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TUYN TP CU HI Vễ C HAY V KHể (PHN 1) Cõu 1: Cho 30 gam hn hp X gm Mg, Al, ZnO v Fe(NO3)2 tan hon ton dung dch cha 0,725 mol H2SO4 loóng Sau cỏc phn ng xy hon ton, thu c dung dch Y ch cha 90,4 gam mui 33 sunfat trung hũa v 3,92 lớt (ktc) hn hp khớ Z gm N2 v H2 Bit t ca Z so vi H2 l Tớnh lng cỏc cht hn hp X? (Trớch thi tuyn HSG bng B tnh Qung Ninh nm hc 2015 2016) Hng dn gii 33 p dng BTKL, ta cú: m H2O = 30 + 0, 725 ì 98 90, 0,175 ì ì = gam H O = 0,5 mol 0,725 ì 0,125 ì 0,5 ì + = 0,05 mol p dng bo ton nguyờn t H, ta cú: NH = 0,05 + 0,05 ì BTNT N BTNT O Fe(NO3 )2 = = 0,075 mol ZnO = 0,5 0,075 ì = 0,05 mol Mg + = a mol 3+ Al = b mol Fe2 + / Fe3+ / (Fe2 + + Fe3+ ) dd Y Mg = a mol 2+ Al = b mol Zn = 0,05 mol H2 O 0,725 mol H2SO4 SO2 = 0,725 mol 30 gam + { 0,5 mol ZnO = 0,05 mol + NH = 0,05 mol 4 Fe(NO3 )2 = 0,075 mol 44244443 90,4 gam N = 0,05 mol Z H = 0,125 p dng bo ton mol electron n e nhận = 0,05 ì 10 + 0,125 ì + 0,05 ì = 1,15 mol Vi format ca tỏc gi thỡ ti õy ngi gii chỳng ta bt buc vo húa thõn thnh cỏc thy búi oỏn xem ý tỏc gi mun dung dch Y ch Fe2+; Fe3+ hay c ion Tht õy l cũn khỏ nhiu tranh lun v vic ó sinh H2 thỡ ỏp t theo dóy in húa thỡ dung dch khụng th tn ti Fe3+ c Theo quan im ca cỏ nhõn mỡnh thỡ vic ỏp t th t phn ng theo dóy in húa ph thụng õy cú nhng cha hp lý nh sau: + Th 1: vic ỏp t hn hp cỏc cht gm kim loi, oxit kim loi, mui ca kim loi phn ng tuõn theo th t nht nh no ú l dng nh khụng n vỡ bn thõn húa hc vụ c khụng cú c ch phn ng nh húa hc hu c nờn vic cỏc hn hp cỏc cht nh trờn tham gia phn ng l rt hn lon (vớ d th hi hn hp Na, Ba cho vo H2O thỡ th t phn ng lm sao???) + Th 2: dóy in húa chng trỡnh ph thụng hin hnh c sp xp da vo th in cc chun Eo (ph thuc vo nng , cỏc bn hc chuyờn s bit c phng trỡnh Nersnt), núi vui l kiu lm bi ny phi thc hin nhit phũng mỏy lnh 250C thỡ mi chun + Th 3: thi ca B ó tng xut hin trng hp nh cú H2 thoỏt dung dch cha c Fe2+, Fe3+ thi Cao ng v minh 2015 ri Chớnh vỡ th cỏ nhõn mỡnh ngh nu l thi CHNH THC ca B s quang minh chớnh i ng ng gii c trng hp tng quỏt nht ch khụng phi mũ th ny! Cũn bi ny, thỡ chỳng ta phi oỏn ý tỏc gi vy! Vi cỏc bi ny thỡ thụng thng hc sinh s tip cn vi vic gi s ln lt ch cha Fe2+, Fe3+ hoc c ú s xut hin trng hp gii nghim, nghim õm v khụng d kin gii t ú dn n kt qu bi toỏn + Trng hp dung dch Y ch cha Fe2+ (yờu tiờn trng hp ny trc vi cỏc thi th vỡ nhiu tỏc gi rt thớch mỏy múc húa Lý thuyt ) 24a + 27b + 0,05 ì 81 + 0,075 ì 180 = 30 gam a = 0,35 mol Khi ú BTĐT dd Y 2a + 3b + 0,05 ì + 0,05 + 0,075 ì = 0,725 ì mol b = 0,15 mol BTE n Kim tra li vi e cho = 2a + 3b = ì 0,35 + ì 0,15 = 1,15 mol = n e nhận (Nghim tha!) Cõu 2: Cho 66,2 gam hn hp X gm Fe3O4, Fe(NO3)2, Al tan hon ton dung dch cha 3,1 mol KHSO4 loóng Sau cỏc phn ng xy hon ton thu c dung dch Y ch cha 466,6 gam mui sunfat trung hũa v 10,08 lớt (ktc) khớ Z gm khớ, ú cú mt khớ húa nõu ngoi khụng khớ Bit t 23 ca Z so vi He l Phn trm lng ca Al hn hp X gn nht vi giỏ tr no sau 18 õy? A 15 B 20 C 25 D 30 ( thi minh k thi THPT Quc Gia nm 2015 B GD&T) Hng dn gii n Z = 0, 45 mol NO = 0,05 mol khí hóa nâu không khí NO Z gm Ta cú: 46 MZ = H = 0, mol NO = 0,05 mol Z 0,4 mol 1H42 = 42 43 46ì0,45 = 2,3 gam Fe3O K + = 3,1 mol KHSO4 = 3,1 mol 3+ 66,2 gam Fe(NO )2 Al Al dd Y Fe? + + H2O SO2 = 3,1 mol NH + 4 4 43 466,6 gam 66,2 + 3,1 ì 136 466,6 2,3 = 1,05 mol Khi ú ỏp dng nh lut bo ton lng, ta cú: H O = 18 3,1 0, ì 1,05 ì + = 0,05 mol p dng bo ton nguyờn t H NH = 0,05 + 0,05 = 0,05 mol p dng bo ton nguyờn t N Fe(NO3 )2 = 4n Fe3O4 + 6n Fe(NO3 )2 = n NO + n H 2O { { Fe O = 0,2 mol (O/ SO trit tiờu 14 43 p dng bo ton nguyờn t O 4 6ì0,05 0,05 1,05 nhau) Khi ú theo lng X, ta cú: m Al = 66,2 0,2 ì 232 180 = 10,8 gam 14 43 0,05 14 2ì 43 Fe3O4 Fe(NO3 )2 10,8 gần ì 100 = 16,31% 15% 66,2 Comment: cõu ny B rt khộo khụng hi v anh Fe trỏnh ng n nhy cm cú phn gõy tranh cói ú l vic ó sinh H2 thỡ dung dch khụng cha Fe3+ Tuy nhiờn nu ta m x thỡ dd Y gm K + = 3,1 mol SO 24 = 3,1 mol 3+ NH +4 = 0,05 mol Al = 0, mol 2+ 3+ Fe , Fe %m Al = Gi s dung dch cha c Fe2+ v Fe3+ BTNT Fe n + n Fe3+ = n Fe X = 0,65 mol Fe2 + = 0,1 mol Fe2 + 3+ BT ĐT Fe = 0,55 mol 2n Fe2 + + 3n Fe3+ = 1,85 mol õy l vớ d mỡnh mun minh cho cỏc bn thy trng hp cú khớ H2 thoỏt dung dch cú th cha c Fe3+ t thi ca B Giỏo dc v o To dng bi kim loi, mui, oxit kim loi phn ng mụi trng