Khi nung X: \(\left\{ \begin{gathered} {\text{2NO}}_3^ - {\text{ }}\xrightarrow{{{{\text{t}}^0}}}{\text{ 2N}}{{\text{O}}_{\text{2}}}{\text{ + }}\frac{1}{2}{{\text{O}}_{\text{2}}}{\text{ + O}}_{{\text{oxit}}}^{2 - } \hfill \\ {\text{x x x/4 x/2}} \hfill \\ {\text{CO}}_3^{2 - }{\text{ }}\xrightarrow{{{{\text{t}}^0}}}{\text{ C}}{{\text{O}}_{\text{2}}}{\text{ + O}}_{{\text{oxit}}}^{2 - } \hfill \\ {\text{x x x }} \hfill \\ \end{gathered} \right.{\text{ }}\)

Giả sử hỗn hợp khí Z gồm có NO₂ (46) và CO₂ (44), lượng khí O₂ sinh ra (x/4 mol) phản ứng hết với Fe và FeO.

- Dùng quy tắc được chéo tính nhanh được \({{\text{n}}_{{\text{N}}{{\text{O}}_{\text{2}}}}}{\text{ = }}{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}{\text{ = x (mol)}}\)

\(\underbrace {({\text{X}})\left\{ \begin{gathered} {\text{Fe}} \hfill \\ {\text{Fe(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}} \hfill \\ {\text{Fe(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}} \hfill \\ {\text{FeC}}{{\text{O}}_{\text{3}}}{\text{ }} \hfill \\ \end{gathered} \right.}_{{\text{m gam}}}{\text{ }}\xrightarrow{{{{\text{t}}^0}}}\left| \begin{gathered} ({\text{Z}})\left\{ \begin{gathered} {\text{C}}{{\text{O}}_{\text{2}}}{\text{ (x)}} \hfill \\ {\text{N}}{{\text{O}}_{\text{2}}}{\text{ (x) }} \hfill \\ \end{gathered} \right. \hfill \\ ({\text{Y}})\left\{ \begin{gathered} {\text{Fe}}{\text{, F}}{{\text{e}}^{2 + }}{\text{, F}}{{\text{e}}^{3 + }} \hfill \\ {{\text{O}}^{2 - }}{\text{ (2}}{\text{.}}\frac{{\text{x}}}{4}{\text{ + }}\frac{{\text{x}}}{2}{\text{ + x = 2x) }} \hfill \\ \end{gathered} \right.{\text{ + }}\left\{ \begin{gathered} {{\text{K}}^{\text{ + }}}{\text{ (0}}{\text{,01)}} \hfill \\ {\text{NO}}_3^ - {\text{ (0}}{\text{,01)}} \hfill \\ {{\text{H}}^ + }{\text{ (0}}{\text{,3)}} \hfill \\ {\text{SO}}_4^{2 - }{\text{ (0}}{\text{,15) }} \hfill \\ \end{gathered} \right.{\text{ }}\xrightarrow{{(2)}}\underbrace {\left\{ \begin{gathered} {{\text{K}}^{\text{ + }}}{\text{ (0}}{\text{,01)}} \hfill \\ {\text{SO}}_4^{2 - }{\text{ (0}}{\text{,15)}} \hfill \\ {\text{F}}{{\text{e}}^{{\text{n + }}}}{\text{ }} \hfill \\ \end{gathered} \right.}_{21,23{\text{ gam}}}{\text{ + hh T}}\left\{ \begin{gathered} {\text{NO (0}}{\text{,01)}} \hfill \\ {{\text{H}}_{\text{2}}}{\text{ (0}}{\text{,01) }} \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.{\text{ }}\)

(M_T = 16 ⇒ trong T có H₂ ⇒ \({\text{NO}}_3^ - \) hết, \({{\text{n}}_{{\text{NO}}}}{\text{ = }}{{\text{n}}_{{\text{NO}}_3^ - }}{\text{ = 0}}{\text{,01}}\).

Dùng quy tắc đường chéo cho hỗn hợp T ta được \({{\text{n}}_{{\text{NO}}}}{\text{ = }}{{\text{n}}_{{{\text{H}}_2}}}{\text{ = 0}}{\text{,01}}\)

Nhận xét: \({\text{m = }}{{\text{m}}_{{\text{Fe}}}}{\text{ + }}{{\text{m}}_{{\text{NO}}_3^ - }}{\text{ + }}{{\text{m}}_{{\text{CO}}_3^{2 - }}}{\text{ }}\)

* Tính m_Fe = 21,32 – (0,01.39) – (0,15.96) = 6,44 (g) (n_Fe = 0,115)

 Số mol ion H⁺ tham gia phản ứng (2): \({\text{m = }}{{\text{m}}_{{\text{Fe}}}}{\text{ + }}{{\text{m}}_{{\text{NO}}_3^ - }}{\text{ + }}{{\text{m}}_{{\text{CO}}_3^{2 - }}}{\text{ }}\)

( kiểm tra lại, nếu Fe chuyển hết thành Fe³⁺ thì mol O^2- cần là (0,115.3):2 = 0,1725 > 0,12 ⇒ O₂ hết (phù hợp với giả sử trên)

⇒  m = 6,44 + (0,06.62) + (0,06.60) = 13,76 (g)

⇒ Chọn D.