Hoà tan hoàn toàn m gam X gồm Mg, Fe, FeCO3, Cu(NO3)2 vào dung dịch chứa H2SO4 và 0,045 mol NaNO3, thu được dung dịch Y c

* Hướng dẫn giải

Đáp án B

\(\underbrace{Mg,Fe,FeC{{O}_{3}},Cu{{(N{{O}_{3}})}_{2}}}_{m(g)\,\,X}+\underbrace{{{H}_{2}}S{{O}_{4}},\overbrace{NaN{{O}_{3}}}^{0,045\,\,mol}}_{dung\,\,d\grave{o}ch\,\,ho\tilde{a}n\,\,h\hat{o}\ddot{i}p}\to \underbrace{M{{g}^{2+}},F{{e}^{a+}},C{{u}^{2+}},\overbrace{N{{a}^{+}}}^{0,045\,\,mol},NH_{4}^{+},SO_{4}^{2-}}_{62,605(g)\,\,Y}+\underbrace{\overbrace{{{H}_{2}}}^{0,02\,\,mol},C{{O}_{2}},NO}_{0,17\,\,mol\,\,ho\tilde{a}n\,\,h\hat{o}\ddot{i}p\,\,Z}\)

• Cho \(\underbrace{M{{g}^{2+}},F{{e}^{a+}},C{{u}^{2+}},NH_{4}^{+},SO_{4}^{2-}}_{62,605(g)\,Y}\xrightarrow{NaOH}\underbrace{Fe{{(OH)}_{a}},Cu{{(OH)}_{2}},Mg{{(OH)}_{2}}}_{31,72(g)\downarrow }+N{{a}_{2}}S{{O}_{4}}\)

\(\Rightarrow a{{n}_{F{{e}^{a+}}}}+2{{n}_{M{{g}^{2+}}}}+2{{n}_{C{{u}^{2+}}}}+{{n}_{NH_{4}^{+}}}={{n}_{NaOH}}=0,865\,\,\,\,\,\,\,\,\left( 1 \right)\)

\(\xrightarrow{BTDT(Y)}{{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{SO_{4}^{2-}}}=0,455\,\,mol\)

\(\Rightarrow {{m}_{\downarrow \max }}=56{{n}_{F{{e}^{a+}}}}+24{{n}_{M{{g}^{2+}}}}+64{{n}_{C{{u}^{2+}}}}+17\left( {{n}_{O{{H}^{-}}}}-{{n}_{NH_{4}^{+}}} \right)\to 56{{n}_{F{{e}^{a+}}}}+24{{n}_{M{{g}^{2+}}}}+64{{n}_{C{{u}^{2+}}}}=17,015+17{{n}_{NH_{4}^{+}}}\)

• Ta có: \({{m}_{Y}}=56{{n}_{F{{e}^{a+}}}}+24{{n}_{M{{g}^{2+}}}}+64{{n}_{C{{u}^{2+}}}}+23{{n}_{N{{a}^{+}}}}+18{{n}_{NH_{4}^{+}}}+96{{n}_{SO_{4}^{2-}}}\)

\(\to 62,605=17,015+17{{n}_{NH_{4}^{+}}}+23.0,045+18{{n}_{NH_{4}^{+}}}+96.0,455\to {{n}_{NH_{4}^{+}}}=0,025(mol)\)

\(\xrightarrow{BT:H}{{n}_{{{H}_{2}}O}}=\frac{2{{n}_{{{H}_{2}}S{{O}_{4}}}}-4{{n}_{NH_{4}^{+}}}-2{{n}_{{{H}_{2}}}}}{2}=0,385\,\,mol\)

\(\xrightarrow{BTKL}{{m}_{X}}={{m}_{Y}}+{{m}_{Z}}+18{{n}_{{{H}_{2}}O}}-85{{n}_{NaN{{O}_{3}}}}-98{{n}_{{{H}_{2}}S{{O}_{4}}}}=27,2(gam)\)

• Khi cho Y tác dụng lần lượt với các dung dịch BaCl₂ và AgNO₃ thì thu được kết tủa gồm:

\(\left\{ \begin{align} & {{n}_{BaS{{O}_{4}}}}={{n}_{SO_{4}^{2-}}}={{n}_{B{{a}^{2+}}}}=0,455\,\,mol \\ & {{n}_{AgCl}}=2{{n}_{BaC{{l}_{2}}}}=0,91\,\,mol \\ \end{align} \right.\Rightarrow {{n}_{Ag}}={{n}_{F{{e}^{2+}}}}=\frac{m\downarrow -233{{n}_{BaS{{O}_{4}}}}-143,5{{n}_{AgCl}}}{108}=0,18\,(mol)\)

• Khí Z chứa các khí H₂ (0,02 mol), CO₂ (0,11 mol), NO (0,04 mol).

\(\xrightarrow{BT:N}{{n}_{Cu{{(N{{O}_{3}})}_{2}}}}=0,5({{n}_{NO}}+{{n}_{NH_{4}^{+}}}-{{n}_{NaN{{O}_{3}}}})=0,01\,\,mol.\)

Từ (1) \(\to 3{{n}_{F{{e}^{3+}}}}+2{{n}_{M{{g}^{2+}}}}=0,46\,\,\,\left( 2 \right)\)

Và \(24{{n}_{M{{g}^{2+}}}}+56{{n}_{F{{e}^{3+}}}}=6,72\,\,\,\,\left( 3 \right).\) Từ (2), (3) ta suy ra:

\(\left\{ \begin{align} & {{n}_{F{{e}^{3+}}}}:0,06 \\ & {{n}_{M{{g}^{2+}}}}:0,14 \\ \end{align} \right.\,\,\,(mol)\)

\(\Rightarrow {{n}_{Fe(X)}}=({{n}_{F{{e}^{3+}}}}+{{n}_{F{{e}^{2+}}}})-\underbrace{{{n}_{FeC{{O}_{3}}}}}_{{{n}_{C{{O}_{2}}}}}=0,06+0,18-0,11=0,13\,\,(mol).\)

Vậy \(%{{m}_{Fe}}=\frac{0,13.56}{27,2}.100%=26,76%.\)