Bồi dưỡng học sinh giỏi hoá 9 nxb Đại học Quốc gia 2014 phần 1

540.76 PGS.TS. CAOCL/GlAC B452D Tai lieu danh cho: Hoc sinh gioi va chuyen hoa hoc • Sinh vien sa pham hoa hoc Giao vien day boi dUSng hoc sinh gioi hoa hoc V 11 1 NHA XUAT BAN DAI HOC QUOC GIA HANOI B6I Dciam Tai lieu danh cho: Hoc sinh gioi va chuyen hoa hpc Sinh vien sU pham hoa hpc Giao vien day boi di/6ng hpc TH 1 NHA JAT BAN D Jji not dAu Chuxrnq 1: MOT S BA T r o n g q u a t r i n h d a y h o c cV T H C S da c 6 n h i e u e m h o c s i n h h o c 16 n a n g k h i e u ve k h a n a n g t i i d u y va h o c t a p m o n hcSa hoc. C h a n g h a n , ccS e m t h i c h k h a m p h a t h e gicvi t u n h i e n n h u t i m h i e u c a u t a o cac chat t h i f c f n g g a p t r o n g c u o c s o n g va d u d o a n , g i a i t h i c h cac t i n h c h a t cua c h u n g . C u n g c o e m thi'ch l a m cac t h i n g h i e m n h o de t u n h i e u va chiVng m i n h cac h i e n tUOng r o i n i t ra q u y l u a t . N h i e u e m c 6 t r i nhci dac biet ve sU p h a n loai cac chat va t i n h chat ciia c h i i n g , t h e h i e n sU sang t a o va t h o n g m i n h t r o n g each g i a i b a i t a p , . . . T u y n h i e n , cac e m m u o n p h a t t r i e n dUOc k h a n a n g t U d u y ciia m i n h v e m o n h o a h o c t h i c a n p h a i duOc t r a n g b i m o t each c 6 he t h o n g t i f l y t h u y e t d e n k l n a n g g i a i b a i t a p va cac ufng d u n g thUc h a n h p h u h o p v 6 i sU p h a t t r i e n n h a n thufc ciia lila t u o i h a o h i i n g va p h a n khcVi nay. D e g i i i p giao v i e n , p h u h u y n h va h o c s i n h bac t r u n g h o c cO -id ccS t h e m t U l i e u b o i dUc^ng h o c s i n h g i o i m o n h o a h o c , c h i i n g t o i b i e n soan b o sach " B o i ditofng h o c s i n h g i o i H o a h o c " t h e o cac n o i d u n g sau: 1. PhiTdng phap tinh theo cdn • Ooi vdi hdp chat c6 hai nguy TacoCTTQ: A,By o/„m^ = — ^ . 1 0 0 % y.M„ %m„=^-^.100% M ^ %m %m B • Doi vcfl hdp chat c6 nhieu ng Ta CO (TTTQ: AxByCz + each 1: Thiet lap cong thiTc d x:y:z=^:^:^^CTDGN M, M 3 M , A . L i t h u y e t t r o n g tarn B. PhUcfng p h a p g i a i b a i t a p Hay: x:y:z = C . B a i t a p ap d u n g A% : B% C% : M3 M , CTPT D . HUcVng d a n g i a i M, Di/a vao Mhdp chat E. Bai t a p t U l u y e n ( t i i l u a n va t r i e n g h i e m ) . + each 2: Thiet lap cong thuTc p H y v o n g vofi k e t cau va n o i d u n g ciia b o sach, p h a n n a o se l a m c h o d o c gia, dac M,.x b i e t la cac e m h o c s i n h t i m t h a y nhufng d i e u b 6 i c h va n i e m say m e t r o n g viec h o c M^.y mg M,.Z_MA,B,C, ^ ^\B^C^ tap h o a h o c . T r a n trong cam d n ! C a c tac gia Nhd sach KJtajtg Vietxitt trdn tronggim thieu toi Quy doc gid vd xin lang nghe moi y kim doug gop, de aion sach ngdy cdng hay horn, bo ich han. Thu xin giti ve: Cty T N H H Mot Thanh Vien - Dich V u Van Hoa Khang Viet Hay: ^ ^ ^ J } o ± ^ A% B% 0% 'V^ 10 Vi du 1: Mot hdp chat cua kim lo Trong do oxi chiem 3 0 % ve kho Hu'dng dan: Ooi vcfi dang na cho hoc sinh theo cac bu'dc sau: + Oat cong thut tong quat (CT CFTQ: R 0 „ ^ Hoa tri ciia R : 71, D i n h T i e n H o a n g , P. D a k a o , Q u a n 1, T P . H C M T e l : (08) 39115694 - 39111969 - 39111968 - 39105797 - F a x : (08) 39110880 + Ap dung cong thCrc:°!