Đáp án đúng: A
Giải chi tiết:Ta có:
\(nX = 0,36\left\{ \begin{gathered} {n_{NO}} = 0,2 \hfill \\ {n_{{H_2}}} = 0,1 \hfill \\ {n_{N{O_2}}} = 0,06 \hfill \\ \end{gathered} \right.\xrightarrow{{BTNT:N}}{n_{N{H_4}^ + }} = 0,34 - 0,2 - 0,06 = 0,08\)
Lại có: nMg(OH)2= 17,4: 58 = 0,3 (mol) =>
\(Y\left\{ \begin{gathered} M{g^{2 + }}:0,3 \hfill \\ N{H_4}^ + :0,08 \hfill \\ A{l^{3 + }}:a \hfill \\ {K^ + }:b \hfill \\ S{O_4}^{2 - }:b \hfill \\ \end{gathered} \right.\xrightarrow{{NaOH\,du:2,28\,mol\,pu}}\left\{ \begin{gathered} N{a^ + }:2,28 \hfill \\ {K^ + }:b \hfill \\ Al{O_2}^ - :a \hfill \\ S{O_4}^{2 - }:b \hfill \\ \end{gathered} \right.\)
\(\begin{gathered} \xrightarrow{{BTDT}}\left\{ \begin{gathered} 2{n_{M{g^{2 + }}}} + {n_{N{H_4}^ + }} + {n_{{K^ + }}} = 2{n_{S{O_4}^{2 - }}} \hfill \\ {n_{N{a^ + }}} + {n_{{K^ + }}} = {n_{Al{O_2}^ - }} + {n_{S{O_4}^{2 - }}} \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} 2.0,3 + 0,08 + 3a + b = 2b \hfill \\ 2,28 + b = a + 2b \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} a = 0,4 \hfill \\ b = 1,88 \hfill \\ \end{gathered} \right. \hfill \\ \xrightarrow{{BTNT:H}}{n_{{H^ + }}} = {n_{HN{O_3}}} + {n_{KHS{O_4}}} = 0,34 + 1,88 = 2,22(mol) \hfill \\ \xrightarrow{{BT\,e}}{n_{{H^ + }}} = 4{n_{NO}} + 2{n_{N{O_2}}} + 10{n_{N{H_4}^ + }} + 2{n_{{H_2}}} + 2{n_{O(trong\,MgO)}} \hfill \\ \Rightarrow 2,22 = 4.0,2 + 2.0,06 + 10.0,08 + 2.0,1 + 2{n_{O(trong\,MgO)}} \hfill \\ \Rightarrow {n_{O(trong\,MgO)}} = 0,15 \hfill \\ \% MgO = \frac{{0,15.40}}{{0,4.27 + 0,3.24 + 0,15.16}}.100\% = 29,41\% \hfill \\ \end{gathered} \)
Đáp án A
