{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" R\u01b0\u1ee3u etylic v\u00e0 axit axetic c\u00f3 c\u00f4ng th\u1ee9c ph\u00e2n t\u1eed l\u1ea7n l\u01b0\u1ee3t l\u00e0 ","select":["A. $C_2H_6O_2, C_2H_4O_2$.","B. $C_3H_6O, C_2H_4O_2$.","C. $C_2H_6O, C_3H_4O_2$.","D. $C_2H_6O, C_2H_4O_2$."],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic: $C_2H_6O$; Axit axetic: $C_2H_4O_2$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2531},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" C\u00e1c ch\u1ea5t \u0111\u1ec1u ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi Na v\u00e0 K l\u00e0","select":["A. r\u01b0\u1ee3u etylic, axit axetic. ","B. benzen, axit axetic","C. r\u01b0\u1ee3u etylic, benzen","D. d\u1ea7u ho\u1ea3, r\u01b0\u1ee3u etylic"],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic, axit axetic \u0111\u1ec1u t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi Na, K<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2532},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi natri cacbonat t\u1ea1o ra kh\u00ed cacbonic l\u00e0 ","select":["A. n\u01b0\u1edbc. ","B. r\u01b0\u1ee3u etylic.","C. axit axetic.","D. r\u01b0\u1ee3u etylic v\u00e0 axit axetic"],"hint":"","explain":"<span class='basic_left'>$2C{{H}_{3}}COOH+N{{a}_{2}}C{{O}_{3}}\\to 2C{{H}_{3}}COONa+C{{O}_{2}}+{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2533},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u01b0\u01a1ng ph\u00e1p l\u00ean men dung d\u1ecbch r\u01b0\u1ee3u etylic lo\u00e3ng d\u00f9ng \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf:","select":["A. Etilen","B. Natri axetat","C. Axit axetic ","D. Etyl axetat"],"hint":"","explain":"<span class='basic_left'>Axit axetic \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p l\u00ean men r\u01b0\u1ee3u etylic<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2534},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ch\u1ea5t h\u1eefu c\u01a1 X \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch cho etilen t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc c\u00f3 axit l\u00e0m x\u00fac t\u00e1c. V\u1eady X l\u00e0:","select":["A. $CH_3COOH$","B. $CH_3OH$","C. $C_2H_5OH$","D. $C_3H_7OH$"],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch cho etilen h\u1ee3p n\u01b0\u1edbc c\u00f3 axit l\u00e0m x\u00fac t\u00e1c.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2535},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Etyl axetat thu\u1ed9c lo\u1ea1i h\u1ee3p ch\u1ea5t n\u00e0o d\u01b0\u1edbi \u0111\u00e2y","select":["A. R\u01b0\u1ee3u","B. Axit","C. Este ","D. Ete"],"hint":"","explain":"<span class='basic_left'>Etyl axetat thu\u1ed9c lo\u1ea1i h\u1ee3p ch\u1ea5t este<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2536},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nh\u00fang qu\u1ef3 t\u00edm v\u00e0o axit axetic hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. Qu\u1ef3 chuy\u1ec3n xanh","B. Qu\u1ef3 kh\u00f4ng chuy\u1ec3n m\u00e0u","C. Qu\u1ef3 chuy\u1ec3n \u0111\u1ecf ","D. Qu\u1ef3 m\u1ea5t m\u00e0u"],"hint":"","explain":"<span class='basic_left'>Axit axetic c\u00f3 t\u00ednh axit n\u00ean l\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n \u0111\u1ecf<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2537},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng kim lo\u1ea1i Na c\u1ea7n l\u1ea5y \u0111\u1ec3 t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi 69 gam r\u01b0\u1ee3u etylic l\u00e0:","select":["A. 34,5gam","B. 44,3gam ","C. 54,3gam ","D. 63,5gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}} \\\\ & {{n}_{Na}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=1,5\\,mol\\Rightarrow {{m}_{Na}}=34,5\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2538},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" S\u1ed1 ml r\u01b0\u1ee3u etylic c\u00f3 trong 500 ml r\u01b0\u1ee3u $40^0$ l\u00e0:","select":["A. 20 ml","B. 200 ml","C. 2 ml ","D. 0,2 ml"],"hint":"","explain":"<span class='basic_left'>${{V}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{500.40}{100}=200\\,ml$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2539},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Vai tr\u00f2 c\u1ee7a $H_2SO_4$ trong ph\u1ea3n \u1ee9ng r\u01b0\u1ee3u etylic t\u00e1c d\u1ee5ng v\u1edbi axit axetic l\u00e0","select":["A. L\u00e0m ch\u1ea5t ph\u1ea3n \u1ee9ng ","B. L\u00e0m ch\u1ea5t x\u00fac t\u00e1c v\u00e0 h\u00fat n\u01b0\u1edbc","C. L\u00e0m ch\u1ea5t h\u00fat n\u01b0\u1edbc","D. Ph\u1ea3n \u1ee9ng v\u1edbi ch\u1ea5t s\u1ea3n ph\u1ea9m."],"hint":"","explain":"<span class='basic_left'>Vai tr\u00f2 c\u1ee7a $H_2SO_4$ trong ph\u1ea3n \u1ee9ng r\u01b0\u1ee3u etylic t\u00e1c d\u1ee5ng v\u1edbi axit axetic l\u00e0: l\u00e0m ch\u1ea5t x\u00fac t\u00e1c v\u00e0 h\u00fat n\u01b0\u1edbc<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2540},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong c\u00e1c ch\u1ea5t sau ch\u1ea5t n\u00e0o l\u00e0 ch\u1ea5t kh\u00ed \u1edf \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng:","select":["A. $C_2H_5OH$","B. $C_2H_4$","C. $CH_3COOH$","D. $CH_3COOC_2H_5$"],"hint":"","explain":"<span class='basic_left'>$C_2H_4$ \u1edf th\u1ec3 kh\u00ed<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2541},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u1ee3p ch\u1ea5t n\u00e0o sau \u0111\u00e2y l\u00e0 ch\u1ea5t l\u1ecfng, \u00edt tan trong n\u01b0\u1edbc","select":["A. $C_2H_5OH$ ","B. $C_2H_4$ ","C. $CH_3COOC_2H_5$","D. $CH_3COOH$ "],"hint":"","explain":"<span class='basic_left'>$CH_3COOC_2H_5$ \u1edf th\u1ec3 l\u1ecfng v\u00e0 \u00edt tan trong n\u01b0\u1edbc.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2542},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 10 gam h\u1ed7n h\u1ee3p g\u1ed3m r\u01b0\u1ee3u etylic v\u00e0 axit axetic tham gia ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi 7,42 gam $Na_2CO_3$. Th\u00e0nh ph\u1ea7n % kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i ch\u1ea5t c\u00f3 trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0 ","select":["A. $CH_3COOH$ (16%), $C_2H_5OH$ (84%).","B. $CH_3COOH$ (58%), $C_2H_5OH$ (42%).","C. $CH_3COOH$ (84%), $C_2H_5OH$ (16%). ","D. $CH_3COOH$ (42%), $C_2H_5OH$ (58%)."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2C{{H}_{3}}COOH+N{{a}_{2}}C{{O}_{3}}\\to 2C{{H}_{3}}COONa+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{C{{H}_{3}}COOH}}=2{{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,14\\,mol\\Rightarrow {{m}_{C{{H}_{3}}COOH}}=8,4\\,gam \\\\ & \\%{{m}_{C{{H}_{3}}COOH}}=\\dfrac{8,4}{10}.100\\%=84\\%\\Rightarrow \\%{{m}_{{{C}_{2}}{{H}_{5}}OH}}=16\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2543},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 60 gam axit axetic t\u00e1c d\u1ee5ng v\u1edbi 55,2 gam r\u01b0\u1ee3u etylic t\u1ea1o ra 55 gam etyl axetat. Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 65,2 %.","B. 62,5 %.","C. 56,2%. ","D. 72,5%."],"hint":"","explain":"<span class='basic_left'>${{n}_{C{{H}_{3}}COOH}}=1\\,mol;\\,\\,\\,{{n}_{{{C}_{2}}{{H}_{5}}OH}}=1,2\\,mol;\\,{{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=0,625\\,mol$ <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai46/lv2/img\/Capture14.PNG' \/><\/center> <br\/>X\u00e9t t\u1ef7 l\u1ec7 <br\/> $\\dfrac{{{n}_{C{{H}_{3}}COOH}}}{1}<\\dfrac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{1}$ => $CH_3COOH$ h\u1ebft, t\u00ednh hi\u1ec7u su\u1ea5t theo $CH_3COOH$ <br\/> $\\begin{aligned} & {{n}_{C{{H}_{3}}COOH\\,\\,(\\text{ph\u1ea3n \u1ee9ng})}}={{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=0,625\\,mol \\\\ & \\Rightarrow H=\\dfrac{{{n}_{C{{H}_{3}}COOH\\,(\\text{ph\u1ea3n \u1ee9ng})}}}{{{n}_{C{{H}_{3}}COOH}}}=\\dfrac{0,625}{0,1}.100\\%=62,5\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2544},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 23 gam r\u01b0\u1ee3u etylic v\u00e0o dung d\u1ecbch axit axetic d\u01b0. Kh\u1ed1i l\u01b0\u1ee3ng etyl axetat thu \u0111\u01b0\u1ee3c l\u00e0 (bi\u1ebft hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng 30%) ","select":["A. 26,4 gam.","B. 13,2 gam. ","C. 36,9 gam.","D. 32,1 gam."],"hint":"","explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai46/lv2/img\/Capture14.PNG' \/><\/center> <br\/> $\\begin{aligned} & {{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,5\\,mol \\\\ & \\Rightarrow {{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\\,(LT)}}=44\\,gam \\\\ & H=30\\%\\Rightarrow {{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\\,\\,(TT)}}=\\dfrac{44.30\\%}{100\\%}=13,2\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2545},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ef1c hi\u1ec7n ph\u1ea3n \u1ee9ng este h\u00f3a ho\u00e0n to\u00e0n m gam $CH_3COOH$ b\u1eb1ng m\u1ed9t l\u01b0\u1ee3ng v\u1eeba \u0111\u1ee7 $C_2H_5OH$ thu \u0111\u01b0\u1ee3c 0,02 mol este. m c\u00f3 gi\u00e1 tr\u1ecb l\u00e0:","select":["A. 2,1 g ","B. 1,1 g","C. 1,2 g ","D. 1,4 g"],"hint":"","explain":"<span class='basic_left'><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai46/lv2/img\/Capture14.PNG' \/><\/center> <br\/> $\\begin{aligned} & {{n}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=0,02\\,mol \\\\ & \\Rightarrow {{m}_{C{{H}_{3}}COOH}}=1,2\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2546},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng axit axetic \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh khi l\u00ean men 14,375 ml r\u01b0\u1ee3u $40^0$ (bi\u1ebft D=0,8g\/ml).","select":["A. 4 gam","B. 5 gam","C. 6 gam ","D. 7 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{V}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{14,375.40}{100}=5,75\\,ml \\\\ & \\Rightarrow {{m}_{{{C}_{2}}{{H}_{5}}OH}}=5,75.0,8=4,6\\,gam \\\\ & {{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,1\\,mol \\\\ & {{C}_{2}}{{H}_{5}}OH+{{O}_{2}}\\xrightarrow{\\text{men}\\,\\text{gi\u1ea5m}}C{{H}_{3}}COOH+{{H}_{2}}O \\\\ & {{n}_{C{{H}_{3}}COOH}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,1\\,mol\\Rightarrow {{m}_{C{{H}_{3}}COOH}}=6\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2547},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 6,72 l\u00edt kh\u00ed etilen h\u1ee3p n\u01b0\u1edbc c\u00f3 axit x\u00fac t\u00e1c thu \u0111\u01b0\u1ee3c bao nhi\u00eau ml dung d\u1ecbch r\u01b0\u1ee3u etylic $60^0$ bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a r\u01b0\u1ee3u l\u00e0 0,8 gam\/ml.","select":["A. 17,25 ml ","B. 25,75 ml","C. 28,75 ml","D. 34,5 ml"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{4}}+{{H}_{2}}O\\xrightarrow{xt}{{C}_{2}}{{H}_{5}}OH \\\\ & {{n}_{{{C}_{2}}{{H}_{4}}}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,3\\,\\,mol\\Rightarrow {{m}_{{{C}_{2}}{{H}_{5}}OH}}=13,8\\,gam \\\\ & {{V}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{13,8}{0,8}=17,25\\,ml\\Rightarrow {{V}_{\\text{dd}\\,{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{17,25.100}{60}=28,75\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2548},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n 9 gam h\u1ee3p ch\u1ea5t h\u1eefu c\u01a1 X ch\u1ee9a C, H v\u00e0 O thu \u0111\u01b0\u1ee3c 19,8 gam kh\u00ed $CO_2$ v\u00e0 10,8 gam $H_2O$. V\u1eady X l\u00e0 ","select":["A. $C_2H_5OH$.","B. $CH_3COOH$.","C. $C_3H_8O$.","D. $CH_4O$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{C}}={{n}_{C{{O}_{2}}}}=0,45\\,mol \\\\ & {{n}_{H}}=2{{n}_{{{H}_{2}}O}}=1,2\\,mol \\\\ & \\Rightarrow {{m}_{O}}=9-{{m}_{C}}-{{m}_{H}}=9-0,45.12-1,2.1=2,4\\,gam \\\\ & \\Rightarrow {{n}_{O}}=0,15\\,mol \\\\ & \\Rightarrow {{n}_{C}}:{{n}_{H}}:{{n}_{O}}=0,45:1,2:0,15=3:8:1 \\\\ \\end{aligned}$ <br\/> Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2549},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ed1t a gam $C_2H_5OH$ thu \u0111\u01b0\u1ee3c 0,2 mol $CO_2$ . \u0110\u1ed1t b gam $CH_3COOH$ thu \u0111\u01b0\u1ee3c 0,2 mol $CO_2$ . Cho a gam $C_2H_5OH$ t\u00e1c d\u1ee5ng v\u1edbi b gam $CH_3COOH$ c\u00f3 x\u00fac t\u00e1c l\u00e0 $H_2SO_4$ \u0111\u1eb7c ( gi\u1ea3 s\u1eed hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0 100%) thu \u0111\u01b0\u1ee3c c gam este. c c\u00f3 gi\u00e1 tr\u1ecb l\u00e0:","select":["A. 4,4 g","B. 8,8 g","C. 13,2 g","D. 17,6 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{5}}OH+3{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O \\\\ & {{n}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{1}{2}{{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & C{{H}_{3}}COOH+2{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+2{{H}_{2}}O \\\\ & {{n}_{C{{H}_{3}}COOH}}=\\dfrac{1}{2}{{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ \\end{aligned}$ <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai46/lv2/img\/Capture14.PNG' \/><\/center> <br\/> X\u00e9t t\u1ec9 l\u1ec7: <br\/> $\\dfrac{{{n}_{C{{H}_{3}}COOH}}}{1}=\\dfrac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{1}$ => C\u1ea3 2 ch\u1ea5t \u0111\u1ec1u ph\u1ea3n \u1ee9ng h\u1ebft. <br\/> $\\begin{aligned} & {{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{C{{H}_{3}}COOH}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,1\\,mol \\\\ & \\Rightarrow {{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=0,1.88=8,8 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2550}],"lesson":{"save":0,"level":2}}