Dẫn từ từ khí CO2 vào dung dịch hỗn hợp gồm NaOH và Ba(OH)2. Sự phụ thuộc của khối lượng kết tủa (y gam) vào số mol CO2 (x mol) được biểu diễn bằng đồ thị sau:
Giá trị của m là
A. 17,73. B. 7,88. C. 14,184. D. 11,82.
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Lời giải:
Báo saiĐoạn 1: CO2 + Ba(OH)2 → BaCO3 + H2O
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % yyaiabg2da9maalaaabaGaamyBaaqaaiaaigdacaaI5aGaaG4naaaa % caGG7aGaamOyaiabg2da9iaaikdacaWGHbGaai4oaiaad6gadaWgaa % WcbaGaamOqaiaadggadaqadaqaaiaad+eacaWGibaacaGLOaGaayzk % aaWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabg2da9iaaiodacaWGHb % aaaa!4BB4! \to a = \frac{m}{{197}};b = 2a;{n_{Ba{{\left( {OH} \right)}_2}}} = 3a\)
Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaaIZaGaamyyaiabgUcaRiaadkgacqGH9aqpcaaI1aGaamyyai % abgkziUkaad6gadaWgaaWcbaGaamOqaiaadggacaWGdbGaam4tamaa % BaaameaacaaIZaaabeaaaSqabaGccqGH9aqpcaaIYaGaamyyaiaacU % dacaWGUbWaaSbaaSqaaiaadkeacaWGHbWaaeWaaeaacaWGibGaam4q % aiaad+eadaWgaaadbaGaaG4maaqabaaaliaawIcacaGLPaaadaWgaa % adbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaG4maiaadggacqGHsisl % caaIYaGaamyyaiabg2da9iaadggacaGG7aGaamOBamaaBaaaleaaca % WGobGaamyyaiaadIeacaWGdbGaam4tamaaBaaameaacaaIZaaabeaa % aSqabaGccqGH9aqpcaaIYaGaamyyaaaa!64FD! {n_{C{O_2}}} = 3a + b = 5a \to {n_{BaC{O_3}}} = 2a;{n_{Ba{{\left( {HC{O_3}} \right)}_2}}} = 3a - 2a = a;{n_{NaHC{O_3}}} = 2a\)
Bảo toàn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaam4qaiabgk % ziUkaadggacqGHRaWkcaaIWaGaaiilaiaaiodacaaI2aGaeyypa0Ja % aGOmaiaadggacqGHRaWkcaaIYaGaamyyaiabgUcaRiaaikdacaWGHb % aaaa!4504! C \to a + 0,36 = 2a + 2a + 2a\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % yyaiabg2da9iaaicdacaGGSaGaaGimaiaaiEdacaaIYaGaeyOKH4Qa % amyBaiabg2da9iaaigdacaaI5aGaaG4naiaadggacqGH9aqpcaaIXa % GaaGinaiaacYcacaaIXaGaaGioaiaaisdaaaa!49DE! \to a = 0,072 \to m = 197a = 14,184\)
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