Đáp án đúng: A
17%.
$\xrightarrow{BT:\,N}{{n}_{N{{H}_{4}}^{+}}}={{n}_{KN{{O}_{3}}}}-{{n}_{NO}}=(x-0,06)\,mol$
$\begin{array}{l}{{m}_{\text{m}\text{.khan}}}-{{m}_{H}}=39{{n}_{{{K}^{+}}}}+18{{n}_{N{{H}_{4}}^{+}}}+35,5{{n}_{C{{l}^{-}}}}-16{{n}_{O(trong\text{ H)}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=39x+18(x-0,06)+35,5.0,725-64a\end{array}$
$\Rightarrow \left\{ \begin{array}{l}{{m}_{\text{m}\text{.khan}}}-{{m}_{H}}=26,23\\10{{n}_{N{{H}_{4}}^{+}}}+2{{n}_{O(trong\text{ H)}}}+4{{n}_{NO}}+2{{n}_{{{H}_{2}}}}={{n}_{HCl}}\end{array} \right.$$\Rightarrow \left\{ \begin{array}{l}57x-64a=1,5725\\10(x-0,06)+8a+0,28=0,725\end{array} \right.\Rightarrow \left\{ \begin{array}{l}x=0,0725\,mol\\a=0,04\,mol\end{array} \right.$
$\to {{m}_{H}}=24.5a+232a=14,08\,(g)\Rightarrow {{m}_{\text{m}\text{.khan}}}=40,31\,(g)$$\Rightarrow {\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}{{m}_{Fe}}=\frac{0,04.3.56}{40,31}.100=16,67$