10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 1 10 PHNG PHP GII NHANH BI TP TRC NGHIM HA HC Phng php 1 P DNG NH LUT BO TON KHI LNGNguyn tc ca phng phpny kh n gin,da vo nhlut bo ton khilng:Tng khi lng cc cht tham gia phn ng bng tng khi lng cc cht to thnh trong phn ng. Cn lu l: khng tnh khi lng ca phn khng tham gia phn ng cng nh phn cht c sn, v d nc c sn trong dung dch. Khi c cn dung dch th khi lng mui thu c bng tng khi lng cc cation kim loi v anion gc axit. V d 1: Hn hp X gm Fe, FeO v Fe2O3. Cho mt lung CO i qua ng s ng m gam hn hp X nung nng. Sau khi kt thc th nghim thu c 64 gam cht rn A trong ng s v 11,2 lt kh B (ktc) c t khi so vi H2 l 20,4. Tnh gi tr m. A. 105,6 gam.B. 35,2 gam.C. 70,4 gam.D. 140,8 gam. Hng dn gii Cc phn ng kh st oxit c th c: 3Fe2O3 + COot2Fe3O4 + CO2 (1) Fe3O4 + COot3FeO + CO2(2) FeO + COotFe + CO2(3) Nh vy cht rn A c th gm 3 cht Fe, FeO, Fe3O4 hoc t hn, iu khng quan trng v vic cnbng cc phng trnh trn cng khngcn thit, quan trngl smol CO phn ngbao gi cng bng s mol CO2 to thnh. B11,2n 0,522,5= = mol. Gi x l s mol ca CO2 ta c phng trnh v khi lng ca B: 44x + 28(0,5 x) = 0,5 20,4 2 = 20,4 nhn c x = 0,4 mol v cng chnh l s mol CO tham gia phn ng. Theo LBTKL ta c:mX + mCO = mA + 2 COmm = 64 + 0,4 44 0,4 28 = 70,4 gam. (p n C) V d 2: un 132,8 gam hn hp 3 ru no, n chc vi H2SO4 c 140oC thu c hn hp cc ete c s mol bng nhau v c khi lng l 111,2 gam. S mol ca mi ete trong hn hp l bao nhiu? 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 2 A. 0,1 mol.B. 0,15 mol. C. 0,4 mol.D. 0,2 mol. Hng dn gii Ta bit rng c 3 loi ru tch nc iu kin H2SO4 c, 140oC th to thnh 6 loi ete v tch ra 6 phn t H2O. Theo LBTKL ta c 2 H O etem m m 132,8 11,2 21,6 = = =ru gam 2 H O21,6n 1, 218= = mol. Mt khc c hai phn t ru th to ra mt phn t ete v mt phn t H2O do s mol H2O lun bng s mol ete, suy ra s mol mi ete l 1,20, 26 = mol. (p n D) Nhn xt: Chng ta khng cn vit 6 phng trnh phn ng t ru tch nc to thnh 6 ete, cngkhngcntmCTPTcaccruvccetetrn.Nuccbnxavovicvitphng trnh phn ngv t nsmol cc ete tnh ton th khngnhng khng gii cm cn tn qu nhiu thi gian. V d 3: Cho 12 gam hn hp hai kim loi Fe, Cu tc dng va vi dung dch HNO3 63%. Sau phn ng thu c dung dchAv 11,2lt kh NO2 duy nht (ktc). Tnhnng % cc cht c trong dung dch A. A. 36,66% v 28,48%.B. 27,19% v 21,12%. C. 27,19% v 72,81%.D. 78,88% v 21,12%. Hng dn gii Fe+6HNO3Fe(NO3)3+3NO2+3H2O Cu+4HNO3Cu(NO3)2 +2NO2+2H2O 2 NOn 0,5 = mol3 2 HNO NOn 2n 1 = = mol. p dng nh lut bo ton khi lng ta c: 22 3NO d HNOm m m m1 63 10012 46 0,5 89 gam.63= + = + =2 2d mui h k.loi tnFe = x mol, nCu = y mol ta c: 56x 64y 123x 2y 0,5+ = + = x 0,1y 0,1= = 3 3 Fe( NO )0,1 242 100%m 27,19%89 = = 3 2 Cu( NO )0,1 188 100%m 21,12%.89 = =(p n B) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 87 A. 10,5.B. 13,5.C. 14,5.D. 16. Hng dn gii Xt 100 gam hn hp X ta c mC = 3,1 gam, 3 Fe Cm= a gamv s gam Fe tng cng l 96 gam. ( ) 3 C trong Fe C12am 100 96 3,1180= =a = 13,5. (p n B) V d 15: Nungm gam Xcha 80% khilng gam CaCO3 (phn cnlil tp cht tr) mt thi gian thu c cht rn Y cha 45,65 % CaO. Tnh hiu sut phn hy CaCO3.A. 50%.B. 75%.C. 80%.D. 70%. Hng dn giiChn mX = 100 gam3 CaCOm 80 gam =v khi lng tp cht bng 20 gam. CaCO3 ot CaO+CO2(hiu sut = h) Phng trnh:100 gam56 gam 44 gam Phn ng:80 gam56.80.h100 44.80. h100 Khi lng cht rn cn li sau khi nung l 2 X CO44.80.hm m 100100 = . 56 80 45,65 44 80 hh 100100 100 100 | | = |\ . h = 0,75hiu sut phn ng bng 75%. (p n B) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 86 V d 13: Trn a gamhnhpX gm 2hirocacbonC6H14 vC6H6 theo tl smol (1:1)vim gammthirocacbonDritchyhontonththuc 2275agam CO82v 94,5a82 gam H2O. a) D thuc loi hirocacbon no A. CnH2n+2. B. CmH2m2. C. CnH2n.D. CnHn. b) Gi tr m lA. 2,75 gam.B. 3,75 gam.C. 5 gam.D. 3,5 gam. Hng dn gii a) Chn a = 82 gam t X v m gam D (CxHy) ta c: 22COH O275n 6, 25 mol4494,5n 5,25 mol18= == = C6H14 + 192O2 6CO2 +7H2O C6H6+ 152O2 6CO2 +3H2O t D: x y 2 2 2y yC H x O xCO H O4 2| |+ + + |\ . t 6 14 6 6 C H C Hn n b mol = =ta c: 86b + 78b = 82b = 0,5 mol. t 82 gam hn hp X thu c: ( )2 COn 0,5 6 6 6 mol = + =( )2 H On 0,5 7 3 5 mol = + = t chy m gam D thu c: 2 COn 6,25 6 0,25 mol = = 2 H On 5,25 5 0, 25 mol = =Do 2 2 CO H On n =D thuc CnH2n. (p n C) b) mD = mC + mH=0,25(12 + 2)=3,5 gam. (p n D) Vd14:Xlhpkimgm(Fe,C,Fe3C),tronghmlngtngcngcaFel96%,hm lng C n cht l 3,1%, hm lng Fe3C l a%. Gi tr a l 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 3 Vd4:Hotanhonton23,8gamhnhpmtmuicacbonatcacckimloihotr(I)v mui cacbonat ca kim loi ho tr (II) trong dung dch HCl. Sau phn ng thu c 4,48 lt kh (ktc). em c cn dung dch thu c bao nhiu gam mui khan? A. 13 gam.B. 15 gam. C. 26 gam.D. 30 gam. Hng dn gii M2CO3+2HCl 2MCl+CO2+H2O R2CO3+2HCl 2MCl2+CO2+H2O 2 CO4,88n 0,222,4= = mol Tng nHCl = 0,4 mol v 2 H On 0,2 mol. =p dng nh lut bo ton khi lng ta c: 23,8 + 0,436,5 = mmui + 0,244 + 0,218 mmui = 26 gam. (p n C) V d 5: Hn hp A gm KClO3, Ca(ClO2)2, Ca(ClO3)2, CaCl2 v KCl nng 83,68 gam. Nhit phn hon ton A ta thu c cht rn B gm CaCl2, KCl v 17,472 lt kh ( ktc). Cho cht rn B tc dng vi 360 ml dung dch K2CO3 0,5M (va ) thu c kt ta C v dung dch D. LngKCltrongdungdchDnhiugp22/3lnlngKClctrongA.%khilng KClO3 c trong A l A. 47,83%.B. 56,72%.C. 54,67%.D. 58,55%. Hng dn gii ooo2t3 2t3 2 2 2t2 2 2 22 2(A) ( A)h B3KClO KCl O (1)2Ca(ClO ) CaCl 3O (2)83,68 gam A Ca(ClO ) CaCl 2O (3)CaCl CaClKCl KCl + + + 2 On 0,78 mol. =p dng nh lut bo ton khi lng ta c:mA = mB + 2 OmmB = 83,68 320,78 = 58,72 gam. Cho cht rn B tc dng vi 0,18 mol K2CO3 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 4 Hn hp B 2 2 3 3( B) ( B)CaCl K CO CaCO 2KCl (4)0,18 0,18 0,36 molKCl KCl+ + + ` )hn hp D ( B) 2 KCl B CaCl ( B)m m m58,72 0,18 111 38,74 gam= = = ( D) KCl KCl ( B) KCl ( pt 4)m m m38,74 0,36 74,5 65,56 gam= += + =

( A) ( D) KCl KCl3 3m m 65,56 8,94 gam22 22= = = (B) (A) KClpt (1) KCl KClm =m m 38,74 8,94 29,8 gam. = =Theo phn ng (1): 3 KClO29,8m 122,5 49 gam.74,5= = 3 KClO (A)49 100%m 58,55%.83,68= =(p n D) Vd6:tchyhonton1,88gamchthucA(chaC,H,O)cn1,904ltO2(ktc)thu c CO2 v hi nc theo t l th tch 4:3. Hy xc nh cng thc phn t ca A. Bit t khi ca A so vi khng kh nh hn 7. A. C8H12O5. B. C4H8O2. C. C8H12O3. D. C6H12O6. Hng dn gii 1,88 gam A + 0,085 mol O2 4a mol CO2 + 3a mol H2O. p dng nh lut bo ton khi lng ta c: 2 2 CO H Om m 1,88 0,085 32 46 gam + = + =Ta c:444a + 183a = 46 a = 0,02 mol. Trong cht A c: nC = 4a = 0,08 mol nH = 3a2 = 0,12 mol nO = 4a2 + 3a 0,0852 = 0,05 molnC : nH : no=0,08 : 0,12 : 0,05=8 : 12 : 5 Vy cng thc ca cht hu c A l C8H12O5 c MA < 203. (p n A) V d 7: Cho 0,1 mol este to bi 2 ln axit v ru mt ln ru tc dng hon ton vi NaOH thu c 6,4 gam ru v mt lng mi c khi lng nhiu hn lng este l 13,56% (so vi lng este). Xc nh cng thc cu to ca este. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 85 V d 12: t chy hon ton a gam hn hp X hai hirocacbon A, B thu c 132.a41 gam CO2 v 245agam H O41. Nu thmvo hnhp Xmt nalngA c trong hnhpX ri t chy hon ton th thu c 2165agam CO41 v 260,75agam H O41. Bit A, B khng lm mt mu nc Br2. a) Cng thc phn t ca A l A. C2H2. B. C2H6. C. C6H12.D. C6H14. b) Cng thc phn t ca B l A. C2H2. B. C6H6. C. C4H4. D. C8H8. c) Phn trm s mol ca A, B trong hn hp X l. A. 60%; 40%.B. 25%; 75%. C. 50%; 50%.D. 30%; 70%. Hng dn gii a) Chn a = 41 gam.t X 2 CO132n 3 mol44= = v 2 H O45n 2,5 mol18= = . t 1X A2| |+ |\ . 2 CO165n 3,75 mol44= =v 2 H O60,75n 3,375 mol18= = . t 1A2 thu c (3,75 3) = 0,75 mol CO2 v(3,375 2,5) = 0,875 mol H2O. t chy A thu c 2 COn 1,5 mol =v 2 H On 1,75 mol = . v 2 2 H O COn n > A thuc loi ankan, do : ( )n 2n 2 2 2 23n 1C H O nCO n 1 H O2+++ + + 22COH On n 1,5n n 1 1,75= =+ n = 6 A l C6H14. (p n D) b) t B thu c (3 1,5) = 1,5 mol CO2 v (2,5 1,75) = 0,75 mol H2O Nh vy CHn 1,5 1n 0,75 2 1= = cng thc tng qut ca B l (CH)n v X khng lm mt mu nc Brom nn B thuc arenB l C6H6. (p n B) c) V A, B c cng s nguyn t C (6C) m lng CO2 do A, B to ra bng nhau (1,5 mol) nA = nB. %nA=%nB=50%. (p n C) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 84 Hn hp kh Z gm x mol CO2 v y10 x4 ( | | + |(\ . mol O2 d. ZM 19 2 38 = = 22coon 1n 1=Vy: yx 10 x4= 8x = 40 y. x = 4, y = 8 tho mn p n C. Vd11:Alhnhpgmmtshirocacbon thkh,Blkhngkh.TrnAviBcng nhit p sut theo tl th tch (1:15) c hnhp kh D. Cho D vo bnh kn dung tch khng i V. Nhit v p sut trong bnh l toC v p atm. Sau khi t chy A trong bnh ch c N2, CO2 v hi nc vi 2 2 CO H OV : V 7 : 4 =a bnh v toC. p sut trong bnh sau khi t l p1 c gi tr l A. 147p p.48= B. p1 = p. C. 116p p.17= D. 13p p.5=Hng dn gii t A:CxHy+2yx O4| |+ |\ . xCO2+2yH O2 V phn ng ch c N2, H2O, CO2 cc hirocacbon b chy ht v O2 va . Chn x y C Hn 1 = nB = 15 mol 2 Oy 15n x 34 5= + = =p.mol. 2 2 N On 4n 12 mol = = yx 34x : y 2 7 : 4 + = = x = 73 ;y = 83 Vnhit v th tch khng i nn p sut t l vi s mol kh, ta c: 1p 7 3 4 3 12 47p 1 15 48+ += =+ 147p p.48=(p n A) Cch 3: CHN GI TR CHO THNG S 22COO( n ) 44 638( n ) 32 610 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 5 A. CH3COO CH3. B. CH3OCOCOOCH3. C. CH3COOCOOCH3. D. CH3COOCH2COOCH3. Hng dn gii R(COOR')2+2NaOH R(COONa)2+2R'OH 0,1 0,20,1 0,2 mol R OH6,4M 320,2'= = Ru CH3OH. p dng nh lut bo ton khi lng ta c: meste + mNaOH=mmui + mru mmui meste=0,240 64=1,6 gam. mmmui meste=13,56100mestemeste = 1,6 10011,8 gam13,56= Meste=118 vC R + (44 + 15)2=118 R = 0. Vy cng thc cu to ca este l CH3OCOCOOCH3. (p n B) V d 8: Thu phn hon ton 11,44 gam hn hp 2 este n chc l ng phn ca nhau bng dung dch NaOH thu c 11,08 gam hn hp mui v 5,56 gam hn hp ru. Xc nh cng thc cu to ca 2 este. A. HCOOCH3 vC2H5COOCH3, B. C2H5COOCH3 v CH3COOC2H5. C. HCOOC3H7 vC2H5COOCH3. D. C B, C u ng. Hng dn gii t cng thc trung bnh tng qut ca hai este n chc ng phn lRCOOR' . RCOOR'+NaOH RCOONa+ R'OH 11,4411,085,56 gam p dng nh lut bo ton khi lng ta c: MNaOH= 11,08 + 5,56 11,44=5,2 gam NaOH5,2n 0,13 mol40= =10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 6 RCOONa11,08M 85,230,13= = R 18, 23 = R OH5,56M 42,770,13'= =R 25,77 ' = RCOOR11,44M 880,13' = =CTPT ca este l C4H8O2 Vy cng thc cu to 2 este ng phn l: HCOOC3H7 vC2H5COOCH3hoc C2H5COOCH3 v CH3COOC2H5. (p n D) V d 9: Chia hn hp gm hai anehit no n chc lm hai phn bng nhau: - Phn 1: em t chy hon ton thu c 1,08 gam H2O. - Phn 2: Tc dng vi H2 d (Ni, to) th thu c hn hp A. em t chy hon ton th th tch kh CO2 (ktc) thu c l A. 1,434 lt.B. 1,443 lt.C. 1,344 lt.D. 0,672 lt. Hng dn gii Phn 1: V anehit no n chc nn 2 2 CO H On n = = 0,06 mol. 2 CO Cn n 0,06(phn 2) (phn 2)= = mol. Theo bo ton nguyn t v bo ton khi lng ta c: C C ( A)n n 0,06(phn 2) = = mol. 2 CO (A)n = 0,06 mol 2 COV = 22,40,06 = 1,344 lt. (p n C) V d 10: Cho mt lung CO i qua ng s ng 0,04 mol hn hp A gm FeO v Fe2O3 t nng. Sau khi kt thc th nghim thu c B gm 4 cht nng 4,784 gam. Kh i ra khi ng s chohpthvodungdchBa(OH)2dththuc9,062gamkt ta.Phntrmkhi lng Fe2O3 trong hn hp A l A. 86,96%.B. 16,04%.C. 13,04%.D.6,01%. Hng dn gii 0,04 mol hn hp A (FeO v Fe2O3) + CO4,784 gam hn hp B + CO2. CO2+Ba(OH)2 d BaCO3 ++H2O 2 3 CO BaCOn n 0,046 mol = =v 2 CO( ) COn n 0,046 molp. = =p dng nh lut bo ton khi lng ta c: 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 83 ( ) 32a 48 1 a 33 + = 215a mol O16= 3 O15 1n 1 mol16 16= = 2 O1 3 3n16 2 32= =b oxi homol Hiu sut phn ng l:3100329,09%3 1532 16=+. (p n B) V d 9: Ho tan hon ton mt lng kim loi R ha tr n bng dung dch H2SO4 long ri c cn dung dch sau phn ng thu c mt lng mui khan c khi lng gp 5 ln khi lng kim loi R ban u em ho tan. Kim loi R l A. Al.B. Ba.C. Zn.D. Mg. Hng dn gii Xt 1 mol kim loi ng vi R (gam) tham gia phn ng. 2R+nH2SO4 R2(SO4)n+nH2 C R (gam) 2R 96ngam mui2+ | | |\ .

