Đáp án A
$\begin{array}{l} \left\{ \begin{array}{l} NO:a\\ {H_2}:b \end{array} \right.\\ \to \left\{ \begin{array}{l} a + b = 0,16\\ 30{\rm{a}} + 2b = 11,5.2.0,16 \end{array} \right.\\ \to \left\{ \begin{array}{l} a = 0,12\\ b = 0,04 \end{array} \right. \end{array}$
$X\left\{ \begin{array}{l} F{{\rm{e}}^{3 + }}:x\\ F{{\rm{e}}^{2 + }}:y\\ M{g^{2 + }}:z\\ NH_4^ + :t\\ C{l^ - }:1,44 \end{array} \right.$ $\xrightarrow{NaOH}$ $47,24\left\{ \begin{array}{l} F{\rm{e}}{\left( {OH} \right)_3}:x\\ F{\rm{e}}{\left( {OH} \right)_2}:y\\ Mg{\left( {OH} \right)_2}:z \end{array} \right.$$\to 107{\rm{x}} + 90y + 58{\rm{z}} = 47,24$ (1)
BTĐT: $3\text{x}+2y+2\text{z}+t=1,44$ (2)
BT N: ${{n}_{F\text{e}{{\left( N{{O}_{3}} \right)}_{2}}}}=\frac{t+0,12}{2}=0,5t+0,06$
BT Fe: ${{n}_{F{{\text{e}}_{3}}{{O}_{4}}}}=\frac{x+y-0,5t-0,06}{3}$
${{n}_{{{H}^{+}}}}=8{{n}_{F{{\text{e}}_{3}}{{O}_{4}}}}+4{{n}_{NO}}+2{{n}_{{{H}_{2}}}}+10{{n}_{NH_{4}^{+}}}$
$\to 8\frac{x+y-0,5t-0,06}{3}+0,12.4+0,04.2+10t=1,44$ (3)
${{m}_{X}}=75,28\to 56\left( x+y \right)+24\text{z}+18t+1,44.35,5=75,28$ (4)
$\begin{array}{l} \to \left\{ \begin{array}{l} x = 0,14\\ y = 0,12\\ z = 0,37\\ t = 0,04 \end{array} \right. \to {n_{Ag}} = {n_{F{{\rm{e}}^{2 + }}}} = 0,12\\ \to {m_ \downarrow } = {m_{Ag}} + {m_{AgCl}}\\ = 0,12.108 + 1,44.143,5 = 219,6\left( {gam} \right) \end{array}$
