Điều chế 23 gam ancol etylic từ xelulozơ, hiệu suất thuỷ phân xelulozơ và lên men glucozơ tương ứng là 90% và 80%. Khối lượng xelulozơ cần dùng là
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaadaqada % qaaiaadoeadaWgaaWcbaGaaGOnaaqabaGccaWGibWaaSbaaSqaaiaa % igdacaaIWaaabeaakiaad+eadaWgaaWcbaGaaGynaaqabaaakiaawI % cacaGLPaaadaWgaaWcbaGaamOBaaqabaGccqGHsgIRcaWGdbWaaSba % aSqaaiaaiAdaaeqaaOGaamisamaaBaaaleaacaaIXaGaaGOmaaqaba % GccaWGpbWaaSbaaSqaaiaaiAdaaeqaaOGaeyOKH4QaaGOmaiaadoea % daWgaaWcbaGaaGOmaaqabaGccaWGibWaaSbaaSqaaiaaiwdaaeqaaO % Gaam4taiaadIeaaeaacaaIWaGaaiilaiaaikdacaaI1aGaaiOlaiaa % c6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaai % Olaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGG % UaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6 % cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOl % aiaac6cacaGGUaGaaiOlaiaaicdacaGGSaGaaGynaaaaaa!6D25! \begin{gathered} {\left( {{C_6}{H_{10}}{O_5}} \right)_n} \to {C_6}{H_{12}}{O_6} \to 2{C_2}{H_5}OH \hfill \\ 0,25.....................................0,5 \hfill \\ \end{gathered} \)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % yBamaaBaaaleaadaqadaqaaiaadoeadaWgaaadbaGaaGOnaaqabaWc % caWGibWaaSbaaWqaaiaaigdacaaIWaaabeaaliaad+eadaWgaaadba % GaaGynaaqabaaaliaawIcacaGLPaaadaWgaaadbaGaamOBaaqabaaa % leqaaOGaeyypa0ZaaSaaaeaacaaIWaGaaiilaiaaikdacaaI1aGaai % OlaiaaigdacaaI2aGaaGOmaaqaaiaaiMdacaaIWaGaaiyjaiaac6ca % caaI4aGaaGimaiaacwcaaaGaeyypa0JaaGynaiaaiAdacaGGSaGaaG % OmaiaaiwdacaWGNbGaamyyaiaad2gacaGGUaaaaa!55C5! \to {m_{{{\left( {{C_6}{H_{10}}{O_5}} \right)}_n}}} = \frac{{0,25.162}}{{90\% .80\% }} = 56,25gam.\)
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