Xu Li Mot Dang Bt Este

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Phm Hng Vng Trng i Hc Dc H Ni

X l mt dng bi tp este

Cu 1 [118133]: Cho 46,4 gam hn hp gm FeO, Fe2O3 , Fe3O4 (trong s mol FeO bng s mol Fe2O3) ho tan hon ton trong V lt dung dch H2SO4 0,2 M v HCl 0,6 M (ly d 20% th tch so vi lng cn thit). V c gi tr l A. 1,80 lt. B. 1,94 lt.

C. 1,92 lt. D. 1,56 lt.

HD: mu cht vn : Fe3O4 = FeO.Fe2O3. Nh vy, vi gi thit s mol FeO bng s mol Fe2O3 th r

ta c th gp chng thnh mt Fe3O4 na. C ngha l 46,4 gam hn hp ca chng ta ch quy i v

mi oxit Fe3O4 na thi.

Do , ta c: nFe3O4 = 0,2 mol nO = 0,8 mol; phn ng vi x mol H2SO4 v 3x mol HCl

(v chng cng chung mt dung dch m 0,6 = 3 0,2).

Bn cht ca phn ng ny l: 2H+ + O H2O. nH+ = 1,6 mol = 5x x = 0,32 mol

Vdng = 0,32 0,2 = 1,6 lt V dng = 1,6 1,2 = 1,92 lt. Chn C. .

Cu 2 [163480]: t chy hon ton 2, 6 gam hn hp gm CxHyCOOH, CxHyCOOCH3, CH3OH thu c 2,6 lt CO2 ( tc) v 1,8 gam H2O. M t hc, cho 2, 6 gam phn ng v a vi 30 ml dung dch NaOH 1M, thu c 0,96 gam CH3OH. C ng thc ca CxHyCOOH l A. C2H5COOH. B. CH3COOH.

C. C2H3COOH. D. C3H5COOH.

HD: vi bi tp ny, chng ta cng thy RCOOH; ROH v RCOOR cng c im gn gn vi

Fe3O4 = FeO. Fe2O3. RCOOR + H2O na th s bng RCOOH + ROH.

y, ta xem tip, phn ng v a vi NaOH l nhm RCOO- (d l ca axit hay este)

nRCOO = 0,03 mol; m nROH = 0,96 32 = 0,03 mol.

Nh vy, tm c mt s im tng ng vi tng ca BT v c trn, ch khc mi H2O.

T chung v ring ta hng n cch gii quyt sau. Quy 2,76 gam hn hp X trn thnh hn hp

gm 0,03 mol CxHyCOOCH3 v a mol H2O. Khi t chy: X + O2 0,12 mol CO2 + 0,1 mol H2O.

s C = 0,12 0,03 = 4 x = 2. nH = 0,2 mol. X gm 3 nguyn t C + H + O.

C bit, H bit, tng khi lng 2, 6 gam nO = 0,0 mol = a + 0,03 2 a = 0,01 mol.

S mol H trong 0,03 mol CxHyCOOCH3 = (0,2 0,01 2 = 0,1 ) y + 3 = 0,18 0,03 y = 3.

Vy cng thc cn tm l C2H3COOH. Chn p n C. .



Cu 3 [108494]: Ha tan hon ton m gam hn hp FeO, Fe2O3 v Fe3O4 bng dung dch H2SO4 long

d thu c dung dch . Chia dung dch thnh hai phn bng nhau:

Phn th nht em c cn thu c 6 ,4 gam mui han.

Phn th hai lm mt mu v a ht 46 ml dung dch KMnO4 0,5M.

Gi tr ca m l:

A. 55,12. B. 58,42.

C. 56,56. D. 60,16.

HD: Vn l hn hp ba oxit ca st: FeO; Fe2O3 v Fe3O4; vn c Fe3O4 = FeO.Fe2O3; nhng h ng c

gi thit s mol FeO v Fe2O3 bng nhau ta c th quy n gin v 1 cht.

Tuy nhin, vn ca BT ny l tch r st 2 v st 3; v th gii quyt bi ny vi iu kin khng

c gi thit s mol hai cht trn bng nhau, ta s quy ngc Fe3O4 thnh hai oxit FeO v Fe2O3.

Nh vy, ta c li, v a lm gn hn hp ch cn hai cht, v a c li trong x l gi thit.

Tht vy: .m gam hn hp FeO v Fe2O3 trong mi phn:

Phn 2: ch c Fe2+ phn ng: bo ton e nhanh: Fe2+ Fe3+ + 1e; Mn+7 + 5e Mn2+.

nFe2+ = 5.nKMnO4 = 5 0,046 0,5 = 0,115 mol. Quay li:

Phn 1: 2 4 2 4 3 2 4 3 2 4 3muoi khan Fe SO Fe (SO ) Fe (SO ) Fe (SO )

m m m 67,48 0,115 152 400n n 0,125 mol

Vy m = 2(0,115 72 + 0,125 160) = 56,56 gam p n ng l C. .

