Đáp án đúng: D
29,0.
– Hướng tư duy 1:
– Tóm tắt quá trình:$\displaystyle \,\underbrace{Fe,Mg,O}_{m\ gam\ X}+\left\langle \begin{array}{l}\xrightarrow{HN{{O}_{3}}}\,\,\underbrace{F{{e}^{3+}},M{{g}^{2+}},NH_{4}^{+},NO_{3}^{-}}_{129,4\,(g)\ \,dd\ Y}+\,\,\underbrace{NO}_{0,26\,mol}\,,\ \underbrace{{{N}_{2}}O}_{0,04\,mol}\ \ (1)\\\xrightarrow{{{H}_{2}}S{{O}_{4}}}\,\underbrace{F{{e}^{3+}},M{{g}^{2+}},SO_{4}^{2-}}_{104\,(g)\,\ dd\ Z}+\,\underbrace{\,S{{O}_{2}}}_{0,7\,mol}\,\ \ (2)\end{array} \right.$
$\xrightarrow{BT:\ e\,\ cho\ (1)\,v\grave{a}\,(2)}{{n}_{N{{H}_{4}}N{{O}_{3}}}}=\frac{2{{n}_{S{{O}_{2}}}}-3{{n}_{NO}}-8{{n}_{{{N}_{2}}O}}}{8}=\,0,0375\,mol$
$\xrightarrow{BTDT\ cho\ (1)\,v\grave{a}\,(2)}2{{n}_{SO_{4}^{2-}}}={{n}_{NO_{3}^{-}}}+{{n}_{NH_{4}^{+}}}$$\xrightarrow{BTKL}\left\{ \begin{array}{l}{{m}_{KL}}+{{m}_{NH_{4}^{+}}}+{{m}_{NO_{3}^{-}}}=129,4\ (1)\\{{m}_{KL}}+{{m}_{SO_{4}^{2-}}}=104\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\end{array} \right.$
$\xrightarrow{(1)\ -\ (2)}62(2{{n}_{SO_{4}^{2-}}}+{{n}_{N{{H}_{4}}^{+}}})-96{{n}_{SO_{4}^{2-}}}=24,725\Rightarrow {{n}_{S{{O}_{4}}^{2-}}}=0,8\ mol$
+ Xét quá trình (2):$\left\{ \begin{array}{l}\xrightarrow{BT:\ e\ +\ BT\text{D}T}2{{n}_{SO_{4}^{2-}}}=2{{n}_{S{{O}_{2}}}}+2{{n}_{O}}\to {{n}_{O}}=0,1\ mol\\{{m}_{KL}}={{m}_{Y}}-{{m}_{SO_{4}^{2-}}}=104-0,8.96=27,2\ gam\end{array} \right.\to m={{m}_{KL}}+{{m}_{O}}=28,8\,gam$
– Hướng tư duy 2:
$\xrightarrow{BT:\ e\,\ cho\ (1)\,v\grave{a}\,(2)}{{n}_{N{{H}_{4}}N{{O}_{3}}}}=\frac{2{{n}_{S{{O}_{2}}}}-3{{n}_{NO}}-8{{n}_{{{N}_{2}}O}}}{8}=\,0,0375\,mol$
+ Gọi T là hỗn hợp muối chứa$\displaystyle Fe{{\left( N{{O}_{3}} \right)}_{2}}$ và$\displaystyle Mg{{\left( N{{O}_{3}} \right)}_{2}}$ suy ra:${{m}_{T}}=\,{{m}_{Y}}-{{m}_{N{{H}_{4}}N{{O}_{3}}}}=126,4\,gam$
+ Áp dụng phương pháp tăng giảm khối lượng + bảo toàn điện tích cho hỗn hợp T và Z ta có
$\to 2{{n}_{N{{O}_{3}}^{-}}}={{n}_{S{{O}_{4}}^{2-}}}=\frac{{{m}_{T}}-{{m}_{Z}}}{2.{{M}_{N{{O}_{3}}^{-}}}-{{M}_{S{{O}_{4}}^{2-}}}}=\frac{126,4-104}{2.62-96}=0,8\,mol$
+ Xét quá trình (2):$\xrightarrow{BT:\ S\ +\ BT:\ H}{{n}_{{{H}_{2}}O}}={{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{S{{O}_{2}}}}\,+{{n}_{S{{O}_{4}}^{2-}}}=1,5\ mol$
$\xrightarrow{BTKL\,}{{m}_{X}}+{{m}_{{{H}_{2}}S{{O}_{4}}}}={{m}_{Z}}+{{m}_{S{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}\Rightarrow m\,=\,28,8\,gam$