( ) 2R 96n5R2+= R = 12n tha mn vi n = 2. Vy: R = 24 (Mg). (p n D) Cch 2: CHN NG T L LNG CHT TRONG U BI CHO V d 10: (Cu 48 - M 182 - khi A - TSH 2007) Hn hp gm hirocacbon X v oxi c t l s mol tng ng l 1:10. t chy hon ton hn hp trn thu c hn hp kh Y. Cho Y qua dung dch H2SO4 c, thu c hn hp kh Z c t khi i vi hiro bng 19. Cng thc phn t ca X l A. C3H8. B. C3H6. C. C4H8. D. C3H4. Hng dn gii t hn hp gm hirocacbon X gm CxHy (1 mol) v O2 (10 mol ). CxHy + yx4| |+ |\ .O2 xCO2+y2H2O 1 mol yx4| |+ |\ .mol x mol y2 mol 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 82 A. 25%.B. 35%.C. 45%.D. 55%. Hng dn gii Xt1molC2H5OH.tamolC2H5OHboxiha.Vyalhiusutcaphnngoxiha ru. C2H5OH+CuO ot CH3CHO+H2O+Cu+ Ban u: 1 mol Oxi ha: a mol a mola mol Sau phn ng:(1 a) mol C2H5OH d a mola mol 46(1 a) 44a 18aM 401 a + += =+ a = 0,25 hay hiu sut l 25%. (p n A) V d 7: Hn hp X gm N2v H2 c XM 12,4 = . Dn X i quabnh ngbt Fe rinungnng bit rng hiu sut tng hp NH3 t 40% th thu c hn hp Y. YMc gi tr l A. 15,12.B. 18,23.C. 14,76.D. 13,48. Hng dn gii Xt 1 mol hn hp XmX = 12,4 gam gm a molN2 v (1 a) mol H2. 28a + 2(1 a) = 12,4 a = 0,4 mol 2 Hn 0,6 mol =N2 + 3H2 oxt, tp 2NH3(vi hiu sut 40%) Ban u:0,4 0,6 Phn ng: 0,08 0,60,4 0,16 mol Sau phn ng: 0,32 0,36 0,16 mol Tng: nY = 0,32 + 0,36 + 0,16 = 0,84 mol; Theo nh lut bo ton khi lng ta c: mX = mY. Y12,4M 14,76 gam0,84= = .(p n C) V d 8: Phng in qua O2 c hn hp kh O2, O3 cM 33 = gam. Hiu sut phn ng l A. 7,09%.B. 9,09%.C. 11,09%. D.13,09%. Hng dn gii 3O2TL2O3 Chn 1 mol hn hp O2, O3 ta c: 2 On a mol = ( )3 On 1 a mol = . 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 7 mA + mCO=mB + 2 COmmA=4,784 + 0,04644 0,04628 = 5,52 gam. t nFeO = x mol, 2 Fe O3n y mol =trong hn hp B ta c: x y 0,0472x 160y 5,52+ = + = x 0,01 moly 0,03 mol= = %mFeO= 0,01 72 10113,04%5,52 =%Fe2O3 = 86,96%. (p n A) MT S BI TP VN DNG GII THEO PHNG PHP S DNG NH LUT BO TON KHI LNG 01. Ha tan 9,14 gam hp kim Cu, Mg, Al bng mt lng va dung dch HCl thu c 7,84 lt kh X (ktc) v 2,54 gam cht rn Yv dung dchZ. Lcb cht rn Y, c cn cn thn dung dch Z thu c lng mui khan l A. 31,45 gam.B. 33,99 gam.C. 19,025 gam.D. 56,3 gam. 02. Cho 15 gam hn hp 3 amin n chc, bc mt tc dng va vi dung dch HCl 1,2 M th thu c 18,504 gam mui. Th tch dung dch HCl phi dng lA. 0,8 lt.B. 0,08 lt. C. 0,4 lt. D. 0,04 lt. 03.Trn8,1gambtAlvi48gambtFe2O3richotinhnhphnngnhitnhmtrongiu kin khng c khng kh, kt thc th nghim lng cht rn thu c l A. 61,5 gam.B. 56,1 gam.C. 65,1 gam.D. 51,6 gam. 04.Hatanhonton10,0gamhnhpXgmhaikimloi(ngtrcHtrongdyinha) bng dung dch HCl d thu c 2,24 lt kh H2 (ktc). C cn dung dch sau phn ng thu c lng mui khan l A. 1,71 gam.B. 17,1 gam.C. 13,55 gam.D. 34,2 gam. 05. Nhit phn hon ton m gam hn hp X gm CaCO3 v Na2CO3 thu c 11,6 gam cht rn v 2,24 lt kh (ktc). Hm lng % CaCO3 trong X l A. 6,25%.B. 8,62%.C. 50,2%.D. 62,5%. 06. Cho 4,4 gam hn hp hai kim loi nhm IA hai chu k lin tip tc dng vi dung dch HCl d thu c 4,48 lt H2 (ktc) v dung dch cha m gam mui tan. Tn hai kim loi v khi lng m l A. 11 gam; Li v Na.B. 18,6 gam; Li v Na. C. 18,6 gam; Na v K.D. 12,7 gam; Na v K. 07. t chy hon ton 18 gam FeS2 v cho ton b lng SO2 vo 2 lt dung dch Ba(OH)2 0,125M. Khi lng mui to thnh l A. 57,40 gam.B. 56,35 gam.C. 59,17 gam.D.58,35 gam. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 8 08. Ha tan 33,75 gam mt kim loi M trong dung dch HNO3 long, d thu c 16,8 lt kh X (ktc) gm hai kh khng mu ha nu trong khng kh c t khi hi so vi hiro bng 17,8. a) Kim loi l A. Cu.B. Zn.C. Fe.D. Al.b) Nu dng dung dch HNO3 2M v ly d 25% th th tch dung dch cn ly l A. 3,15 lt.B. 3,00 lt.C. 3,35 lt.D. 3,45 lt. 09. Ho tan hon ton 15,9 gamhnhp gm 3kimloiAl, MgvCu bng dung dch HNO3 thu c6,72ltkhNOvdungdchX.emccndungdchXthucbaonhiugammui khan? A. 77,1 gam.B. 71,7 gam.C. 17,7 gam.D. 53,1 gam. 10. Ha tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500 ml axit H2SO4 0,1M (va ). Sau phn ng, hn hp mui sunfat khan thu c khi c cn dung dch c khi lng l A. 6,81 gam. B. 4,81 gam. C. 3,81 gam. D. 5,81 gam. p n cc bi tp vn dng: 1. A2. B3. B4. B5. D 6. B7. D8. a-D, b-B9. B10. A 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 81 a = 0,2 2 Nn 0, 2 mol =v 2 Hn 0,8 mol = H2 d. N2 + 3H2 oxt, tp 2NH3 Ban u: 0,2 0,8 Phn ng: x 3x2x Sau phn ng: (0,2 x) (0,8 3x)2x nY = (1 2x) mol p dng nh lut bo ton khi lng ta c mX = mY YYYmnM=( )7, 21 2x8 = x = 0,05. Hiu sut phn ng tnh theo N2 l 0,05 10025%0,2= . (p n D) Vd5:HnhpAgmmtAnkenvhiroc tkhisoviH2bng6,4.ChoAiquaniken nung nng c hn hp B c t khi so vi H2 bng 8 (gi thit hiu sut phn ng xy ra l 100%). Cng thc phn t ca anken l A. C2H4. B. C3H6. C. C4H8. D. C5H10. Hng dn gii Xt 1 mol hn hp A gm (a mol CnH2n v (1a) mol H2) Ta c: 14.n.a + 2(1 a) = 12,8 (1) Hn hp B cM 16 14n = 2) trong hn hp B c H2 d CnH2n + H2 oNi , t CnH2n+2 Ban u: a mol(1a) mol Phn ng:a a a mol Sau phn ng hn hp B gm (1 2a) mol H2 d v a mol CnH2n+2. tng nB = 1 2a. p dng nh lut bo ton khi lng ta c mA = mB BBBmnM= ( )12,81 2a16 = a = 0,2 mol. Thay a = 0,2 vo (1) ta c140,2n + 2(1 0,2) = 12,8 n = 4 anken l C4H8. (p n C) Vd6:OxihaC2H5OHbngCuOnungnng,thuchnhpchtlnggmCH3CHO, C2H5OH d v H2O cM = 40 vC. Hiu sut phn ng oxi ha l 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 80 A. 20%.B. 16%.C. 15%. D.13%. Hng dn gii Xt 1 mol CH3COOH: CH3COOH+NaOH CH3COONa+H2O 60 gam40 gam82 gam 3 dd CH COOH60 100m gamx= ddNaOH40 100m 400 gam10= = 60 100 82 100m 400x 10,25 = + =dd mui gam. x = 15%. (p n C). V d 3: (Cu 1 - M 231 - Khi A - TSC 2007) Khiha tanhiroxit kimloiM(OH)2 bngmtlngva dung dch H2SO4 20% thu c dung dch mui trung ho c nng 27,21%. Kim loi M l A. Cu. B. Zn. C. Fe. D. Mg. Hng dn gii Xt 1 mol M(OH)2 tham gia phn ng M(OH)2 + H2SO4MSO4 + 2H2O C (M + 34) gam 98 gam (M + 96) gam 2 4 dd H SO98 100m 490 gam20= =( )( )4 dd MSOM 96 100m M 34 49027, 21+ = + + =M = 64 M l Cu. (p n A) V d 4: Hn hp X gm N2 v c H2 c t khi hi so vi H2 bng 3,6. Sau khi tin hnh phn ng tng hp c hn hp Y c t khi hi so vi H2 bng 4. Hiu sut phn ng tng hp l A. 10%.B. 15%.C. 20%.D. 25%. Hng dn gii Xt 1 mol hn hp X, ta c: mx = XM = 7,2 gam. t 2 Nn a mol = , ta c: 28a + 2(1 a) = 7,2 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 9 Phng php 2 BO TON MOL NGUYN T C rt nhiu phng php gii ton ha hc khc nhau nhng phng php bo ton nguyn t v phng php bo ton s mol electron cho php chng ta gp nhiu phng trnh phn ng li lm mt, qui gn vic tnh ton v nhm nhanh p s. Rt ph hp vi vic gii cc dng bi tonha hc trc nghim. Cch thc gp nhng phng trnhlmmt v cch lp phng trnh theo phng php bo ton nguyn t s c gii thiu trong mt s v d sau y. V d 1: kh hon ton 3,04 gam hn hp X gm FeO, Fe3O4, Fe2O3 cn 0,05 mol H2. Mt khc hatanhonton3,04gamhnhpXtrongdungdchH2SO4c thucthtchkh SO2 (sn phm kh duy nht) iu kin tiu chun l A. 448 ml.B. 224 ml.C. 336 ml.D. 112 ml. Hng dn gii Thc cht phn ng kh cc oxit trn l H2+O H2O 0,05 0,05 mol t s mol hn hp X gm FeO, Fe3O4, Fe2O3 ln lt l x, y, z. Ta c: nO=x + 4y + 3z=0,05 mol(1) Fe3,04 0,05 16n 0,04 mol56 = =x + 3y + 2z= 0,04 mol(2) Nhn hai v ca (2) vi 3 ri tr (1) ta c: x + y=0,02 mol. Mt khc: 2FeO + 4H2SO4Fe2(SO4)3 + SO2 + 4H2O xx/2 2Fe3O4 + 10H2SO43Fe2(SO4)3 + SO2 + 10H2O yy/2 tng: SO2x y 0,2n 0,01 mol2 2+= = =Vy: 2 SOV 224 ml. =(p n B) V d 2: Thi t t V lt hn hp kh (ktc) gm CO v H2 i qua mt ng ng 16,8 gam hn hp 3 oxit:CuO,Fe3O4,Al2O3nungnng,phnnghonton.Sauphnngthucmgam cht rn v mt hn hp kh v hi nng hn khi lng ca hn hp V l 0,32 gam. Tnh V v m. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 10 A. 0,224 lt v 14,48 gam.B. 0,448 lt v 18,46 gam. C. 0,112 lt v 12,28 gam. D. 0,448 lt v 16,48 gam. Hng dn gii Thc cht phn ng kh cc oxit trn l CO+O CO2 H2+O H2O. Khilnghnhpkhtothnhnnghnhnhpkhbanuchnhlkhilngca nguyn t Oxi trong cc oxit tham gia phn ng. Do vy: mO = 0,32 gam. O0,32n 0,02 mol16= =( )2 CO Hn n 0,02 mol + = . p dng nh lut bo ton khi lng ta c: moxit = mcht rn + 0,32 16,8 = m + 0,32 m = 16,48 gam. 2 hh (CO H )V 0,02 22, 4 0,448+= = lt. (p n D) V d 3: Thi rt chm 2,24 lt (ktc) mt hn hp kh gm CO v H2 qua mt ng s ng hn hp Al2O3,CuO,Fe3O4,Fe2O3ckhilngl24gamdangcunnng.Saukhikt thc phn ng khi lng cht rn cn li trong ng s l A. 22,4 gam.B. 11,2 gam.C. 20,8 gam.D. 16,8 gam. Hng dn gii 2 hh (CO H )2, 24n 0,1 mol22, 4+= =Thc cht phn ng kh cc oxit l: CO+O CO2 H2+O H2O. Vy: 2 O CO Hn n n 0,1 mol = + = . mO = 1,6 gam. Khi lng cht rn cn li trong ng s l: 24 1,6 = 22,4 gam. (p n A) V d 4: Cho m gam mt ancol (ru) no, n chc X qua bnh ng CuO (d), nung nng. Sau khi phn ng hon ton, khi lng cht rn trong bnh gim 0,32 gam. Hn hp hi thu c c t khi i vi hiro l 15,5. Gi tr ca m l 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 79 Phng php 10 T CHN LNG CHT Trongmtscuhivbitptrcnghimchngtac thgpmts trnghpcbit sau: - C mt s bi ton tng nh thiu d kin gy b tc cho vic tnh ton. - C mt s bi ton ngi ta cho di dng gi tr tng qut nh a gam, V lt, n mol hoc cho t l th tch hoc t l s mol cc cht...Nh vy kt qu gii bi ton khng ph thuc vo cht cho. Trong cc trng hp trn tt nht ta t chn mt gi tr nh th no cho vic gii bi ton tr thnh n gin nht. Cch 1: Chn mt mol nguyn t, phn t hoc mt mol hn hp cc cht phn ng. Cch 2: Chn ng t l lng cht trong u bi cho. Cch 3: Chn cho thng s mt gi tr ph hp chuyn phn s phc tp v s n gin tnh ton. Sau y l mt s v d in hnh: Cch 1: CHN 1 MOL CHT HOC HN HP CHT PHN NG V d 1: Ho tan mt mui cacbonat kim loi M ha tr n bng mt lng va dung dch H2SO4 9,8% ta thu c dung dch mui sunfat 14,18%. M l kim loi g? A. Cu.B. Fe.C. Al.D. Zn. Hng dn gii Chn 1 mol mui M2(CO3)n. M2(CO3)n+ nH2SO4M2(SO4)n + nCO2| + nH2O C (2M + 60n) gam 98n gam (2M + 96n) gam 2 4 dd H SO98n 100m 1000n gam9,8= = 2 3 n 2 4 2 M (CO ) dd H SO COm m m m = + dd mui = 2M + 60n + 1000.n 44.n = (2M + 1016.n) gam. ( ) + = =+ddmui2M 96 100C% 14,182M 1016n M = 28.n n = 2 ;M = 56 l ph hp vy M l Fe. (p n B) V d 2: Cho dung dch axit axetic c nng x% tc dng va vi dung dch NaOH 10% th thu c dung dch mui c nng 10,25%. Vy x c gi tr no sau y? 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 78 - Phn 2: Oxi ha bng Oxi thnh HCOOH vi hiu sut 40% thu c dung dch A. Cho A tc dng vi dung dch AgNO3 /NH3 thu c m' gam Ag. T s mm' c gi tr bng A. 0,2.B. 0,4.C. 0,6.D. 0,8. 07. A l axit cha ba nguyn t cacbon trong phn t. Cho 0,015 mol A tc dng vi dung dch cha a mol Ba(OH)2thu c dung dch B. Ngi ta nhn thy: Nu a = 0,01 mol th dung dch B lm qu tm. Nu a = 0,02 mol th dung dch B lm xanh qu tm. B c cng thc cu to: A. CH3CH2COOH.B. CH2=CHCOOH. C. CHCCOOH.D. HOOCCH2COOH. 08. C 2 axit hu c no: (A) l axit n chc v (B) l axit a chc. Hn hp (X) cha x mol (A) v y mol (B). t chy hon ton (X) th thu c 11,2 lt CO2 (ktc). Cho x + y = 0,3 v MA < MB. Vy cng thc phn t ca (A) l: A. CH3COOH.B. C2H5COOH. C. HCOOH.D. C3H7COOH. 09. Hn hp A gm Al v Fe2O3 c khi lng trung bnh l AM . Tin hnh phn ng nhit nhm, saumt thigianthuchnhpBckhilngphnt trungbnhl BM .Quanhgia AM v BMl A. AM = BM .B. AM> BM . C. AM< BM .D. AM> BM . 10. Kh hon ton mt lng oxit st cn V lt H2. ha tan hon ton lng st sinh ra trn trong dung dch HCl thy to ra V' lt H2. Bit V > V' (cc kh o cng iu kin). Cng thc oxit st l A. Fe2O3.B. FeO. C. Fe3O4. D. Fe2O3 v Fe3O4. p n cc bi tp vn dng: 1. B2. C3. C4. B5. A 6. D7. D8. C9. A10. D 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 11 A. 0,92 gam.B. 0,32 gam.C. 0,64 gam.D. 0,46 gam. Hng dn gii CnH2n+1CH2OH + CuO ot CnH2n+1CHO + Cu+ + H2O Khi lng cht rn trong bnh gim chnh l s gam nguyn t O trong CuO phn ng. Do nhn c: mO = 0,32 gam O0,32n 0,02 mol16= =Hn hp hi gm: n 2n 12C H CHO : 0,02 molH O : 0,02 mol.