Lời giải chi tiết:

Đáp án: C

$$\eqalign{ & 4Fe{(N{O_3}{\rm{)}}_3}{\rm{ }}\buildrel {{t^0}} \over \longrightarrow {\rm{ 2F}}{{\rm{e}}_2}{{\rm{O}}_3}{\rm{ + 12N}}{{\rm{O}}_2} + {\rm{ 3}}{O_2} \uparrow \cr & 2Cu{(N{O_3}{\rm{)}}_2}{\rm{ }}\buildrel {{t^0}} \over \longrightarrow {\rm{ 2CuO + 4}}N{O_2} \uparrow {\rm{ }} + {\rm{ }}{O_2} \uparrow \cr & {n_{{O_2}}} = {\rm{ }}0,2{\rm{ }} = > {\rm{ }}{n_{N{O_2}}} = {\rm{ }}4{n_{{O_2}}} = {\rm{ }}0,8 \cr & = > {m_{r\raise.5ex\hbox{$\scriptstyle 3$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 3$} n}} = {\rm{ }}67,2{\rm{ }} - {\rm{ }}0,2.32{\rm{ }} - {\rm{ }}0,8.46{\rm{ }} = {\rm{ }}24{\rm{ }}(g) \cr} $$