A. 60,5%
B. 75,0%
C. 72,5%
D. 67,5%
B
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbw5usTq % vATv2CaeXatLxBI9gBaerbd9wDYLwzYbItLDharuavP1wzZbItLDhi % s9wBH5garqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8 % qqaqFr0xc9pk0xbba9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9 % pgeaYRXxe9vr0-vr0-vqpWqaaeaabaGaaiaacaqabeaadaabauaaaO % qaamaayaaabaWaaiqaaqaabeqaaiaadgeacaWGSbaabaGaam4qaiaa % dkhaaeaacaWGdbGaamOCamaaBaaaleaacaaIYaaabeaakiaad+eada % WgaaWcbaGaaG4maaqabaaaaOGaay5EaaaaleaacaaI0aGaaGynaiaa % cYcacaaIWaGaaGOnaiaadEgaaOGaayjo+dWaa4ajaSqaaiaadshada % ahaaadbeqaaiaaicdaaaaaleqakiaawkziaiaadIfadaGabaabaeqa % baGaamyqaiaadYgadaWgaaWcbaGaaGOmaaqabaGccaWGpbWaaSbaaS % qaaiaaiodaaeqaaaGcbaGaamyqaiaadYgaaeaacaWGdbGaamOCaaqa % aiaadoeacaWGYbWaaSbaaSqaaiaaikdaaeqaaOGaam4tamaaBaaale % aacaaIZaaabeaaaaGccaGL7baadaWabaabaeqabaGaey4kaSIaamOt % aiaadggacaWGpbGaamisaiabgkziUkaadIeadaWgaaWcbaGaaGOmaa % qabaGccaGG6aGaaGimaiaacYcacaaIXaGaaG4maiaaiwdacqGHRaWk % daagaaqaamaaceaaeaqabeaacaWGdbGaamOCaiaacQdacaWG4baaba % Gaam4qaiaadkhadaWgaaWcbaGaaGOmaaqabaGccaWGpbWaaSbaaSqa % aiaaiodaaeqaaOGaaiOoaiaadMhaaaGaay5EaaaaleaacaaIYaGaaG % ioaiaacYcacaaI4aGaaGOnaaGccaGL44paaeaacqGHRaWkcaWGibGa % am4qaiaadYgacaGG6aGaaGOmaiaacYcacaaIYaGaeyOKH46aaiqaaq % aabeqaaiaadgeacaWGSbGaam4qaiaadYgadaWgaaWcbaGaaG4maaqa % baaakeaacaWGdbGaamOCaiaadoeacaWGSbWaaSbaaSqaaiaaikdaae % qaaOGaaiOoaiaadIhaaeaacaWGdbGaamOCaiaadoeacaWGSbWaaSba % aSqaaiaaiodaaeqaaOGaaiOoaiaaikdacaWG5baaaiaawUhaaaaaca % GLBbaaaaa!9411! \underbrace {\left\{ \begin{gathered} Al \hfill \\ Cr \hfill \\ C{r_2}{O_3} \hfill \\ \end{gathered} \right.}_{45,06g}\xrightarrow{{{t^0}}}X\left\{ \begin{gathered} A{l_2}{O_3} \hfill \\ Al \hfill \\ Cr \hfill \\ C{r_2}{O_3} \hfill \\ \end{gathered} \right.\left[ \begin{gathered} + NaOH \to {H_2}:0,135 + \underbrace {\left\{ \begin{gathered} Cr:x \hfill \\ C{r_2}{O_3}:y \hfill \\ \end{gathered} \right.}_{28,86} \hfill \\ + HCl:2,2 \to \left\{ \begin{gathered} AlC{l_3} \hfill \\ CrC{l_2}:x \hfill \\ CrC{l_3}:2y \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.\)
\(\begin{gathered} \xrightarrow{{BT.e}}nAl = 0,135.2:3 = 0,09\xrightarrow{{mX = 45,06}}nA{l_2}{O_3} = (45,06 - 28,86 - 0,09.27):102 = 0,135 \hfill \\ \xrightarrow{{BT.Al}}nA{l_{{\text{ ban đầu}}}} = 0,36{\text{ }}(1) \to \left\{ \begin{gathered} 52x + 152y = 28,86 \hfill \\ \xrightarrow{{BT.Cl}}2x + 6y = 2,2 - 0,36.3 \hfill \\ \end{gathered} \right. \to y = 0,13 \hfill \\ \xrightarrow{{BT.O}}nC{r_2}{O_3}{{\text{ }}_{{\text{ban đầu}}}} = 0,135 + 0,13 = 0,265{\text{ (2)}} \hfill \\ {\text{Từ (1) và (2) }} \to {\text{tính hiệu suất theo Al }} \to {\text{H = 0,135}}{\text{.2:0,36}}{\text{.100% = 75% }} \hfill \\ \end{gathered} \)
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