+ Vy hn hp hi c tng s mol l 0,04 mol. CM= 31mhh hi = 31 0,04 = 1,24 gam. mancol + 0,32=mhh hi mancol=1,24 0,32=0,92 gam. (p n A) Ch : Vi ru bc (I) hoc ru bc (II) u tha mn u bi. V d 5: t chy hon ton 4,04 gam mt hn hp bt kim loi gm Al, Fe, Cu trong khng kh thu c 5,96 gam hn hp 3 oxit. Ha tan ht hn hp 3 oxit bng dung dch HCl 2M. Tnh th tch dung dch HCl cn dng. A. 0,5 lt.B. 0,7 lt.C. 0,12 lt.D. 1 lt. Hng dn gii mO=moxit mkl=5,96 4,04=1,92 gam. O1,92n 0,12 mol16= = . Ha tan ht hn hp ba oxit bng dung dch HCl to thnh H2O nh sau: 2H+ +O2 H2O 0,240,12 mol HCl0, 24V 0,122= = lt. (p n C) V d 6: t chyhon ton 0,1molmt axit cacbonxylic n chc cnva Vlt O2 ( ktc), thu c 0,3 mol CO2 v 0,2 mol H2O. Gi tr ca V l A. 8,96 lt.B. 11,2 lt.C. 6,72 lt.D. 4,48 lt. Hng dn gii Axit cacbonxylic n chc c 2 nguyn t Oxi nn c th t l RO2. Vy: 2 2 2 2 O ( RO ) O (CO ) O (CO ) O ( H O)n n n n + = +10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 12 0,12 + nO (p.)=0,32 + 0,21 nO (p.)=0,6 mol 2 On 0,3 mol = 2 OV 6,72 = lt. (p n C) V d 7: (Cu 46 - M 231 - TSC Khi A 2007) Cho 4,48 lt CO ( ktc) t t i qua ng snung nng ng 8 gammt oxit st n khi phnngxyrahonton.Khthucsauphnngctkhisovihirobng20. Cngthc ca oxit st v phn trm th tch ca kh CO2 trong hn hp kh sau phn ng l A. FeO; 75%.B. Fe2O3; 75%. C. Fe2O3; 65%. D. Fe3O4; 65%. Hng dn gii FexOy+yCOxFe + yCO2 Kh thu c cM 40 =gm 2 kh CO2 v CO d 2 COCOn 3n 1=2 CO%V 75% = . Mt khc: 2 CO ( ) CO75n n 0, 2 0,15100p. = = = molnCO d = 0,05 mol. Thc cht phn ng kh oxit st l do CO+O(trong oxit st) CO2 nCO = nO = 0,15 mol mO = 0,1516 = 2,4 gam mFe = 8 2,4 = 5,6 gamnFe = 0,1 mol. Theo phng trnh phn ng ta c: 2FeCOn x 0,1 2n y 0,15 3= = = Fe2O3. (p n B) V d 8: Cho hn hp A gm Al, Zn, Mg. em oxi ho hon ton 28,6 gam A bng oxi d thu c 44,6gamhnhpoxitB.Ho tanhtBtrongdungdchHClthucdungdchD.C cn dung dch D c hn hp mui khan l A. 99,6 gam.B. 49,8 gam. C. 74,7 gam. D. 100,8 gam. Hng dn gii 2 COCOn 44 1240n 28 410 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 77 Phn ng:0,03 0,08 0,020,02 mol V1 tng ng vi 0,02 mol NO. TN2:nCu = 0,06 mol ;3 HNOn 0,08 mol = ;2 4 H SOn 0,04 mol. = Tng Hn + = 0,16 mol ;3 NOn = 0,08 mol. 3Cu+8H++2NO3 3Cu2++2NO|+4H2O u bi:0,06 0,160,08Cu v H+ phn ng ht Phn ng:0,06 0,16 0,040,04 mol V2 tng ng vi 0,04 mol NO. Nh vy V2 = 2V1. (p n B)MT S BI TP VN DNG GII THEO PHNG PHP CC I LNG DNG TNG QUT 01. Dung dch A c a mol NH4+, b mol Mg2+, c mol SO42 v d mol HCO3. Biu thc no biu th s lin quan gia a, b, c, d sau y l ng? A. a + 2b =c + d.B. a + 2b = 2c + d. C. a + b = 2c + d.D. a + b =c+ d. 02. Cho a mol Fe vo dung dch cha b mol dung dch AgNO3. a v b c quan h nh th no thu c dung dch Fe(NO3)3 duy nht sau phn ng? A. b =2a.B. b>a.C. b=3a.D. b>a. 03. Dung dch A cha cc ion Na+: a mol; HCO3: b mol; CO32: c mol; SO42: d mol. to ra kt ta ln nht ngi ta dng 100 ml dung dch Ba(OH)2 nng x mol/l. Lp biu thc tnh x theo a v b. A. x = a + b.B. x = a b.C. x = a b0,2+.D. x = a b0,1+. 04. Dung dch X cha a mol NaAlO2. Khi thm vo dung dch X b mol hoc 2b mol dung dch HCl th lng kt ta sinh ra u nh nhau. T s ab c gi tr bng A. 1. B. 1,25.C. 1,5.D. 1,75. 05. Oxi ha mt lng Fe thnh hn hp X gm FeO, Fe3O4, Fe2O3cn a mol Oxi. Kh hon ton hn hp X thnh Fe cn b mol Al. T s ab c gi tr bng A. 0,75.B. 1.C. 1,25.D. 1,5. 06. C mt lng anehit HCHO c chia lm 2 phn bng nhau, mi phn cha a mol HCHO. - Phn 1: Cho tc dng vi dung dch AgNO3 /NH3 thu c m gam Ag. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 76 A. a.M10d. B. d.M10a.C. 10aM.d.D. a.M1000d. Hng dn gii Xt 1 lt dung dch cht X: nX = a mol mX = a.M mdd X = a.M.100C% = 1000d C% = a.M10d. (p n A) V d 17: Hn hp X c mt s ankan. t chy 0,05 mol hn hp X thu c a mol CO2 v b mol H2O. Kt lun no sau y l ng? A. a = b.B. a = b 0,02. C. a = b 0,05.D. a = b 0,07. Hng dn gii t cng thc tng qut ca 1 s ankan l x 2x 2C H+ x 2x 2C H+ + 23x 1O2+ x CO2 +(x 1) + H2O 0,50,05x 0,05(x 1) + mol 0,05x a0,05(x 1) b= + = a = b 0,05.(p n C) V d 18: (Cu 40 - M 285 - Khi B - TSH 2007) Thc hin hai th nghim: 1) Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO3 1M thot ra V1 lt NO. 2) Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO4 0,5 M thot ra V2 lt NO. Bit NO l sn phm kh duy nht, cc th tch kh o cng iu kin. Quan h gia V1 v V2 l A. V2 = V1.B. V2 = 2V1.C. V2 = 2,5V1.D. V2 = 1,5V1. Hng dn gii TN1:3CuHNO3,84n 0,06 mol64n 0,08 mol= = = 3HNOn 0,08 moln 0,08 mol+= = 3Cu+8H++2NO3 3Cu2++2NO|+4H2O u bi:0,060,08 0,08 H+ phn ng ht 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 13 Gi M l kim loi i din cho ba kim loi trn vi ho tr l n. M+n2O2 M2On(1) M2On+2nHCl 2MCln+nH2O(2) Theo phng trnh (1) (2) 2 HCl On 4.n = . p dng nh lut bo ton khi lng 2 Om 44,6 28,6 16 = = gam 2 On 0,5 = mol nHCl = 40,5 = 2 mol Cln 2 mol = mmui = mhhkl + Clm = 28,6 + 235,5 = 99,6 gam. (p n A) V d 9: Cho mt lung kh CO i qua ng ng 0,01 mol FeO v 0,03 mol Fe2O3 (hn hp A) t nng. Sau khi kt thc th nghim thu c 4,784 gam cht rn B gm 4 cht. Ho tan cht rn Bbng dung dch HCl d thy thot ra 0,6272lt H2 ( ktc). Tnh s mol oxit st t trong hn hp B. Bit rng trong B s mol oxit st t bng 1/3 tng s mol st (II) oxit v st (III) oxit. A. 0,006.B. 0,008.C. 0,01.D. 0,012. Hng dn gii Hn hp A 2 3FeO : 0,01 molFe O : 0,03 mol + CO4,784 gam B (Fe, Fe2O3, FeO, Fe3O4) tng ng vi s mol l: a, b, c, d (mol). Ho tan B bng dung dch HCl d thu c 2 Hn 0,028 = mol. Fe+2HClFeCl2 +H2 a = 0,028 mol. (1) Theo u bi: ( )3 4 2 3 Fe O FeO Fe O1n n n3= +( )1d b c3= + (2) Tng mB l: (56.a + 160.b + 72.c + 232.d) = 4,78 gam.(3) S mol nguyn t Fe trong hn hp A bng s mol nguyn t Fe trong hn hp B. Ta c:nFe (A) = 0,01 + 0,032 = 0,07 molnFe (B) = a + 2b + c + 3d a + 2b + c + 3d = 0,07(4) T (1, 2, 3, 4) b = 0,006 mol c = 0,012 mol d = 0,006 mol. (p n A) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 14 V d 10: Khhon ton 24 gam hn hp CuOv FexOybng H2 d nhit cao thu c 17,6 gam hn hp 2 kim loi. Khi lng H2O to thnh l A. 1,8 gam.B. 5,4 gam.C. 7,2 gam. D. 3,6 gam. Hng dn gii mO (trong oxit)=moxit mkloi = 24 17,6 = 6,4 gam. ( ) 2 O HOm 6, 4 = gam ;2 H O6, 4n 0,416= = mol. 2 H Om 0, 4 18 7,2 = = gam. (p n C) V d 11: Kh ht m gam Fe3O4 bng CO thu c hn hp A gm FeO v Fe. A tan va trong 0,3 lt dung dch H2SO4 1M cho ra 4,48 lt kh (ktc). Tnh m? A. 23,2 gam.B. 46,4 gam.C. 11,2 gam.D. 16,04 gam. Hng dn gii Fe3O4(FeO, Fe)3Fe2+ n mol ( )24 4Fe trong FeSO SOn n 0,3 = = mol p dng nh lut bo ton nguyn t Fe: ( ) ( ) 4 3 4 Fe FeSO Fe Fe On n =3n = 0,3 n = 0,1 3 4 Fe Om 23,2 =gam (p n A) Vd12:unhairunchcviH2SO4c,140oCchnhpbaete.Ly0,72gammt trong ba ete em t chy hon ton thu c 1,76 gam CO2 v 0,72 gam H2O. Hai ru l A. CH3OH v C2H5OH.B. C2H5OH v C3H7OH. C. C2H5OH v C4H9OH. D. CH3OH v C3H5OH. Hng dn gii t cng thc tng qut ca mt trong ba ete l CxHyO, ta c: C0,72m 12 0, 4844= = gam ;H0,72m 2 0,0818= = gam mO = 0,72 0,48 0,08 = 0,16 gam. 0, 48 0,08 0,16x : y :1 : :12 1 16==4 : 8 : 1. Cng thc phn t ca mt trong ba ete l C4H8O. Cng thc cu to l CH3OCH2CH=CH2. Vy hai ancol l CH3OH v CH2=CHCH2OH. (p n D) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 75 Cc kh o cng iu kin. Quan h gia V1 v V2 l A. V1 = V2.B. V1 > V2.C. V1 < V2.D. V1 s V2.Hng dn gii Cc phng trnh phn ng khi ha tan hn hp Na v Al vi H2O v vi dung dch NaOH d: Na+H2O NaOH+ 12H2(1) 2Al+6H2O+2NaOH Na[Al(OH)4]+ 3H2(2) t s mol Na v Al ban u ln lt l x v y (mol). TN1:x>ynNaOHvahocdkhihatanAlchaithnghimcngtothnh x 3x2 2| |+ |\ . mol H2. V1 = V2. TN2: x < ytrong TN1 (1) Al d, TN2 (2) Al tan ht2 2 H (TN2) H (TN2)n n . >V2 > V1. Nh vy (x,y > 0) th V2 > V1. (p n D) Vd15:MtbnhknchaVltNH3vV' ltO2cngiukin.Nungnngbnhcxctc NH3chuynhtthnhNO,sauNOchuynht thnhNO2.NO2vlngO2cnli trong bnh hp th va vn ht trong nc thnh dung dch HNO3. T s VV' l A. 1.B. 2.C. 3.D. 4. Hng dn gii Cc phng trnh phn ng: 4NH3+5O2 oxtt4NO+6H2O V 5V/4 V 2NO+O2 2NO2 V V/2V 4NO2 + O2+2H2O 4HNO3 V 5V VV4 2| |' |\ . V = 45V VV4 2| |' |\ .VV' = 2. (p n B) V d 16: Cht X c khi lng phn t l M. Mt dung dch cht X c nng a mol/l, khi lng ring d gam/ml. Nng C% ca dung dch X l 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 74 A. a = b.B. a = 2b.C. b = 5a.D. a < b < 5a. Hng dn gii Phng trnh phn ng: NaOH+HCl NaCl+H2O(1) a mola mol NaAlO2+HCl+H2O Al(OH)3++NaCl (2) Al(OH)3+3HCl AlCl3+3H2O(3) NaAlO2+4HCl AlCl3+NaCl+2H2O(4) a mol 4a mol iu kin khng c kt ta khinHCl > 2 NaAlO4n + nNaOH = 5a. Vy suy ra iu kin c kt ta: nNaOH < nHCl < 2 NaAlO4n + nNaOH a < b < 5a. (p n D) V d 13: Dung dch cha a mol NaOH tc dng vi dung dch cha b mol H3PO4 sinh ra hn hp Na2HPO4 + Na3PO4. T s ab l A. 1 < ab < 2.B. ab > 3. C. 2 < ab < 3.D. ab > 1. Hng dn gii Cc phng trnh phn ng: NaOH+H3PO4 NaH2PO4+H2O(1) 2NaOH+H3PO4 Na2HPO4+2H2O(2) 3NaOH+H3PO4 Na3PO4+3H2O(3) Ta c:nNaOH = a mol ;3 4 H POn = b mol. thu c hn hp mui Na2HPO4 + Na3PO4 th phn ng xy ra c hai phng trnh (2 v 3), do : 2 < 3 4NaOHH POnn < 3, tc l 2 < ab < 3. (p n C) V d 14: Hn hp X gm Na v Al. - Th nghim 1: Nu cho m gam X tc dng vi H2O d th thu c V1 lt H2. - Thnghim 2:nu chom gam X tc dngvidung dch NaOH d th thu c V2lt H2. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 15 MT S BI TP VN DNG GII THEO PHNG PHP BO TON MOL NGUYN T 01. Ha tan hon ton hn hp X gm 0,4 mol FeO v 0,1mol Fe2O3 vo dung dch HNO3 long, d thu c dung dch A v kh B khng mu, ha nu trong khng kh. Dung dch A cho tc dng vidungdchNaOHdthuckt ta.Lytonbkt tanungtrongkhngkhnkhi lng khng i thu c cht rn c khi lng l A. 23,0 gam.B. 32,0 gam.C. 16,0 gam.D. 48,0 gam. 02. Cho kh CO i qua ng s cha 16 gam Fe2O3 un nng, sau phn ng thu c hn hp rn X gm Fe, FeO, Fe3O4, Fe2O3. Ha tan hon ton X bng H2SO4 c, nng thu c dung dch Y. C cn dung dch Y, lng mui khan thu c l A. 20 gam.B. 32 gam.C. 40 gam.D. 48 gam. 03. Kh hon ton 17,6 gam hnhp X gmFe,FeO, Fe2O3 cn 2,24lt CO ( ktc). Khilng st thu c l A. 5,6 gam.B. 6,72 gam.C. 16,0 gam.D. 11,2 gam. 04. t chy hn hp hirocacbon X thu c 2,24 lt CO2 (ktc) v 2,7 gam H2O. Th tch O2 tham gia phn ng chy (ktc) l A. 5,6 lt.B. 2,8 lt.C. 4,48 lt.D. 3,92 lt. 05. Ho tan hon ton a gam hn hp X gm Fe v Fe2O3 trong dung dch HCl thu c 2,24 lt kh H2 ktc v dung dch B. Cho dung dch B tc dng dung dch NaOH d, lc ly kt ta, nung trong khng kh n khi lng khng i thu c 24 gam cht rn. Gi tr ca a l A. 13,6 gam.B. 17,6 gam.C. 21,6 gam.D. 29,6 gam. 06. Hn hp X gm Mg v Al2O3. Cho 3 gam X tc dng vi dung dch HCl d gii phng V lt kh (ktc). Dung dch thu c cho tc dng vi dung dch NH3 d, lc v nung kt ta c 4,12 gam bt oxit. V c gi tr l: A. 1,12 lt.B. 1,344 lt.C. 1,568 lt.D. 2,016 lt. 07. Hn hpA gmMg,Al, Fe, Zn. Cho 2 gamA tc dngvi dung dch HCl d gii phng 0,1 gam kh. Cho 2 gamA tc dngvikh clo d thu c 5,763 gamhnhpmui. Phn trm khi lng ca Fe trong A l A. 8,4%.B. 16,8%.C. 19,2%.D. 22,4%. 08. (Cu 2 - M 231 - TSC - Khi A 2007) tchyhontonmt thtchkhthinnhingmmetan,etan,propanbngoxikhngkh (trong khng kh Oxi chim 20% th tch), thu c 7,84 lt kh CO2 (ktc) v 9,9 gam H2O. Th tch khng kh (ktc) nh nht cn dng t chy hon ton lng kh thin nhin trn l A. 70,0 lt.B. 78,4 lt.C. 84,0 lt.D. 56,0 lt. 09. Ho tan hon ton 5 gam hn hp 2 kim loi X v Y bng dung dch HCl thu c dung dch A v kh H2. C cn dung dch A thu c 5,71 gam mui khan. Hy tnh th tch kh H2 thu c ktc. