Đáp án đúng: D
13,76.
Khi nung X:
$\displaystyle \left\{ \begin{array}{l}\text{2NO}_{3}^{-}\text{ }\xrightarrow{{{\text{t}}^{0}}}\text{ 2N}{{\text{O}}_{\text{2}}}\text{ + }\frac{1}{2}{{\text{O}}_{\text{2}}}\text{ + O}_{\text{oxit}}^{2-}\\\text{x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x/4\,\,\,\,\,\,\,\,\,\,x/2}\\\text{CO}_{3}^{2-}\text{ }\xrightarrow{{{\text{t}}^{0}}}\text{ C}{{\text{O}}_{\text{2}}}\text{ + O}_{\text{oxit}}^{2-}\\\text{x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x }\end{array} \right.\text{ }$
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$\displaystyle {{\text{n}}_{\text{N}{{\text{O}}_{\text{2}}}}}\text{ = }{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{ = x (mol)}$
$\displaystyle \underbrace{(\text{X})\left\{ \begin{array}{l}\text{Fe}\\\text{Fe(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\\\text{Fe(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}\\\text{FeC}{{\text{O}}_{\text{3}}}\text{ }\end{array} \right.}_{\text{m gam}}\text{ }\xrightarrow{{{\text{t}}^{0}}}\left| \begin{array}{l}(\text{Z})\left\{ \begin{array}{l}\text{C}{{\text{O}}_{\text{2}}}\text{ (x)}\\\text{N}{{\text{O}}_{\text{2}}}\text{ (x) }\end{array} \right.\\(\text{Y})\left\{ \begin{array}{l}\text{Fe, F}{{\text{e}}^{2+}}\text{, F}{{\text{e}}^{3+}}\\{{\text{O}}^{2-}}\text{ (2}\text{.}\frac{\text{x}}{4}\text{ + }\frac{\text{x}}{2}\text{ + x = 2x) }\end{array} \right.\text{ + }\left\{ \begin{array}{l}{{\text{K}}^{\text{+}}}\text{ (0,01)}\\\text{NO}_{3}^{-}\text{ (0,01)}\\{{\text{H}}^{+}}\text{ (0,3)}\\\text{SO}_{4}^{2-}\text{ (0,15) }\end{array} \right.\text{ }\xrightarrow{(2)}\underbrace{\left\{ \begin{array}{l}{{\text{K}}^{\text{+}}}\text{ (0,01)}\\\text{SO}_{4}^{2-}\text{ (0,15)}\\\text{F}{{\text{e}}^{\text{n+}}}\text{ }\end{array} \right.}_{21,23\text{ gam}}\text{ + hh T}\left\{ \begin{array}{l}\text{NO (0,01)}\\{{\text{H}}_{\text{2}}}\text{ (0,01) }\end{array} \right.\end{array} \right.\text{ }$
(MT = 16 ⇒ trong T có H2 ⇒$\displaystyle \text{NO}_{3}^{-}$ hết,$\displaystyle {{\text{n}}_{\text{NO}}}\text{ = }{{\text{n}}_{\text{NO}_{3}^{-}}}\text{ = 0,01}$.
Dùng quy tắc đường chéo cho hỗn hợp T ta được$\displaystyle {{\text{n}}_{\text{NO}}}\text{ = }{{\text{n}}_{{{\text{H}}_{2}}}}\text{ = 0,01}$
Nhận xét:$\displaystyle \text{m = }{{\text{m}}_{\text{Fe}}}\text{ + }{{\text{m}}_{\text{NO}_{3}^{-}}}\text{ + }{{\text{m}}_{\text{CO}_{3}^{2-}}}\text{ }$
* Tính mFe = 21,32 – (0,01.39) – (0,15.96) = 6,44 (g) (nFe = 0,115)
* Tính số mol$\displaystyle \text{NO}_{3}^{-}$,$\displaystyle \text{CO}_{3}^{2-}$.
Số mol ion$\displaystyle {{\text{H}}^{+}}$ tham gia phản ứng (2):$\displaystyle {{\text{n}}_{{{\text{H}}^{+}}}}\text{ = 2}\text{.}{{\text{n}}_{{{\text{O}}^{2-}}}}\text{ + 4}\text{.}{{\text{n}}_{\text{NO}}}\text{ + 2}\text{.}{{\text{n}}_{{{\text{H}}_{\text{2}}}}}\text{ }\Rightarrow \text{ 0,3 = 2}\text{.2x + 4}\text{.0,01 + 2}\text{.0,01 }\Leftrightarrow \text{ x = 0,06 }$
( kiểm tra lại, nếu Fe chuyển hết thành$\displaystyle \text{F}{{\text{e}}^{3+}}$ thì mol$\displaystyle {{\text{O}}^{2-}}$ 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)
