Dẫn 0,55 mol hỗn hợp X (gồm hơi nước và khí CO2) qua cacbon nung đỏ thu được 0,95 mol hỗn hợp Y gồm CO, H2 và CO2. Cho Y hấp thụ vo dung dịch chứa 0,1 mol Ba(OH)2 sau khi phản ứng xảy ra hoàn toàn, thu được m gam kết tủa. Giá trị của m là
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Lời giải:
Báo sainC phản ứng = nY – nX
Bảo toàn electron: 4nC phản ứng = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaad6 % gadaWgaaWcbaGaam4qaiaad+eaaeqaaOGaey4kaSIaamOBamaaBaaa % leaacaWGibWaaSbaaWqaaiaaikdaaeqaaaWcbeaaaaa!3D35! 2{n_{CO}} + {n_{{H_2}}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGdbGaam4taaqabaGccqGHRaWkcaWGUbWaaSba % aSqaaiaadIeadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaG % imaiaacYcacaaI4aGaeyOKH4QaamOBamaaBaaaleaacaWGdbGaam4t % amaaBaaameaacaaIYaaabeaalmaabmaabaGaamywaaGaayjkaiaawM % caaaqabaGccqGH9aqpcaWGUbWaaSbaaSqaaiaadMfaaeqaaOGaeyOe % I0YaaeWaaeaacaWGUbWaaSbaaSqaaiaadoeacaWGpbaabeaakiabgU % caRiaad6gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaaSqa % baaakiaawIcacaGLPaaacqGH9aqpcaaIWaGaaiilaiaaigdacaaI1a % aaaa!59AD! \to {n_{CO}} + {n_{{H_2}}} = 0,8 \to {n_{C{O_2}\left( Y \right)}} = {n_Y} - \left( {{n_{CO}} + {n_{{H_2}}}} \right) = 0,15\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGcbGaamyyaiaadoeacaWGpbWaaSbaaWqaaiaa % iodaaeqaaaWcbeaakiabg2da9iaaikdacaWGUbWaaSbaaSqaaiaadk % eacaWGHbWaaeWaaeaacaWGpbGaamisaaGaayjkaiaawMcaamaaBaaa % meaacaaIYaaabeaaaSqabaGccqGHsislcaWGUbWaaSbaaSqaaiaado % eacaWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabg2da9iaaicda % caGGSaGaaGimaiaaiwdacqGHsgIRcaWGTbWaaSbaaSqaaiaadkeaca % WGHbGaam4qaiaad+eadaWgaaadbaGaaG4maaqabaaaleqaaOGaeyyp % a0JaaGyoaiaacYcacaaI4aGaaGynaiaabccacaqGNbGaaeyyaiaab2 % gaaaa!5D39! \to {n_{BaC{O_3}}} = 2{n_{Ba{{\left( {OH} \right)}_2}}} - {n_{C{O_2}}} = 0,05 \to {m_{BaC{O_3}}} = 9,85{\text{ gam}}\)
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