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 16 A. 0,56 lt.B. 0,112 lt. C. 0,224 ltD. 0,448 lt 10. t chy hon ton m gam hn hp Y gm C2H6, C3H4 v C4H8 th thu c 12,98 gam CO2 v 5,76 gam H2O. Vy m c gi tr l A. 1,48 gam.B. 8,14 gam.C. 4,18 gam. D. 16,04 gam. p n cc bi tp vn dng: 1. D2. C3. C4. D5. C 6. C7. B8. A9. C10. C 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 73 Ta c:b = a + c ax = a + a.y2 y = 2x 2. Cng thc tng qut ca anehit n chc X l CxH2x2O c dng Cx1H2(x1)1CHO l anehit khng no c mt lin kt i, n chc. (p n C) V d 9: Cng thc phn t ca mt ancol A l CnHmOx. cho A l ancol no th m phi c gi tr A. m = 2n.B. m = 2n + 2. C. m = 2n 1.D. m = 2n + 1. Hng dn gii Theophngphpngnhths:CngthctngqutcaancolnolCnH2n+2-x(OH)xhay CnH2n+2Ox. Vy m = 2n+2. (p n B) V d 10: Hi t l th tch CO2 v hi nc (T) bin i trong khong no khi t chy hon ton cc ankin. A. 1 < T s 2.B. 1 s T < 1,5. C. 0,5 < T s 1.D. 1 < T < 1,5. Hng dn gii CnH2n-2 nCO2+(n 1)H2O iu kin: n > 2 v n e N. T = 22COH Onn = n 1.1 n 11n= Vi mi n > 2T > 1; mt khc n tngT gim. n = 2 T = 2 l gi tr ln nht. Vy: 1 < T s 2. (p n A) V d 11: t chy 1 mol aminoaxitNH2(CH2)nCOOH phi cn s mol O2 l A. 2n 3.2+B. 6n 3.2+C. 6n 3.4+D. 2n 3.4+ Hng dn gii Phng trnh t chy amino axit l H2N(CH2)nCOOH+6n 34+O2 (n + 1)CO2+2n 32+H2O (p n C) V d 12: Mt dung dch hn hp cha a mol NaAlO2 v a mol NaOH tc dng vi mt dung dch cha b mol HCl. iu kin thu c kt ta sau phn ng l 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 72 inphndungdchchaamolCuSO4vbmolNaCl(viincctr,cmngngn xp). dung dch sau in phn lm phenolphtalein chuyn sang mu hng th iu kin ca a v b l (bit ion SO42 khng b in phn trong dung dch) A. b > 2a.B. b = 2a.C. b < 2a.D. 2b = a. Hng dn gii Phng trnh in phn dung dch CuSO4+2NaClpdd Cu++Cl2|+Na2SO4 (1) a2a mol Dungdchsauinphnlmphenolphtaleinchuynsangmuhngsauphnng(1)th dung dch NaCl cn d v tip tc b in phn theo phng trnh 2NaCl+2H2Opddmng ngn 2NaOH+H2+Cl2 (2) Vy:b > 2a. (p n A) Ch : Tng t cng cu hi trn chng ta c th hi: + dung dch sau in phn c mi trng axit th iu kin ca a v b l. A. b > 2a.B. b = 2a.C. b < 2a.D. a = 2b. + dung dch sau in phn c kh nng ha tan kt ta Al(OH)3 th iu kin ca a, b l A. b > 2a.B. b < 2a.C. b = 2a.D. b > 2a. V d 8: t chy hon ton a mol mt anehit X (mch h) to ra b mol CO2 v c mol H2O (bit b = a + c). Trong phn ng trng gng, mt phn t X ch cho 2 electron. X thuc dy ng ng anehit A. no, n chc. B. khng no c hai ni i, n chc. C. khng no c mt ni i, n chc. D. no, hai chc. Hng dn gii Trong phn ng trng gng mt anehit X ch cho 2e X l anehit n chc bi v: 1RCHO+34RCOONH+ trong : C+1 2eC+3. t cng thc phn t ca anehit n chc X l CxHyO ta c phng trnh CxHyO+2y 1x O4 2| |+ |\ . xCO2+y2H2O aa.xa.y2 mol (b mol)(c mol)10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 17 Phng php 3 BO TON MOL ELECTRON Trchtcnnhnmnhykhngphilphngphpcnbngphnngoxiha-kh, mc d phng php thng bng electron dng cn bng phn ng oxi ha - kh cng da trn s bo ton electron. Nguyn tc ca phng phpnhsau: khic nhiu cht oxiha, cht kh trongmt hn hp phn ng (nhiu phn ng hoc phn ng qua nhiu giai on) th tng s electron ca cc cht kh cho phi bng tng s electron m cc cht oxi ha nhn. Ta ch cn nhn nh ng trng thi u vtrngthicuicaccchtoxihahocchtkh,thmchkhngcnquantmnviccn bng cc phng trnh phn ng. Phng php ny c bit l th i vi cc bi ton cn phi bin lun nhiu trng hp c th xy ra. Sau y l mt s v d in hnh. V d 1: Oxi ha hon ton 0,728 gam bt Fe ta thu c 1,016 gam hn hp hai oxit st (hn hp A). 1. Ha tan hn hp A bng dung dch axit nitric long d. Tnh th tch kh NO duy nht bay ra ( ktc). A. 2,24 ml.B. 22,4 ml. C. 33,6 ml.D. 44,8 ml. 2.CnghnhpAtrntrnvi5,4gambtAlritinhnhphnngnhitnhm(hiusut 100%). Ha tan hn hp thu c sau phn ng bng dung dch HCl d. Tnh th tch bay ra ( ktc). A. 6,608 lt.B. 0,6608 lt. C. 3,304 lt.D. 33,04. lt Hng dn gii 1. Cc phn ng c th c: 2Fe + O2ot2FeO(1) 2Fe + 1,5O2otFe2O3(2) 3Fe + 2O2otFe3O4 (3) Cc phn ng ha tan c th c: 3FeO + 10HNO33Fe(NO3)3 + NO| + 5H2O(4) Fe2O3 + 6HNO32Fe(NO3)3 + 3H2O(5) 3Fe3O4 + 28HNO39Fe(NO3)3 + NO| + 14H2O(6) Ta nhn thy tt c Fe t Fe0 b oxi ha thnh Fe+3, cn N+5 b kh thnh N+2, O20 b kh thnh 2O2 nn phng trnh bo ton electron l: 0,7283n 0,009 4 3 0,03956+ = = mol. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 18 trong ,nl s mol NO thot ra. Ta d dng rt ra n = 0,001 mol; VNO = 0,00122,4 = 0,0224 lt = 22,4 ml. (p n B) 2. Cc phn ng c th c: 2Al + 3FeOot3Fe + Al2O3(7) 2Al + Fe2O3ot2Fe + Al2O3(8) 8Al + 3Fe3O4ot9Fe + 4Al2O3(9) Fe + 2HClFeCl2 + H2|(10) 2Al + 6HCl2AlCl3 + 3H2| (11) Xt cc phn ng (1, 2, 3, 7, 8, 9, 10, 11) ta thy Fe0 cui cng thnh Fe+2, Al0 thnh Al+3, O20 thnh 2O2 v 2H+ thnh H2 nn ta c phng trnh bo ton electron nh sau: 5,4 30,013 2 0,009 4 n 227 + = + Fe0 Fe+2 Al0 Al+3O20 2O2 2H+ H2 n = 0,295 mol 2 HV 0, 295 22, 4 6,608 = = lt. (p n A) Nhn xt: Trong bi ton trn cc bn khng cn phi bn khon l to thnh hai oxit st (hn hp A) gm nhng oxit no v cng khng cn phi cn bng 11 phng trnh nh trn m ch cn quan tm ti trng thi u v trng thi cui ca cc cht oxi ha v cht kh ri p dng lut bo ton electron tnh lc bt c cc giai on trung gian ta s tnh nhm nhanh c bi ton. V d 2: Trn 0,81 gam bt nhmvibt Fe2O3 v CuO ri t nng tinhnh phn ngnhit nhm thu c hn hp A. Ho tan hon ton A trong dung dch HNO3 un nng thu c V lt kh NO (sn phm kh duy nht) ktc. Gi tr ca V l A. 0,224 lt.B. 0,672 lt.C. 2,24 lt.D. 6,72 lt. Hng dn gii Tm tt theo s : o2 3 tNOFe O0,81 gam Al V ?CuO3ha tan hon tondung dchHNOhn hp A+ = Thc cht trong bi ton ny ch c qu trnh cho v nhn electron ca nguyn t Al v N. Al Al+3 + 3e 0,81270,09 mol vN+5 +3e N+2 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 71 DungdchHClvdungdchCH3COOHccngnngmol/l,pHcahaidungdch tng ng l x v y. Quan h gia x v y l (gi thit, c 100 phn t CH3COOH th c 1 phn t in li) A. y = 100x.B. y = 2x.C. y = x 2.D. y = x + 2. Hng dn gii pHHCl = x [H+]HCl = 10x 3 CH COOHpH y = 3yCH COOH[H ] 10+ = Ta c:HCl H++Cl 10x10x (M) CH3COOH H++CH3COO 100.10y 10y (M). Mt khc:[HCl] = [CH3COOH] 10x = 100.10y y = x + 2. (p n D) V d 6: (Cu 53 - M 182 - Khi A - TSH 2007) thu ly Ag tinh khit t hn hp X (gm a mol Al2O3, b mol CuO, c mol Ag2O), ngi ta ho tan X bi dung dch cha (6a + 2b + 2c) mol HNO3 c dung dch Y, sau thm (gi thit hiu sut cc phn ng u l 100%) A. c mol bt Al vo Y.B. c mol bt Cu vo Y. C. 2c mol bt Al vo Y.D. 2c mol bt Cu vo Y. Hng dn gii Ha tan hn hp X trong dung dch HNO3 Al2O3+6HNO3 2Al(NO3)3+3H2O a6a2a mol CuO+2HNO3 Cu(NO3)2+H2O b 2b b mol Ag2O+2HNO3 2AgNO3+H2O c 2c 2c mol Dung dch HNO3 va . Dung dch Y gm 2a mol Al(NO3)3, b mol Cu(NO3)2, 2c mol AgNO3. thu Ag tinh khit cn cho thm kim loi Cu vo phng trnh Cu+2AgNO3 Cu(NO3)2+2Ag c mol 2c Vy cn c mol bt Cu vo dung dch Y. (p n B) V d 7: (Cu 32 - M 285 - Khi B - TSH 2007) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 70 Vy 35,5 (n k) 35,5 2 k27 (n k) 26 k + + = 63,9636,04 nk = 3. (p n A). V d 3: (Cu 21 - M 182 - Khi A - TSH 2007) Trn dung dch cha a mol AlCl3 vi dung dch cha b mol NaOH. thu c kt ta th cn c t l A. a : b = 1 : 4.B. a : b < 1 : 4. C. a : b = 1 : 5.D. a : b > 1 : 4. Hng dn gii Trn a mol AlCl3 vi b mol NaOH thu c kt ta th 33 2 232 2Al3 3OH Al(OH)Al(OH) OH AlO 2H OAl 4OH AlO 2H Oa 4 mol+ + + + ++ + + + kt ta tan hon ton th 3OHAlnn+ > 4 ba > 4. Vy c kt ta th ba < 4 a : b > 1 : 4. (p n D) V d 4: (Cu 37 - M 182 - Khi A - TSH 2007) t chy hon ton a mol axit hu c Y c 2a mol CO2. Mt khc, trung ha a mol Y cn va 2a mol NaOH. Cng thc cu to thu gn ca Y l A. HOOCCH2CH2COOH.B. C2H5COOH. C. CH3COOH.D. HOOCCOOH. Hng dn gii - t a mol axit hu c Y c 2a mol CO2 axit hu c Y c hai nguyn t C trong phn t. - Trung ha a mol axit hu c Y cn dng 2a mol NaOH axit hu c Y c 2 nhm chc cacboxyl (COOH). Cng thc cu to thu gn ca Y l HOOCCOOH. (p n D) V d 5: (Cu 39 - M 182 - Khi A - TSH 2007) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 19 0,09 mol 0,03 mol VNO = 0,0322,4 = 0,672 lt. (p n D) Nhn xt: Phn ng nhit nhm cha bit l hon ton hay khng hon ton do hn hp A khng xc nh c chnh xc gm nhng cht no nn vic vit phng trnh ha hc v cn bng phngtrnhphctp.KhihatanhontonhnhpAtrongaxitHNO3thAl0tothnhAl+3, nguyn t Fe v Cu c bo ton ha tr. C bn s thc mc lng kh NO cn c to bi kim loi Fe v Cu trong hn hp A. Thc cht lng Al phn ng b li lng Fe v Cu to thnh. V d 3: Cho 8,3 gam hn hp X gm Al, Fe (nAl = nFe) vo 100 ml dung dch Y gm Cu(NO3)2 v AgNO3. Sau khi phn ng kt thc thu c cht rn A gm 3 kim loi. Ha tan hon ton cht rn A vo dung dch HCl d thy c 1,12 lt kh thot ra (ktc) v cn li 28 gam cht rn khng tan B. Nng CM ca Cu(NO3)2 v ca AgNO3 ln lt l A. 2M v 1M.B. 1M v 2M. C. 0,2M v 0,1M.D. kt qu khc. Tm tt s : Al Fe8,3 gam hn hp X(n = n )AlFe + 100 ml dung dch Y 33 2AgNO : x molCu(NO ) :y mol Cht rn A(3 kim loi) 2HCl d1,12 lt H2,8 gam cht rn khng tan B+

Hng dn gii Ta c:nAl = nFe = 8,30,1 mol.83 =t 3 AgNOn x mol =v 3 2 Cu( NO )n y mol =X + Y Cht rn A gm 3 kim loi. Al ht, Fe cha phn ng hoc cn d. Hn hp hai mui ht.Qu trnh oxi ha: AlAl3+ + 3eFeFe2+ + 2e 0,10,30,10,2 Tng s mol e nhng bng 0,5 mol. Qu trnh kh: Ag+ + 1eAgCu2+ + 2eCu 2H+ +2eH2 x xxy 2y y 0,1 0,05 Tng s e mol nhn bng (x + 2y + 0,1). 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 20 Theo nh lut bo ton electron, ta c phng trnh: x + 2y + 0,1=0,5 hay x + 2y=0,4(1) Mt khc, cht rn B khng tan l:Ag: x mol ;Cu: y mol. 108x + 64y =28(2) Gii h (1), (2) ta c: x = 0,2 mol ;y = 0,1 mol. 3 MAgNO0,2C0,1== 2M;3 2 MCu( NO )0,1C0,1== 1M. (p n B) Vd4:Ha tan15gamhnhpXgmhaikimloiMgvAlvodungdchYgmHNO3v H2SO4 c thu c 0,1 mol mi kh SO2, NO, NO2, N2O. Phn trm khi lng ca Al v Mg trong X ln lt l A. 63% v 37%.B. 36% v 64%. C. 50% v 50%.D. 46% v 54%. Hng dn gii t nMg = x mol ;nAl = y mol. Ta c:24x + 27y = 15. (1) Qu trnh oxi ha: MgMg2+ + 2eAl Al3+ + 3e x2xy3y Tng s mol e nhng bng (2x + 3y). Qu trnh kh: N+5 + 3eN+22N+5 + 24e2N+1 0,3 0,10,80,2 N+5 + 1eN+4S+6 + 2eS+4 0,1 0,1 0,2 0,1 Tng s mol e nhn bng 1,4 mol. Theo nh lut bo ton electron: 2x + 3y=1,4(2) Gii h (1), (2) ta c: x = 0,4 mol ;y = 0,2 mol. 27 0,2%Al 100% 36%.15= =%Mg = 100% 36% = 64%. (p n B) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 69 Phng php 9 CC I LNG DNG KHI QUT Trongcckimtravthituynsinhtheophngphptrcnghimchngtathyrngs lng cu hi v bi tp kh nhiu v a dng bao trm ton b chng trnh ha hc ph thng. Rt nhiu cc phng php, cc dng bi c bn c bit n. Sau y l mt s v d v dng bi tm mi lin h khi qut gia cc i lng thng xut hin trong trong cc thi tuyn sinh i hc. V d 1: (Cu 11 - M 182 - Khi A - TSH 2007) Cho t t dung dch cha amolHClvo dung dch chabmol Na2CO3 ng thi khuy u, thu c V lt kh ( ktc) v dung dch X. Khi cho d nc vi trong vo dung dch X thy c xut hin kt ta. Biu thc lin h gia V vi a, b l A. V = 22,4(a b).B. V = 11,2(a b). C. V = 11,2(a + b).D. V = 22,4(a + b). Hng dn gii Cho t t dung dch HCl vo dung dch Na2CO3 ta c phng trnh: HCl + Na2CO3 NaHCO3+NaCl(1) b b b mol HCl + NaHCO3 NaCl+CO2|+H2O(2) (a b)(a b) mol Dung dch X cha NaHCO3 d do HCl tham gia phn ng ht, NaHCO3+Ca(OH)2 d CaCO3++NaOH+H2O Vy:V = 22,4(a b). (p n A) V d 2: (Cu 13 - M 182 - Khi A - TSH 2007) Cloho PVC thu cmt polime cha 63,96% clo v khilng, trung bnh 1 phn t clo phn ng vi k mt xch trong mch PVC. Gi tr ca k l A. 3.B. 6.C. 4.D. 5. Hng dn gii Mt phn t Clo phn ng vi k mt xch trong mch PVC theo phng trnh: 2nCH CH|Cl | | | |\ .