H+, NO3 D nhiờn bi vit trờn mỡnh ó nhn mnh l xột trờn quan im cỏ nhõn cng nh kin thc hn ch ca mỡnh kim chng iu ny cú l cn lm thớ nghim thc nghim, nhiờn vi iu kin hc v mc ph thụng ca nc ta thỡ rt khú cho c cỏc giỏo viờn v hc sinh kim chng Thụi thỡ l hc sinh, vi i a s ụng hc sinh hin thỡ u HC THI vỡ th gp nhng cõu hi kiu ny thỡ ụi thc dng li hay (tỡm cỏch gii ỏp s ca bi toỏn thay vỡ ln tn gỡ ú hay ỏp t rng buc lý thuyt Dóy in húa vo, cú lm tr nờn muụn trựng khú khn) Cõu 3: Cho mt lng d Mg vo 500 ml dung dch gm H2SO4 1M v NaNO3 0,4M Sau kt thỳc cỏc phn ng thu c Mg d, dung dch Y cha m gam mui v thy ch bay 2,24 lớt khớ NO (kc) Giỏ tr ca m l: A 61,32 B 71,28 C 64,84 D 65,52 Hng dn gii Mg 2+ + Na + H 2SO 0,5 mol NH Mg + NaNO3 0, mol SO NO x 0, y + H 2O 0,5 z NO 0,1mol [ +,] 2x + y z = 0,8 [e] 2x 8y = 0,1.3 [N] y + z = 0,1 x = 0,39 y = 0, 06 m = 65,52 gam z = 0, 04 Cõu 4: Cho gam bt Mg vo dung dch hn hp KNO3 v H2SO4, un nh, iu kin thớch hp, n phn ng xy hon ton thu c dung dch A cha m gam mui, 1,792 lớt hn hp khớ B (ktc) gm hai khớ khụng mu, ú cú mt khớ húa nõu ngoi khụng khớ v cũn li 0,44 gam cht rn khụng tan Bit t hi ca B i vi H2 l 11,5 Giỏ tr ca m l: A 27,96 B 29,72 C 31,08 D 36,04 Hng dn gii Mg 2+ 0,19 + K NH + KNO3 0,19 mol Mg + 24 H 2SO SO NO 0, 06 mol H 0, 02 mol + H 2O + [e] S mol NH = 0,02 (mol) BTNT Nito S mol KNO3 = 0,08 (mol) [K ] S mol K+ = 0,08 (mol) [ +,] S mol SO4 = 0,24 (mol) => m = 31,08 gam Cõu 5: Cho 3,9 gam hn hp Al, Mg t l mol : tan ht dung dch cha KNO3 v HCl Sau phn ng thu c dung dch A ch cha m gam hn hp cỏc mui trung hũa v 2,24 lớt hn hp khớ B gm NO v H2 Khớ B cú t so vi H2 bng Giỏ tr ca m gn giỏ tr no nht? A 24 B 26 C 28 D 30 Hng dn gii KNO3 Al 0,1 + Mg 0, 05 HCl [e] n NH+ = Al3+ 0,1 2+ Mg 0, 05 K + NH +4 Cl NO 0, 05 H 0, 05 + H 2O 0,1.3 + 0, 05.2 0, 05.3 0, 05.2 = 0, 01875 (mol) [N] n KNO3 = n K + = 0, 01875 + 0, 05 = 0, 06875(mol) [ + , ] n Cl = 0,1.3 + 0, 05.2 + 0, 06875 + 0, 01875 = 0, 4875 (mol) m = 24, 225(gam) Cõu 6: Cho 4,32 gam Mg vo dung dch hn hp NaNO3 v H2SO4, un nh n phn ng xy hon ton thu c dung dch A ; 0,896 lớt (ktc) hn hp khớ B cú lng 0,92 gam gm khớ khụng mu cú mt khớ húa nõu khụng khớ v cũn li 2,04 gam cht rn khụng tan Cụ cn cn thn dung dch A thu c m gam mui khan Giỏ tr ca m l: A 18,27 B 14,90 C 14,86 D 15,75 Hng dn gii NaNO3 Mg 0, 095 + H 2SO [e] n NH+ = Mg 2+ + Na + H 2O NH + SO NO 0, 03 H 0, 01 0, 095.2 0, 03.3 0, 01.2 = 0, 01 (mol) [N] n NaNO3 = n Na + = 0, 01 + 0, 03 = 0, 04 (mol) [ +, ] n SO2 = 0, 095.2 + 0, 04 + 0, 01 = 0,12 (mol) => m = 14,9 (gam) Cõu 7: Cho 31,15 gam hn hp bt Zn v Mg (t l mol 1:1) tan ht dung dch hn hp gm NaNO3 v NaHSO4 thu c dung dch A ch cha m gam hn hp cỏc mui v 4,48 lớt (ktc) hn hp khớ B gm N2O v H2 Khớ B cú t so vi H2 bng 11,5 m gn giỏ tr no nht? A 240 B 255 C 132 D 252 Hng dn gii Zn 0,35 Mg 0,35 + [e] n NH+ = NaNO3 0, 25 NaHSO x Zn 2+ 0,35 2+ Mg 0,35 dd A Na + x + 0, 25 NH + 0, 05 SO x N 2O 0,1 H 0,1 + H 2O 0,35.2 + 0,35.2 0,1.8 0,1.2 = 0, 05 (mol) [N] n NaNO3 = 0, 05 + 0,1.2 = 0, 25 (mol) [ + , ] 0,35.2 + 0,35.2 + x + 0, 25 + 0, 05 = 2x t x l s mol NaHSO4 => x = 1, (mol) => m = 240,1 (gam) Cõu 8: Cho 5,6 gam hn hp X gm Mg v MgO cú t l mol tng ng l : tan va dung dch hn hp cha HCl v KNO3 Sau phn ng thu c 0,224 lớt khớ N2O (ktc) v dung dch Y ch cha mui clorua Bit cỏc phn ng hon ton Cụ cn cn thn Y thu c m gam mui Giỏ tr ca m l: A 20,51 B 18,25 C 23,24 D 24,17 Hng dn gii HCl Mg 0,1 + MgO 0, 08 KNO3 [e] n NH+ = Mg 2+ 0,18 + K NH +4 Cl N O 0, 01 + H 2O 0,1.2 0, 01.8 = 0, 015 (mol) [N] n KNO3 = n K + = 0, 015 + 0, 01.2 = 0, 035 (mol) [+, ] n Cl = 0,18.2 + 0, 035 + 0, 015 = 0, 41 (mol) => m = 20,51 (gam) Cõu 9: Cho 12,56 gam hn hp gm Mg v Mg(NO3)2 tan va dung dch hn hp cha 0,98 mol HCl v x mol KNO3 Sau phn ng thu c dung dch Y ch cha mui clorua v 0,04 mol khớ N2 Cụ cn cn thn Y thu c m gam mui khan Bit cỏc phn ng hon ton Giỏ tr ca m l: A 46,26 B 52,12 C 49,28 D 42,23 Hng dn gii: y Mg 12,56 g + Mg(NO3 ) z Mg 2+ y + z + x K HCl 0,98 NH +4 x + 2z 0, 08 KNO x Cl 0,98 N 0, 04 + H 2O t y, z ln lt l s mol ca Mg v Mg(NO3)2 [N] n NH+ = x + 2z 0, 08 (mol) [ + , ] 2(y + z) + x + x + 2z 0, 08 = 0, 98 (1) [e] 2y = 8(x + 2z 0, 08) + 0, 04.10 (2) Mt khỏc : 12,56 = 24y + 148z (3) x = 0, 09 => y = 0, z = 0, 02 => m = 49,28 (gam) Cõu 10: Cho Zn ti d vo dung dch gm HCl, 0,05 mol NaNO3, 0,1 mol KNO3 Sau kt thỳc phn ng thu