Các phương trình hóa học: $8\text{Al}+3{\text{Fe}}_3{\text{O}}_4\stackrel{t^o}{\to }{\text{4Al}}_2{\text{O}}_3+9\text{Fe}$

Hòa tan X trong HCl dư:

$$\begin{gathered} 2\text{Al}+6\text{HCl}\to 2{\text{AlCl}}_3+3{\text{H}}_2 \\ \text{Fe}+2\text{HCl}\to {\text{FeCl}}_2+{\text{H}}_2 \\ {\text{Al}}_2{\text{O}}_3+6\text{HCl}\to 2{\text{AlCl}}_3+3{\text{H}}_2\text{O} \\ {\text{Fe}}_3{\text{O}}_4+8\text{HCl}\to {\text{FeCl}}_2+3{\text{FeCl}}_3+4{\text{H}}_2\text{O} \end{gathered}$$

Bảo toàn O: ${\text{n}}_{\text{O(ox}it\text{) }}={\text{n}}_{H_{2}O}=0,04.4=0,16$ mol

Bảo toàn H: ${\text{n}}_{\text{HCl}}=2.{\text{n}}_{{\text{H}}_2}+2.{\text{n}}_{{\text{H}}_2\text{O}}=2.0,15+2.0,16=0,62$ mol

Bảo toàn khối lượng ta có:

${\text{m}}_{\text{x}}+{\text{m}}_{\text{HCl}}={\text{m}}_{\text{muoi}}+{\text{m}}_{{\text{H}}_2}+{\text{m}}_{{\text{H}}_2\text{O}}$

$\to {\text{m}}_{muoi}=0,12.27+0,04.232+0,62.36,5-0,15.2-0,16.18=31,97$ gam