Nung nóng hỗn hợp gồm FexOy và 8,64 gam Al trong khí trơ, thu được hỗn hợp rắn X. Chia X làm hai phần bằng nhau. Phần một cho vào dung dịch NaOH loãng, dư thấy thoát ra a mol khí H...

* Hướng dẫn giải

* ⁿAl = 0,16 → Muối chứa \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqqrFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabaGaaiaacaqabeaadaqaaqaaaOqaamaaceaaeaqabe % aacaWGbbGaamiBamaaBaaaleaacaaIYaaabeaakiaacIcacaWGtbGa % am4tamaaBaaaleaacaaI0aaabeaakiaacMcadaqhaaWcbaGaaG4maa % qaaaaakiaacQdacaaIWaGaaiilaiaaicdacaaI4aaabaGaamOraiaa % dwgadaWgaaWcbaGaaGOmaaqabaGccaGGOaGaam4uaiaad+eadaWgaa % WcbaGaaGinaaqabaGccaGGPaWaaSbaaSqaaiaaiodaaeqaaOGaaiOo % aiaaicdacaGGSaGaaGimaiaaisdaaaGaay5Eaaaaaa!4DC3! \left\{ \begin{gathered} A{l_2}(S{O_4})_3^{}:0,08 \hfill \\ F{e_2}{(S{O_4})_3}:0,04 \hfill \\ \end{gathered} \right.\) 

* X chứa \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqqrFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabaGaaiaacaqabeaadaqaaqaaaOqaamaaceaaeaqabe % aacaWGbbGaamiBaiaacQdacaaIWaGaaiilaiaaigdacaaI2aGaeyOe % I0IaaGOmaiaadkgaaeaacaWGbbGaamiBamaaBaaaleaacaaIYaaabe % aakiaad+eadaWgaaWcbaGaaG4maaqabaGccaGG6aGaaiOyaaqaaiaa % dAeacaWGLbGaaiOoaiaaicdacaGGSaGaaGimaiaaiIdaaaGaay5Eaa % aaaa!4AAD! \left\{ \begin{gathered} Al:0,16 - 2b \hfill \\ A{l_2}{O_3}:b \hfill \\ Fe:0,08 \hfill \\ \end{gathered} \right.\)  → 2.3.(0,16 – 2b) =  3.(0,16 – 2b) + 3.0,08 → b = 0,04

* Tỉ lệ x : y = 0,08 : 0,12 = 2 : 3