c dung dch X cha m gam mui, 0,125 mol hn hp khớ Y gm hai khớ khụng mu, ú cú mt khớ húa nõu ngoi khụng khớ T hi ca Y so vi H2 l 12,2 Giỏ tr ca m l: A 64,05 gam B 49,775 gam C 57,975 gam D 61,375 gam Hng dn gii: HCl Zn + NaNO3 0, 05 KNO 0,1 Na + 0, 05 + K 0,1 Zn 2+ NH +4 Cl NO 0,1 H 0, 025 + H 2O [N] n NH+ = 0, 05 (mol) [e] n Zn = 0, 05.8 + 0,1.3 + 0, 025.2 = 0,375 (mol) [ + , ] n Cl = 0, 05 + 0,1 + 0,375.2 + 0, 05 = 0,95 (mol) Vy m = 64,05 (gam) Cõu 11: Cho 50,82 gam hn hp X gm NaNO3, Fe3O4, Fe(NO3)2 v Mg tan hon ton dung dch cha 1,8 mol KHSO4 loóng Sau cỏc phn ng xy hon ton thu c dung dch Y ch cha 275,42 gam mui sunfat trung hũa v 6,272 lớt khớ (ktc) Z gm khớ ú cú mt khớ húa nõu khụng khớ Bit t ca Z so vi H2 l 11 Phn trm lng Mg hn hp X l: A 25,5% B 20,2% C 19,8% D 22,6% Hng dn gii: Chỳ ý: Kinh nghim cho thy cú khớ cú H2 thỡ 60% - 70% Fe2+ NaNO3 Fe O 50,82 g X + KHSO 1,8 Fe(NO3 ) Mg Mg 2+ + Na 2+ 275, 42 g Fe + K NH + SO NO 0, H 0, 08 [m] 50,82 + 1,8.136 = 275, 42 + 0, 2.30 + 0, 08.2 + 18.n H 2O [H] n NH + = + H2O => n H 2O = 0, 78 (mol) 1,8 0, 78.2 0, 08.2 = 0, 02 (mol) 0,98 0, 22.3 = 0, 08 (mol) => n Fe3O4 = = 0, 78 + 0, = 0,98 (mol) [N] n NO (X ) = n N = 0, 22 (mol) [O] n O (X) [e] 2.n Mg = 0, 08.2 + 0, 2.3 + 0, 08.2 + 0, 02.8 => n Mg = 0,54 (mol) Vy %Mg = 25,5% Bi ny cú th t x, y, z, t ln lt l s mol cỏc cht hn hp X gii cng c Cõu 12: X l hn hp rn gm Mg, NaNO3 v FeO (trong ú oxi chim 26,4% v lng) Hũa tan ht m gam X 2107 gam dung dch H2SO4 loóng, nng 10% thu c dung dch Y ch cha mui sunfat trung hũa v 11,2 lớt (kc) hn hp NO, H2 cú t so vi H2 l 6,6 Cụ cn dung dch sau phn ng c rn khan Z v 1922,4 gam H2O Phn trm lng FeO X gn vi giỏ tr no nht di õy? A 50% B 12% C 33% D 40% (Thy Nguyn ỡnh - 2015) Hng dn gii: Mg m (g) X NaNO3 + H 2SO 2,15 FeO [H] n NH + = Mg 2+ + Na Fe n + + H 2O 1,45 NH +4 SO4 NO 0, H 0,3 2,15.2 1, 45.2 0,3.2 = 0, (mol) [N] n NaNO3 = 0, + 0, = 0, (mol) [O] n O X = 1, 45 + 0, = 1, 65 (mol) => mX = 100 (gam) n FeO = n O(FeO) = 1, 65 0, 4.3 = 0, 45 (mol) => %FeO = 32,4% Chỳ ý: Bi ny khụng cn xỏc nh st cú s oxi húa bao nhiờu dung dch Y Cõu 13: Hũa tan hon ton 15,76 gam hn hp X gm Mg, MgO v Mg(NO3)2 bng dung dch hn hp cha 1,14 mol HCl v x mol NaNO3 va Sau phn ng thu c 0,04 mol N2 v dung dch Y ch cha mui Cho NaOH d vo Y thỡ thy cú a mol NaOH tham gia phn ng Bit X phn trm lng ca MgO l 20,30457% Giỏ tr ca a l: A 1,0 B 1,05 C 1,10 D 0,98 Hng dn gii: Mg 2+ y + z + 0, 08 + Na x y Mg NH + x + 2z 0, 08 HCl 1,14 15, 76g MgO 0, 08 + Cl 1,14 NaNO3 x Mg(NO ) z H O N 0, 04 + a mol NaOH ? [N] n NH+ = x + 2z 0, 08 (mol) [ + , ] 2(y + z + 0, 08) + x + x + 2z 0, 08 = 1,14 => x + y + 2z = 0,53 (1) [e] 2y = 8(x + 2z 0, 08) + 0, 04.10 => 8x 2y + 16z = 0, 24 (2) Mt khỏc : 24y + 148z = 15,76 0,08.40 => 24y + 148z = 12,56 (3) x = 0, 09 => y = 0, => a = 1,05 (mol) z = 0, 02 Cõu 14: Hũa tan hon ton 7,028 gam hn hp rn X gm: Zn, Fe3O4, ZnO (s mol Zn bng s mol ZnO) vo 88,2 gam dung dch HNO3 20% thu c dung dch Y v 0,2688 lớt khớ NO nht (ktc) Cho t t V lớt dung dch NaOH 1M vo dung dch Y cho n phn ng ht vi cỏc cht Y thu c lng kt ta cc i, nung lng kt ta ny khụng khớ n lng khụng i thu c 7,38 gam rn Giỏ tr ca V l: A 0,267 lớt B 0,257 lớt C 0,266 lớt D 0,256 lớt Hng dn gii: Zn 2+ ZnO 2x (1) + NaOH 3+ Fe 3y (2) to Fe2 O3 x Zn + NH 7, 028gam Fe3O y + HNO3 (0, 28 mol) + + H 2O ZnO x H NO3 NO 0, 012 (mol) 65x + 232y + 81x = 7, 028 x = 0, 01 => Ta cú : 3y y = 0, 024 81.2x + 160 = 7,38 [e] n NH + = 0, 01.2 + 0, 024 0, 012.3 = 0, 001 (mol) 8 [N] n NO dd = 0, 28 0, 012 0, 001 = 0, 267 (mol) [ + , ] n H + = 0, 01 (mol) 2+ 3+ + + NaOH tỏc dng vi Zn , Fe , NH , H => V = 0,267 lớt Cõu 15: Hũa tan hon ton hn hp X gm Mg, MgO, Fe3O4 v Fe(NO3)2 (trong ú oxi chim 29,68% theo lng) dung dch HCl d thy cú 4,61 mol HCl phn ng Sau cỏc phn ng xy xong thu c dung dch Y ch cha 231,575 gam mui clorua v 14,56 lớt (kc) khớ Z gm NO, H2 Z cú t so 69 vi H2 l Thờm dung dch NaOH d vo Y, sau phn ng thu c kt ta Z Nung Z khụng khớ 13 n lng khụng i c 102,2 gam cht rn T Phn trm lng MgO X gn nht vi giỏ tr no sau õy? A 13,33% B 33,33% C 20,00% D 6,80% (thy Nguyn ỡnh 2015) Hng dn gii: Mg 2+ 2+ Fe MgO (x + y) 3+ x Mg (1) + NaOH 231,575gam Fe (3z + t) MgO (2) t o C Fe O3 y + 4,61 mol HCl m (gam) Fe O z + NH + H O1, 655 mol Fe(NO3 )2 t Cl mO (X) = 0,2968m ; [O] => n H2O NO 0, mol H 0, 45 mol 0, 2968m = 0, 16 0, 2968m 0, 2).18 => m = 