Đáp án đúng: D

Phương pháp giải:

\(7,65(g)X\left\{ \begin{array}{l}Al\\A{l_2}{O_3}\\\% {m_{A{l_2}{O_3}}} = 40\% \end{array} \right. + Y\left\{ \begin{array}{l}{H_2}S{O_4}\\KN{O_3}\end{array} \right. \to \left\{ \begin{array}{l}m(g)hhT:\;0,015\;mol\;{H_2}\\Z\left\{ \begin{array}{l}{K^ + }\\A{l^{3 + }}\\NH_4^ + \\SO_4^{2 - }\end{array} \right. + \left[ \begin{array}{l}BaC{l_2} \to BaS{O_4}:0,4\\KOH:0,935\end{array} \right.\\{H_2}O\end{array} \right.\)- Ta có: %mAl2O3 = 40% ⟹ mAl2O3  ⟹ nAl2O3 ⟹ nAl- Từ nBaSO4 ⟹ nSO42-  ⟹ nH2SO4- Từ nKOH ⟹ nNH4+- BTĐT cho dung dịch Z ⟹ nK+ ⟹ nKNO3- BTNT H ⟹ nH2O- BTKL ⟹ mT.Giải chi tiết:\(7,65(g)X\left\{ \begin{array}{l}Al\\A{l_2}{O_3}\\\% {m_{A{l_2}{O_3}}} = 40\% \end{array} \right. + Y\left\{ \begin{array}{l}{H_2}S{O_4}\\KN{O_3}\end{array} \right. \to \left\{ \begin{array}{l}m(g)hhT:\;0,015\;mol\;{H_2}\\Z\left\{ \begin{array}{l}{K^ + }\\A{l^{3 + }}\\NH_4^ + \\SO_4^{2 - }\end{array} \right. + \left[ \begin{array}{l}BaC{l_2} \to BaS{O_4}:0,4\\KOH:0,935\end{array} \right.\\{H_2}O\end{array} \right.\)- Ta có: %mAl2O3 = 40% ⟹ mAl2O3 = 7,65.40/100 = 3,06 g ⟹ mAl =  7,65 - 3,06 = 4,59 g⟹ nAl2O3 = 3,06/102 = 0,03 mol;    nAl =  4,59/27 = 0,17 mol.⟹ nAl3+ = 2. nAl2O3 + nAl = 0,23 mol- Có: nBaSO4 = 93,2/233 = 0,4 mol ⟹ nH2SO4 = nSO42- = nBaSO4 = 0,4 mol.- Ta có: nKOH = (93,5.56/100)/56 = 0,935 mol⟹ nKOH = 4nAl3+ + nNH4+ = 4.0,23 + nNH4+ = 0,935  ⟹ nNH4+ = 0,015 mol.- BTĐT dung dịch Z: nK+ = 2nSO42- – 3nAl3+ - nNH4+ = 0,095 mol.- BTNT H: 2nH2SO4 = 2nH2 + 4nNH4+ + 2nH2O ⟹ nH2O = 0,355 mol- BTKL: mX + mY = mT + mZ + mH2O ⟹ 7,65 + (0,4.98 + 0,095.101) = mT + (0,095.39 + 0,23.27 + 0,015.18 + 0,4.96) + 0,355.18⟹ mT = 1,47 gam.