Đáp án đúng: A
Giải chi tiết:\(\begin{gathered} x\,gam\left\{ \begin{gathered} Fe:x + y \hfill \\ Cu:z \hfill \\ O:t \hfill \\ \end{gathered} \right.\xrightarrow[{(1)}]{{ + HNO3}}\left\{ \begin{gathered} T\left\{ \begin{gathered} F{e^{3 + }}:x \hfill \\ F{e^{2 + }}:y \hfill \\ C{u^{2 + }}:z \hfill \\ \end{gathered} \right.\left\langle \begin{gathered} \xrightarrow[{(2)}]{{{t^o},kk}}(x + 3,84)gam\,ran\left\{ \begin{gathered} F{e_2}{O_3} \hfill \\ CuO \hfill \\ \end{gathered} \right. \hfill \\ \xrightarrow[{(3)}]{{ + 0,57\,mol\,KOH}}\left\{ \begin{gathered} Fe{(OH)_3}:x \hfill \\ Fe{(OH)_2}:y \hfill \\ Cu{(OH)_2}:z \hfill \\ \end{gathered} \right.\xrightarrow[{(4)}]{{{t^o}}}19,76g\left\{ \begin{gathered} F{e_2}{O_3}:0,5x \hfill \\ FeO:y \hfill \\ CuO:z \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \hfill \\ NO:0,15 \hfill \\ \end{gathered} \right. \hfill \\ + Coi\,qua\,trinh\,(2):x\,gam\left\{ \begin{gathered} Fe:x + y \hfill \\ Cu:z \hfill \\ O:t \hfill \\ \end{gathered} \right.\xrightarrow{{ + O}}(x + 3,84)gam\,ran\left\{ \begin{gathered} F{e_2}{O_3} \hfill \\ CuO \hfill \\ \end{gathered} \right. \hfill \\ = > {m_O} = 3,84g = > {n_O} = 3,84/16 = 0,24mol \hfill \\ BTe:3(x + y) + 2z = 2t + 2.0,24(1) \hfill \\ + {n_{KOH}} = 3x + 2y + 2z = 0,57(2) \hfill \\ + BT\,e\,(1):3x + 2y = 2z + 0,15.3(3) \hfill \\ + {m_{cran(4)}} = 160.0,5x + 72y + 80z = 19,76(4) \hfill \\ (1,2,3,4) = > x = 0,07;y = 0,03;z = 0,15;t = 0,06 \hfill \\ BTNT\,N:{n_{HN{O_3}}} = 3{n_{Fe{{(N{O_3})}_3}}} + 2{n_{Fe{{(N{O_3})}_2}}} + 2{n_{Cu{{(N{O_3})}_2}}} + {n_{NO}} \hfill \\ = 3.0,07 + 2.0,03 + 2.0,15 + 0,15 = 0,72mol \hfill \\ {m_{dd\,HN{O_3}}} = 0,72.63.\frac{{100}}{{20}} = 226,8g \hfill \\ BTKL:{m_{dd\,sau\,pu}} = 0,1.56 + 0,15.64 + 0,06.16 + 226,8 - 0,15.30 = 238,46g \hfill \\ C{\% _{Fe{{(N{O_3})}_2}}} = 0,03.180/238,46 = 2,26\% \hfill \\ \end{gathered} \)
Đáp án A