+ kCl2oxtt 2k n kCH CH CH CH| | |Cl Cl Cl | | | | || ||\ . \ . Do:%mCl = 63,96%%mC,H cn li = 36,04%. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 68 01. Ha tan hon ton m gam Na2O nguyn cht vo 40 gam dung dch NaOH 12% thu c dung dch NaOH 51%. Gi tr ca m (gam) l A. 11,3.B. 20,0.C. 31,8.D. 40,0. 02.Thtchncnguynchtcnthmvo1ltdungdchH2SO498%(d=1,84g/ml)c dung dch mi c nng 10% l A. 14,192 ml.B. 15,192 ml.C. 16,192 ml.D. 17,192 ml. 03. Nguyn t khi trung bnh ca ng 63,54. ng c hai ng v bn: 6329Cuv 6529CuThnh phn % s nguyn t ca 6529Cul A. 73,0%.B. 34,2%.C.32,3%.D. 27,0%. 04.CnlyV1ltCO2vV2ltCOcc24lthnhpCO2vCOctkhihiivi metan bng 2. Gi tr ca V1 (lt) l A. 2.B. 4.C. 6.D. 8. 05. Thm 150ml dung dchKOH 2Mvo 120ml dung dch H3PO4 1M. Khilng ccmui thu c trong dung dch l A. 10,44 gam KH2PO4 ;8,5 gam K3PO4. B. 10,44 gam K2HPO4 ;12,72 gam K3PO4. C. 10,44 gam K2HPO4 ;13,5 gam KH2PO4. D. 13,5 gam KH2PO4;14,2 gam K3PO4. 06. Ha tan 2,84 gam hn hp 2 mui CaCO3 v MgCO3 bng dung dch HCl (d) thu c 0,672 lt kh iu kin tiu chun. Thnh phn % s mol ca MgCO3 trong hn hp l A. 33,33%.B. 45,55%.C. 54,45%.D. 66,67%. 07. Lng SO3 cn thm vo dung dch H2SO4 10% c 100 gam dung dch H2SO4 20% l A. 2,5 gam.B. 8,88 gam.C. 6,66 gam.D. 24,5 gam. 08. Dung dch ru etylic 13,8o c d (g/ml) =?. Bit 2 5 CHOH(ng.cht)d= 0,8 g/ml ; 2 H Od 1 g ml = . A. 0,805.B. 0,8 55.C. 0,972. D. 0,915. 09. Ha tan m gam Al bng dung dch HNO3 long thu c hn hp kh NO v N2O c t khi so vi H2 bng 16,75. T l th tch kh trong hn hp l A. 2 : 3.B. 1 : 2.C. 1 : 3.D. 3 : 1. 10. T1 tn qung hematit A iu ch c 420 kg Fe. T 1 tn qung manhetit B iu ch c 504 kg Fe. Hi phi trn hai qung trn vi t l khi lng (mA : mB) l bao nhiu c 1 tn qung hn hp m t 1 tn qung hn hp ny iu ch c 480 kg Fe. A. 1 : 3.B. 2 : 5.C. 2 : 3.D. 1 : 1. p n cc s bi tp vn dng: 1. B2. C3. D4. C5. B 6. A7. B8. C9. D10. B 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 21 V d 5: Trn 60 gam bt Fe vi 30 gam bt lu hunh ri un nng (khng c khng kh) thu c cht rn A. Ho tan A bng dung dch axit HCl d c dung dch B v kh C. t chy C cn V lt O2 (ktc). Bit cc phn ng xy ra hon ton. V c gi tr l A. 11,2 lt.B. 21 lt.C. 33 lt.D. 49 lt. Hng dn gii VFe S30n n32> =nn Fe d v S ht. Kh C l hn hp H2S v H2. t C thu c SO2 v H2O. Kt qu cui cng ca qu trnh phn ng l Fe v S nhng e, cn O2 thu e. Nhng e:Fe Fe2+ + 2e 60mol56 60256 mol S S+4 +4e

30mol32 30432mol Thu e: Gi s mol O2 l x mol. O2 +4e 2O-2

x mol 4x Ta c: 60 304x 2 456 32= + gii ra x = 1,4732 mol. 2 OV 22,4 1,4732 33 = = lt. (p n C) Vd6:HnhpAgm2kimloiR1,R2cho trx,ykhngi(R1,R2khngtcdngvi ncv ng trc Cu trong dyhot ng hahc ca kimloi). Cho hnhpA phn ng hon ton vi dung dch HNO3 d thu c 1,12 lt kh NO duy nht ktc. Nu cho lng hn hp A trn phn ng hon ton vi dung dch HNO3 th thu c bao nhiu lt N2. Cc th tch kh o ktc. A. 0,224 lt.B. 0,336 lt.C. 0,448 lt.D. 0,672 lt. Hng dn gii Trong bi ton ny c 2 th nghim: TN1: R1 v R2 nhng e cho Cu2+ chuyn thnh Cu sau Cu li nhng e cho 5N+ thnh 2N+(NO). S mol e do R1 v R2 nhng ra l 5N+ + 3e 2N+ 0,1505 , 04 , 2212 , 1= 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 22 TN2: R1 v R2 trc tip nhng e cho 5N+ to ra N2. Gi x l s mol N2, th s mol e thu vo l 25N+ + 10e 02N 10x x mol Ta c: 10x = 0,15 x = 0,015 2 NV = 22,4.0,015 = 0,336 lt. (p n B) V d 7: Cho 1,35 gamhnhp gm Cu,Mg,Al tc dnght vi dung dch HNO3thu chn hp kh gm 0,01 mol NO v 0,04 mol NO2. Tnh khi lng mui to ra trong dung dch. A. 10,08 gam.B. 6,59 gam.C. 5,69 gam.D. 5,96 gam. Hng dn gii Cch 1: t x, y, z ln lt l s mol Cu, Mg, Al. Nhng e: Cu = 2Cu+ + 2e Mg = 2Mg+ + 2e Al = 3Al+ + 3e x x 2xy y2y z z 3z Thu e:5N+ + 3e =2N+ (NO)5N+ + 1e =4N+ (NO2) 0,03 0,010,04 0,04 Ta c: 2x + 2y + 3z = 0,03 + 0,04 = 0,07 v 0,07 cng chnh l s mol NO3

Khi lng mui nitrat l:1,35 + 620,07 = 5,69 gam. (p n C) Cch 2:Nhn nh mi: Khi cho kim loi hoc hn hp kim loi tc dng vi dung dch axit HNO3 to hn hp 2 kh NO v NO2 th 3 2 HNO NO NOn 2n 4n = + 3 HNOn 2 0,04 4 0,01 0,12 = + = mol 2 HOn 0,06 = mol p dng nh lut bo ton khi lng: 3 2 2 KL HNO mui NO NO HOm m m m m m + = + + +1,35 + 0,1263 = mmui + 0,0130 + 0,0446 + 0,0618 mmui=5,69 gam. V d 8: (Cu 19 - M 182 - Khi A - TSH - 2007) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 67 C. 60 gam v 220 gam.D. 40 gam v 240 gam. Hng dn gii 4 2160250CuSO .5H O Ta coi CuSO4.5H2O nh l dung dch CuSO4 c: C% = 160 100250=64%. Gi m1 l khi lng ca CuSO4.5H2O v m2 l khi lng ca dung dch CuSO4 8%. Theo s ng cho: 12m 8 1m 48 6= = . Mt khc m1 + m2=280 gam. Vy khi lng CuSO4.5H2O l: m1 = 28011 6 + = 40 gam v khi lng dung dch CuSO4 8% l: m2=280 40=240 gam. (p n D) V d 10: Cn bao nhiu lt axit H2SO4 (D = 1,84) v bao nhiu lt nc ct pha thnh 9 lt dung dch H2SO4 c D = 1,28 gam/ml? A. 2 lt v 7 lt. B. 3 lt v 6 lt. C. 4 lt v 5 lt.D. 6 lt v 3 lt. Hng dn gii Ta c s ng cho: 22 4H OH SOV 0,56 2V 0, 28 1= = . Cn phi ly 19 31 2 =+lt H2SO4 (d = 1,84 g/ml) v 6 lt H2O.(p n B) MT S BI TP VN DNG GII THEO PHNG PHP S NG CHO 12(m ) 64 8 16 816(m ) 8 64 16 48 = =22 4H O: 1 |1,84 1,28| 0,561,28HSO : 1,84 |1,28 1| 0,28 = =10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 66 Hng dn gii C: 3 4NaOHH POn 0,25 2 51 2n 0, 2 1,5 3< = = < to ra hn hp 2 mui: NaH2PO4, Na2HPO4. S ng cho: 2 42 4Na HPONaH POn 2n 1= 2 4 2 4 Na HPO NaH POn 2n =M: 2 4 2 4 3 4 Na HPO NaH PO H POn n n 0,3 + = = mol 2 42 4Na HPONaH POn 0,2 moln 0,1 mol= = 2 42 4Na HPONaH POm 0,2 142 28,4 gamn 0,1 120 12 gam= = = = (p n C) V d 8: Ha tan 3,164 gamhnhp 2muiCaCO3vBaCO3bng dung dch HCld, thu c 448 ml kh CO2 (ktc). Thnh phn % s mol ca BaCO3 trong hn hp l A. 50%.B. 55%.C. 60%.D. 65%. Hng dn gii 2 CO0,488n22, 4== 0,02 mol 3,164M0,02== 158,2. p dng s ng cho: 3 BaCO58,2%n58,2 38,8=+100% = 60%.(p n C) V d 9: Cnlybao nhiu gam tinh th CuSO4.5H2O vbao nhiu gam dung dch CuSO4 8% pha thnh 280 gam dung dch CuSO4 16%? A. 180 gam v 100 gam.B. 330 gam v 250 gam. 2 4 12 4 25 2Na HPO n 2 13 35n35 1NaH PO n 1 23 3= === =3 13 2BaCO (M 197) 100 158,2 58,2M 158,2CaCO (M 100) 197 158,2 38,8= === =10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 23 Ha tan hon ton 12 gam hn hp Fe, Cu (t l mol 1:1) bng axit HNO3, thu c V lt ( ktc) hn hp kh X (gm NO v NO2) v dung dch Y (ch cha hai mui v axit d). T khi ca X i vi H2 bng 19. Gi tr ca V l A. 2,24 lt. B. 4,48 lt. C. 5,60 lt. D. 3,36 lt. Hng dn gii t nFe = nCu = a mol56a + 64a = 12a = 0,1 mol. Cho e:FeFe3++3eCu Cu2+ +2e 0,1 0,30,1 0,2 Nhn e:N+5+3e N+2N+5+1e N+4 3x x yy Tng ne cho bng tng ne nhn. 3x + y=0,5 Mt khc:30x + 46y=192(x + y). x = 0,125 ;y = 0,125. Vhh kh (ktc)=0,125222,4=5,6 lt. (p n C) V d 9: Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ha tan ht hn hp X trong dung dch HNO3 (d), thot ra 0,56 lt ( ktc) NO (l sn phm kh duy nht). Gi tr ca m l A. 2,52 gam.B. 2,22 gam. C. 2,62 gam. D. 2,32 gam. Hng dn gii m gam Fe + O23 gam hn hp cht rn X 3 HNO d 0,56 lt NO. Thc cht cc qu trnh oxi ha - kh trn l: Cho e:Fe Fe3+ + 3e m563m56mol e Nhn e:O2+4e2O2N+5 +3e N+2 3 m32 4(3 m)32mol e 0,075 mol 0,025 mol 3m56 = 4(3 m)32 + 0,075 m = 2,52 gam. (p n A) V d 10: Hn hp X gm hai kim loi A v B ng trc H trong dy in ha v c ha tr khng i trong cc hp cht. Chia m gam X thnh hai phn bng nhau: - Phn 1: Ha tan hon ton trong dung dch cha axit HCl v H2SO4 long to ra 3,36 lt kh H2. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 24 - Phn 2: Tc dng hon ton vi dung dch HNO3 thu c V ltkh NO (sn phm kh duy nht). Bit cc th tch kh o iu kin tiu chun. Gi tr ca V l A. 2,24 lt.B. 3,36 lt.C. 4,48 lt.D. 6,72 lt. Hng dn gii t hai kim loi A, B l M. - Phn 1:M+nH+ Mn++2nH2(1) - Phn 2:3M+4nH++nNO3 3Mn+ +nNO+2nH2O(2) Theo (1): S mol e ca M cho bng s mol e ca 2H+ nhn; Theo (2): S mol e ca M cho bng s mol e ca N+5 nhn. Vy s mol e nhn ca 2H+ bng s mol e nhn ca N+5. 2H++2e H2 v N+5+ 3eN+2 0,3 0,15 mol0,3 0,1 mol VNO = 0,122,4 = 2,24 lt. (p n A) V d 11: Cho m gam bt Fe vo dung dch HNO3 ly d, ta c hn hp gm hai kh NO2 v NO c VX = 8,96 lt (ktc) v t khi i vi O2 bng 1,3125. Xc nh %NO v %NO2 theo th tch trong hn hp X v khi lng m ca Fe dng? A. 25% v 75%; 1,12 gam.B. 25% v 75%; 11,2 gam. C. 35% v 65%; 11,2 gam.D. 45% v 55%; 1,12 gam. Hng dn gii Ta c:nX = 0,4 mol;MX = 42. S ng cho: 22NO NONO NOn : n 12: 4 3n n 0,4 mol= = + = 2NONOn 0,1 moln 0,3 mol= = 2NONO%V 25%%V 75%= = vFe3e Fe3+ N+5+ 3eN+2N+5+1eN+4 3xx 0,3 0,1 0,3 0,3 Theo nh lut bo ton electron: 3x = 0,6 molx = 0,2 mol 2NO : 46 42 30 1242NO : 30 46 42 4 = =10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 65 81357935% Br 0,319% Br 1,681= 81350,319% Br1,681 0,319=+100% = 15,95%. (p n D) V d 5: Mt hn hp gm O2, O3 iu kin tiu chun c t khi hi vi hiro l 18. Thnh phn % v th tch ca O3 trong hn hp l A. 15%.B. 25%.C. 35%.D. 45%. Hng dn gii p dng s ng cho: 32OOV 4 1V 12 3= = 3 O1%V3 1=+ 100% = 25%.(p n B) V d 6: Cn trn hai th tch metan vi mt th tch ng ng X ca metan thu c hn hp kh c t khi hi so vi hiro bng 15. X l A. C3H8.B. C4H10.C. C5H12.D. C6H14. Hng dn gii p dng s ng cho: 42CH 2MV M 30 2V 14 1= = M2 30 = 28 M2 = 58 14n + 2 = 58 n = 4. Vy: X l C4H10.(p n B) V d 7: Thm 250 ml dung dch NaOH 2M vo 200 ml dung dch H3PO4 1,5M. Mui to thnh v khi lng tng ng l A. 14,2 gam Na2HPO4 ;32,8 gam Na3PO4. B. 28,4 gam Na2HPO4 ;16,4 gam Na3PO4. C. 12 gam NaH2PO4 ;28,4 gam Na2HPO4. D. 24 gam NaH2PO4;14,2 gam Na2HPO4. 32OOV M 48 32 36M 18 2 36V M 32 48 36= = == 42CH 2M 2V M 16 M 30M 15 2 30V M M 16 30= = == 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 64 1245 25 m 20 2m 15 25 10 1= = =.(p n C) V d 2: pha c 500 ml dung dch nc mui sinh l (C = 0,9%) cn ly V ml dung dch NaCl 3% pha vi nc ct. Gi tr ca V l A. 150 ml.B. 214,3 ml.C. 285,7 ml.D. 350 ml. Hng dn gii Ta c s : V1 = 0,95002,1 0,9+ = 150 ml.(p n A) V d 3: Ha tan 200 gam SO3 vo m2 gam dung dch H2SO4 49% ta c dung dch H2SO4 78,4%. Gi tr ca m2 l A. 133,3 gam.B. 146,9 gam.C. 272,2 gam.D. 300 gam. Hng dn gii Phng trnh phn ng: SO3+H2O H2SO4 100 gam SO398 10080 = 122,5 gam H2SO4. Nng dung dch H2SO4 tng ng 122,5%. Gi m1, m2 ln lt l khi lng ca SO3 v dung dch H2SO4 49% cn ly. Theo (1) ta c: 1249 78,4 m 29, 4m 122,5 78,4 44,1= = 244,1m 20029, 4= = 300 gam.(p n D) V d 4: Nguyn t khi trungbnh cabroml79,319. Brom chai ngvbn: 7935Brv 8135Br . Thnh phn % s nguyn t ca 8135Brl A. 84,05.B. 81,02.C. 18,98.D. 15,95. Hng dn gii Ta c s ng cho: V1 (NaCl)V2 (H2O)0,930| 0,9 - 0 || 3 - 0,9 |81357935Br (M 81) 79, 319 79 0, 319A 79, 319Br (M 79) 81 79, 319 1, 681= === =10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 25 mFe = 0,256 = 11,2 gam. (p p B). V d 12: Cho 3 kim loi Al, Fe, Cu vo 2 lt dung dch HNO3 phn ng va thu c 1,792 lt kh X (ktc) gm N2 v NO2 c t khihi so vi Hebng 9,25. Nng mol/lt HNO3 trong dung dch u l A. 0,28M.B. 1,4M.C. 1,7M.D. 1,2M. Hng dn gii Ta c: ( )2 2 N NOXM MM 9, 25 4 372+= = =l trung bnh cng khi lng phn t ca hai kh N2 v NO2 nn: 2 2XN NOnn n 0,04 mol2= = =v NO3+10eN2 NO3 +1e NO2 0,080,40,04 mol0,04 0,04 0,04 mol MMn++n.e 0,04 mol 3 HNO (b kh)n 0,12 mol. =Nhn nh mi: Kim loi nhng bao nhiu electron th cng nhn by nhiu gc NO3 to mui. 3 HNO ( ) ( ) ( )n n.e n.e 0,04 0, 4 0,44 mol.to mui nhng nhn= = = + =Do : 3 HNO ( )n 0,44 0,12 0,56 molphn ng = + =| |30,56HNO 0, 28M.2= =(p n A) V d 13: Khi cho 9,6 gam Mg tc dng ht vi dung dch H2SO4 m c, thy c 49 gam H2SO4 tham gia phn ng, to mui MgSO4, H2O v sn phm khX. X l A. SO2B. SC. H2S D. SO2, H2SHng dn gii Dung dch H2SO4 m c va l cht oxi ha va l mi trng. Gi a l s oxi ha ca S trong X.Mg Mg2++ 2eS+6 + (6-a)e S a 0,4 mol0,8 mol0,1 mol0,1(6-a) mol Tng s mol H2SO4 dng l : 490,598 = (mol) S mol H2SO4 dng to mui bng s mol Mg = 9,6 : 24 = 0,4 mol. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 26 S mol H2SO4 dng oxi ha Mg l: 0,5 0,4 = 0,1 mol. Ta c:0,1(6 a) = 0,8 x = 2. Vy X l H2S. (p n C) V d 14: a gam bt st ngoi khng kh, sau mt thi gian s chuyn thnh hn hp A c khi lng l 75,2 gam gm Fe, FeO, Fe2O3 v Fe3O4. Cho hn hp A phn ng ht vi dung dch H2SO4 m c, nng thu c 6,72 lt kh SO2 (ktc). Khi lng a gam l: A. 56 gam.B. 11,2 gam.C. 22,4 gam.D. 25,3 gam. Hng dn gii S mol Fe ban u trong a gam: Fean56= mol. S mol O2 tham gia phn ng: 2 O75, 2 an32= mol. Qu trnh oxi ha: 3Fe Fe 3ea 3amol mol56 56+ + (1) S mol e nhng: e3an mol56=Qu trnh kh:O2+4e 2O2(2) SO42 + 4H+ + 2eSO2 + 2H2O(3) T (2), (3)cho 2 2 e O SOn 4n 2n = +

