$\underbrace{Fe{{(N{{O}_{3}})}_{3}},\,Al,Cu,\,MgC{{O}_{3}}\,}_{28,4\,(g)\,A}\xrightarrow{{{H}_{2}}S{{O}_{4}}}\underbrace{NO,\,{{N}_{2}}O,\,{{N}_{2}},\,\overbrace{{{H}_{2}}}^{0,02\,mol}}_{\text{hn h }\!\!\hat{\mathrm{i}}\!\!\text{ p kh }\!\!\acute{\mathrm{Y}}\!\!\text{  Z}}+\underbrace{F{{e}^{n+}},\,A{{l}^{3+}},\,M{{g}^{2+}},C{{u}^{2+}},N{{H}_{4}}^{+},\,S{{O}_{4}}^{2-}\,}_{65,48\,(g)\,B}+{{H}_{2}}O$

- Khi cho B tác dụng với BaCl₂ thì $\xrightarrow{BT:\,S{{O}_{4}}^{2-}}{{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{BaS{{O}_{4}}}}=0,53$

- Khi cho B tác dụng với NaOH thì $\xrightarrow{BT:\,Na}{{n}_{NaOH}}=2{{n}_{N{{a}_{2}}S{{O}_{4}}}}=1,06$và ${{n}_{N{{H}_{3}}}}={{n}_{{{H}_{2}}O}}$

$\xrightarrow{BTKL}18{{n}_{{{H}_{2}}O}}+17{{n}_{N{{H}_{3}}}}={{m}_{B}}+40{{n}_{NaOH}}-{{m}_{\downarrow }}-142{{n}_{N{{a}_{2}}S{{O}_{4}}}}\Rightarrow {{n}_{N{{H}_{4}}^{+}}}={{n}_{N{{H}_{3}}}}={{n}_{{{H}_{2}}O}}=0,02$

- Khi cho A tác dụng với H₂SO₄ thì :

$\xrightarrow{BT:\,H}{{n}_{{{H}_{2}}O}}=\frac{2{{n}_{{{H}_{2}}S{{O}_{4}}}}-4{{n}_{N{{H}_{4}}^{+}}}-2{{n}_{{{H}_{2}}}}}{2}=0,47\xrightarrow{BTKL}{{m}_{Z}}={{m}_{A}}+98{{n}_{{{H}_{2}}S{{O}_{4}}}}-{{m}_{B}}-18{{n}_{{{H}_{2}}O}}=6,4$

$\Rightarrow {{n}_{Z}}=\frac{{{m}_{Z}}}{{{M}_{Z}}}=\frac{6,4}{16.2}=0,2\to {{V}_{Z}}=4,48\,(l)$