^I!1KL Hoac Emaih [email protected] va hoa tri cua R. Bdi dudng hgc sink gidi Hda hgc 9 — Cao Cxi Gidc M„ = Ll^ do hoa tri cue R la " 3.x ^. 70.y_ 30.x 16 Ta c6: " /x %mg. nen 56 2y 3 x va hoa trj cua R. 63.64.y^M, 36,36.x + Cuoi cung ta bien luan hoa trj de xac djnh ten kim loai R (Li/u y: kim loai c6 hoa tri tif 1 den 4, phi kim hoa tri tif 1 den 7) 32 => M , = 2y/ 1 / X 28 MR MR 1 2 3 4 18,7 37,3 Loai 56 74,7 2. Phu'G^ng p h a p bao toan kh (Fe) Loai - Nguyen tac cua phUdng p ling bang tong khoi luWng cac ch - Trong mot phan trtig hoa h thanh), neu biet khoi lu'dng cua khoi lu'dng. Loai Vay nguyen to R la Fe. Ket luan: Kim loai c6 hoa tri I I I va M = 56 la Fe Vi dM 2: Mot oxit c6 chu^a 6 0 % oxi ve khoi lydng. Xac djnh cong thCrc Oxit tren. Hitaing dan: + Dat cong thiTc tong quat (CTTQ) va suy ra hoa trj cua nguyen to con iai CTTQ: R^Oy ^ Hoa tri cua R : + Ap dung cong thCrc: %m3.x =— MB de thiet lap bieu thirc hien he giiJa M R .y K va hoa tri cua R. Ta c6: 4?^ =^ 60.x 16 »/. " 16 2y 3 x M. = ^ 3.x nen do hoa tri cua R la Vi du 1: Cho luong khf CO di q gam chat ran, va 17,6 gam CO Hitdng dan: Phan tich dC k ngay vi so chat tham gia va cha chat. Mat khac chu'a biet khoi lu du" nen khong tinh theo kieu bai Do do ta di/a vao phu'dng trin Fe304 + CO — FeO+ CO Cuoi cung ta bien luan hoa trj de xac djnh ten nguyen to R ^ 3Fe0 -> Fe + Qua phu'dng trinh nhan thay: 17,6 2y/ / x 1 2 3 4 5 6 7 MR 5,3 10,6 16 21,3 26,7 32 37,3 Vay nguyen to R la lull huynh => CTPT Oxit: SO3 V I du 3: Mot hcJp chat gom Kim loai chu-a biet hoa trj vdi luU huynh (biet S c6 hoa tri II), trong do lu'u huynh chiem 36,36% ve khoi luWng. Xac dinh cong thirc phan tCr hdp chat tren? Hu-dngdan: ' ^ CTTQ: R^Sy ^ Hoa trj cua R : ^v/ .| = 0,4m 44 Khi du du' kien ta ap dung dun "2 V i d u 2: Cho luong khf CO du' d MgO, ZnO, CuO nung nong, 23,6 gam chat ran Z. Cho Y Ip thay CO 40 gam ket tua xuat h Hitdng dan: n CaCOg 40 .| QQ = COz + Ca(0H)2 ^ CaCOj i + H2O (1) 0,4 <- 0,4 X + CO, t°: AI2O3, MgO khong bi khu". CuO + CO — ^ Fe304 + CO — FeO + CO — ^ ^ (2) (3) (4) (5) ^ 3FeO + CO2 (3) Fe + CO2 (4) n,o Sddo phan iTng : RCOOH + So mol O: 0,1.2 Di/a vao sd do tren ta ap dung n Cu + CO2 (2) Ta CO no(RCOOH) + no(02 pu") = no(02 pu')= 0,6mol - > n Vi d u 2: Cho 13,36g hon hdp gom dung dich HNO3 bang du', thu d Zn + CO2 (5) ZnO + CO Hi/dng dan: Ta nhan thay kho biet cong thiTc cu the axit ma chi bi ta thIet lap sd do phan iTng : =n,o^=0,4mol (dktc) va dung djch X. Co can du gia tri m thu du'dc? Theo dinh luat bao toan khoi li/dng: mx + mco = MrSnz + nn^^^ => mx = mz + m(,Q^ - mco = 23,6 + 44.0,4 - 28.0,4 = 30 gam Vi du 3: Hoa tan het 7,74 gam hon hdp bot Mg, Al bang 500ml dung dich HCI IM va H2SO4 0,28M thu du-dc dung dich X va 8,7361 lit H2 (dktc). Co can ' ' dung dich X thu du'dc l^dng muoi khan la bao nhieu? $ HWdngdan: Hitdng dan: Ta c6: n^,„ = 1^11 22,4 Sd do phan iTng: Fe, FeO, Fe203 va Fe304 + H Gpi x la so mol Fe(N03)3 Ap dung dinh luat bao toan ngu %HN03) Taco: n„ = M ^ = 0,39mol H2 22,4 HHCI n^, Di/a vao sd do ta thay: n^^^ = ^ = 0,5.1 =0,5mol H2SO4 Mac khac: m.^ + m^^^^ = m,^, =0,28.0,5 = 0,14mol Ap dung OLBTKL: mhh+ mHci + m^^ => 11,36 + (3x+0,06).63 = 24 = mmua + m^^ =^mmu6i = 7,74 + 0,5.3,05 + 0,14.0,8 - 0,39.2 = 38,93 gam. 3. Phu'dng phap bao toan nguyen to Trong mot cac phu'dng trinh hoa hoc, cac nguyen to luon du'dc bao toan ->T6ng so mol cua mot nguyen to A tru'dc phan iTng hoa hoc luon bang tong so mol cua mot nguyen to A do sau phan lirig. 001 vdi dang bai tap nay cac em khong can viet phu'dng trinh (hdac c6 tru'dng hdp khong the viet phu'dng trinh, hoac phu'dng trinh phtfc tap) ma chi lap sd do phan (fng giCa chat tham gia va san pham. Sau do thiet lap moi quan he bao toan cho nguyen to c6 lien quan den du' lieu can tim va giai Vi du 1: Dot chay hoan toan O,lmol axit Cacboxylic ddn chiTc, can vCra du V lit 02 (dktc) thu du'dc 0,3 mol CO2 va 0,2 mol H2O. Tinh gia tri cua V ? 6 = "N(Fe(N03)3 + ^NINO) =^ X = 0,16 mol => m,^,^o^,^ = Vi d u 3: Hoa tan hoan toan hon axit HNO3 vLTa du, thu di/dc d duy nhat NO. Tim so mol CU2S Hitdngdan: Sd do phan iTng: 2Fe2S ^ ^ ^ ^ ^ Fe2(S04) 0,14mol CU2S a mol 0,7nriol > 2CuS0'4 2a mol Ap dung dinh luat bao toan ngu 0,14.2 + a = 0,07.3 + 2a ^ 4. Phu'dng phap tang — giam khoi lu'^ng N2(C03)3 - Nguyen tac cua phu'dng phap: Khi chuyen tCr chat nay sang chat khac thl khoi lu'cJng tang hay giam mot lu'dngAm( h a y A V d o i vdi chat khi), d o cac chat khac nhau c 6 khoi lu'dng mol khac nhau (hay doi vdi chat khi: ti le mol khac nhau). Di/a v a o su" tu'dng quan ti le thuan cua s i / tang - giam, tinh du'dc khoi lu'dng (hay the tich) chat tham gia hay tao thanh sau phan (ing. Bai toan giai du'dc theo phu'dng phap bao toan khoi lu'dng se a p dung du'dc cho phu'dng phap nay. Nhu'ng vcfi phu'dng phap khong can biet het (n-1) TGKL dai lu'dng ta van giai du'dc neu biet du'dc SLT bien thien A m h a y A V . 6HCI -> (2N+180)g ( T u ' ( l ) va (2), ta thay: CiTtao ra 1 mol CO2 thi khoi lu'd Vay de tao ra 0,03mol CO2 thi k Vay khoi lu'dng muoi sau phan C V i d u 2: N h u n g mot thanh C u ' s 0 4 0,5M, sau phan uTig la VI d u : Khi day ve tinh chat hoa hoc cua axit tac dung vdi muoi : la 51gam. a) Tinh khoi lu'dng dong tao tha Phu'dng trinh : C a C O j + b) Tinh CM cac chat trong dung 2HCI CaCb + C O 2 + H 2 O Nhan thay cu" I m o l CaCOs tham gia thi tao ra I m o l muoi CaCb Ve khoi lu'dng lOOgam lllgam Neu X mol CaCOs phan iTng thi sau phan uCng : A m = agam , dt/a v