75, 2 a 3a4 2 0,332 56= + =a = 56 gam. (p n A) Vd15:Cho1,35gamhnhpAgmCu,Mg,AltcdngviHNO3dc1,12ltNOv NO2 (ktc) c khi lng mol trung bnh l 42,8. Tng khi lng mui nitrat sinh ra l: A. 9,65 gamB. 7,28 gamC. 4,24 gamD. 5,69 gamHng dn gii Da vo s ng cho tnh c s mol NO v NO2 ln lt l 0,01 v 0,04 mol. Ta c cc bn phn ng: NO3 + 4H+ + 3eNO + 2H2O NO3 + 2H+ + 1eNO2 + H2O Nh vy, tng electron nhn l 0,07 mol. Gix,y, zlnltl smol Cu, Mg,Al c trong 1,35 gamhnhp kimloi. Ta c ccbn phn ng:CuCu2++ 2e MgMg2++ 2eAlAl3++ 3e 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 63 Phng php 8 S NG CHO Bi ton trn ln cc cht vi nhau l mt dng bi tp hay gp trong chng trnh ha hc ph thng cngnh trong cc thi kim tra v thi tuyn sinh ihc, cao ng. Ta c th giibi tp dng ny theo nhiu cch khcnhau, songvic giiloi dng bi tp ny theo phng phps ng cho theo tc gi l tt nht. Nguyn tc: Trn ln hai dung dch: Dung dch 1: c khi lng m1, th tch V1, nng C1 (nng phn trm hoc nng mol), khi lng ring d1. Dung dch 2: c khi lng m2, th tch V2, nng C2 (C2 > C1 ), khi lng ring d2. Dung dch thu c: c khi lng m = m1 + m2, th tch V = V1 + V2, nng C (C1 < C < C2) v khi lng ring d. S ng cho v cng thc tng ng vi mi trng hp l: a. i vi nng % v khi lng: 2 12 1C C mm C C=(1) b. i vi nng mol/lt: 2 12 1C C VV C C=(2) c. i vi khi lng ring: 2 12 1C C VV C C=(3) Khi s dng s ng cho cn ch : - Cht rn coi nh dung dch c C = 100% - Dung mi coi nh dung dch c C = 0% - Khi lng ring ca H2O l d = 1g/ml. Sau y l mt s v d s dng phng php s ng cho trong tnh ton cc bi tp. V d 1: thu c dung dch HCl 25% cn ly m1 gam dung dch HCl 45% pha vi m2 gam dung dch HCl 15%. T l m1/m2 l A. 1:2.B. 1:3.C. 2:1.D. 3:1. Hng dn gii p dng cng thc (1): C1C2C| C2 - C || C1 - C |C| C2 - C || C1 - C |` CM1CM2d1d2| d2 - d || d1 - d |d10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 62 V d 7: Nung 8,96 gam Fe trong khng kh c hn hp A gm FeO, Fe3O4, Fe2O3. A ha tan va vn trong dung dch cha 0,5 mol HNO3, bay ra kh NO l sn phm kh duy nht. S mol NO bay ra l. A. 0,01.B. 0,04.C. 0,03.D. 0,02. Hng dn gii Fe8, 96n 0,1656= = mol Quy hn hp A gm (FeO, Fe3O4, Fe2O3) thnh hn hp (FeO, Fe2O3) ta c phng trnh: 2Fe+O22FeO x x 4Fe+3O2 2Fe2O3 y y/2 3FeO+ 10HNO33Fe(NO3)3+NO +2H2O x10x/3x/3 Fe2O3+6HNO3 2Fe(NO3)3+3H2O y/2 3y H phng trnh: x y 0,1610x3y 0,53+ = + =x 0,06 moly 0,1 mol= = NO0,06n 0,023= = mol. (p n D) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 27 2x + 2y + 3z = 0,07. Khi lng mui nitrat sinh ra l: m= 3 2 Cu( NO )m + 3 2 Mg( NO )m + 3 3 Al( NO )m = 1,35 + 62(2x + 2y + 3z) = 1,35 + 62 0,07 = 5,69 gam. MT S BI TP VN DNG GIAI THEO PHNG PHP BO TOM MOL ELECTRON 01. Ho tan hon ton m gam Al vo dung dch HNO3 rt long th thu c hn hp gm 0,015 mol kh N2O v 0,01mol kh NO (phn ng khng to NH4NO3). Gi tr ca m l A. 13,5 gam.B. 1,35 gam.C.0,81 gam.D. 8,1 gam. 02. Chomt lung CO i qua ngs ng 0,04molhnhpA gmFeOvFe2O3 t nng. Sau khi kt thc thnghim thu c cht rn B gm4 cht nng 4,784 gam. Kh i ra khi ng s hpthvodungdchCa(OH)2d,ththuc4,6gamktta.PhntrmkhilngFeO trong hn hp A l A. 68,03%.B. 13,03%.C. 31,03%.D. 68,97%. 03. Mt hn hp gm hai bt kim loi Mg v Al c chia thnh hai phn bng nhau: - Phn 1: cho tc dng vi HCl d thu c 3,36 lt H2.-Phn2:hotanhttrongHNO3longdthucVltmtkhkhngmu,honutrong khng kh (cc th tch kh u o ktc). Gi tr ca V l A. 2,24 lt.B. 3,36 lt.C. 4,48 lt.D. 5,6 lt. 04. Dung dch X gm AgNO3 v Cu(NO3)2 c cng nng . Ly mt lng hn hp gm 0,03 mol Al; 0,05 mol Fe cho vo 100 ml dung dch X cho ti kh phn ng kt thc thu c cht rn Y cha 3 kim loi.Cho Y vo HCl d gii phng 0,07 gam kh. Nng ca hai mui lA. 0,3M.B. 0,4M.C. 0,42M.D. 0,45M. 05. Cho 1,35 gam hn hp Cu, Mg, Al tc dng vi HNO3 d c 896 ml hn hp gm NO v NO2 c M 42 = . Tnh tng khi lng mui nitrat sinh ra (kh ktc). A. 9,41 gam.B. 10,08 gam. C. 5,07 gam.D. 8,15 gam. 06. Ha tan ht 4,43 gamhnhpAlv Mg trong HNO3long thu c dung dchAv 1,568lt (ktc)hnhphaikh(ukhngmu)ckhilng2,59gamtrongcmtkhbha thnh mu nu trong khng kh. Tnh s mol HNO3 phn ng. A. 0,51 mol.B. A. 0,45 mol.C. 0,55 mol.D. 0,49 mol. 07. Ha tan hon ton m gam hn hp gm ba kim loi bng dung dch HNO3 thu c 1,12 lt hn hp kh D (ktc) gm NO2 v NO. T khi hi ca D so vi hiro bng 18,2. Tnh th tch ti thiu dung dch HNO3 37,8% (d = 1,242g/ml) cn dng. A. 20,18 ml.B. 11,12 ml.C. 21,47 ml.D. 36,7 ml. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 28 08. Ha tan 6,25 gam hn hp Zn v Al vo 275 ml dung dch HNO3 thu c dung dch A, cht rn B gm cc kim loi cha tan ht cn nng 2,516 gam v 1,12 lt hn hp kh D ( ktc) gm NO v NO2. T khi ca hn hp D so vi H2 l 16,75. Tnh nng mol/l ca HNO3 v tnh khi lng mui khan thu c khi c cn dung dch sau phn ng. A. 0,65M v 11,794 gam.B. 0,65M v 12,35 gam. C. 0,75M v 11,794 gam.D. 0,55M v 12.35 gam. 09. t chy 5,6 gam bt Fe trong bnh ng O2 thu c 7,36 gam hn hp A gm Fe2O3, Fe3O4 v Fe. Ha tan hon ton lng hn hp A bng dung dch HNO3 thu c V lt hn hp kh B gm NO v NO2. T khi ca B so vi H2 bng 19. Th tch V ktc l A. 672 ml.B. 336 ml.C. 448 ml.D. 896 ml. 10. Cho a gam hn hp A gm oxit FeO, CuO, Fe2O3 c s mol bng nhau tc dng hon ton vi lng va l 250 ml dung dch HNO3 khi un nng nh, thu c dung dch B v 3,136 lt (ktc) hn hp kh C gm NO2 v NO c t khi so vi hiro l 20,143. Tnh a. A. 74,88 gam.B. 52,35 gam.C. 61,79 gam.D. 72,35 gam. p n cc bi tp vn dng 1. B2. B3. A4. B5. C 6. D7. C8. A9. D10. A 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 61 V d 5: Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ha tan ht hn hp X trong dung dch HNO3 (d) thot ra 0,56 lt NO ( ktc) (l sn phm kh duy nht). Gi tr ca m l A. 2,52 gam.B. 2,22 gam.C. 2,62 gam.D. 2,32 gam. Hng dn gii Quy hn hp cht rn X v hai cht Fe, Fe2O3: Fe + 4HNO3Fe(NO3)3 + NO+ 2H2O 0,025 0,025 0,025 mol 2 3 Fe Om = 3 560,025 = 1,6 gam 2 3 Fe ( trong Fe O )1,6m 2160= = 0,02 mol mFe = 56(0,025 + 0,02) = 2,52 gam. (p n A) V d 6: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol, ha tan ht vo dungdchYgm(HClvH2SO4long)dthucdungdchZ.Nhttdungdch Cu(NO3)21MvodungdchZchotikhingngthotkhNO.Thtchdungdch Cu(NO3)2 cn dng v th tch kh thot ra ktc thuc phng n no? A. 25 ml; 1,12 lt.B. 0,5 lt; 22,4 lt. C. 50 ml; 2,24 lt. D. 50 ml; 1,12 lt. Hng dn gii Quy hn hp 0,1 mol Fe2O3 v 0,1 mol FeO thnh 0,1 mol Fe3O4.Hn hp X gm:Fe3O4 0,2 mol; Fe 0,1 mol + dung dch Y Fe3O4 + 8H+Fe2++2Fe3++4H2O 0,2 0,2 0,4 mol Fe+2H+ Fe2++H2|

0,10,1 mol Dung dch Z: (Fe2+: 0,3 mol; Fe3+: 0,4 mol) + Cu(NO3)2: 3Fe2++NO3+4H+ 3Fe3++NO|+2H2O 0,3 0,1 0,1 mol VNO =0,122,4 = 2,24 lt. 3 2 3Cu( NO ) NO1n n2 = = 0,05 mol. 23 2 d Cu( NO )0,05V1== 0,05 lt(hay 50 ml). (p n C) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 60 V d 3: Ha tan hon ton 49,6 gam hn hp X gm Fe, FeO, Fe2O3, Fe3O4 bng H2SO4 c nng thu c dung dch Y v 8,96 lt kh SO2 (ktc). a) Tnh phn trm khi lng oxi trong hn hp X. A. 40,24%.B. 30,7%.C. 20,97%.D. 37,5%. b) Tnh khi lng mui trong dung dch Y. A. 160 gam.B.140 gam.C. 120 gam.D. 100 gam. Hng dn gii Quy hn hp X v hai cht FeO, Fe2O3, ta c: 2 4 2 4 3 2 22 3 2 4 2 4 3 22FeO 4H SO Fe (SO ) SO 4H O0,8 0,4 0,4 mol49,6 gamFe O 3H SO Fe (SO ) 3H O0,05 0,05 mol + + + + + 2 3 Fe Om = 49,6 0,872 = 8 gam (0,05 mol) nO (X) = 0,8 + 3(0,05) = 0,65 mol. Vy: a)O0,65 16 100%m49,9 == 20,97%. (p n C) b)2 4 3 Fe (SO )m = [0,4 + (-0,05)]400 = 140 gam. (p n B) V d 4: kh hon ton 3,04 gam hnhp X gm FeO, Fe2O3, Fe3O4 th cn 0,05 mol H2. Mt khc ha tan hon ton 3,04 gam hn hp X trong dung dch H2SO4 c nng th thu c th tch kh SO2 (sn phm kh duy nht ktc) l. A. 224 ml.B. 448 ml.C. 336 ml.D. 112 ml. Hng dn gii Quy hn hp X v hn hp hai cht FeO v Fe2O3 vi s mol l x, y, ta c: FeO+H2 ot Fe+H2O x y Fe2O3+3H2 ot 2Fe+3H2O x3y x 3y 0,0572x 160y 3,04+ = + = x 0,02 moly 0,01 mol= = 2FeO+4H2SO4 Fe2(SO4)3+SO2+4H2O 0,02 0,01 mol Vy: 2 SOV = 0,0122,4 = 0,224 lt(hay 224 ml). (p n A) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 29 Phng php 4 S DNG PHNG TRNH ION - ELETRON lmttccbitonbngphngphpioniuutinccbnphinmchcphng trnh phn ng di dngcc phn t t suy ra cc phng trnhion, i khi cmt s bi tp khng th gii theo cc phng trnh phn t c m phi gii da theo phng trnh ion. Vic gii bi ton ha hc bng phng php ion gip chng ta hiu k hn v bn cht ca cc phng trnh ha hc. T mt phng trnhion c th ngvi rt nhiu phng trnh phn t. V d phn ng gia hn hp dung dch axit vi dung dch baz u c chung mt phng trnh ion l H++OH H2O hoc phn ng ca Cu kim loi vi hn hp dung dch NaNO3 v dung dch H2SO4 l 3Cu+8H++2NO3 3Cu2++2NO|+4H2O... Sau y l mt s v d: V d 1: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol, ha tan ht vo dungdchYgm(HClvH2SO4long)dthucdungdchZ.Nhttdungdch Cu(NO3)21MvodungdchZchotikhingngthotkhNO.Thtchdungdch Cu(NO3)2 cn dng v th tch kh thot ra ktc thuc phng n no? A. 25 ml; 1,12 lt.B. 0,5 lt; 22,4 lt. C. 50 ml; 2,24 lt.D. 50 ml; 1,12 lt. Hng dn gii Quy hn hp 0,1 mol Fe2O3 v 0,1 mol FeO thnh 0,1 mol Fe3O4.Hn hp X gm: (Fe3O4 0,2 mol; Fe 0,1 mol) tc dng vi dung dch Y Fe3O4 + 8H+Fe2+ +2Fe3++4H2O 0,2 0,2 0,4 mol Fe+2H+ Fe2++H2|

