$\underbrace{Al,\,A{{l}_{2}}{{O}_{3}}}_{7,65\,(g)\text{ hn h }\!\!\hat{\mathrm{i}}\!\!\text{ p X}}+\underbrace{NaN{{O}_{3}},\,{{H}_{2}}S{{O}_{4}}}_{dung\text{ dch Y}}\xrightarrow{{}}\underbrace{A{{l}^{3+}},\,N{{a}^{+}},\,N{{H}_{4}}^{+},\,S{{O}_{4}}^{2-}}_{dung\text{ dch Z}}+\underbrace{{{N}_{{\bar{x}}}}{{O}_{{\bar{y}}}},\,{{H}_{2}}}_{\text{m (g)}\,\text{kh }\!\!\acute{\mathrm{Y}}\!\!\text{  T}}+{{H}_{2}}O$

- Khi cho dung dịch Z tác dụng với BaCl₂ dư thì : ${{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{BaS{{O}_{4}}}}=0,4\,mol$

- Xét quá trình hỗn hợp X tác dụng với dung dịch Y ta có :

+ ${{n}_{Al\text{ (trong}\,\text{X)}}}=0,17\,mol\Rightarrow {{n}_{A{{l}_{2}}{{O}_{3}}}}=\frac{{{m}_{X}}-27{{n}_{Al}}}{102}=0,03\,mol\xrightarrow{BT:Al}{{n}_{A{{l}^{3+}}(trong\,Z)}}={{n}_{Al}}+2{{n}_{A{{l}_{2}}{{O}_{3}}}}=0,23\,mol$

+ Khi dung dịch Z tác dụng với 0,935 mol NaOH thì :

 ${{n}_{N{{H}_{4}}^{+}}}={{n}_{NaOH}}-4{{n}_{A{{l}^{3+}}}}=0,015\,mol\xrightarrow{BT:\,H}{{n}_{{{H}_{2}}O}}=\frac{2{{n}_{{{H}_{2}}S{{O}_{4}}}}-2{{n}_{{{H}_{2}}}}-4{{n}_{N{{H}_{4}}^{+}}}}{2}=0,355\,mol$

$\xrightarrow[(Z)]{BTDT}{{n}_{N{{a}^{+}}}}=2{{n}_{S{{O}_{4}}^{2-}}}-{{n}_{N{{H}_{4}}^{+}}}-3{{n}_{A{{l}^{3+}}}}=0,095\,mol\Rightarrow {{m}_{Z}}=23{{n}_{N{{a}^{+}}}}+27{{n}_{A{{l}^{3+}}}}+18{{n}_{N{{H}_{4}}^{+}}}+96{{n}_{S{{O}_{4}}^{2-}}}=47,065\,(g)$$\xrightarrow{BTKL}{{m}_{T}}={{m}_{X}}+98{{n}_{{{H}_{2}}S{{O}_{4}}}}+85{{n}_{NaN{{O}_{3}}}}-18{{n}_{{{H}_{2}}O}}-{{m}_{Z}}=1,47\,(g)$.