100 (gam) 16 => S mol H2O = 1,655 mol ; S mol O (X) = 1,855 mol + [H] => nNH = 0,1 24x + 40y + 232z + 180t = 100 x = y + 4z + 6t = 1,855 y = 0,355 Ta cú: => => %MgO = 14,2% 40(x + y) + 80(3z + t) = 102, z = 0,15 2t = 0, + 0,1 [N] t = 0,15 Cõu 16: Cho 24,06 gam hn hp X gm Zn, ZnO v ZnCO3 cú t l s mol : : theo th t trờn, tan hon ton dung dch Y gm H2SO4 v NaNO3, thu c dung dch Z ch cha mui trung hũa v V 218 lớt hn hp khớ T (ktc) gm NO, N2O, CO2, H2 (Bit t ca T so vi H2 l ) Cho dung dch 15 BaCl2 d vo Z n cỏc phn ng xy hon ton, thu c 79,22 gam kt ta Cũn nu cho Z phn ng vi NaOH thỡ lng NaOH phn ng ti a l 1,21 mol Giỏ tr ca V gn nht vi: [m] => m + 4, 61.36,5 = 231,575 + 0, 2.30 + 0, 45.2 + ( A 3,0 B 4,0 C 5,0 D 2,6 (Nguyn Anh Phong ln 2016) Hng dn gii: 0,18 Zn H 2SO ZnO 0, 06 + NaNO3 0, 07 ZnCO 0, 06 Zn 2+ 0,3 + Na 0, 07 + NH 0, 01 SO 0,34 NO 436 N O M= 15 H CO + H 2O x y z 0, 06 BaSO n SO2 = 0, 34 (mol) n NH + = 1, 21 0,3.4 = 0, 01(mol) 1,21mol NaOH [ + , ] n Na + = n NaNO3 = 0,34.2 0, 01 0,3.2 = 0, 07 (mol) [N] 0, 07 = x + 2y + 0, 01 (1) 0,18.2 = 0, 08 + 3x + 8y + 2z (2) [e] x = 0, 04 => y = 0, 01 z = 0, 04 => V = 3,36 lớt 436 15 30x + 44y + 2z + 44.0, 06 436 = (3) x + y + z + 0, 06 15 Cõu 17: Hũa tan ht 35,4 gam hn hp gm Mg v FeCO3 dung dich HCl loóng d thu c 20,16 lớt khớ (ktc) Mt khỏc cng hũa tan ht 35,4 gam hn hp trờn cn dựng dung dch hn hp cha H2SO4 0,25M v HNO3 0,75M un núng Sau kt thỳc phn ng thu c dung dch Y cha m gam mui v hn hp khớ Z gm khớ khụng mu, ú cú mt khớ húa nõu ngoi khụng khớ T ca Z so vi He bng 8,8125 Giỏ tr ca m l: M= A 152,72 gam Hng dn gii: 0, 75 Mg FeCO3 0,15 [e] n NH+ = B 172,42 gam + H 2SO x HNO3 3x C 142,72 gam D 127,52 gam Mg 2+ 0, 75 3+ Fe 0,15 dd Y NH + 0,1125 SO x NO 3x 0,3625 H 2O CO 0,15 NO 0, 25 0, 75.2 + 0,15 0, 25.3 = 0,1125(mol) [N] n NO = 3x 0,1125 0, 25 = 3x 0,3625 (mol) [ +, ] 0, 75.2 + 0,15.3 + 0,1125 = 2x + 3x 0,3625 => x = 0, 485(mol) 10 x = 0,3.27 + 23x + 18y + z + 96x = 127,88 => y = 0, 04 y + z = 0,1 z = 0, 06 BTDT x + y + z + 0,3.3 = 2.x BTKL 10,92 + 1.120 + 0, 09.63 = 127,88 + 0, 08.20 + 18.n H 2O BTNT H n H2 = => nH2O = 0,395 mol 0, 09 + 0, 04.4 0, 06 0,395.2 = 0, 04 (mol) a + b = 0, 04 a = 0, 015 => 28a + 44b + 0, 04.2 = 20.0, 08 b = 0, 025 => %N2O = 68,75% Cõu 28: Hũa tan hon ton m gam hn hp X gm Mg, Fe, FeCO3, Cu(NO3)2 vo dung dch cha NaNO3 (0,045 mol) v H2SO4, thu c dung dch Y ch cha 62,605 gam mui trung hũa (khụng cú ion Fe3+) v 19 3,808 lớt (ktc) hn hp khớ Z (trong ú cú 0,02 mol H2) T ca Z so vi O2 bng Thờm dung dch 17 NaOH 1M vo Y n thu c lng kt ta ln nht l 31,72 gam thỡ va ht 865 ml Mt khỏc, cho Y tỏc dng va vi BaCl2 c hn hp T Cho lng d dung dch AgNO3 vo T thu c 256,04 gam kt ta Giỏ tr ca m l A 34,6 B 32,8 C 27,2 D 28,4 Hng dn gii: Na + 2+ Mg Fe 2+ 62,605 g 2+ Cu Mg NH + Fe NaNO3 0, 045(mol) X + SO H 2SO FeCO3 Cu(NO3 ) H O â ê Mg(OH) ê ê Fe(OH) + NaOH 0,865(mol) êê ê Cu(OH) ê ê êôdd : Na 2SO 0, 455(mol) â ê AgCl 0,91 ê (1) BaCl2 256, 04 g ê Ag (2) AgNO3 ê ê BaSO 0, 455 ôê N x Oy CO H 0, 02 (mol) n Ag = 256, 04 0, 91.143,5 0, 455.233 = 0,18(mol) => n Fe2+ = n Ag = 0,18 108 17 Na + 0, 045 2+ Mg x Fe2+ 0,18 [ +, ] 2x + 2y + z = 0, 505 x = 0, 62,605 g 2+ Cu y => y = 0, 04 Ta cú: 24x + 64y + 18z = 7,81 58x + 98y = 31, 72 0,18.90 z = 0, 025 NH + z SO 0, 455 H 2O [H] n H2O = 0, 455.2 0, 025.4 0, 02.2 = 0,385(mol) [m] m X + 0, 045.85 + 0, 455.98 = 62, 605 + 0,17 608 + 0,385.18 => mX = 27,2 gam 17 Cõu 29: Hũa tan hon ton 18,025 gam hn hp bt rn gm Fe2O3, Fe(NO3)2, Zn bng 480 ml dung dch HCl 1M sau phn ng thu c dung dch X cha 30,585 gam cht tan v 1,12 lớt (ktc) hn hp khớ gm (N2O, NO, H2) cú t vi He l 6,8 Cho AgNO3 d vo dung dch X trờn thy thu c 0,112 lớt khớ NO (ktc) (sn phm kh nht ) v 72,66 gam kt ta Phn trm lng ca Fe(NO 3)2 hn hp ban u l A 29,96% B 39,89% C 17,75% D 62,32% ( thi th THPT Quc Giỏ ln BOOKGOL 2016) Hng dn gii: H 0,05 mol N O NO { 1,36 gam Zn 2+ 2+ NO = 0,005 mol Zn = x mol 3+ Fe , Fe HCl = 0,48 mol AgNO3 AgCl = 0,48 mol Fe O3 = y mol dd X Cl = 0, 48 mol Fe(NO ) = z mol = 0,035 mol 44 1Ag H + = 0, 02 mol 4 43 44 4 43 18,025 mol 72,66 gam NH + 4 4 30,585 gam H2O 18,025 + 0, 48 ì 36,5 1,36 30,585 = 0,2 mol 18 AgNO3 + Do dung dch X cú sinh khớ NO X cú H d 4H + + NO3 + 3e NO + 2H O H+ X = 0,005 ì = 0,02 mol Do mol : 0,005 72,66 0, 48 ì 143, = 0,035 mol Mt khỏc, ta cú: Ag = 108 p dng bo ton mol electron, ta cú: Fe2 + = 0,035 + 0,005 ì = 0,05 mol p dng nh lut bo ton lng, ta cú: H O = 18 p dng bo ton nguyờn t O n N O NO = (3y + 6z 0,2) mol H2 = 0,05 (3y + 6z 0,2) = 0,25 3y 6z p dng bo ton nguyờn t H NH4+ = 0, 48 0,02 0,2 ì ì ( 0,25 3y 6z ) = 6y + 12 z 0, 44 mol Theo gi thuyt, lng cht tan X v bo ton in tớch dung dch X, ta cú: 65 x + 160 y + 180 z = 18, 025 x = 0,145 mol y + 12 z 0, 44 18 + 0, 48.35,5 = 30,585 y = 0, 02 mol 65 x + 56.