2<\/sub><\/p>\n– Khi gi\u1ea3i b\u00e0i t\u1eadp v\u1ec1 ph\u1ea3n \u1ee9ng c\u1ee7a axit cacboxylic t\u00e1c d\u1ee5ng v\u1edbi kim lo\u1ea1i c\u1ea7n l\u01b0u \u00fd:<\/p>\n
\u00a0 \u00a0 \u00a0+ B\u1ea3o to\u00e0n nguy\u00ean t\u1ed1 H: nH2<\/sub>\u00a0= 1\/2nCOOH<\/sub><\/p>\n\u00a0 \u00a0 \u00a0+ B\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng: mmu\u1ed1i<\/sub>\u00a0= mkim lo\u1ea1i<\/sub>\u00a0+ maxit<\/sub>\u00a0– mH2<\/sub><\/p>\n\u00a0 \u00a0 \u00a0+ T\u0103ng gi\u1ea3m kh\u1ed1i l\u01b0\u1ee3ng: mmu\u1ed1i<\/sub>\u00a0– maxit<\/sub>\u00a0= mkim lo\u1ea1i<\/sub>\u00a0– mH2<\/sub><\/p>\n* V\u00cd D\u1ee4 MINH H\u1eccA<\/p>\n
V\u00ed d\u1ee5<\/strong>\u00a01:<\/strong> Cho\u00a0Na\u00a0d\u01b0 t\u00e1c d\u1ee5ng v\u1edbi\u00a0a\u00a0gam dung d\u1ecbch\u00a0CH3<\/sub>COOH\u00a0. K\u1ebft th\u00fac ph\u1ea3n \u1ee9ng, th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng\u00a0H2<\/sub>sinh ra l\u00e0\u00a011a\/240\u00a0. T\u00ednh n\u1ed3ng \u0111\u1ed9\u00a0C% c\u1ee7a dung dich axit<\/p>\nA.\u00a010% \u00a0 \u00a0 B.\u00a025% \u00a0 \u00a0 C.\u00a04,58% \u00a0 \u00a0 D.\u00a036%<\/p>\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\nCh\u1ecdn a=240 gam
\nnH2<\/sub>=11\/2=5,5 mol ; nCH3COOH <\/sub>= 240.C\/60 = 0,04C mol
\nnH2O<\/sub> = (240\u20132,4C)\/18 mol
\nPTP\u01af
\n2CH3<\/sub>COOH\u00a0 + 2NaOH \u2192 2CH3<\/sub>COONa + H2<\/sub>
\n2H2<\/sub>O + 2Na \u2192 2NaOH + H2<\/sub>
\n=> nCH3<\/sub>COOH + nH2<\/sub>O = 2nH2<\/sub><\/p>\n=> 0,04C + (240\u20132,4C)\/18 = 2.5,5 => C=25
\n=> \u0110\u00e1p \u00e1n B<\/p>\n
V\u00ed d\u1ee5<\/strong>\u00a02:<\/strong> Cho\u00a013,4\u00a0gam h\u1ed7n h\u1ee3p\u00a0X\u00a0g\u1ed3m hai axit no \u0111\u01a1n ch\u1ee9c , m\u1ea1ch h\u1edf , k\u1ebf ti\u1ebfp nhau trong c\u00f9ng d\u00e3y \u0111\u1ed3ng \u0111\u1eb3ng t\u00e1c d\u1ee5ng v\u1edbi\u00a0Na\u00a0d\u01b0, thu \u0111\u01b0\u1ee3c\u00a017,8\u00a0gam mu\u1ed1i . Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a axit c\u00f3 s\u1ed1 nguy\u00ean t\u1eed cacbon \u00edt h\u01a1n c\u00f3 trong\u00a0X\u00a0l\u00e0<\/p>\nA.\u00a03\u00a0gam \u00a0 \u00a0 B.\u00a04,6\u00a0gam \u00a0 \u00a0 C.\u00a07,4\u00a0gam \u00a0 \u00a0 D.\u00a06\u00a0gam<\/p>\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\u0110\u1eb7t c\u00f4ng th\u1ee9c ph\u00e2n t\u1eed trung b\u00ecnh c\u1ee7a hai axit l\u00e0\u00a0Cn<\/sub>H2n+1<\/sub>COOH<\/p>\n2 Cn<\/sub>H2n+1<\/sub>COOH \u00a0+ 2Na\u00a0 \u2192 \u00a02Cn<\/sub>H2n+1<\/sub>COONa+H2<\/sub><\/p>\nx\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x<\/p>\n
Theo pt v\u00e0 gi\u1ea3 thi\u1ebft ta c\u00f3<\/p>\n
(14n+67) . x \u2013 (14n+45) . x=17,8\u201313,4<\/p>\n
=>\u00a0x=0,2<\/p>\n
=> \u00a00,2(14n+46)=13,4\u00a0=>\u00a0n=1,5<\/p>\n
V\u1eady\u00a02\u00a0axit l\u00e0\u00a0CH3<\/sub>COOH\u00a0v\u00e0\u00a0C2<\/sub>H5<\/sub>COOH<\/p>\nDo\u00a0n=1,5\u00a0l\u00e0 trung b\u00ecnh c\u1ed9ng c\u1ee7a\u00a01\u00a0v\u00e0\u00a02\u00a0n\u00ean\u00a02\u00a0axit c\u00f3 s\u1ed1 mol b\u1eb1ng nhau v\u00e0 l\u00e0\u00a00,1\u00a0mol<\/p>\n
=>\u00a0mCH3<\/sub>COOH=0,1.60=6\u00a0gam<\/p>\n=> \u0110\u00e1p \u00e1n\u00a0D<\/p>\n
V\u00ed d\u1ee5 3<\/strong>: Hh X g\u1ed3m axit Y \u0111\u01a1n ch\u1ee9c v\u00e0 axit Z hai ch\u1ee9c (Y, Z c\u00f3 c\u00f9ng s\u1ed1 nguy\u00ean t\u1eed cacbon). Chia X th\u00e0nh hai ph\u1ea7n b\u1eb1ng nhau. Cho ph\u1ea7n m\u1ed9t t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi Na, sinh ra 4,48 l\u00edt kh\u00ed H2<\/sub> (\u1edf \u0111ktc).\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n ph\u1ea7n hai, sinh ra 26,4 gam CO2<\/sub>. CTCT thu g\u1ecdn v\u00e0 ph\u1ea7n tr\u0103m v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Z trong hh X l\u1ea7n l\u01b0\u1ee3t l\u00e0
\nA. HOOC-COOH v\u00e0 42,86%.<\/p>\nB. HOOC-CH2<\/sub>-COOH v\u00e0 70,87%.
