$\rm n_{Al} = \dfrac{m}{27}(mol)$

$\rm n_{O_2} = \dfrac{m-m+2,88}{32} = 0,09$ mol

$\rm 4Al + 3O_2 \xrightarrow[]{t^o}2Al_2O_3(1)$

Theo PTHH $\rm (1)$

$\rm n_{Al_2O_3} = \dfrac{2}{3}n_{O_2} = 0,06$ mol

Ta có

$\rm n_{Al} = 2n_{Al_2O_3} + n_{Al(dư)}$

$\rm \to \dfrac{m}{27} = 0,12 + n_{Al(dư)}$

$\rm \to n_{Al(dư)} = \dfrac{m}{27} - 0,12(mol)$

$\rm X: Al_2O_3, Al$ dư

Cho $\rm X + HCl$ dư

$\rm 2Al + 6HCl \to 2AlCl_3+3H_2(2)$

$\rm Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O(3)$

$\rm \to Y: AlCl_3, HCl$

Theo PTHH $\rm (2)$

$\rm n_{H_2} = \dfrac{3}{2}n_{Al}$

$\rm \to 2a = \dfrac{m}{9} - 0,36$

$\rm \to \dfrac{m}{9} - 2a = 0,36(I)$

Bảo toàn nguyên tố $\rm Al$

$\rm n_{AlCl_3} = n_{Al} = \dfrac{m}{27}$ mol

Xét $\rm n_{NaOH} = 0,5a$

Hết $\rm HCl$

$\rm NaOH + HCl \to NaCl + H_2O(4)$

$\rm \to n_{HCl} = n_{NaOH} = 0,5a$ (mol)

Xét $\rm n_{NaOH} = \dfrac{29a}{6}$

Kết tủa bị hòa tan 1 phần

Xảy ra các phản ứng:

$\rm HCl + NaOH \to NaCl + H_2O(4)$

$\rm AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl(5)$

$\rm Al(OH)_3 + NaOH \to NaAlO_2 + 2H_2O(6)$

Theo PTHH $\rm (5)$

$\rm n_{Al(OH)_3(max)} = n_{AlCl_3} = \dfrac{m}{27}$(mol)

$\rm \to n_{Al(OH)_3} = n_{Al(OH)_3(max)} - n_{Al(OH)_3(còn)}$

$\rm \to n_{Al(OH)_3(6)} = \dfrac{m}{27} - a(mol)$

Theo PTHH $\rm (4),(5),(6)$

$\rm n_{NaOH} = n_{HCl} + 3n_{AlCl_3} + n_{Al(OH)_3}$

$\rm \to \dfrac{29a}{6} = 0,5a + \dfrac{3m}{27} + \dfrac{m}{27} - a$

$\rm \to \dfrac{16a}{3} - \dfrac{4m}{27} = 0(II)$

Từ $\rm (I),(II)$

$\rm \to \begin{cases} \rm a = 0,18\\\rm m = 6,48\end{cases}$