(2 y + z) + 0, 02.1 + z = 0, 03 mol y + 12 z 0, 44 x + 0, 05.2 + (2 y + z 0, 05).3 + 0, 02 + = 0, 48 0,03 ì 180 ì 100 = 29,96% %m Fe(NO3 )2 = 18,025 [Hay] Cõu 30: Cho m gam hn hp H gm FexOy, Fe, Cu tỏc dng ht vi 200 gam dung dch cha HCl 32,85% v HNO3 9,45%, sau phn ng thu c 5,824 lớt khớ NO (ktc) l sn phm kh nht v dung dch X cha (m + 60,24) gam cht tan Cho a gam Mg vo dung dch X, kt thỳc phn ng thy thoỏt khớ Y gm khớ, ú cú khớ húa nõu khụng khớ; t ca Y i vi He bng 4,7 v (m - 6,04) gam cht rn T Giỏ tr ca a l A 21,48 B 21,84 C 21,60 D 21,96 Hng dn gii: (m + 60,24)g T : Cu, Fe, Mg (m 6, 04) g 3+ Fe Mg 2+ Cu 2+ + + NH y Mg H Fe a gam HCl 1,8 Cl 1,8 m (g) H Cu + NO H O z HNO3 0,3 O Cl NO 3x H 2x H 2O NO 0, 26 1,8.36,5 + 0,3.63 60, 24 0, 26.30 BTKL n H2O = = 0,92 (mol) 18 Fe3+ 2+ Cu BTNT (m + 60, 24) H + 0, 26 => m(Cu + Fe) = (m 6,4) gam < mT Vy T cú thờm Mg d = 0,36 g NO 0, 04 Cl 1,8 3x + y = 0, 04 x = 0, 01 4x + 4y + 2z = 0, 26 => y = 0, 01 3x + z = 0, 04.3 z = 0, 09 1,8 0, 01 BTDT n Mg2+ = = 0,895(mol) => a = 0,895.24 + 0,36 = 21,84 (gam) BTNT ( N,H,O) 19 Cõu 31: Hũa tan ht 13,52 gam hn hp X gm Mg(NO3)2, Al2O3, Mg v Al vo dung dch NaNO3 v 1,08 mol HCl (un núng) Sau kt thc phn ng thu c dung dch Y ch cha cỏc mui v 3,136 lớt (ktc) hn hp khớ Z gm N2O v H2 T ca Z so vi He bng Dung dch Y tỏc dng ti a vi dung dch cha 1,14 mol NaOH, ly kt ta nung ngoi khụng khớ ti lng khụng i thu c 9,6 gam rn Phn trm lng ca Al cú hn hp X l A 31,95% B 19,97% C 23,96% D 27,96% (Thy To Mnh c) Hng dn gii: Mg 2+ 0, 24 3+ Al x Na + y 1,14mol NaOH MgO Mg(NO3 ) 0,24 Al O + NH z NaNO3 y 13,52g X + HCl 1, 08 Mg Cl 1, 08 Al H O 0, 46 2z N O 0, 06 H 0, 08 BTDT 3x + y + z = 0, 06 x = 0,16 y = 0,1 4x + z = 0, 66 z = 0, 02 BTKL 13,52 + 85y + 39, 42 = 27x + 23y + 18z + 46,9 + 18(0, 46 2z) 0, 02 + 0, 06.2 0,1 BTNT N n Mg(NO3 )2 = = 0, 02 (mol) => nMg = 0,22 (mol) 8.0, 02 + 8.0, 06 + 2.0, 08 0, 22.2 BTe n Al = = 0,12 (mol) => %Al = 23,96% Cõu 32: Hũa tan ht 14,88 gam hn hp gm Mg , Fe3O4 , Fe(NO3)2 vo dung dch cha 0,58 mol HCl, sau cỏc phn ng kt thỳc thu c dung dch X cha 30,05 gam cht tan v thy thoỏt 1,344 lớt (ktc) hn hp khớ Y gm H2, NO, NO2 cú t so vi H2 bng 14 Cho dung dch AgNO3 (d) vo dung dch X , sau cỏc phn ng xy hon ton thu c dung dch Z; 84,31 gam kt ta v thy thoỏt 0,224 lớt (ktc) khớ NO l sn phm kh nht ca NO3- Phn trm lng ca Mg hn hp ban u gn nht vi giỏ tr no sau õy? 1,14mol NaOH A 16% B 17% C 18% D 19% Hng dn gii: 20 x Mg 14,88g Fe3O y + HCl 0,58 Fe(NO ) z Mg 2+ x 2+ Fe 0, 04 Fe3+ 3y + z 0, 04 AgNO3 + NH t + H 0, 04 Cl 0,58 H O 0, 24 H 0, 06 molY NO NO Mg 2+ 3+ Fe NH + NO3 H O AgCl 0,58 Ag 0, 01 NO 0, 01 M = 28 4H + + NO3 + 3e NO + 2H 2O 0, 04 0, 01 BTe n Fe2+ = 0, 01 + 0, 01.3 = 0, 04 (mol) BTKL n H2O = BTNT H n H2 = 14,88 + 0,58.36,5 0, 06.28 30, 05 = 0, 24 (mol) 18 0,58 0, 24.2 0, 04 4t = 0, 03 2t (mol) x = 0,105 2x + 3(3y + z 0, 04) + t = 0, 46 y = 0, 03 => BTNT N 2z = t + 0, 06 (0, 03 2t) z = 0, 03 t = 0, 01 24x + 56(3y + z 0, 04) + 18t = 7,18 24x + 232y + 180z = 14,88 BTDT => %Mg = 16,9% Cõu 33: Cho 7,65 gam hn hp X gm Al v Al2O3 (trong ú Al chim 60% v lng) tan hon ton dung dch Y gm H2SO4 v NaNO3, thu c dung dch Z ch cha mui trung hũa v m gam hn hp khớ T (trong T cú 0,015 mol H2) Cho dung dch BaCl2 d vo Z n cỏc phn ng xy hon ton, thu c 93,2 gam kt ta Cũn nu cho Z phn ng vi NaOH thỡ lng NaOH phn ng ti a 0,935 mol Giỏ tr ca m gn giỏ tr no nht sau õy ? A 2,5 B C 1,5 D ( thi THPT Quc Gia nm 2015-B GD & T) Al 0, 23 + 0,935 mol NaOH Na NH + 0, 015 BaCl2 BaSO 0, 4 0,17 H 2SO 0, Al SO 24 0, 7, 65g X + Al O 0, 03 NaNO 3 H 2O H 0, 015 T N x Oy BTDT n NaNO3 = n Na + = 0, 4.2 0, 23.3 0, 015 = 0, 095 (mol) 3+ 21 0, 4.2 0, 015.4 0, 015.2 = 0,355 (mol) BTKL m T = 7, 65 + 0, 4.98 + 0, 095.85 0, 355.18 0, 23.27 0, 095.23 0, 015.18 0, 4.96 = 1, 47 (g) Cõu 34: Hũa tan 17,32 gam hn hp X gm Mg, Fe3O4 v Fe(NO3)2 cn va ỳng dung dch hn hp gm 1,04 mol HCl v 0,08 mol HNO3, un nh thu c dung dch Y