\nC. HOOC-COOH v\u00e0 60,00%.<\/p>\nD. HOOC-CH2<\/sub>-COOH v\u00e0 54,88%.<\/p>\nL\u1eddi gi\u1ea3i<\/strong><\/p>\nTa c\u00f3 x + 2y = 0,4 (*)(s\u1ed1 mol HCOOH) => 2(x + y) > 0,4 (1)<\/p>\n
Ta c\u00f3 A(x+y) = 0,6 (s\u1ed1 mol C) (2)<\/p>\n
Chia t\u1eebng v\u1ebf c\u1ee7a (2) cho (1) ta \u0111\u01b0\u1ee3c:<\/p>\n
=> A\/2 < 0,6\/0,4 = 1,5 -> A = 2 v\u00ec A > 1.<\/p>\n
=> CH3<\/sub>-COOH, HOOC-COOH<\/p>\nT\u1eeb (*) (2) v\u00e0 A = 2 ta c\u00f3 h\u1ec7<\/p>\n
x + 2y = 0,4; 2(x+y) = 0,3 => y = 0,1; x = 0,2.<\/p>\n
C%(Y) = 0,1 * 90 \/ (0,1 *90 + 0.2 * 60) = 9\/21 = 42,86%<\/p>\n
\u0110\u00e1p \u00e1n A.<\/strong><\/p>\nV\u00ed d\u1ee5 4:<\/strong> Cho h\u1ed7n h\u1ee3p X g\u1ed3m ancol metylic v\u00e0 hai axit cacboxylic (no, \u0111\u01a1n ch\u1ee9c, k\u1ebf ti\u1ebfp nhau trong d\u00e3y \u0111\u1ed3ng \u0111\u1eb3ng) t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi Na, gi\u1ea3i ph\u00f3ng ra 6,72 l\u00edt kh\u00ed H2<\/sub>(\u0111ktc). N\u1ebfu \u0111un n\u00f3ng h\u1ed7n h\u1ee3p X (c\u00f3 H2<\/sub>SO4<\/sub>\u0111\u1eb7c l\u00e0m x\u00fac t\u00e1c) th\u00ec c\u00e1c ch\u1ea5t trong h\u1ed7n h\u1ee3p ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi nhau t\u1ea1o th\u00e0nh 25 gam h\u1ed7n h\u1ee3p este (hi\u1ec7u su\u1ea5t 100%). Hai axit trong h\u1ed7n h\u1ee3p X l\u00e0
\nA. HCOOH v\u00e0 CH3<\/sub>COOH.<\/p>\nB. CH3<\/sub>COOH v\u00e0 C2<\/sub>H5<\/sub>COOH.<\/p>\nC. C2<\/sub>H5<\/sub>COOH v\u00e0 C3<\/sub>H7<\/sub>COOH.<\/p>\nD. C3<\/sub>H7<\/sub>COOH v\u00e0 C4<\/sub>H9<\/sub>COOH<\/p>\nL\u1eddi gi\u1ea3i<\/strong><\/p>\nnX<\/sub>=2nH2<\/sub>=2.0,3=0,6 (mol)
\n\u0110un hh X v\u1edbi H2<\/sub>SO4\u0111<\/sub> th\u00ec c\u00e1c ch\u1ea5t trong hh ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi nhau t\u1ea1o este v\u1edbi H=100% –> nCH3OH<\/sub> = naxit<\/sub>= 0,3 (mol)
\nneste<\/sub> \u00a0= \u00a0nCH3OH<\/sub>=0,3 mol
\nG\u1ecdi CT axit: RCOOH –> CT este: RCOOCH3<\/sub>
\nMeste<\/sub> = 25\/0,3 = 83,33g<\/p>\n–> R=24,33g<\/p>\n
–> R1=15g, R2=29g (2 axit no \u0111\u1ed3ng \u0111\u1eb3ng k\u1ebf ti\u1ebfp)<\/p>\n
V\u1eady 2 axit l\u00e0 CH3<\/sub>COOH v\u00e0 C2<\/sub>H5<\/sub>COOH –>\u00a0\u0110\u00e1p \u00e1n B<\/p>\nV\u00ed d\u1ee5 5:<\/strong> A l\u00e0 h\u1ed7n h\u1ee3p g\u1ed3m r\u01b0\u1ee3u Etylic v\u00e0 2 axit h\u1eefu c\u01a1 k\u1ebf ti\u1ebfp nhau c\u00f3 d\u1ea1ng Cn<\/sub>H2n+1<\/sub>COOH v\u00e0 Cn+1<\/sub>H2n+3<\/sub>COOH. Cho 1\/2 h\u1ed7n h\u1ee3p A t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi Na