$\underbrace{{{H}_{2}}N{{C}_{x}}{{H}_{y}}{{(COOH)}_{2}}(X)}_{0,1\,mol}+\underbrace{{{H}_{2}}S{{O}_{4}}}_{0,1\,mol}\to \underbrace{X{{H}^{+}},{{H}^{+}},\,S{{O}_{4}}^{2-}\,}_{dung\text{ dch X}}\xrightarrow{NaOH,\,KOH}\underbrace{\overbrace{{{X}^{2-}}}^{0,1\,mol},\overbrace{S{{O}_{4}}^{2-}}^{0,1\,mol},\,\overbrace{N{{a}^{+}}}^{x\,mol},\,\overbrace{{{K}^{+}}}^{3x\,mol}}_{\text{36,7}\,\text{(g) Y}}$

$\xrightarrow[(Y)]{BTDT}{{n}_{N{{a}^{+}}}}+{{n}_{{{K}^{+}}}}=2{{n}_{S{{O}_{4}}^{2-}}}+2{{n}_{{{X}^{2-}}}}\to 4x=0,4\Rightarrow x=0,1$

$\to {{m}_{{{X}^{2-}}}}={{m}_{Y}}-96{{n}_{S{{O}_{4}}^{2-}}}-39{{n}_{{{K}^{+}}}}-23{{n}_{N{{a}^{+}}}}=13,1\Rightarrow {{M}_{{{X}^{2-}}}}=\frac{13,1}{0,1}=131\to {{M}_{X}}=133\Rightarrow %N=10,52$