Đáp án đúng: B
14,90.
$\displaystyle \,\left\{ \begin{array}{l}{{\overline{M}}_{2\,\uparrow }}=23\\\,NO\,\end{array} \right.\Rightarrow \,\,\left\{ \begin{array}{l}{{H}_{2}}:\,\,0,01\\NO:\,\,0,03\end{array} \right.$⟹ X không có$NO_{3}^{-}$
$\displaystyle \begin{array}{l}\,\,\,\,\underbrace{Mg}_{0,095\,\,mol}+\left\{ \begin{array}{l}NaN{{O}_{3}}\\{{H}_{2}}S{{O}_{4}}\end{array} \right\}\xrightarrow{{}}\underbrace{\left\{ \begin{array}{l}M{{g}^{2+}};\,\,N{{a}^{+}}\\S{{O}_{4}}^{2-};\,\,N{{H}_{4}}^{+}\end{array} \right\}}_{dd\,\,Y}+\,\,\left\{ \begin{array}{l}{{H}_{2}}\\NO\end{array} \right\}\\\,\left\{ \begin{array}{l}BT\,\,E:\,\,2{{n}_{Mg\,\,pu}}=2{{n}_{{{H}_{2}}}}+3{{n}_{NO}}+8{{n}_{N{{H}_{4}}^{+}}}\\BT\,\,N:\,\,{{n}_{N{{O}_{3}}^{-}}}={{n}_{NO}}+{{n}_{N{{H}_{4}}^{+}}}\\BTNT:\,\,{{n}_{N{{a}^{+}}}}={{n}_{N{{O}_{3}}^{-}}}\end{array} \right.\Rightarrow \left\{ \begin{array}{l}{{n}_{N{{H}_{4}}^{+}}}=0,01\\{{n}_{N{{a}^{+}}}}={{n}_{N{{O}_{3}}^{-}}}=0,04\end{array} \right.\\\left\{ \begin{array}{l}\,trong\,\,Y:\,\,{{n}_{S{{O}_{4}}^{2-}}}=0,12\\{{m}_{muoi\,\,trong\,\,X}}={{m}_{M{{g}^{2+}}}}+{{m}_{N{{a}^{+}}}}+{{m}_{N{{H}_{4}}^{+}}}+{{m}_{S{{O}_{4}}^{2-}}}=14,9\,gam\end{array} \right.\end{array}$