tho\u00e1t ra 3,92 l\u00edt H2<\/sub>\u00a0(\u0111ktc). \u0110\u1ed1t 1\/2 h\u1ed7n h\u1ee3p A ch\u00e1y ho\u00e0n to\u00e0n, s\u1ea3n ph\u1ea9m ch\u00e1y \u0111\u01b0\u1ee3c h\u1ea5p th\u1ee5 h\u1ebft v\u00e0o dung d\u1ecbch Ba(OH)2<\/sub>\u00a0d\u01b0 th\u00ec c\u00f3 147,75g k\u1ebft t\u1ee7a v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng b\u00ecnh Ba(OH)2<\/sub>\u00a0t\u0103ng 50,1 g.<\/p>\na, T\u00ecm c\u00f4ng th\u1ee9c 2 axit tr\u00ean.<\/p>\n
b, T\u00ecm th\u00e0nh ph\u1ea7n h\u1ed7n h\u1ee3p A.<\/p>\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 nH2<\/sub>\u00a0=\u00a0\u00a0= 0,175 (mol)<\/p>\nPT ph\u1ea3n \u1ee9ng:<\/p>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2C2<\/sub>H5<\/sub>OH + 2Na\u00a0\u00a02C2<\/sub>H5<\/sub>ONa + H2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(1)<\/em><\/p>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2Cn<\/sub>H2n+1<\/sub>\u00a0COOH +2Na\u00a0\u00a02Cn<\/sub>H\u00a02n+1<\/sub>COONa + H2<\/sub>\u00a0\u00a0\u00a0\u00a0(2)<\/em><\/p>\n2Cn+1<\/sub>H2n+3<\/sub>\u00a0COOH +2Na\u00a0\u00a02Cn+1<\/sub>H2n+3<\/sub>COONa + H2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(3)<\/em><\/p>\nBi\u1ec7n lu\u1eadn theo tr\u1ecb s\u1ed1 trung b\u00ecnh.<\/p>\n
T\u1ed5ng s\u1ed1 mol 3 ch\u1ea5t trong 1\/2 h\u1ed7n h\u1ee3p = 0,175.2= 0,35 (mol)<\/p>\n
C2<\/sub>H6<\/sub>O + 3O2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02CO2<\/sub>\u00a0+ 3H2<\/sub>O\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(4)<\/em><\/p>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Cx<\/sub>H2x<\/sub>O2<\/sub>\u00a0+\u00a0O2<\/sub>\u00a0\u00a0xCO2<\/sub>\u00a0+ xH2<\/sub>O\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(5)<\/em><\/p>\nCh\u1ea5t k\u1ebft t\u1ee7a l\u00e0 BaCO3<\/sub>\u00a0\u00a0nBaCO3<\/sub>\u00a0=\u00a0\u00a0= 0,75 (mol)<\/p>\nPT: CO2<\/sub>\u00a0+ Ba(OH)2<\/sub>\u00a0\u00a0BaCO3<\/sub>\u00a0+ H2<\/sub>O (6)<\/p>\nTheo PT (6) ta c\u00f3: nCO2<\/sub>\u00a0= nBaCO3<\/sub>\u00a0= 0,75 (mol)<\/p>\n\u00a0mCO2<\/sub>\u00a0= 0,75 x44 = 33(g)<\/p>\n\u00a0mH2<\/sub>O = m t\u0103ng – mCO2<\/sub><\/p>\n\u00a0mH2<\/sub>O = 50,1 – 33 = 17,1 (g)<\/p>\n\u00a0nH2<\/sub>O =\u00a0\u00a0= 0,95 (mol)<\/p>\nT\u1eeb PT\u00a0 (4) ta th\u1ea5y ngay:<\/p>\n
