Đáp án đúng: C
Giải chi tiết:Kết tủa Y : BaSO4, Mg(OH)2
Coi hỗn hợp X gồm Mg, S
nN2 = V/22,4 =0,13 mol
Mg →Mg+2 +2e
x 2x
\(\begin{gathered} {S^0}\; \to \mathop {{\text{ }}S}\limits^{ + 6} {O_4}^{2 - } + 6{\text{e}} \hfill \\ {\text{y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{6y}} \hfill \\ {\text{2}}{{\text{N}}^{ + 5}} + 10{\text{e}} \to {\mathop N\limits^0 _2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,3\,\,\,\,\,\,0,13 \hfill \\ \end{gathered} \)
Kết tủa Y : BaSO4, Mg(OH)2
Bảo toàn nguyên tố S, Mg => nBaSO4 = y, nMg(OH)2 = x
\(\left\{ \begin{gathered} 58{\text{x}} + 233y = 46,55 \hfill \\ 2{\text{x}} + 6y = 1,3 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x = 0,2 \hfill \\ y = 0,15 \hfill \\ \end{gathered} \right.\)
=> m= mMg + mS = 0,2.24+ 0,15.32= 9,6 gam
Đáp án C