v 2,24 lớt hn hp khớ Z (ktc) cú t hi i vi H2 l 10,8 gm hai khớ khụng mu ú cú mt khớ húa nõu ngoi khụng khớ Cho dung dch Y tỏc dng vi mt lng dung dch AgNO3 va thu c m gam kt ta v dung dch T Cho dung dch T tỏc dng vi mt lng d dung dch NaOH, lc kt ta nung n n lng khụng i thu c 20,8 gam cht rn Cỏc phn ng xy hon ton Giỏ tr ca m l BTNT H n H 2O = A 150,32 B 151,40 C 152,48 D 153,56 Hng dn gii: Mg HCl 1, 04 17,32g X Fe3O + HNO3 0, 08 Fe(NO ) AgCl 1, 04 Mg 2+ x Ag y 2+ Fe y Mg 2+ AgNO3 Fe3+ z MgO x Fe3+ + NaOH (1) + NH t y+z t o , kk (2) + Fe 2O3 NH Cl 1, 04 NO H O NO 0, 07 H 0, 03 BTNT H n H2 O = 1, 04 + 0, 08 0, 03.2 4t = 0,53 2t (mol) BTNT N n Fe(NO3 )2 = BTNT O n Fe3O4 = t + 0, 07 0, 08 t 0, 01 = (mol) 2 0,53 2t + 0, 07 3.(t 0, 01) 0, 08.3 0,39 5t = (mol) 4 20,8g x = 0, 40.x + 80.(y + z) = 20,8 y = 0, 01 0,39 5t t 0, 01 17,32g 24x + 232 + 180 = 17,32 => z = 0, 05 t = 0, 07 0,39 5t t 0, 01 BTNT Fe + = y+z BTDT 2x + 2y + 3z + t = 1, 04 ==> m = 150,32 (gam) Chỳ ý: Cú th dựng phng trỡnh bo ton electron c quỏ trỡnh cng c Cõu 35: Hũa tan 35,04 gam hn hp X gm Mg, MgCO3 v Al(NO3)3 vo dung dch cha 1,68 mol NaHSO4 Sau kt thỳc phn ng thu c dung dch Y ch cha cỏc mui trung hũa v 0,2 mol hn hp khớ Z gm CO2, N2O, N2 v H2 tỏc dng ti a cỏc cht tan dung dch Y cn dựng dung dch cha 1,75 mol NaOH, thu c 40,6 gam kt ta Phn trm lng ca N2O cú hn hp Z gn ỳng nht l A 49 B 46 C 48 D 47 Hng dn gii: 22 c Mg 35, 04g X MgCO d + NaHSO 1,68 Al(NO ) 3 Mg 2+ 0,7 3+ Al a NH + b NaOH Mg(OH) 1,75 0,7 Na + 1, 68 SO 1, 68 H 2O CO N O x 0, (mol) N2 y H z a = 0, 07 => 1,75mol NaOH 4a + b + 0, 7.2 = 1, 75 b = 0, 07 BTDT 3a + b = 0, 28 n Al( NO3 )3 = 0, 07 (mol) c = 0, 6445 => 35,04g 24c + 84d + 0, 07.213 = 35, 04 d = 0, 0555 BTNT Mg c + d = 0, x = 0, 06 BTNT N 2x + 2y + 0, 07 = 0, 07.3 => y = 0, 01 BTe 8x + 10y + 2z + 8.0, 07 = 0, 6445.2 z = 0, 0745 n CO2 = n MgCO3 = 0, 0555(mol) x + y + z + 0, 0555 = 0, => %N 2O = 47,9% Cõu 36: Ngi ta hũa 216,55 gam hn hp mui KHSO4 v Fe(NO3)3 vo nc d thu c dung dch A Sau ú cho m gam hn hp B gm Mg, Al, Al2O3 v MgO vo dung dch A ri khuy u ti cỏc phn ng xy hon ton thy B tan ht, thu c dung dch C ch cha cỏc mui v cú 2,016 lớt hn hp khớ D cú tng lng l 1,84 gam gm khớ (ktc) thoỏt ú v th tớch H2, N2O, NO2 ln lt 1 chim , , Cho BaCl2 d vo C thy xut hin 356,49 gam kt ta trng Bit B oxi chim 9 64 v lng Giỏ tr ca m gn nht vi giỏ tr no sau õy? 205 A 18 B 20 C 22 D 24 Hng dn gii: 23 Mg 2+ 3+ Al Fe n + BaCl2 BaSO4 + K 1,53 + NH SO H O Mg KHSO4 1,53 + 216,55g Quy i Al Fe(NO ) 0, 035 3 64m O 205 H 0, 04 N O 0, 01 1,84g NO 0, 01 N 0, 02 = x NO 0, 01 = y 14444442 4444443 0,09 mol n KHSO4 = n BaSO4 = 1,53(mol) n Fe(NO3 )3 = 0, 035(mol) BTNT N n NH+ = 0, 035.3 0, 01.2 0, 01 0, 02.2 0, 01 = 0, 025(mol) 1,53 0, 025.4 0, 04.2 = 0, 675 (mol) 64m BTNT O 0, 035.9 + = 0, 675 + 0, 01 + 0, 01.2 + 0, 01 => m = 20,5 (gam) 205.16 BTNT H n H2 O = Cõu 37: Hũa tan hon ton hn hp X gm Al, Mg, FeO, CuO cn dựng lớt dung dch HNO3 0,35M, thu c dung dch Y ch cha mui nitrat (khụng cha ion Fe2+) v 3,36 lớt NO (ktc, sn phm kh nht) Mt khỏc, cho X tỏc dng ht vi dung dch HCl va , thờm AgNO3 (d) vo hn hp phn ng, thu c 77,505 gam kt ta Tng lng ca oxit kim loi X l A 7,68 gam B 3,84 gam C 3,92 gam D 3,68 gam Hng dn gii: Al Mg FeO CuO x y TH1 Al3+ x 2+ Mg y Fe3+ z Cu 2+ t NO3 0,55 H 2O HNO3 0,7 z t NO 0,15 TH AgCl 3x + 2y + 2z + 2t HCl z Ag 3x + 2y = 0, 41 BT e 3x + 2y + z = 0,15.3 => m oxit = 3, 68(gam) => z = 0, 04 77,505g 143,5.(3x + 2y + 2z + 2t) + 108z = 77,505 t = 0, 01 BTDT 3x + 2y + 3z + 2t = 0,55 24 Cõu 38: Cho m gam hn hp X gm Fe, Fe3O4 v Fe(NO3)2 tan ht 320 ml dung dch KHSO4 1M Sau phn ng thu c dung dch Y ch cha 59,04 gam mui trung hũa v 0,896 lớt khớ NO (ktc, sn phm kh nht) Cho dung dch NaOH d vo Y thỡ cú 0,44 mol NaOH phn ng Bit cỏc phn ng xy hon ton Phn trm lng ca Fe X gn giỏ tr no nht sau õy? A 3,5% B 2,0% C 3,0% D 2,5% Hng dn gii: Fe X Fe3O + KHSO4 0,32 Fe(NO ) BTNT H n H2O = Fe 2+ x 3+ Fe y 59, 04g K + 0,32 NaOH 0,44 SO 0,32 NO3 z H O 0,16 NO 0, 04 0,32 = 0,16 (mol) x = 0, 01 56x + 56y + 62z = 15,84 => y = 0,14 0,44mol NaOH 2x + 3y = 0, 44 z = 0,12 BTDT 2x + 3y z = 0,32 59,04g BTNT N n Fe(NO3 )2 = BTNT O n Fe3O4 = 0,12 + 0, 04 = 0, 08 (mol) 0,12.3 + 0,16 + 0, 04 0, 08.6 = 0, 02 (mol) BTNT Fe n Fe = 0, 01 + 0,14 0, 08 0, 02.3 = 0, 01 (mol) => %Fe = 2,857% Cõu 39: Nung m gam hn hp gm Mg v Cu(NO3)2 iu kin khụng cú khụng khớ, sau mt thi gian thu c cht rn X v 10,08 lớt (ktc) hn hp khớ gm NO2 v O2 Hũa tan hon ton X bng 650 ml dung dch HCl 2M, thu c dung dch Y ch cha 71,87 gam mui clorua v 0,05 mol hn hp khớ Z gm N2 v H2 T ca Z so vi He bng 5,7 Giỏ tr ca m gn giỏ tr no nht sau õy? A 50 B 55 C 45 D 60 Hng dn gii: x Mg m (g) Cu(NO ) y Mg MgO X + HCl 1,3 CuO Cu(NO3 ) Mg 2+ x 2+ Cu y 71,87g NH + z Cl 1, H O 6y 0,9 N 0, 04 H 0, 01 NO 0, 45 O 25 BTNT O n H2O = 6y 0,9 (mol) x = 0,39 24x + 64y + 18z + 35,5.1,3 = 71,87 => y = 0, 25 z = 0, 02 BTNT H 2.