S\u1ed1 mol r\u01b0\u1ee3u C2<\/sub>H5<\/sub>OH = 0,95 – 0,75 = 0,2 ( mol)<\/p>\nTheo PT (4) ta th\u1ea5y s\u1ed1 mol CO2<\/sub>\u00a0t\u1ea1o ra l\u00e0<\/p>\nnCO2\u00a0<\/sub>\u00a0<\/sub>=\u00a0 2.nC2<\/sub>H5<\/sub>OH \u00a0= 2.0,2 = 0,4 (mol)<\/p>\nSuy ra: 2 a x\u00edt ch\u00e1y t\u1ea1o ra\u00a0 0,75 – 0,4 = 0,35\u00a0 (mol CO2<\/sub>)<\/p>\nT\u1eeb PT\u00a0(4)<\/em>\u00a0ta th\u1ea5y nH2<\/sub>O = 3.nC2<\/sub>H5<\/sub>OH = 3.0,2 = 0,6 (mol)<\/p>\nSuy ra 2 axit ch\u00e1y t\u1ea1o ra: 0,95 – 0,6 = 0,35 mol H2<\/sub>O<\/p>\nV\u1edbi s\u1ed1 mol 2axit = 0,35 – 0,2 = 0,15\u00a0\u00a0x = 0,35 : 0,15 = 2,33<\/p>\n
(x l\u00e0 s\u1ed1 mol trung b\u00ecnh gi\u1eefa n+1 v\u00e0 n+2)<\/p>\n
\u00a02 axit l\u00e0 CH3<\/sub>COOH v\u00e0 C2<\/sub>H5<\/sub>COOH. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/em><\/p>\n\u00a0G\u1ecdi s\u1ed1 mol CH3<\/sub>COOH, C2<\/sub>H5<\/sub>COOH trong 1\/2 A l\u00e0 a, b.<\/p>\nTheo ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1t ch\u00e1y ta c\u00f3:<\/p>\n
S\u1ed1 mol c\u1ee7a 2 axit = 0,15mol = a + b.<\/p>\n
nCO2<\/sub>\u00a0sinh ra = 2a + 3 b = 0,35.\u00a0Gi\u1ea3i ra ta c\u00f3: a = 0,1; b = 0,05.<\/p>\nV\u1eady h\u1ed7n h\u1ee3p c\u00f3 0,2 mol CH3<\/sub>COOH l\u00e0 12 g v\u00e0 0,10 mol C2<\/sub>H5<\/sub>COOH l\u00e0\u00a0 7,4g<\/p>\n","protected":false},"excerpt":{"rendered":"Axit cacboxylic ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi kim lo\u1ea1i \u0111\u1ee9ng tr\u01b0\u1edbc H trong d\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a kim lo\u1ea1i t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 gi\u1ea3i ph\u00f3ng kh\u00ed H2 (th\u01b0\u1eddng g\u1eb7p l\u00e0 ph\u1ea3n \u1ee9ng v\u1edbi kim lo\u1ea1i ki\u1ec1m). B\u00e0i vi\u1ebft h\u01b0\u1edbng d\u1eabn b\u1ea1n \u0111\u1ecdc pp ph\u00e1p gi\u1ea3i v\u00e0 c\u00e1c ch\u00fa \u00fd khi l\u00e0m d\u1ea1ng b\u00e0i t\u1eadp […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[24],"tags":[],"yoast_head":"\n
B\u00e0i t\u1eadp axit cacboxylic t\u00e1c d\u1ee5ng v\u1edbi kim lo\u1ea1i<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/bai-tap-axit-cacboxylic-tac-dung-voi-kim-loai\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"B\u00e0i t\u1eadp axit cacboxylic t\u00e1c d\u1ee5ng v\u1edbi kim lo\u1ea1i\" \/>\n<meta property=\"og:description\" content=\"Axit cacboxylic ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi kim lo\u1ea1i \u0111\u1ee9ng tr\u01b0\u1edbc H trong d\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a kim lo\u1ea1i t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 gi\u1ea3i ph\u00f3ng kh\u00ed H2 (th\u01b0\u1eddng g\u1eb7p l\u00e0 ph\u1ea3n \u1ee9ng v\u1edbi kim lo\u1ea1i ki\u1ec1m). 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