0,1 0,1 mol Dung dch Z: (Fe2+: 0,3 mol; Fe3+: 0,4 mol) + Cu(NO3)2: 3Fe2++NO3+4H+ 3Fe3++NO|+2H2O 0,30,10,1 mol VNO =0,122,4 = 2,24 lt. 3 2 3Cu( NO ) NO1n n 0,052 = = mol 3 2 dd Cu( NO )0,05V 0,051= = lt(hay 50 ml).(p n C) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 30 V d 2: Ha tan 0,1molCu kimloi trong 120 ml dung dch X gmHNO3 1Mv H2SO4 0,5M. Sau khi phn ng kt thc thu c V lt kh NO duy nht (ktc). Gi tr ca V l A. 1,344 lt.B. 1,49 lt.C. 0,672 lt.D. 1,12 lt. Hng dn gii 3 HNOn 0,12 = mol ;2 4 H SOn 0,06 = mol Tng: Hn 0,24 + = mol v 3 NOn 0,12 = mol. Phng trnh ion: 3Cu + 8H+ +2NO3 3Cu2++2NO|+4H2O Ban u: 0,1 0,240,12 mol Phn ng:0,09 0,240,060,06 mol Sau phn ng: 0,01 (d)(ht)0,06 (d) VNO=0,0622,4=1,344 lt. (p n A) Vd3:DungdchXchadungdchNaOH0,2MvdungdchCa(OH)20,1M.Sc7,84ltkh CO2 (ktc) vo 1 lt dung dch X th lng kt ta thu c l A. 15 gam.B. 5 gam.C. 10 gam.D. 0 gam. Hng dn gii 2 COn = 0,35 mol ;nNaOH = 0,2 mol;2 Ca(OH)n = 0,1 mol. Tng: OHn = 0,2 + 0,12 = 0,4 mol v2Can + = 0,1 mol. Phng trnh ion rt gn: CO2+2OH CO32+H2O 0,35 0,4 0,2 0,4 0,2 mol 2 CO ( )nd= 0,35 0,2 = 0,15 mol tip tc xy ra phn ng: CO32+CO2+H2O 2HCO3 Ban u:0,2 0,15 mol Phn ng:0,15 0,15 mol 23 COn cn li bng 0,15 mol 3CaCOn+= 0,05 mol 3 CaCOm = 0,05100 = 5 gam. (p n B) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 59 ta c: 22 2 32Fe O 2FeO0,1 0,1 mol0,15 mol4Fe 3O 2Fe O0,05 0,025 mol + + 2h Xm = 0,172 + 0,025160 = 11,2 gam. (p n A) Ch : Vn c th quyhnhp Xvhai cht (FeO vFe3O4) hoc (Fev FeO), hoc (Fev Fe3O4) nhng vic gii tr nn phc tp hn (c th l ta phi t n s mol mi cht, lp h phng trnh, gii h phng trnh hai n s). - Quy hn hp X v mt cht l FexOy: FexOy + (6x2y)HNO3Fe(NO3)3+ (3x2y) NO2 + (3xy)H2O 0,13x 2y mol0,1 mol. Fe8,4 0,1.xn56 3x 2y= =x 6y 7=mol. Vy cng thc quy i l Fe6O7 (M = 448) v 6 7 Fe O0,1n3 6 2 7= = 0,025 mol. mX = 0,025448 = 11,2 gam. Nhn xt: Quy ihnhp gm Fe,FeO, Fe2O3, Fe3O4 vhnhphai chtlFeO, Fe2O3l n gin nht. V d 2: Ha tan ht m gam hn hp X gm FeO, Fe2O3, Fe3O4 bng HNO3 c nng thu c 4,48 lt kh NO2 (ktc). C cn dung dchsau phn ng thu c 145,2 gammui khan gi tr ca m l A. 35,7 gam.B. 46,4 gam.C. 15,8 gam.D. 77,7 gam. Hng dn gii Quy hn hp X v hn hp hai cht FeO v Fe2O3 ta c FeO + 4HNO3Fe(NO3)3 + NO2 + 2H2O 0,2 mol0,2 mol 0,2 mol Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O 0,2 mol 0,4 mol 3 3 Fe( NO )145, 2n242== 0,6 mol. mX = 0,2(72 + 160) = 46,4 gam. (p n B) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 58 Phng php 7 QUI I HN HP NHIU CHT V S LNG CHT T HN Mtsbitonhahcc thgiinhanhbngccphngphpbo tonelectron,bo ton nguynt,botonkhilngsongphngphpquyicngtmrapsrtnhanhvl phng php tng i u vit, c th vn dng vo cc bi tp trc nghim phn loi hc sinh. Cc ch khi p dng phng php quy i: 1. Khi quy i hn hp nhiu cht (hn hp X) (t ba cht tr ln) thnh hn hp hai cht hay ch cn mt cht ta phi bo ton s mol nguyn t v bo ton khi lng hn hp. 2. C th quy i hn hp X v bt k cp cht no, thm ch quy i v mt cht. Tuy nhin ta nn chn cp cht no n gin c t phn ng oxi ha kh nht n gin vic tnh ton. 3. Trong qu trnh tnh ton theo phng php quy i i khi ta gp s m l do s b tr khilng ca cc cht tronghnhp. Trong trnghpny tavn tnh tonbnh thngv kt qu cui cng vn tha mn. 4.KhiquyihnhpXvmtchtlFexOythoxitFexOytmcchloxitginh khng c thc. Vd1:Nung8,4gamFetrongkhngkh,sauphnngthucmgamchtrnXgmFe, Fe2O3, Fe3O4, FeO. Ha tanm gamhn hp Xvo dung dch HNO3 d thu c 2,24lt kh NO2 (ktc) l sn phm kh duy nht. Gi tr ca m l A. 11,2 gam.B. 10,2 gam.C. 7,2 gam.D. 6,9 gam. Hng dn gii - Quy hn hp X v hai cht Fe v Fe2O3: Ha tan hn hp X vo dung dch HNO3 d ta c Fe + 6HNO3Fe(NO3)3+ 3NO2+ 3H2O 0,13 0,1 mol S mol ca nguyn t Fe to oxit Fe2O3 l Fe8,4 0,1 0,35n56 3 3= = 2 3 Fe O0,35n3 2= Vy: 2 3 X Fe Fe Om m m = + X0,1 0,35m 56 1603 3= + = 11,2 gam. - Quy hn hp X v hai cht FeO v Fe2O3: FeO + 4HNO3Fe(NO3)3+NO2+ 2H2O 0,1 0,1 mol 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 31 V d 4: Ha tan ht hn hp gm mt kim loi kim v mt kim loi kim th trong nc c dung dch A v c 1,12 lt H2 bay ra ( ktc). Cho dung dch cha 0,03 mol AlCl3 vo dung dch A. khi lng kt ta thu c l A. 0,78 gam.B. 1,56 gam.C. 0,81 gam.D. 2,34 gam. Hng dn gii Phn ng ca kim loi kim v kim loi kim th vi H2O: M+nH2O M(OH)n+2nH2 T phng trnh ta c: 2 H OHn 2n = = 0,1mol. Dung dch A tc dng vi 0,03 mol dung dch AlCl3: Al3+ + 3OH Al(OH)3+ Ban u:0,03 0,1 mol Phn ng:0,03 0,09 0,03 mol OH ( )nd = 0,01mol tip tc ha tan kt ta theo phng trnh: Al(OH)3+OH AlO2+2H2O 0,010,01 mol Vy: 3 Al(OH)m = 780,02 = 1,56 gam. (p n B) V d 5: Dung dch A cha 0,01 mol Fe(NO3)3 v 0,15 mol HCl c kh nng ha tan ti a bao nhiu gam Cu kim loi? (Bit NO l sn phm kh duy nht) A. 2,88 gam.B. 3,92 gam.C. 3,2 gam.D. 5,12 gam. Hng dn gii Phng trnh ion: Cu + 2Fe3+ 2Fe2+ + Cu2+ 0,005 0,01 mol 3Cu + 8H+ +2NO3 3Cu2++2NO|+4H2O Ban u: 0,150,03 mol H+ d Phn ng:0,045 0,12 0,03 mol mCu ti a=(0,045 + 0,005) 64=3,2 gam. (p n C) Vd6:ChohnhpgmNaClvNaBrtcdngvidungdchAgNO3dthuckt tac khilng ngbng khilngAgNO3 phnng. Tnh phn trm khilng NaCl trong hn hp u. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 32 A. 23,3%B. 27,84%.C. 43,23%.D. 31,3%. Hng dn gii Phng trnh ion:Ag++Cl AgCl+ Ag++Br AgBr+ t:nNaCl = x mol ;nNaBr = y mol mAgCl + mAgBr=3( ) AgNOmp. 3 Cl Br NOm m m + =35,5x + 80y=62(x + y) x : y=36 : 53 Chn x = 36, y = 53NaCl58,5 36 100%m58,5 36 103 53 = + =27,84%. (p n B) V d 7: Trn 100 ml dung dch A (gm KHCO3 1M v K2CO3 1M) vo 100 ml dung dch B (gm NaHCO3 1M v Na2CO3 1M) thu c dung dch C. Nh t t 100 ml dung dch D (gm H2SO4 1M v HCl 1M) vo dung dch C thu c V lt CO2 (ktc) v dung dch E. Cho dung dch Ba(OH)2 ti d vo dung dch E th thu c m gam kt ta. Gi tr ca m v V ln lt l A. 82,4 gam v 2,24 lt.B. 4,3 gam v 1,12 lt. C. 43 gam v 2,24 lt.D. 3,4 gam v 5,6 lt. Hng dn gii Dung dch C cha:HCO3 : 0,2 mol ; CO32 : 0,2 mol. Dung dch D c tng: Hn + = 0,3 mol. Nh t t dung dch C v dung dch D: CO32+H+ HCO3 0,2 0,2 0,2 mol HCO3+H+ H2O+CO2 Ban u:0,40,1 mol Phn ng:0,1 0,1 0,1 mol D:0,3 mol Tip tc cho dung dch Ba(OH)2 d vo dung dch E: Ba2++ HCO3 + OH BaCO3+ + H2O 0,3 0,3 mol 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 57 07. Nhng mt thanh st nng 12,2 gam vo 200 ml dung dch CuSO4 0,5M. Sau mt thi gian ly thanh kim loi ra, c cn dung dch c 15,52 gam cht rn khan. a)Vitphngtrnhphnngxyra,tmkhilngtngchtctrong15,52gamchtrn khan. b)Tnhkhilngthanhkimloisauphnng.Hatanhontonthanhkimloinytrong dung dch HNO3 c nng, d thu c kh NO2 duy nht, th tch V lt (o 27,3 oC, 0,55 atm). Vit cc phng trnh phn ng xy ra. Tnh V. 08.Ngmmtthanhngckhilng140,8gamvodungdchAgNO3saumtthigianly thanh ng em cnli thynng 171,2 gam. Tnh thnh phn khilngca thanh ngsau phn ng. 09. Ngm mt l km nh trong mt dung dch c cha 2,24 gam ion kim loi c in tch 2+. Phn ng xong, khi lng l km tng thm 0,94 gam. Hy xc nh tn ca ion kim loi trong dung dch. 10. C hai l kim loi cng cht, cng khi lng, c kh nng to ra hp cht c s oxi ha +2. Mt l c ngm trong dung dch Pb(NO3)2 cn l kia c ngm trong dung dch Cu(NO3)2. Sau mt thi gian ngi ta ly l kim loi ra khi dung dch, ra nh. Nhn thy khi lng l kim loi c ngm trong mui ch tng thm 19%, khi lng l kim loi kia gim 9,6%. Bit rng, trong hai phn ng trn, khi lng cc kim loi b ha tan nh nhau. Hy xc nh tn ca hai l kim loi ang dng. p n cc bi tp vn dng: 01. B02. D.03. B.04. A. 05. Fe2O3. 06. VCO = 8,512 lt ;%nFe = 46,51% ;%nFeO = 37,21% ;2 3 Fe O%n 16,28%. =07. a) 6,4 gam CuSO4 v 9,12 gam FeSO4. b) mKL = 12,68 gam ;2 NOV 26,88 = lt. 08. Thanh Cu sau phn ng c mAg (bm) = 43,2 gam v mCu (cn li) = 128 gam. 09. Cd2+ 10. Cd 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 56 tng thm 0,8 gam. C cn dung dch sau phn ng thu c m gam mui khan. Gi tr m l A. 4,24 gam.B. 2,48 gam.C. 4,13 gam.D. 1,49 gam. Hng dn gii p dng nh lut bo ton khi lng: Sau mt khong thi gian tng khi lng ca thanh Fe bng gim khi lng ca dung dch mui. Do : m = 3,28 0,8=2,48 gam. (p n B) MT S BI TP VN DNG GII THEO PHNG PHP TNG GIM KHI LNG 01. Cho 115 gam hn hp gm ACO3, B2CO3, R2CO3 tc dng ht vi dung dch HCl thy thot ra 22,4 lt CO2 (ktc). Khi lng mui clorua to ra trong dung dch l A. 142 gam.B. 126 gam.C. 141 gam.D. 132 gam. 02. Ngm mt l st trong dung dch CuSO4. Nu bit khi lng ng bm trn l st l 9,6 gam th khi lng l st sau ngm tng thm bao nhiu gam so vi ban u? A. 5,6 gam.B. 2,8 gam.C. 2,4 gam.D. 1,2 gam. 03. Cho hai thanh st c khi lng bng nhau. - Thanh 1 nhng vo dung dch c cha a mol AgNO3. - Thanh 2 nhng vo dung dch c cha a mol Cu(NO3)2. Sau phn ng, ly thanh st ra, sy kh v cn li thy s cho kt qu no sau y? A. Khi lng hai thanh sau nhng vn bng nhau nhng khc ban u. B. Khi lng thanh 2 sau nhng nh hn khi lng thanh 1 sau nhng. C. Khi lng thanh 1 sau nhng nh hn khi lng thanh 2 sau nhng. D. Khi lng hai thanh khng i vn nh trc khi nhng. 04.ChoVltdungdchAchangthiFeCl31MvFe2(SO4)30,5Mtcdngvidungdch Na2CO3 c d, phn ng kt thc thy khi lng dung dch sau phn ng gim 69,2 gam so vi tng khi lng ca cc dung dch ban u. Gi tr ca V l: A. 0,2 lt. B. 0,24 lt.C. 0,237 lt.D.0,336 lt. 05. Cho lung kh CO i qua 16 gam oxit st nguyn cht cnung nng trong mt ci ng.Khi phn ng thc hin hon ton v kt thc, thy khi lng ng gim 4,8 gam. Xc nh cng thc v tn oxit st em dng. 06. Dng CO kh 40 gam oxit Fe2O3 thu c 33,92 gam cht rn B gm Fe2O3, FeO v Fe. Cho 1B2 tc dng vi H2SO4 long d, thu c 2,24 lt kh H2 (ktc). Xc nh thnh phn theo s mol cht rn B, th tch kh CO (ktc) ti thiu c c kt qu ny. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 33 Ba2+ + SO42 BaSO4 0,1 0,1 mol 2 COV = 0,122,4 = 2,24 lt. Tng khi lng kt ta: m = 0,3197 + 0,1233 = 82,4 gam.(p n A) V d 8: Ha tan hon ton 7,74 gam mt hn hp gm Mg, Al bng 500 ml dung dch gm H2SO4 0,28M v HCl 1M thu c 8,736 lt H2 (ktc) v dung dch X. ThmVltdungdchchangthiNaOH1MvBa(OH)20,5MvodungdchXthu c lng kt ta ln nht. a) S gam mui thu c trong dung dch X l A. 38,93 gam.B. 38,95 gam. C. 38,97 gam.D. 38,91 gam. b) Th tch V l A. 0,39 lt.B. 0,4 lt. C. 0,41 lt.D. 0,42 lt. c) Lng kt ta l A. 54,02 gam.B. 53,98 gam. C. 53,62 gam.D. 53,94 gam. Hng dn gii a) Xc nh khi lng mui thu c trong dung dch X: 2 4 H SOn = 0,280,5 = 0,14 mol 24 SOn = 0,14 molv Hn + = 0,28 mol. nHCl = 0,5 mol Hn + = 0,5 molvCln = 0,5 mol. Vy tng Hn + = 0,28 + 0,5 = 0,78 mol. M2 Hn = 0,39 mol. Theo phng trnh ion rt gn: Mg0+2H+ Mg2++H2|(1) Al+3H+ Al3++32H2|(2) Ta thy 2 H H ( p)n 2n + = H+ ht. mhh mui=mhh k.loi +24 SO Clm m + =7,74 + 0,1496 + 0,535,5 = 38,93gam. (p n A) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 34 b) Xc nh th tch V: 2NaOHBa(OH)n 1V moln 0, 5V mol= `=) Tng OHn = 2V mol v2Ban + = 0,5V mol. Phng trnh to kt ta: Ba2+ +SO42BaSO4+(3) 0,5V mol0,14 mol Mg2++2OH Mg(OH)2+(4) Al3++3OH Al(OH)3+(5) kt ta t ln nht th s mol OH kt ta ht cc ion Mg2+ v Al3+. Theo cc phng trnh phn ng (1), (2), (4), (5) ta c: Hn + =OHn = 0,78 mol 2V = 0,78 V = 0,39 lt. (p n A) c) Xc nh lng kt ta: 2Ban + = 0,5V = 0,50,39 = 0,195 mol > 0,14 mol Ba2+ d. 4 BaSOm = 0,14233 = 32,62 gam. Vy mkt ta=4 BaSOm + m2 k.loi + OHm =32,62 + 7,74 + 0,78 17=53,62 gam. (p n C) V d 9: (Cu 40 - M 182 - TS i Hc - Khi A 2007) Chom gamhnhp Mg,Alvo 250mldung dch X chahnhp axitHCl 1Mv axit H2SO4 0,5M, thu c 5,32lt H2 ( ktc) v dung dch Y (coi th tch dung dch khng i). Dung dch Y c pH l A. 1. B. 6. C. 7. D. 2. Hng dn gii nHCl = 0,25 mol ;2 4 H SOn = 0,125. Tng: Hn + = 0,5 mol ; 2 H ( )nto thnh= 0,2375 mol. Bit rng: c 2 mol ion H+1 mol H2 vy 0,475 mol H+ 0,2375 mol H2 H ( )nd+ = 0,5 0,475 = 0,025 mol 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 55 V d 12: Cho 3,78 gam bt Al phn ng va vi dung dch mui XCl3 to thnh dung dch Y. Khilng cht tan trong dung dch Y gim 4,06 gam so vi dung dch XCl3. xc nh cng thc ca mui XCl3. A. FeCl3.B. AlCl3.C. CrCl3.D. Khng xc nh. Hng dn gii Gi A lnguyn t khi ca kim loi X. Al +XCl3 AlCl3+X

