Đáp án đúng: C
Giải chi tiết:\(\begin{gathered} mO = {\text{20}}{\text{.}}\frac{{16,8}}{{100}} = 3,36{\text{ }}gam \to nO = 0,21{\text{ }}mol \hfill \\ 20\,gam\,X\left\{ \begin{gathered} Cu:x \hfill \\ Fe:y + z \hfill \\ O:0,21 \hfill \\ \end{gathered} \right.\xrightarrow{\begin{subarray}{l} + HCl:b \\ + HN{O_3}:0,2 \end{subarray} }\left\{ \begin{gathered} dd\,Y\left\{ \begin{gathered} C{u^{2 + }}:x \hfill \\ F{e^{2 + }}:y \hfill \\ F{e^{3 + }}:z \hfill \\ \end{gathered} \right.\xrightarrow{{ + AgN{O_3}(vd)}}\left\{ \begin{gathered} \left\{ \begin{gathered} C{u^{2 + }}:x \hfill \\ F{e^{3 + }}:y + z \hfill \\ \end{gathered} \right. \hfill \\ Ket\,tua:m\, = ? \hfill \\ \end{gathered} \right.\xrightarrow{{ + NaOH\,du}}22,4\,gam\left\{ \begin{gathered} CuO:x \hfill \\ F{e_2}{O_3}:0,5(y + z) \hfill \\ \end{gathered} \right. \hfill \\ NO:0,06 \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} {m_X} = 64x + 56(y + z) + 0,21.16 = 20 \hfill \\ BTe:2x + 2y + 3z = 0,21.2 + 0,06.3 \hfill \\ 80x + 160.0,5(y + z) = 22,4 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x = 0,12 \hfill \\ y = 0,12 \hfill \\ z = 0,04 \hfill \\ \end{gathered} \right. \hfill \\ BTNT\,N:{n_{N{O_3}^ - (muoi)}} = {n_{HN{O_3}}} - {n_{NO}} = 0,2 - 0,06 = 0,14mol \hfill \\ BTDT\,dd\,Y:{n_{C{l^ - }}} = 2{n_{C{u^{2 + }}}} + 2{n_{F{e^{2 + }}}} + 3{n_{F{e^{3 + }}}} - {n_{N{O_3}^ - }} = 0,46mol \hfill \\ C{l^ - } + A{g^ + } \to AgCl \downarrow \hfill \\ F{e^{2 + }} + A{g^ + } \to F{e^{3 + }} + Ag \downarrow \hfill \\ {n_{Ag}} = {n_{F{e^{2 + }}}} = 0,12\,mol;nAgCl = nCl - = 0,46mol \hfill \\ \to m \downarrow = mAg + mAgCl = 0,12.108 + 0,46.143,5 = 78,97gam \hfill \\ \end{gathered} \)
Đáp án C
