Đáp án đúng: A
Giải chi tiết:
\(\begin{gathered} {n_N} = \frac{{17,44.6,422}}{{14}} = 0,08\,\,\,(mol) \hfill \\ \xrightarrow{{BTNT:\,N}}{n_{Fe{{(N{O_3})}_2}}} = \frac{1}{2}{n_N} = 0,04\,(mol) \hfill \\ \end{gathered} \)
\(17,44\,g\left\{ \begin{gathered} FeS:\,a\,mol \hfill \\ C{u_2}S:\,b\,mol \hfill \\ Fe{(N{O_3})_2}\,:\,0,04\,mol \hfill \\ \end{gathered} \right. + HN{O_3}\,du\left\langle \begin{gathered} Y\left\{ \begin{gathered} N{O_2} \hfill \\ S{O_2} \hfill \\ \end{gathered} \right. \hfill \\ Z\left\{ \begin{gathered} F{e^{3 + }} \hfill \\ C{u^{2 + }} \hfill \\ N{O_3}^ - \hfill \\ {H^ + } \hfill \\ S{O_4}^{2 - } \hfill \\ \end{gathered} \right.\xrightarrow{{ + Ba{{(OH)}_{2\,}}du}}T\underbrace {\left\{ \begin{gathered} BaS{O_4} \hfill \\ Fe{(OH)_3} \hfill \\ Cu{(OH)_2} \hfill \\ \end{gathered} \right.}_{35,4\,g}\xrightarrow{{}}E\underbrace {\left\{ \begin{gathered} BaS{O_4} \hfill \\ F{e_2}{O_3} \hfill \\ CuO \hfill \\ \end{gathered} \right.}_{31,44\,g\,} + {H_2}O \hfill \\ \end{gathered} \right.\)
\(\xrightarrow{{BTKL}}{n_{{H_2}O}} = \frac{{{m_T} - {m_E}}}{{18}} = \frac{{34,5 - 31,44}}{{18}} = 0,22(mol)\)
Khi nhiệt phân T thì BaSO4 không bị nhiệt phân
2Fe(OH)3 \(\xrightarrow{{{t^0}}}\)Fe2O3 + 3H2O
(a + 0,04 ) → 1,5. (a + 0,04 ) (mol)
Cu(OH)2 \(\xrightarrow{{{t^0}}}\)CuO + H2O
b → b (mol)
\(\begin{gathered} \left\{ \begin{gathered} {m_X} = 88a + 160b + 0,04.180 = 17,44 \hfill \\ {n_{{H_2}O}} = 1,5(a + 0,04) + 2b = 0,22 \hfill \\ \end{gathered} \right. = > \left\{ \begin{gathered} a = 0,08\,mol \hfill \\ b = \,0,02\,mol \hfill \\ \end{gathered} \right. \hfill \\ = > {n_{BaS{O_4}}} = \frac{{{m_E} - {m_{F{e_2}{O_3}}} - {m_{CuO}}}}{{233}} = \frac{{31,44 - 0,06.160 - 0,04.80}}{{233}} = 0,08\,(mol) \hfill \\ = > \% mO = \frac{{(4.0,08 + 3.0,06 + 0,04).16}}{{31,44}}.100\% = 27,48\% \hfill \\ \end{gathered} \)
Gần nhất với 27,5%
Đáp án A
