Đáp án đúng: B
Giải chi tiết:\(\begin{gathered} X\underbrace {\left\{ \begin{gathered} Al \hfill \\ MgC{O_3} \hfill \\ Fe \hfill \\ FeC{O_3} \hfill \\ \end{gathered} \right.}_{18,32(g)}\xrightarrow[{HNO3:0,25mol}]{{NaHSO4:1,22mol}}KhiY\underbrace {\left\{ \begin{gathered} C{O_2}:b \hfill \\ {N_2}:a \hfill \\ NO:2a \hfill \\ {H_2}:0,025 \hfill \\ \end{gathered} \right.}_{7,97(g)} + {H_2}O + {\text{dd}}Z\left\{ \begin{gathered} F{e^{2 + }}:x \hfill \\ F{e^{3 + }}:y \hfill \\ A{l^{3 + }}:z \hfill \\ M{g^{2 + }}:t \hfill \\ N{H_4}^ + :c \hfill \\ N{a^ + }:1,22 \hfill \\ S{O_4}^{2 - }:1,22 \hfill \\ \end{gathered} \right.\xrightarrow{{ + NaOH:1,54}}\left\{ \begin{gathered} Mg{(OH)_2} \hfill \\ Fe{(OH)_2} \hfill \\ Fe{(OH)_3} \hfill \\ \end{gathered} \right.\xrightarrow{{{t^o}}}\underbrace {\left\{ \begin{gathered} MgO:t \hfill \\ F{e_2}{O_3}:0,5(x + y) \hfill \\ \end{gathered} \right.}_{8,8(g)} \hfill \\ {m_Y} = 44b + 28a + 2a.30 + 0,025.2 = 7,97 \hfill \\ {n_{{H^ + }}} = 2b + 12a + 8a + 0,025.2 + 10c = 1,22 + 0,25 \hfill \\ BTN:c + 2a + 2a = 4a + c = 0,25 \hfill \\ \to \left\{ \begin{gathered} a = 0,06 \hfill \\ b = 0,06 \hfill \\ c = 0,01 \hfill \\ \end{gathered} \right. \to X\underbrace {\left\{ \begin{gathered} Al:z \hfill \\ MgC{O_3}:t \hfill \\ Fe:x + y + t - 0,06 \hfill \\ FeC{O_3}:0,06 - t \hfill \\ \end{gathered} \right.}_{18,32(g)} \hfill \\ BTDT\,{\text{dd}}Z:2x + 3y + 3z + 2t + 0,01 + 1,22 = 1,22.2 \to 2x + 3y + 3z + 2t = 1,21 \hfill \\ {n_{O{H^ - }}} = 2x + 3y + 4z + 2t + 0,01 = 1,54 \to 2x + 3y + 4z + 2t = 1,53 \hfill \\ {m_{cran}} = 40t + 160(0,5x + 0,5y) = 8,8 \to 80x + 80y + 40t = 8,8 \hfill \\ {m_X} = 27z + 84t + 56(x + y + t - 0,06) + 116(0,06 - t) = 18,32 \to 56x + 56y + 27z + 24t = 14,72 \hfill \\ \to \left\{ \begin{gathered} x = 0,09 \hfill \\ y = 0,01 \hfill \\ z = 0,32 \hfill \\ t = 0,02 \hfill \\ \end{gathered} \right. \to {m_{Fe}} = x + y + t - 0,06 = 0,06 \to {m_{Fe}} = 3,36(g) \to \% {m_{Fe}} = \frac{{3,36}}{{18,32}}.100\% = 18,34\% \hfill \\ \end{gathered} \)
Đáp án B
