Đáp án đúng: A
Giải chi tiết:\(\begin{gathered} Ta\,co:\left\{ \begin{gathered} NO:0,001 \hfill \\ {N_2}O:0,0015 \hfill \\ \end{gathered} \right.\,va\,\left\{ \begin{gathered} Mg\,du:0,0025 \hfill \\ Ag:0,001 \hfill \\ \end{gathered} \right. \hfill \\ Mg + X\left\{ \begin{gathered} {H^ + }: \hfill \\ N{O_3}^ - : \hfill \\ A{g^ + }:0,001 \hfill \\ \end{gathered} \right. \to Y\left\{ \begin{gathered} M{g^{2 + }}:x \hfill \\ N{H_4}^ + :y \hfill \\ \xrightarrow{{BTDT}}N{O_3}^ - :2x + y \hfill \\ \end{gathered} \right. + \left\{ \begin{gathered} NO:0,001 \hfill \\ {N_2}O:0,0015 \hfill \\ Ag:0,001 \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} 24x + 18y + 62(2x + y) = 1,52 \hfill \\ \xrightarrow{{BTe}}2x = 0,001.3 + 0,0015.8 + 8y + 0,001 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x = 0,01 \hfill \\ y = 0,0005 \hfill \\ \end{gathered} \right. \hfill \\ \xrightarrow{{BTN}}{n_{N{O_3}^ - (X)}} = {n_{N{O_3}^ - (Y)}} + {n_{N{H_4}^ + }} + {n_{NO}} + 2{n_{{N_2}O}} = 0,025\,mol \hfill \\ \xrightarrow{{BTDT\,dd\,X}}{n_{{H^ + }}} = {n_{N{O_3}^ - }} - {n_{A{g^ + }}} = 0,024\,mol \hfill \\ Anot: \hfill \\ {H_2}O - 2e\,\,\,\,\,\, \to 2{H^ + } + 0,5{O_2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,0,024 \leftarrow 0,024 \hfill \\ \to t = \frac{{{n_e}.F}}{I} = \frac{{0,024.96500}}{2} = 1158(s) \hfill \\ \end{gathered} \)
Đáp án A
