Nung m gam hỗn hợp X gồm Fe, Fe(N...

C

- Hướng dẫn giải

Quá trình:  

\(X\left\{ \begin{array}{l}
Fe;\;\\
Fe{(N{O_3})_2}\\
Fe{(N{O_3})_3}\;\\
FeC{O_3}
\end{array} \right. \to \left\langle \begin{array}{l}
\underbrace {N{O_2},C{O_2}}_{Hon{\rm{ hop khi Z}}}({{\bar M}_Z} = 45,5)\\
Y + \underbrace {\overbrace {NaN{O_3}}^{0,04\,mol},\,\overbrace {KHS{O_4}}^{0,92\,mol}}_{{\rm{dung dich hon hop}}} \to \underbrace {F{e^{n + }};\overbrace {N{a^ + }}^{0,04};\overbrace {{K^ + }}^{0,92}:\overbrace {SO_4^{2 - }}^{0,92}}_{21,23\;gam} + \underbrace {{H_2},\,NO}_{{\rm{hon hop khi}}}({{\bar M}_{khi}} = 13,2)
\end{array} \right.\)

Ta có:  \({m_{F{e^{n + }}}} + {m_{SO_4^{2 - }}} + {m_{{K^ + }}} + {m_{N{a^ + }}} = 143,04 \to {m_{F{e^{n + }}}} = l7,92\;(g)\)

\( \to {n_{NaN{O_3}}} = {n_{NO}} = 0,04\;mol\) mà \(M = \frac{{{M_{{H_2}}} + {M_{NO}}}}{2} = 13,2 \to {n_{{H_2}}} = 0,06\;mol\) 

\(\begin{array}{l}
 \to {n_{{H_2}O}} = 0,5{n_{KHS{O_4}}} - {n_{{H_2}}} = 0,4\;mol\\
 \to {n_{O(Y)}} + 3{n_{NaN{O_3}}} = {n_{NO}} + {n_{{H_2}O}} \to {n_{O(Y)}} = 0,32\;mol\\
{{\bar M}_X} = \frac{{{M_{N{O_2}}} + {M_{C{O_2}}}}}{2} = 45 \to \left\{ \begin{array}{l}
{n_{N{O_2}}} = 0,24\\
{n_{C{O_2}}} = 0,08
\end{array} \right.\; \Rightarrow {m_X} = {m_{Fe}} + 62{n_{NO_3^ - }} + 60{n_{CO_3^{2 - }}} = 37,6\;(g)
\end{array}\)