{"id":32211,"date":"2018-06-30T09:33:08","date_gmt":"2018-06-30T02:33:08","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=32211"},"modified":"2021-03-21T23:27:21","modified_gmt":"2021-03-21T16:27:21","slug":"chuong-6-bai-31-luyen-tap-anken-va-ankadien","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-6-bai-31-luyen-tap-anken-va-ankadien\/","title":{"rendered":"Ch\u01b0\u01a1ng 6: B\u00e0i 31 : Luy\u1ec7n t\u1eadp : Anken v\u00e0 anka\u0111ien"},"content":{"rendered":"

B\u00e0i 1 (trang 137 SGK H\u00f3a 11):<\/b> <\/span>Vi\u1ebft c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc minh h\u1ecda:<\/p>\n

a. \u0110\u1ec3 t\u00e1ch metan t\u1eeb h\u1ed7n h\u1ee3p v\u1edbi m\u1ed9t l\u01b0\u1ee3ng nh\u1ecf etilen, ng\u01b0\u1eddi ta d\u1eabn h\u1ed7n h\u1ee3p kh\u00ed \u0111i qua dung d\u1ecbch brom d\u01b0<\/p>\n

b. S\u1ee5c kh\u00ed propilen v\u00e0o dung d\u1ecbch KMnO_{4<\/sub>, th\u1ea5y m\u00e0u c\u1ee7a dung d\u1ecbch nh\u1ea1t d\u1ea7n c\u00f3 k\u1ebft t\u1ee7a m\u00e0u n\u00e2u \u0111en xu\u1ea5t hi\u1ec7n<\/p>\n}

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a. Khi d\u1eabn h\u1ed7n h\u1ee3p kh\u00ed (g\u1ed3m CH_{4<\/sub> v\u00e0 C_{2<\/sub>H_{4<\/sub>) qua dung d\u1ecbch brom d\u01b0 th\u00ec C_{2<\/sub>H_{4<\/sub> s\u1ebd t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch n\u01b0\u1edbc brom, CH_{4<\/sub> kh\u00f4ng t\u00e1c d\u1ee5ng s\u1ebd \u0111i ra kh\u1ecfi b\u00ecnh \u0111\u01b0\u1ee3c dung d\u1ecbch n\u01b0\u1edbc brom.<\/p>\n}}}}}}

$\"Gi\u1ea3i$ <\/p>\n

B\u00e0i 2 (trang 138 SGK H\u00f3a 11):<\/b><\/span> Tr\u00ecnh b\u00e0y ph\u01b0\u01a1ng ph\u00e1p h\u00f3a h\u1ecdc \u0111\u1ec3 ph\u00e2n bi\u1ec7t ba b\u00ecnh \u0111\u1ef1ng ba kh\u00ed ri\u00eang bi\u1ec7t l\u00e0 metan, etilen v\u00e0 cacbonic. Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc minh h\u1ecda<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

$\"Gi\u1ea3i$ <\/p>\n

B\u00e0i 3 (trang 138 SGK H\u00f3a 11):<\/b> <\/span>Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a c\u00e1c ph\u1ea3n \u1ee9ng th\u1ef1c hi\u1ec7n s\u01a1 \u0111\u1ed3 chuy\u1ec3n h\u00f3a sau:<\/p>\n

$\"Gi\u1ea3i$ <\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

$\"Gi\u1ea3i$ <\/p>\n

B\u00e0i 4 (trang 138 SGK H\u00f3a 11):<\/b> <\/span>Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a c\u00e1c ph\u1ea3n \u1ee9ng \u0111i\u1ec1u ch\u1ebf: 1,2 \u2013 \u0111icloetan; 1,1 \u2013 \u0111icloetan t\u1eeb etan v\u00e0 c\u00e1c ch\u1ea5t v\u00f4 c\u01a1 c\u1ea7n thi\u1ebft<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

$\"Gi\u1ea3i$ <\/p>\n

B\u00e0i 5 (trang 138 SGK H\u00f3a 11):<\/b> <\/span>Cho 4,48 l\u00edt kh\u00ed g\u1ed3m metan v\u00e0 etilen \u0111i qua dung d\u1ecbch brom d\u01b0, th\u1ea5y dung d\u1ecbch nh\u1ea1t m\u00e0u v\u00e0 c\u00f2n 1,12 l\u00edt kh\u00ed tho\u00e1t ra. C\u00e1c th\u1ec3 t\u00edch kh\u00ed \u0111o \u1edf \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n. Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch c\u1ee7a kh\u00ed metan trong h\u1ed7n h\u1ee3p l\u00e0:<\/p>\n

A. 25% ; B. 50% ; C. 60% ; D. 37,5%<\/p>\n

H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– \u0110\u00e1p \u00e1n A<\/b><\/p>\n

– Khi cho h\u1ed7n h\u1ee3p kh\u00ed (g\u1ed3m CH_{4<\/sub> v\u00e0 C_{2<\/sub>H_{4<\/sub> ) \u0111i qua dung d\u1ecbch brom th\u00ec C_{2<\/sub>H_{4<\/sub> s\u1ebd ph\u1ea3n \u1ee9ng h\u1ebft v\u1edbi dung d\u1ecbch brom, CH_{4<\/sub> kh\u00f4ng t\u00e1c d\u1ee5ng s\u1ebd \u0111i ra kh\u1ecfi b\u00ecnh.<\/p>\n}}}}}}

B\u00e0i 7 (trang 138 SGK H\u00f3a 11):<\/b> <\/span>\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n 5,40g anka\u0111ien li\u00ean h\u1ee3p X thu \u0111\u01b0\u1ee3c 8,96 l\u00edt kh\u00ed CO_{2<\/sub> (\u0111ktc). C\u00f4ng th\u1ee9c n\u00e0o sau \u0111\u00e2y l\u00e0 c\u00f4ng th\u1ee9c c\u1ea5u t\u1ea1o c\u1ee7a X ?<\/p>\n}

$\"Gi\u1ea3i$ <\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– \u0110\u00e1p \u00e1n A<\/b><\/p>\n

– Ta c\u00f3:<\/p>\n

$\"Gi\u1ea3i$ <\/p>\n

=> m_{C<\/sub> = 12.0,4 = 4,8 (gam)<\/p>\n}

=> m_{H<\/sub> = 5,4 – 4,8 = 0,6 (gam)<\/p>\n}

=> n_{H<\/sub> = 0,6 (mol)<\/p>\n}

X\u00e9t n_{C<\/sub> : n_{H<\/sub> = 0,4 : 0,6 = 4 : 6 = 2 : 3 (\u0111\u00e2y ch\u00ednh l\u00e0 t\u1ec9 l\u1ec7 s\u1ed1 nguy\u00ean t\u1eed C so v\u1edbi s\u1ed1 nguy\u00ean t\u1eed H)<\/p>\n}}

=> \u0110\u00e1p \u00e1n A ho\u1eb7c D \u0111\u00fang<\/p>\n

V\u00ec X l\u00e0 anka\u0111ien li\u00ean h\u1ee3p \u0111\u00e1p \u00e1n A \u0111\u00fang<\/p>\n

(B\u00e0i n\u00e0y kh\u00f4ng c\u1ea7n vi\u1ebft ptp\u01b0)<\/p>\n\n\n