Đáp án A
$24,04\,\,A\left\{ \begin{array}{l} FeO\\ Fe{(N{O_3})_2}:0,07\\ Fe \end{array} \right.(\frac{1}{2}{n_{N{O_2}}})$$\xrightarrow{{{t}^{o}}}$$X\left\{ \begin{array}{l} Fe\\ O \end{array} \right.$$\xrightarrow[{{H}_{2}}S{{O}_{4}}:0,36]{NaN{{O}_{3}}}$$\left\{ \begin{array}{l} NO:0,05\\ {\rm{muoi}} \end{array} \right.$
BTKL: ${{m}_{X}}={{m}_{A}}-{{m}_{N{{O}_{2}}}}=17,6$
$\begin{array}{l} 4{H^ + } + NO_3^ - + 3e \to NO + 2{H_2}O\\ 2{H^ + } + O_2^ - \to {H_2}O\\ \to {n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 4{n_{NO}} + 2{n_{O\,\left( X \right)}} \to {n_{O\,\left( X \right)}} = 0,26\\ \to {n_{Fe\,\left( X \right)}} = \frac{{17,6 - 0,26.16}}{{56}} = 0,24\\ \to A\left\{ \begin{array}{l} FeO:{n_{FeO}} = {n_{O\,\left( X \right)}} + 2{n_{N{O_2}}} - 6{n_{Fe{{(N{O_3})}_2}}} = 0,12\,\,(BT\,\,O)\\ Fe{(N{O_3})_2}:0,07\\ Fe:\,0,05\,\,(BTFe) \end{array} \right.\\ {n_{HN{O_3}}} = 4{n_{NO}} + 2{n_{FeO}} = 0,64\\ \to {m_{dd\,\,HN{O_3}}} = 200 \end{array}$
BTKL: ${{m}_{A}}+{{m}_{dd\,\,HN{{O}_{3}}}}={{m}_{dd\,\,sau}}+{{m}_{NO}}\to {{m}_{dd\,sau}}=221,04$
$\begin{array}{l} Y\left\{ \begin{array}{l} Fe{(N{O_3})_2}:a\\ Fe{(N{O_3})_2}:b \end{array} \right.\\ \to \left\{ \begin{array}{l} a + b = 0,24\\ 3a + 2b = 0,07.2 + 0,64 - 0,1\,\,(BTN) \end{array} \right.\\ \to \left\{ \begin{array}{l} a = 0,04\\ b = 0,2 \end{array} \right.\\ \to C{\% _{Fe{{(N{O_3})}_3}}} = 22\% \end{array}$