(6y 0,9) + 4z + 2.0, 01 = 1,3 BTDT 2x + 2y + z = 1,3 71,87g => m = 56,36 (gam) Cõu 40: Trn 10,17 gam hn hp X gm Fe(NO3)2 v Al vi 4,64 gam FeCO3 c hn hp Y Cho Y vo lng va dung dch cha 0,56 mol KHSO4 c dung dch Z cha 83,41 gam mui sunfat trung hũa v m gam hn hp khớ T ú cú cha 0,01 mol H2 Thờm NaOH vo Z n ton b mui st chuyn ht thnh hidroxit v ngng khớ thoỏt thỡ cn 0,57 mol NaOH, lc kt ta nung khụng khớ n lng khụng i thu c 11,5 gam cht rn Giỏ tr m l: A 3,22 B 2,52 C 3,42 D 2,70 Hng dn gii: x Al 10,17g Fe(NO3 ) y + KHSO 0,56 FeCO 0, 04 Al3+ 2+ Fe a Al(OH)3 x 0, 01 Fe3+ b Al2 O3 + NaOH t o ,kk 83, 41g + + H O Fe(OH) 0,57 K 0,56 Fe(OH) Fe O3 y + 0, 04 NH + c SO 24 0,56 H 0, 01 T CO N O x y 3+ 2+ 3+ + ý ta thy lng OH dựng tỏc dng ht vi Al , Fe , Fe , NH l 0,56 mol, nhng lng OH cho vo dung dch l 0,57 mol Vy lng kt ta Al(OH)3 ó tan 0,01 mol 27x + 180y = 10,17 x = 0,11 => Ta cú: x 0, 01 (y+ 0, 04) + 160 = 11,5 y = 0, 04 102 2 [ Fe ] a + b = 0, 08 [ + , ] 2a + 3b + c = 0, 23 56a + 56b + 18c = 4,84 [H] n H2O = => a = 0,03 ; b = 0,05 ; c = 0,02 0,56 0, 02.4 0, 01.2 = 0, 23(mol) [m] 10,17 + 4, 64 + 0,56.136 = 83, 41 + 0, 23.18 + m T => m T = 3, 42 gam Cõu 41: Hn hp X gm Al, Fe3O4 v CuO, ú oxi chim 25% lng hn hp Cho 1,344 lớt khớ CO (ktc) i qua m gam X nung núng, sau mt thi gian thu c cht rn Y v hn hp khớ Z cú t so vi H2 bng 18 Hũa tan hon ton Y dung dch HNO3 loóng (d), thu c dung dch cha 3,08m gam mui v 0,896 lớt khớ NO ( ktc, l sn phm kh nht) Giỏ tr m gn giỏ tr no nht sau õy ? A 9,5 B 8,5 C 8,0 D 9,0 ( THPT Quc Gia A 2014) Hng dn gii: 26 CO 0, 03 CO 0, 03 KL :Al, Fe, Cu 0, 75m CO Quy i: KL :0, 75m O 0, 25m + HNO3 O :0, 25m 0, 03.16 KL : 0, 75m NO3 : 2,33m H O NO 0, 04 Ta cú: n NO = 3n NO + 2n O 2,33m 0, 25m = 3.0, 04 + 2.( 0, 03) 62 16 m ; 9, 477 (gam) Cõu 42: Cho m gam hn hp X gm MgO, CuO, MgS v Cu2S (Oxi chim 30% lng) tan ht dung dch H2SO4 v NaNO3, thu c dung dch Y ch cha 4m gam mui trung hũa v 0,672 lớt (ktc) hn hp khớ gm NO2, SO2 (khụng cũn sn phm kh khỏc) Cho Y tỏc dng va vi dung dch Ba(NO3)2 c dung dch Z v 9,32 gam kt ta Cụ cn Z c cht rn T, nung T n lng khụng i thu c 2,688 lớt (ktc) hn hp khớ (cú t so vi H2 bng 19,5) Giỏ tr ca m gn giỏ tr no nht sau õy? A 3,0 B 2,5 C 3,5 D 4,0 Hng dn gii: Mg Cu m (g) + S O 0,3m H 2SO a NaNO 0,09 BaSO 0, 04 Mg 2+ x to 2+ Mg(NO ) MgO + 2NO2 + O Cu y x Ba ( NO3 )2 4m (g)Y Na + 0, 09 o t Cu(NO3 ) CuO + 2NO + O y NO3 SO 24 0, 04 to NaNO3 NaNO + O z H 2O a NO 0, 03mol SO x = 0, 06 = n NO2 x + y = 0,12 => 46x + 32y = 4, 68 y = 0, 06 = n O2 0,06mol NO2 2x + 2y = 0, 06 2x + 2y = 0,06 => x y z 0,06molO2 + + = 0, 06 z = 0, 09 = n Na + = n NaNO3 2 BTDT dd Y n YNO = 2x 2y + 0, 09 0, 04.2 = 0, 07 (mol) 1442+ 443 0,06 BTNT N n NO = 0, 09 0, 07 = 0, 02 (mol) n SO2 = 0, 03 0, 02 = 0, 01 (mol) BTKL m + 98a + 7, 65 = 4m + 18a + 1,56 (1) m 2,959 => 0,3m BTNT O + 4a + 0, 09.3 = 0, 04.4 + 0, 07.3 + 0, 06 (2) a 0, 035 16 27 Cõu 43: Cho hn hp X cha 56,9 gam gm Fe, Al, FeO, Fe3O4, Al2O3 v CuO Hũa tan ht X dung dch HNO3 d thy cú 2,825 mol HNO3 tham gia phn ng thu c 208,7 gam mui v 2,24 lớt khớ NO nht Mt khỏc, t hn hp X ta cú th iu ch c ti a m gam kim loi Giỏ tr ca m cú th l A 39,75 gam B 46,2 gam C 48,6 gam D 42,5 gam Hng dn gii: Fe KL Al 208, 7g NH +4 FeO 56,9g + HNO NO3 2,825 Fe3O4 H O Al O3 NO 0,1 CuO 56,9 + 2,825.63 208, 0,1.30 BTKL n H2O = = 1, 2875(mol) 18 2,825 2.1, 2875 BTNT H n NH+ = = 0, 0625 (mol) 4 BTNT N n NO = 2,825 0,1 0, 0625 = 2, 6625 (mol) m KL = 208, 0, 0625.18 2, 6625.62 = 42,5(gam) Cõu 44: Hũa tan ht hn hp Q gm Mg, Al, MgO v Al2O3 (trong ú oxi chim 16,72% v lng) bng dung dch cha 0,4 mol HNO3 v 0,709 mol H2SO4, sau kt thỳc phn ng thu c dung dch X ch cha cỏc mui trung hũa cú lng 95,36 gam v 4,4 gam hn hp khớ Y gm NO, N2O v N2 Cho t t dung dch NaOH vo dung dch X n kt ta t cc i, lc ly kt ta, nung ngoi khụng khớ n lng khụng i thu c 28,96 gam rn khan Nu tỏc dng ti a cỏc cht tan cú dung dch X cn dựng a mol NaOH Giỏ tr gn nht ca a l A.1,60 B.1,75 Mg HNO3 0, Quy i: Q Al + H 2SO 0, 709 O C.1,80 D 1,85 Mg 2+ x 3+ Al y MgO x 95,36g NH + z (1) + NaOH 28,96g y (2) t o Al2 O3 SO 0, 709 NO3 t H 2O NO N Y N 2O 4, 4g O N 0, + 0, 709.2 4z = 0,909 z (mol) BTKL m Q = 95,36 + 4, + 18(0,909 2z) 0, 4.63 0, 709.98 = 21, 44 36z (gam) BTNT H n H 2O = 0,1672.