a o ti le nay khi biet a ta xac djnh du'dc x va ngu'dc lai. Nang cao hdn muoi kirn loai chu'a biet hoa t n II hay hon hdp muoi kirn loai chu'a b i e t : Phu'dng trinh : MCO3 + Nhan thay CLT 2HCI MCI2 + Ve khoi lu'dng M + 6 0 g a m -> MCI2 Cur I m o l MCO3 phan iTng thi sau phan uTig : A m = ( 7 1 - 6 0 ) g a m X= 71-60 b gam Doi vdi nhuTig loai bai tap khong c 6 it so lieu ma phu'dng trinh phdc tap c 6 the giai hoac khong giai du'dc thong qua he phu'dng trinh. Nhung ta thay c6 s i / bien thien ve A m h a y A V thi c6 the ap dung theo phu'dng phap tang giam khoi lu'dng. V I d u 1: C h o l O g h o n h d p 2 m u o i c a c b o n a t k i m loai h o a t r i II v a I I I t a c dung vdi axit HCI vu^ d u , thu du'dc dung djch A v a 672ml khi (dktc). CO can dung djch A thu du'dc bao nhieu g a m muoi? MCO3 + 2HCI X = ^ = 0,125mol -» npe = nc ^ CM CUSO4 c o n lai = 0,5 - 0,3 V i du 3: Khi lay 14,25g muoi cloru la 7,95g. Hay tim cong thuTc cu Hifdng dan: T a c 6 cong thC X(N03)2 (M = X + 124). Ta thay CLT I m o l muoi nitrat cu la: (X + 124) - (X + 71) = 53g. T h e o gia thiet khoi lu'dng khac hdn goc CI" nen tCr gia thiet ta c6: mX(N03)2 - mXCl2 = 7,95g. MCO3 ddA N2(C03)3 672mlkhi r i g muoi Cac phu'dng trinh hoa hoc: (M+60)g Vay X mol sat phan uTng se tao lu'dng muoi nitrat cua X vdi so , difa v a o ti le nay khi biet b ta xac dinh du'dc x va ngu'dc lai Hi/dng dan: l O g a m hon hdp \ Imol Nong d p M cua dung dich CUSO M + 7 1 gam Neu X mol MCO3 phan (fng thi sau phan Crng : Am = CUSO4 Nong d o M cua dung dich FeS0 CO2 + H2O I m o l MCO3 tham gia thi tao ra I m o l muoi + Imol CLT I m o l sat (56g sat) phan ung a1 => X = Hitdng dan: Phu'dng trinh hoa Fe GUI' I m o l CaCOa phan uYig thi sau phan ling : A m = 1 1 1 - 1 0 0 = l l g a m Hoi + Vay so mol muoi cua kim loai X => Khoi lu'dng mol cua muoi cl -> MCI2 + (M+71)g CO2 Imol + H H 2 0 (1) M =^-^^= 9 5 0,15 X + 71 = 5. Phu'dng phap dung khoi lu'dng mol trung binh ( M ) Nguyen tac cua phu'dng phap: M la khoi lu'dng cue 1 mol hon hdp. M n.|M, + n^Mj + njMg + ... = Va rdi vao be tac khong giai du'dc. f V2M2 + V3M3 + ... Vi 2 kim loai cung nhom I nen cung v.. Oat ky hieu chung ciia 2 kim loai la Phu'dng trinh phan iCng : 2 R + 2 H Hoac : M = XiM] + X2M2 + X3M3 + ... Vdi Ml, M2, M3...: 0,2 mol khoi lu'dng mdl va ni, n2, n3...la so mol cac chat trong hon hdp. Vi, V2, V3...; the ti'ch cac khi trong hon hdp khi. Xi, X2, X3...: so phan mol ciia cac chat trong 1 mol hon hdp. , 0,2 Do 2kim loai cung nhom I va thuoc Khi hdn hdp g S m 2 chat: Mi < M < M2. -n^M^ X + y = n,,, ^ - 0,1 - each giai theo trung binh: n, + n^ + ng + ... M hhkhi Ax + B y = 5,4 Sau d o lap he +(n-n^)Mg MNa = 23 < 27 < MK = 39. -_3M,+(V-V,)M, n Vay hai kim loai d o la Na, K. V Vi du 2 : Hon hdp X g o m hai muoi cac chu ky ke tiep nhau trong bang t M = XiMi + (1 -X2)M2. bang dung dich HNO3 du, thu du'dc Co the ti'nh di/a vao M theo s d d d cheo: Ml ^ M - M2 ni, Vi, Xi. het