Cu 4 [34544]: Cho hn hp gm mt axit cacboxylic n chc, mt ancol n chc v este to bi axit v ancol . t chy hon ton 1,55 gam hn hp thu c 1, 36 lt CO2 ( tc) v 1,26 gam H2O. M t hc hi cho 1,55 gam hn hp tc dng v a ht vi 125 ml dung dch NaOH 0,1 M to ra m gam mui. Sau phn ng tng s gam ancol thu c l 0, 4 gam v ng vi 0,01 mol. Vy gi tr ca m l: A. 1,175 gam B. 1,205 gam

C. 1,275 gam D. 1,305 gam

HD: Mt s hng gii trc y:



Trn l mt s li gii ph bin ca cc bn hsmoon cng nh ngi ra . Tuy nhin, i

theo nhng phn tch nh trn, tng cho TH ny l tch este ra thnh ancol v axit.



Thm na, sau khi kt thc bui moontv s 8 T3 ngy 14/10/2014 ta thy rng, vic gii v x

l bi ton ny cn n gin v kh t nhin nh sau:

HD: X gm:

2

0,0125 mol 0,01 mol mol

0,14 mol0,0775 mol 0,03 mol

RCOOH + R'OH H O

= 1,55 gam C + H + O

x

Gii thch s trn: X nu nhn theo nguyn t C, H, O th da vo phn ng t chy, xc

nh c C, H s mol O; X nu nhn theo hng quy i s gm RCOOH (s mol xc nh

c t gc quy i RCOO = nNaOH = 0,0125 mol); ROH (s mol ancol cho l 0,01 mol).

Cng l X nn bo ton O: 0,01 0,0125 2 0,03 0,005 x x mol

Vy mRCOOH = 1,55 0,05 18 0,74 = 0,09 gam mRCOONa = 0,9 + 0,0125 22 = 1,175 gam.

Nh vy, gii quyt bi ny n gin bng mt s v mt php tnh l ok!.



Tng kt: qua mt s v d trn, ta tng kt li c vn g ?

X l bi ton t chy hp cht hu c ta xem xt n cc nguyn t cu thnh: C + H + O:

gi tr CO2 cho s C, H2O cho s H,... (phng php nh lng).

Cch nhn quy i, cc vn v c & hu c. Ci cui cng ta hng n l mt cch

nhn n gin ri c mt hng gii n gin, gn g v nhanh chng, r rng. Nhng phn

tch, suy din cui cng cng hng n ci nhn n gin, c bn .

[...] Cn kh nhiu khai thc th v cho dng ton ny. Cc bn c gng + th tm ti thm +

vn dng gii nhng BT khc. Tm mi lin h v t duy chung cho nhiu bi tp, dng bi

tp khc nhau.

Hi vng cc bn s c nhng pht hin th v sau bui moontv s 8 ny. !

Pht trin ln vi hn hp ancol ta c:

V d: Hn hp X gm mt axit cacboxylic no, n chc; hai ancol n chc (hn m nhau mt

nguyn t cacbon) v hai este to bi chnh axit v hai ancol trn. em t chy hon ton 5,476 gam X

th thu c 10,692 gam CO2 v 4,320 gam H2O. M t hc, hi cho cng lng X trn tc dng v a

vi 50 ml dung dch KOH 1 M th thu c dung dch Y v 2,926 gam hn hp ancol c t khi so vi

H2 l 26,600. em c cn dung dch Y th khi lng cht rn thu c l

A. 4,900 gam. B. 4,100 gam. C. 3,690 gam. D. 4,125 gam.

Pht trin thm ru ria, long ngong lm nhiu: ( i hc khi A 2014):

Cho X, Y l hai cht thuc dy ng ng ca axit acrylic v MX < MY; Z l ancol c cng s

nguyn t cacbon vi X; T l este hai chc to bi , Y v Z. t chy hon ton 11,16 gam hn hp

E gm X, Y, Z, T cn v a 13,216 lt kh O2 ( tc), thu c kh CO2 v 9,36 gam nc. M t khc

11,16 gam E tc dng ti a vi dung dch cha 0,04 mol Br2. Khi lng mui thu c khi cho cng

lng E trn tc dng ht vi dung dch KOH d l

A. 4,68 gam. B. 5,04 gam. C. 5,80 gam. D. 5,44 gam.

D nh th no nhng nu ta nhn v nhng vn c bn trn th s gii quyt c ht; khng

c vn g y c. V ng nhin, bi tp hay kh ca dng ny khng d ng li y. .

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Xu Li Mot Dang Bt Este