\rm [OH-]=\dfrac{10^{-14}}{10^{-12}}=0,01(M)$

$\rm n_{OH-}=0,01.1=0,01(mol)$

gọi CTTQ: $\rm R_2O_x$

$\rm PTHH:R_2O_x+xH_2O \to 2R(OH)_x$

$\rm n_{R(OH)_x}=\dfrac{0,01}{x}(mol)$

$\rm n_{R_2O_x}=\dfrac{1}{2}n_{R(OH)_x}=\dfrac{0,005}{x}(mol)$

$\rm M_{R_2O_x}=\dfrac{0,31}{\dfrac{0,005}{x}}=62x(amu)$

cho $\rm x=1 \to M_R=23 \to chọn$

$\rm x=2 \to M_R=46 \to loại$

vậy oxide kim loại là $\rm Na_2O$