(21, 44 36z) (mol) 16 BTNT N Y m = 14(0, z t) (gam) N n Otrong Q = n Otrong Y = 4, 14(0, z t) (mol) 16 28 0,1672.(21, 44 36z) 4, 14(0, z t) + 0, 4.3 = 3t + 0,909 2z + 16 16 0, 7488z + 3,875t = 0,590048 (1) BTNT O BTDT 2x + 3y + z t = 1, 418 (2) 95,36g 24x + 27y + 18z + 62t + 96.0, 709 = 95,36 (3) 28,96g 40x + 51y = 28,96 (4) x = 0, 469 y = 0, => a = 1, 778(mol) z = 0, 04 t = 0,16 Cõu 45: Hũa tan hon ton 11,6 gam hn hp A gm Fe v Cu vo 87,5 gam dung dch HNO3 50,4%, sau kim loi tan ht thu c dung dch X Cho 500 ml dung dch KOH 1M vo dung dch X thu c kt ta Y v dung dch Z Lc ly Y ri nung khụng khớ n lng khụng i thu c 16 gam cht rn Cụ cn dung dch Z c cht rn T Nung T n lng khụng i thu c 41,05 gam cht rn Nng phn trm ca mui Fe(NO3)3 cú dung dch X gn nht vi giỏ tr no di õy? A 14% B 28% C 12% D 37% ( Nguyn Khuyn ln 2016) Hng dn gii: 56x + 64y = 11, x = 0,15 => 80x + 80y = 16 y = 0, 05 Gi s KOH ht thỡ T cú 0,5 mol KNO3 => m KNO2 = 0,5.85 = 42,5 g > 41, 05 g (Vụ lý) => KOH d KNO3 a KNO a to T 41, 05g KOH b KOH b a + b = 0,5 a = 0, 45 => 85a + 56b = 41, 05 b = 0, 05 Fe3+ 0,15 2+ Cu 0, 05 Gi s HNO3 d, dung dch X gm + phn ng vi 0,45 mol KOH l vụ lý Suy HNO3 ht H NO 3+ Fe z 2+ z + t = 0,15 z = 0, 05 Fe t => Dung dch X: 2+ 3z + 2t + 0, 05.2 = 0, 45 t = 0,1 Cu 0, 05 NO 87,5g dd Fe 11, 6g + HNO3 0,7 mol Cu Fe(OH)3 3+ Fe2 O3 Fe 0, 05 to Fe(OH) 16,0g 2+ CuO 0,5molKOH Fe 0,1 Cu(OH) Cu 2+ 0, 05 KNO3 t o KNO 41, 05g NO3 dd KOH KOH H O NxOy BTDT n NO = 0, 05.3 + 0,1.2 + 0, 05.2 = 0, 45(mol) 29 0, = 0,35 (mol) = 0, 7.63 0, 45.62 0,35.18 = 9,9 (gam) BTNT H n H2O = BTKL m N x Oy 0, 05.242 ; 13,565% 89, Cõu 46: Dung dch X cha m gam cht tan gm Cu(NO3)2 ( m Cu (NO3 )2 > gam ) v NaCl in phõn dung dch X vi in cc tr, mng ngn xp vi cng dũng in khụng i Sau thi gian t giõy thỡ thu c dung dch Y cha m -18,79 gam cht tan v cú khớ thoỏt catot Nu thi gian in phõn l 2t giõy thỡ thu c dung dch Z cha a gam cht tan v hn hp khớ T cha khớ v cú t hi so vi hidro l 16 Cho Z vo dung dch cha 0,1 mol FeCl2 v 0,2 mol HCl thỡ thu c dung dch cha a + 16,46 gam cht tan v cú khớ NO thoỏt Tng giỏ tr m + a l: A 73,42 B 72,76 C 74,56 D 76,24 ( Nguyn Khuyn ln 2016) Hng dn gii: n Cu ( NO3 )2 = x =y Gi n NaCl n electron (t giay) = z in phõn 2t giõy m ddX = 87,5 + 11, 9,9 = 89, (gam) => C% Fe( NO3 )3 = + Anot catot 2Cl 2e Cl 2+ Cu + 2e Cu x 2x y y y/2 2H 2O + 2e 2OH + H 2H 2O 4e 4H + + O 2z 2x zx 2z y (2z y) / y 2z y 2(z x) + 71 + 32 M = 32 = 32 => 30x + 19,5y 30z = (1) y 2z y (z x) + + + Na y FeCl2 0,1 Dung dch Z a (gam) NO3 2x + HCl 0, OH t + H + OH H 2O t 3Fe2+ 0,1 t + 4H + 0,2 t t NO 3Fe3+ + NO + 2H 2O > 0,053 0,1 0, t < => t < TH1 : Fe2+ ht trc (iu kin ) 15 0,1 0, mc tan =16,46g 20 18.t 30 18 = 16, 46 t ; 0, 074 (loi) 3 0,1 0, t > => t > ) TH2 : H+ Ht trc (iu kin 15 0, t 0, t mc tan =16,46g 20 18.t 30 18 = 16, 46 t = 0,16 (tha) BTDT y 2x = 0,16 (2) in phõn t giõy + 30 TH1 : Bờn anot H2O cha in phõn (iu kin y > z ) (+) Anot () catot 2Cl 2e Cl2 Cu 2+ + 2e Cu x 2x x y > z z/2 2H O + 2e 2OH + H z 2x z 2x mc tan =18,79g 64x + 35,5z 17(z 2x) = 18, 79 98x + 18,5z = 18, 79 (3) x = 0,12 (1),(2),(3) y = 0, (tha y > z) z = 0, 38 TH2 : Bờn anot H2O ó in phõn (iu kin y < z ) (+) Anot () catot Cu 2+ + 2e Cu x 2x x 2Cl 2e Cl2 y y y/2 2H O + 2e 2OH + H 2H 2O 4e 4H + + O z 2x z 2x z y z y (z y) / mc tan =18,79g 64x + 35,5y 17[(z 2x) (z y) ] = 18, 79 98x + 18,5y = 18, 79 (4) x ; 0,117 (1),(2),(4) y ; 0,394 (khụng tha y < z) z ; 0,373 => m + a = 72, 76 (gam) a = 2.0,12.62 + 0, 4.23 + 0,16.17 = 26,8(gam) m = 0,12.188 + 0, 4.58,5 = 45,96 (gam) 31 ... sinh hin thỡ u HC THI vỡ th gp nhng cõu hi kiu ny thỡ ụi thc dng li hay (tỡm cỏch gii ỏp s ca bi toỏn thay vỡ ln tn gỡ ú hay ỏp t rng buc lý thuyt Dóy in húa vo, cú lm tr nờn muụn trựng khú... 05.2 + (2 y + z 0, 05).3 + 0, 02 + = 0, 48 0,03 ì 180 ì 100 = 29,96% %m Fe(NO3 )2 = 18,025 [Hay] Cõu 30: Cho m gam hn hp H gm FexOy, Fe, Cu tỏc dng ht vi 200 gam dung dch cha HCl 32,85% v