Hydrocarbon không no

Home Lớp 11 Hoá học Lớp 11 - Sách Kết nối tri thức Bài 16: Hydrocarbon không noBài tập nâng cao

{"common":{"save":0,"post_id":"9638","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"8283","post_id":"9638","mon_id":"1159892","chapter_id":"1160152","question":"\u0110\u1ed1t cháy hoàn toàn a gam h\u1ed7n h\u1ee3p eten, propen, but-2-en c\u1ea7n dùng v\u1eeba \u0111\u1ee7 b lít oxi (\u1edf \u0111ktc) thu \u0111\u01b0\u1ee3c 2,4 mol CO2. Giá tr\u1ecb c\u1ee7a b là:<\/p>","options":["A. 92,4 lít.","B. 94,2 lít.","C. 80,64 lít.","D. 24,9 lít."],"correct":"3","level":"3","hint":"","answer":"\u0110áp án: C. 80,64 lít.<\/strong><\/p>G\u1ecdi s\u1ed1 mol: etilen = x, propen = y, but-2-en = z PT t\u1ea1o CO\u2082: 2x + 3y + 4z = 2.4 (1) PT O\u2082: 3x + 4.5y + 6z → b = (3x + 4.5y + 6z) × 22.4 (2)<\/p>T\u1eeb (1) ⇒ z = (2.4 – 2x – 3y)\/4 Th\u1ebf vào (2): b = (3x + 4.5y + 6 × (2.4 – 2x – 3y)\/4) × 22.4 = (3x + 4.5y + (14.4 – 12x – 18y)\/4) × 22.4 = (3x + 4.5y + 3.6 – 3x – 4.5y) × 22.4 = 3.6 × 22.4 = 80.64 lít<\/strong><\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-07-12 00:00:13","option_type":"txt","len":2},{"id":"8284","post_id":"9638","mon_id":"1159892","chapter_id":"1160152","question":"H\u1ed7n h\u1ee3p khí X g\u1ed3m H2 và C2H4 có t\u1ec9 kh\u1ed1i so v\u1edbi He là 3,75. D\u1eabn X qua Ni nung nóng, thu \u0111\u01b0\u1ee3c h\u1ed7n h\u1ee3p khí Y có t\u1ec9 kh\u1ed1i so v\u1edbi He là 5. Tính hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng hi\u0111ro hoá?<\/p>","options":["A. 20%.","B. 25%.","C. 50%.","D. 40%"],"correct":"3","level":"3","hint":"","answer":"\u0110áp án: C. 50%.<\/strong><\/p>G\u1ecdi H\u2082 = a, C\u2082H\u2084 = a (vì dX\/He = 3.75 ⇒ M = 15 ⇒ 2a + 28a = 15×2a ⇒ a = b)<\/p>Hi\u0111ro hóa: C\u2082H\u2084 + H\u2082 → C\u2082H\u2086, g\u1ecdi s\u1ed1 mol ph\u1ea3n \u1ee9ng là x<\/p>Sau ph\u1ea3n \u1ee9ng:<\/p><ul data-end=\"401\" data-is-last-node=\"\" data-is-only-node=\"\" data-start=\"151\"> <li data-end=\"168\" data-start=\"151\">H\u2082 còn: a – x<\/p><\/li> <li data-end=\"188\" data-start=\"169\">C\u2082H\u2084 còn: a – x<\/p><\/li> <li data-end=\"401\" data-is-last-node=\"\" data-start=\"189\">C\u2082H\u2086: x → dY\/He = 5 ⇒ M = 20 Kh\u1ed1i l\u01b0\u1ee3ng Y = 2(a – x) + 28(a – x) + 30x = 30a Mol Y = (a – x) + (a – x) + x = 2a – x ⇒ 30a \/ (2a – x) = 20 ⇒ x = 0.5a Hi\u1ec7u su\u1ea5t = x\/a × 100% = 50%<\/strong><\/p><\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-07-12 00:01:19","option_type":"txt","len":0},{"id":"8285","post_id":"9638","mon_id":"1159892","chapter_id":"1160152","question":"\u0110\u1ec3 chuy\u1ec3n hoá ankin thành anken ta th\u1ef1c hi\u1ec7n ph\u1ea3n \u1ee9ng c\u1ed9ng H2 trong \u0111i\u1ec1u ki\u1ec7n có xúc tác :<\/p>","options":["A. Ni, tº.","B. Mn, tº.","C. Pd\/ PbCO3, tº.","D. Fe, tº."],"correct":"3","level":"3","hint":"","answer":"\u0110áp án: C. Pd\/ PbCO3, tº.<\/strong><\/p>\u0110\u1ec3 chuy\u1ec3n ankin → anken<\/strong>, c\u1ea7n c\u1ed9ng H\u2082 m\u1ed9t ph\u1ea7n<\/strong> (không c\u1ed9ng hoàn toàn thành ankan).<\/p>→ Xúc tác ch\u1ecdn l\u1ecdc<\/strong> là Pd\/PbCO\u2083 (xúc tác Lindlar)<\/strong> \u0111\u1ec3 d\u1eebng l\u1ea1i \u1edf anken<\/strong> (d\u1ea1ng cis).<\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-07-12 00:03:59","option_type":"txt","len":2}]}