$$\begin{gathered} \text{Z}+{\text{O}}_2\stackrel{{\text{t}}^0}{\to }{\text{CO}}_2+{\text{H}}_2\text{O}+{\text{Na}}_2{\text{CO}}_3 \\ {\text{n}}_{{\text{Na}}_2{\text{CO}}_3}=\frac{6,36}{106}=0,06 \text{mol}\to {\text{n}}_{\text{NaOH}}={\text{n}}_{\text{Muôí }}=0,12 \text{mol} \end{gathered}$$

Đặt số mol CO₂ và H₂O lần lượt là a, b mol.

Bảo toàn nguyên tố oxi, ta có:

${\text{n}}_{\text{O}\left(\text{Z}\right)}+2\times {\text{n}}_{{\text{O}}_2}=2\times {\text{n}}_{{\text{CO}}_2}+{\text{n}}_{{\text{H}}_2\text{O}}\to 0,12\times 2+2\times 0,285=2\text{a}+\text{b}+0,06\times 3\left(1\right)$

Mặt khác, ta có: ${\text{n}}_{{\text{CO}}_2}+{\text{n}}_{{\text{H}}_2\text{O}}=\text{a}+\text{b}=0,405\left(2\right)$

Từ (1) và (2), giải hệ ta được: $\left\{\begin{array}{l}\text{a}=0,225\\ \text{b}=0,18\end{array}\text{X}\right.$ gồm ${\text{C}}_{\text{n}}{\text{H}}_{2\text{n}}{\text{O}}_2\left(\text{xmol}\right)$ và ${\text{C}}_{\text{m}}{\text{H}}_{2 \text{m}-2}{\text{O}}_2(\text{y\hspace{0.17em}mol)}$

Số mol muối không no $=\text{y}={\text{n}}_{{\text{CO}}_2}-{\text{n}}_{{\text{H}}_2\text{O}}=0,225-0,18=0,045( \text{mol})$

-> Số mol muối no $=x=0,12-0,045=0,075 \text{mol}$

Ta có: ${\text{m}}_{\text{x}}=11,1=0,075\times (14\text{n}+32)+0,045\times (14 \text{m}+30)\to 5n+3 \text{m}=35\to \left\{\begin{array}{l}\text{n}=4\\ \text{m}=5\end{array}\right.$

Vậy X gồm: ${\text{C}}_4{\text{H}}_8{\text{O}}_2(0,075 \text{mol})$ và ${\text{C}}_5{\text{H}}_8{\text{O}}_2(0,045 \text{mol})$