Cho 4,26 gam P2O5 vào dung dịch chứa x mol KOH và 0,04 mol K3PO4. Sau khi các phản ứng xảy ra hoàn toàn, thu được dung dịch chứa 16,64 gam hai chất tan. Giá trị của x là
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGqbWaaSbaaWqaaiaaikdaaeqaaSGaam4tamaaBaaameaa % caaI1aaabeaaaSqabaGccqGH9aqpcaaIWaGaaiilaiaaicdacaaIXa % GaaGynaiabgkziUkaad6gadaWgaaWcbaGaamisamaaBaaameaacaaI % ZaaabeaaliaadcfacaWGpbWaaSbaaWqaaiaaisdaaeqaaaWcbeaaki % abg2da9iaaicdacaGGSaGaaGimaiaaiodaaaa!4AB1! {n_{{P_2}{O_5}}} = 0,015 \to {n_{{H_3}P{O_4}}} = 0,03\)
Nếu kiềm còn dư thì trong 2 chất tan chứa \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaaIZaaabeaakiaadcfacaWGpbWaaSbaaSqaaiaaisdaaeqa % aOWaaeWaaeaacaaIWaGaaiilaiaaicdacaaIZaGaey4kaSIaaGimai % aacYcacaaIWaGaaGOmaiabg2da9iaaicdacaGGSaGaaGimaiaaiwda % aiaawIcacaGLPaaacqGHsgIRcaWGTbWaaSbaaSqaaiaadUeadaWgaa % adbaGaaG4maaqabaWccaWGqbGaam4tamaaBaaameaacaaI0aaabeaa % aSqabaGccqGH9aqpcaaIXaGaaGimaiaacYcacaaI2aGaeyOpa4JaaG % ioaiaacYcacaaIZaGaaGOmaiaacQdaaaa!567C! {K_3}P{O_4}\left( {0,03 + 0,02 = 0,05} \right) \to {m_{{K_3}P{O_4}}} = 10,6 > 8,32:\) vô lý
Vậy KOH phản ứng hết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaSGaam4taaqa % baGccqGH9aqpcaWGUbWaaSbaaSqaaiaadUeacaWGpbGaamisaaqaba % GccqGH9aqpcaWG4baaaa!4240! \to {n_{{H_2}O}} = {n_{KOH}} = x\)
Bảo toàn khối lượng:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaaIWa % GaaiilaiaaicdacaaIZaGaaiOlaiaaiMdacaaI4aGaey4kaSIaaGyn % aiaaiAdacaWG4bGaey4kaSIaaGimaiaacYcacaaIWaGaaGOmaiaac6 % cacaaIYaGaaGymaiaaikdacqGH9aqpcaaI4aGaaiilaiaaiodacaaI % YaGaey4kaSIaaGymaiaaiIdacaWG4baabaGaeyOKH4QaamiEaiabg2 % da9iaaicdacaGGSaGaaGimaiaaiodaaaaa!5334! \begin{gathered} 0,03.98 + 56x + 0,02.212 = 8,32 + 18x \hfill \\ \to x = 0,03 \hfill \\ \end{gathered} \)
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