3,7827 = (0,14 mol) 0,140,14 mol. Ta c :(A + 35,53)0,14 (133,50,14) = 4,06 Gii ra c: A = 56. Vy kim loi X l Fe v mui FeCl3. (p n A) V d 13: Nung 100 gam hn hp gm Na2CO3 v NaHCO3 cho n khi khi lng hn hp khng i c 69 gam cht rn. Xc nh phn trm khi lng ca mi cht tng ng trong hn hp ban u. A. 15,4% v 84,6%.B. 22,4% v 77,6%. C. 16% v 84%.D. 24% v 76%. Hng dn gii Ch c NaHCO3 b phn hy. t x l s gam NaHCO3.2NaHCO3 ot Na2CO3 + CO2| + H2O C nung 168 gamkhi lng gim: 44 + 18 = 62 gam x khi lng gim: 100 69 = 31 gam Ta c:168 62x 31= x = 84 gam. Vy NaHCO3 chim 84% v Na2CO3 chim 16%. (p n C) V d 14: Ha tan 3,28 gam hn hp mui CuCl2 v Cu(NO3)2 vo nc c dung dch A. Nhng Mg vo dung dch A cho n khi mt mu xanh ca dung dch. Ly thanh Mg ra cn li thytngthm0,8gam.Ccndungdchsauphnngthucmgammuikhan. Tnh m? A. 1.28 gam.B. 2,48 gam.C. 3,1 gam.D. 0,48 gam. Hng dn gii Ta c: mtng = mCu mMg phn ng = ( ) 2 2 2Cu Mg Mgm m 3, 28 m m 0,8gc axit + + + = + =m = 3,28 0,8 = 2,48 gam. (p n B) V d 15: Ha tan 3,28 gam hn hp mui MgCl2 v Cu(NO3)2 vo nc c dung dch A. Nhng vodungdchAmtthanhst.Saumtkhongthigianlythanhstracnlithy 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 54 Gi khi lng thanh km ban u l a gam th khi lng tng thm l 2,35a100 gam. Zn+CdSO4ZnSO4 +Cd 65 1 mol112, tng (112 65) = 47 gam 8,32208 (=0,04 mol) 2,35a100 gam Ta c t l: 1 472,35a 0,04100= a = 80 gam. (p n C) V d 11: Nhng thanh kim loi M ho tr 2 vo dung dch CuSO4, sau mt thi gian ly thanh kim loi ra thy khilng gim 0,05%. Mt khcnhng thanh kimloi trnvo dung dch Pb(NO3)2,saumtthigianthykhilngtng7,1%.XcnhM,bitrngsmol CuSO4 v Pb(NO3)2 tham gia 2 trng hp nh nhau. A. Al.B. Zn.C. Mg.D. Fe. Hng dn gii Gi m l khi lng thanh kim loi, M l nguyn t khi ca kimloi, x l s mol mui phn ng. M+ CuSO4MSO4 + Cu+ M (gam) 1 mol 64 gam, gim (M 64)gam. x mol gim 0,05. m100 gam. x = 0,05.m100M 64 (1) M + Pb(NO3)2 M(NO3)2 + Pb+ M (gam) 1 mol 207, tng (207 M) gam x mol tng 7,1.m100 gam x = 7,1. m100207 M (2) T (1) v (2) ta c: 0,05.m100M 64 = 7,1. m100207 M (3) T (3) gii ra M = 65. Vy kim loi M l km. (p n B) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 35 0,025H0,25+ ( = = 0,1 = 101M pH = 1. (p n A) V d 10: (Cu 40 - M 285 - Khi B - TSH 2007) Thc hin hai th nghim: 1) Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO3 1M thot ra V1 lt NO. 2) Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO4 0,5 M thot ra V2 lt NO. Bit NO l sn phm kh duy nht, cc th tch kh o cng iu kin. Quan h gia V1 v V2 l A. V2 = V1.B. V2 = 2V1.C. V2 = 2,5V1.D. V2 = 1,5V1. Hng dn gii TN1:3CuHNO3,84n 0,06 mol64n 0,08 mol= = = 3HNOn 0,08 moln 0,08 mol+= = 3Cu+8H++2NO3 3Cu2++2NO|+4H2O Ban u:0,060,080,08 molH+ phn ng ht Phn ng:0,03 0,08 0,020,02 mol V1 tng ng vi 0,02 mol NO. TN2:nCu = 0,06 mol ;3 HNOn = 0,08 mol ;2 4 H SOn = 0,04 mol. Tng:Hn + = 0,16 mol ; 3 NOn = 0,08 mol. 3Cu+ 8H++2NO3 3Cu2++2NO|+4H2O Ban u:0,060,160,08 mol Cu v H+ phn ng ht Phn ng:0,06 0,16 0,04 0,04 mol V2 tng ng vi 0,04 mol NO. Nh vy V2 = 2V1. (p n B) V d 11: (Cu 33 - M 285 - Khi B - TSH 2007) Trn 100 ml dung dch (gm Ba(OH)2 0,1M v NaOH 0,1M) vi 400 ml dung dch (gm H2SO4 0,0375M v HCl 0,0125M), thu c dung dch X. Gi tr pH ca dung dch X l A. 7. B. 2. C. 1. D. 6. Hng dn gii 2 Ba(OH)NaOHn 0,01 moln 0,01 mol= `=) Tng OHn = 0,03 mol. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 36 2 4 H SOHCln 0,015 moln 0,005 mol= `=) Tng Hn + = 0,035 mol. Khi trn hn hp dung dch baz vi hn hp dung dch axit ta c phng trnh ion rt gn: H+ +OH H2O Bt u0,035 0,03 mol Phn ng:0,030,03 Sau phn ng: H ( )nd+= 0,035 0,03 = 0,005 mol. Tng:Vdd (sau trn) = 500 ml(0,5 lt). 0,005H0,5+ ( = = 0,01 = 102 pH = 2. (p n B) V d 12: (Cu 18 - M 231 - TS Cao ng - Khi A 2007) Cho mt mu hp kim Na-Ba tc dng vi nc (d), thu c dung dch X v 3,36 lt H2 ( ktc). Th tch dung dch axit H2SO4 2M cn dng trung ho dung dch X l A. 150 ml. B. 75 ml. C. 60 ml. D. 30 ml. Hng dn gii Na+H2O NaOH+12H2 Ba+2H2O Ba(OH)2+H2 2 Hn = 0,15 mol, theo phng trnhtng s22 H OH (d X)n 2n = = 0,3 mol. Phng trnh ion rt gn ca dung dch axit vi dung dch baz l H++OH H2O Hn + = OHn = 0,3 mol2 4 H SOn = 0,15 mol 2 4 H SO0,15V2== 0,075 lt(75 ml). (p n B) V d 13: Ha tan hn hp X gm hai kim loi A v B trong dung dch HNO3 long. Kt thc phn ngthuchnhpkhY(gm0,1molNO,0,15molNO2v0,05molN2O).Bit rng khng c phn ng to mui NH4NO3. S mol HNO3 phn ng l:A. 0,75 mol. B. 0,9 mol.C. 1,05 mol. D. 1,2 mol. Hng dn gii Ta c bn phn ng: NO3+2H++1e NO2+H2O(1) 2 0,15 0,15 NO3+4H++ 3e NO+2H2O(2) 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 53 A. 12,8 gam; 32 gam.B. 64 gam; 25,6 gam. C. 32 gam; 12,8 gam.D. 25,6 gam; 64 gam. Hng dn gii V trong cng dung dch cn li (cng th tch) nn: [ZnSO4] = 2,5 [FeSO4] 4 4 ZnSO FeSOn 2,5n =Zn + CuSO4 ZnSO4+Cu+(1) 2,5x 2,5x 2,5x mol Fe + CuSO4 FeSO4+Cu+(2) x x xx mol T (1), (2) nhn c gim khi lng ca dung dch l mCu (bm) mZn (tan) mFe (tan) 2,2 = 64(2,5x + x) 652,5x 56x x = 0,4 mol. Vy:mCu (bmln thanh km) = 642,50,4 = 64 gam; mCu (bmln thanh st) = 640,4= 25,6 gam.(p n B) V d 9: (Cu 15 - M 231 - TSC - Khi A 2007) Cho5,76gamaxithucXnchc,mchhtcdnghtviCaCO3thuc7,28 gam mui ca axit hu c. Cng thc cu to thu gn ca X l A. CH2=CHCOOH. B. CH3COOH. C. HCCCOOH.D. CH3CH2COOH. Hng dn gii t CTTQ ca axit hu c X n chc l RCOOH. 2RCOOH+CaCO3 (RCOO)2Ca+CO2|+H2O C 2 mol axit phn ng to mui th khi lng tng (40 2) = 38 gam. x mol axit (7,28 5,76) = 1,52 gam. x = 0,08 molRCOOH5,76M 720,08= = R = 27 Axit X: CH2=CHCOOH. (p n A) V d 10: Nhng thanh km vo dung dch cha 8,32 gam CdSO4. Sau khi kh hon ton ion Cd2+ khi lng thanh km tng 2,35% so vi ban u. Hi khi lng thanh km ban u. A. 60 gam.B. 70 gam.C. 80 gam.D. 90 gam. Hng dn gii 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 52 Vy: x (gam) = 0,52. M216 M khi lng kim loi tng 0,52 gam. Ta c: 0, 24. MM 64 = 0,52. M216 M M = 112 (kim loi Cd). (p n B) V d 6: Ho tan hon ton 104,25 gam hn hp X gm NaCl v NaI vo nc c dung dch A. Sc kh Cl2 d vo dung dch A. Kt thc th nghim, c cn dung dch thu c 58,5 gam mui khan. Khi lng NaCl c trong hn hp X l A. 29,25 gam.B. 58,5 gam. C. 17,55 gam.D. 23,4 gam. Hng dn gii Kh Cl2 d ch kh c mui NaI theo phng trnh 2NaI+Cl2 2NaCl+I2 C 1 mol NaI to thnh 1 mol NaCl Khi lng mui gim 127 35,5 = 91,5 gam. Vy: 0,5 mol Khi lng mui gim 104,25 58,5 = 45,75 gam. mNaI = 1500,5 = 75 gam mNaCl = 104,25 75 = 29,25 gam. (p n A) V d 7: Ngm mt vt bng ng c khi lng 15 gam trong 340 gam dung dch AgNO3 6%. Sau mtthigianlyvtrathykhilngAgNO3trongdungdchgim25%.Khilng ca vt sau phn ng l A. 3,24 gam.B. 2,28 gam.C. 17,28 gam. D. 24,12 gam. Hng dn gii 3 AgNO ( )340 6n= 170 100ban u = 0,12 mol; 3 AgNO ( )25n= 0,12100ph.ng= 0,03 mol.Cu + 2AgNO3 Cu(NO3)2+2Ag+ 0,015 0,03 0,03 mol mvt sau phn ng=mvt ban u + mAg (bm) mCu (tan) =15 + (1080,03) (640,015) = 17,28 gam. (p n C) V d 8: Nhng mt thanh km v mt thanh st vo cng mt dung dch CuSO4. Sau mt thi gian ly hai thanh kim loi ra thy trong dung dch cn li c nng mol ZnSO4 bng 2,5 ln nng mol FeSO4. Mt khc, khi lng dung dch gim 2,2 gam. Khi lng ng bm ln thanh km v bm ln thanh st ln lt l 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 37 4 0,1 0,1 2NO3+10H++8e N2O+5H2O(3) 10 0,05 0,05 T (1), (2), (3)nhn c: 3 HNO Hn np+ ==2 0,15 4 0,1 10 0,05 + + = 1,2 mol. (p n D) V d 14: Cho 12,9 gam hn hp Al v Mg phn ng vi dung dch hn hp hai axit HNO3 v H2SO4 (c nng) thu c 0,1 mol mi kh SO2, NO, NO2. C cn dung dch sau phn ng khi lng mui khan thu c l: A. 31,5 gam.B. 37,7 gam.C. 47,3 gam.D. 34,9 gam. Hng dn gii Ta c bn phn ng: 2NO3+2H++1e NO2+H2O+NO3(1) 0,10,1 4NO3+ 4H++3e NO+2H2O+3NO3 (2) 0,1 3 0,1 2SO42+4H++2e SO2+H2O+SO42 (3) 0,1 0,1 T (1), (2), (3) s mol NO3 to mui bng 0,1 + 3 0,1 = 0,4 mol; s mol SO42 to mui bng 0,1 mol. mmui = mk.loi + 3 NOm +24 SOm = 12,9 + 62 0,4 + 96 0,1 = 47,3. (p n C) V d 15: Ha tan 10,71 gamhnhp gmAl,Zn, Fe trong 4lt dung dch HNO3 aMva thu c dung dch Av 1,792 lthn hp kh gmN2 v N2O c tlmol 1:1. C cn dung dch A thu c m (gam.) mui khan. gi tr ca m, a l: A. 55,35 gam. v 2,2MB. 55,35 gam. v 0,22M C. 53,55 gam. v 2,2MD. 53,55 gam. v 0,22M Hng dn gii 2 2 N O N1,792n n 0,042 22,4= = =mol. Ta c bn phn ng: 2NO3+12H++10e N2+6H2O 0,080,480,04 2NO3+10H++8e N2O+5H2O 0,080,4 0,04 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 38 3 HNO Hn n 0,88 + = = mol. 0,88a 0,224= = M. S mol NO3 to mui bng 0,88 (0,08 + 0,08) = 0,72 mol. Khi lng mui bng 10,71 + 0,72 62 = 55,35 gam. (p n B) V d 16: Ha tan 5,95 gam hn hp Zn, Al c t lmol l 1:2 bng dung dch HNO3long d thu c 0,896 lt mt sn shm kh X duy nht cha nit. X l: A. N2OB. N2C. NOD. NH4+ Hng dn gii Ta c:nZn = 0,05 mol; nAl = 0,1 mol. Gi a l s mol ca NxOy, ta c: ZnZn2++2eAlAl3++3e 0,05 0,10,10,3 xNO3 + (6x 2y)H++(5x 2y)eNxOy + (3x 2y)H2O 0,04(5x 2y)0,04 0,04(5x 2y) = 0,45x 2y = 10 VyX l N2. (p n B) V d 17: Cho hn hp gm 0,15mol CuFeS2 v 0,09 mol Cu2FeS2tc dng vi dung dch HNO3 d thu c dung dch Xvhn hp kh Y gmNO v NO2. ThmBaCl2 d vo dung dchX thu cmgam kt ta. Mt khc,nu thmBa(OH)2 dvo dung dch X,ly kt ta nung trong khng kh n khi lng khng i thu c a gam cht rn. Gi tr ca m v a l: A. 111,84g v 157,44gB. 111,84g v 167,44g C. 112,84g v 157,44gA. 112,84g v 167,44g Hng dn gii Ta c bn phn ng: CuFeS2 + 8H2O 17eCu2+ + Fe3+ + 2SO42 + 16+ 0,150,150,15 0,3 Cu2FeS2 + 8H2O 19e2Cu2+ + Fe3+ + 2SO42 + 16+ 0,09 0,180,09 0,18 24 SOn 0, 48 = mol;Ba2++SO42BaSO4 0,48 0,48 m = 0,48 233 = 111,84 gam. 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 51 A. HCOOHB. C3H7COOH C. CH3COOHD. C2H5COOH. Hng dn gii C 1 mol axit n chc to thnh 1 mol mui th khi lng tng (23 1) = 22 gam, m theo u bi khi lng mui tng (4,1 3) = 1,1 gam nn s mol axit l naxit = 1,122 = 0,05 mol. Maxit = 30,05 = 60 gam. t CTTQ ca axit no, n chc A l CnH2n+1COOH nn ta c: 14n + 46 = 60 n = 1. Vy CTPT ca A l CH3COOH. (p n C) V d 4: Cho dung dch AgNO3 d tc dng vidung dch hnhp c ha tan 6,25 gamhaimui KClvKBrthuc10,39gamhnhpAgClvAgBr.Hyxcnhsmolhnhp u. A. 0,08 mol.B. 0,06 mol.C. 0,03 mol.D. 0,055 mol. Hng dn gii C 1 mol mui halogen to thnh 1 mol kt ta khi lng tng: 108 39 = 69 gam; 0,06 molkhi lng tng: 10,39 6,25 = 4,14 gam. Vy tng s mol hn hp u l 0,06 mol. (p n B) V d 5: Nhng mt thanh graphit c ph mt lp kim loi ha tr (II) vo dung dch CuSO4 d. Sau phn ng khi lng ca thanh graphit gim i 0,24 gam. Cng thanh graphit ny nu cnhngvodungdchAgNO3thkhiphnngxongthykhilngthanhgraphit tng ln 0,52 gam. Kim loi ha tr (II) l kim loi no sau y? A. Pb.B. Cd.C. Al.D. Sn. Hng dn gii t kim loi ha tr (II) l M vi s gam l x (gam). M+CuSO4 d MSO4+Cu C M gam kimloi tan ra ths c 64 gam Cubmvo. Vy khilng kimloi gim (M 64) gam; Vy: x (gam) = 0, 24. MM 64 khi lng kim loi gim 0,24 gam. Mt khc:M+2AgNO3 M(NO3)2+2Ag C M gam kim loi tan ra th s c 216 gam Ag bm vo. Vy khi lng kim loi tng (216 M) gam; 10 phng php gii nhanh mn Hahttp://phuongphaphoctap.tk 50 Na2CO3 2Na++CO32 (NH4)2CO3 2NH4++CO32 BaCl2 Ba2++2Cl CaCl2 Ca2++2Cl Cc phn ng: Ba2++CO32 BaCO3+ (1) Ca2++CO32 CaCO3+ (2) Theo (1) v (2) c 1 mol BaCl2, hoc CaCl2 bin thnh BaCO3 hoc CaCO3 th khi lng mui gim (71 60) = 11 gam. Do tng s mol hai mui BaCO3 v CaCO3 bng: 43 39,711 = 0,3 mol m tng s mol CO32 = 0,1 + 0,25 = 0,35, iu chng t d CO32. Gi x, y l s mol BaCO3 v CaCO3 trong A ta c: x y 0,3197x 100y 39,7+ = + = x = 0,1 mol ;y = 0,2 mol. Thnh phn ca A: 3 BaCO0,1 197%m 10039,7= = 49,62%; 3 CaCO%m = 100 49,6 = 50,38%. (p n C) Vd2:Ho tanhonton23,8gamhnhpmtmuicacbonatcakimloihotr(I)vmt mui cacbonat ca kimloiho tr (II) bng dung dch HCl thy thot ra 4,48 lt kh CO2 (ktc). C cn dung dch thu c sau phn ng th khi lng mui khan thu c l bao nhiu? A. 26,0 gam.B. 28,0 gam.C. 26,8 gam.D. 28,6 gam. Hng dn gii C 1 mol mui cacbonat to thnh 1 mol mui clorua cho nn khi lng mui khan tng (71 60) = 11 gam, m 2 COn = nmui cacbonat = 0,2 mol. Suy ra khi lng mui khan tng sau phn ng l 0,211 = 2,2 gam. Vy tng khi lng mui khan thu c l 23,8 + 2,2 = 26