Đáp án đúng: D
64,12%.
Đặt a, b, c là số mol Mg, MgCO3 và FeCO3
$\displaystyle \Rightarrow {{m}_{X}}=12a+84b+116c=17,6\,gam\,\,\,\,\left( 1 \right)$
${{M}_{Y}}=\frac{44.\left( {{n}_{C{{O}_{2}}}}+{{n}_{{{N}_{2}}O}} \right)+2.0,08}{{{n}_{C{{O}_{2}}}}+{{n}_{{{N}_{2}}O}}+0,08}=6.8,4\Rightarrow {{n}_{C{{O}_{2}}}}+{{n}_{{{N}_{2}}O}}=0,12\text{mol}$
$\Rightarrow {{n}_{{{N}_{2}}O}}=0,12-\left( b+c \right)$
$\xrightarrow[{}]{BTNT\text{ N}}{{n}_{NH_{4}^{+}}}={{n}_{HN{{O}_{3}}}}-2{{n}_{{{N}_{2}}O}}=0,16-2.\left( 0,12-b-c \right)=2b+2c-0,08$
${{n}_{{{H}^{+}}}}=10{{n}_{{{N}_{2}}O}}+10{{n}_{NH_{4}^{+}}}+2{{n}_{{{H}_{2}}}}$
$\Rightarrow 10.\left( 0,12-b-c \right)+10.\left( 2b+2c-0,08 \right)+2.0,08=1,12+0,16\,\,\,\,\left( 2 \right)$
${{m}_{\text{chat}\,\text{ran}}}=40.\left( a+b \right)+80c=22,8\,gam\,\,\left( 3 \right)$
Từ (1), (2), (3) suy ra:
$\left\{ \begin{array}{l}a=0,47\\b=0,02\\c=0,04\end{array} \right.\Rightarrow {\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}{{m}_{Mg}}=\frac{24.0,47}{17,6}.100{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}=64,09{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}$
Gần nhất với giá trị 64,12%