Nung m gam hỗn hợp X gồm Fe,

Chọn đáp án C.

Quá trình: $X\left\{\begin{array}{l}Fe;\\ Fe{\left(N{O}_3\right)}_2\\ Fe{\left(N{O}_3\right)}_3\\ FeC{O}_3\end{array}\right.\stackrel{t^o}{\to }〈\begin{array}{l}\underset{Hon\text{ hop khi Z}}{\underbrace{N{O}_2,C{O}_2}}({\overline{M}}_Z=\mathrm{45,5})\\ Y+\underset{\text{dung dich hon hop}}{\underbrace{\stackrel{\mathrm{0,04}mol}{\overbrace{NaN{O}_3}},\stackrel{\mathrm{0,92}mol}{\overbrace{KHS{O}_4}}}}\to \underset{\mathrm{21,23}gam}{\underbrace{F{e}^{n+};\stackrel{\mathrm{0,04}}{\overbrace{N{a}^{+}}};\stackrel{\mathrm{0,92}}{\overbrace{K^{+}}}:\stackrel{\mathrm{0,92}}{\overbrace{S{O}_4^{2-}}}}}+\underset{\text{hon hop khi}}{\underbrace{H_2,NO}}({\overline{M}}_{khi}=\mathrm{13,2})\end{array}$

- Ta có: $m_{F{e}^{n+}}+m_{S{O}_4^{2-}}+m_{K^{+}}+m_{N{a}^{+}}=\mathrm{143,04}\to m_{F{e}^{n+}}=\mathrm{17,92}\left(g\right)$

 $\stackrel{BT:N}{\to }{n}_{NaN{O}_3}=n_{NO}=\mathrm{0,04}mol$mà  $M=\frac{M_{H_2}+M_{NO}}{2}=\mathrm{13,2}\to n_{H_2}=\mathrm{0,06}mol$

$\stackrel{BT:\text{\hspace{0.17em}}H}{\to }{n}_{H_{2}O}=\mathrm{0,5}{n}_{KHS{O}_4}-n_{H_2}=\mathrm{0,4}mol$

$\stackrel{BT:O}{\to }{n}_{O\left(Y\right)}+3n_{NaN{O}_3}=n_{NO}+n_{H_{2}O}\to n_{O\left(Y\right)}=\mathrm{0,32}mol$

${\overline{M}}_X=\frac{M_{N{O}_2}+M_{C{O}_2}}{2}=45\to \left\{\begin{array}{l}{n}_{N{O}_2}=\mathrm{0,24}\\ n_{C{O}_2}=\mathrm{0,08}\end{array}\right.\Rightarrow m_X=m_{Fe}+62n_{N{O}_3^{-}}+60n_{C{O}_3^{2-}}=\mathrm{37,6}\left(g\right)$