bdi dung dich Ba(0H)2 du' thu cong thiTc cua hai muoi va tinh thanh HWdngdan: Ml - M - n2, V2, X2. Oat C r Chung cua 2 muoi cacbonat ci PTTHH : RCO3 + 2HNO3 Phu'dng phap nay t h a d n g ap dung giai bai toan hon hdp hai hay nhieu chat R(N03)2 CO2 + Ba(OH)2 -> B a C O g CO cung mpt ti'nh chat tu'dng du'dng nad d d . Sai lam nhat khi giai bai tap nay hdc sinh thu'dng du'a v'e dang bai toan hdn hdp. Nhu'ng khi tien hanh giai '^BaCOg thu'dng rdi vao be tac la an so nhieu hdn so phu'dng trinh nen khong tim du'dc '^'BaCOg 7,88 197 - 0.04mol triet de cac an sd. Dan den mat thdi gian va chan nan trdng qua trinh giai bai tap hda hoc. VI d u 1 : Hoa tan 5,4g hon hdp 2 kim b a i cung nhdm I 6 2 chu ky lien tiep nhau vad nu'dc thu du'dc thu du'dc 2,24 lit khi (dktc). Xac dinh ten 2 kim b a i . Hitdng dan: Dat an so cho ten va s5 mol cho kim b a i can tim 2 B + 2 H2O y 10 2 BOH + H2 T y 3 ' 3,6 ^ = ^ — = 90 ^ MR nRC03 "RCO3 0,04 0,04 Vi 2 kim b a i hpa tri II lai thupc 2 chu Ml < M < M2 r:> Ml < MR - Hdc sinh thu'dng sai lam nhu' sau: 2 A + 2 H 2 0 - > 2 A O H + H2T MRCC, = ^'^'^^ Vay 2 kim b a i dp la Mg va Ca CTHH cua 2 mupi: MgCOa va Ca Gpi a, b Tan lu'dt la so mol cua MgCO , 184a + 100b = 3,6 Ta co: ' a + b = 0,04 < ^ a = 0, b-0,0 o/ ' M9CO3 = !!Wo^ . 1 ooo/o rn^^ 3,6 . 1 ooo/o = 58.33% = 100% - 58,33% = 41,67% =^ % " i c 3 C 0 3 Ta CO he phu'dng trinh: ^ [ 95x + 133 Vi dv 3: Hoa tan vao niTdc 7,14g hon hdp muoi cacbonat trung hoa va cacbonat axit cua mot kim loai hoa tri I, roi do them lu'dng dung dich HCI vLTa du thi thu du'dc 0,672 I khi d dktc. Xac dinh ten kim loai tao muoi. Ht/dng dan: Oat kf hieu kim loai la M, x, y Ian lu'dt la so mol cua M2CO3 va MHCO3. Ta CO phu'dng trinh phan Cfng : M2CO3 + 2 HCI 2 MCI + H2O + CO2 T X X MHCO3 + HCI ^ MCI + H2O + CO2 t y y muoi = X + y = M muoi = — 0,03 o 89 < = = 238. Vi M + 6 1 < M < 177 r:> 0,03 (mol) Goi a, b, c, d Ian lu'dt la so mol cua M 2Mg + O2 ^ ^ 2 MgO M muon = 238 < 2M + 60 a M la Cs. Mg + 2 HCI ^ MgCl2 + H2r x ' 2 Al + 6 H C I 2 AICI3 + 3H2T X y X y i,5y Fe + 2 HCI -> FeCb + H2t z ». z z 0,5a ' b 0,75b a 0,5b 2Zn + O2 2ZnO c c 0,5c 2 Cu + O2 d 0,5d (3) 2 CuO d B gom (MgO, AI2O3, ZnO, MgO + 2HCI a 2a (4) CuO) MgCl2 + H2O (5) AI2O3 + 6HCI ->2AICl3 + 3 H2O (6) 0,5b 3b ZnO + 2Ha -^ZnCl2 + H2O c 2c (7) CuO + 2HCICUCI2 + H2O (8) d 2d Ap dung dinh luat bao toan khoi lu'dng => "^02 = 12 (1) 4 A I + 3 O 2 — ^ 2 A l 2 0 3 (2) 6. Phu'dng phap ghep an so ( Nguyen tac cua phu'dng phap: Dung thu thuat toan hoc ghep an so de giai cac bai toan c6 an so Idn hdn so phu'dng trinh toan hoc lap du'dc ma yeu cau bai ra khong can giai chi tiet, day du cac an. Dang nay thu'cJng gap khi tinh toan khoi lu'dng chung cua hon hdp cac chat (hon hdp kim bai, hon hdp muoi, ...) tru'dc hoac sau phan (inq ma khong can tinh chi'nh xac khoi lu'dng tiTng chat trong hon hdp. Vi du 1: Chd hon hdp X gom Al, Fe, Mg tac dung v6i dung dich HCI du" thu dutJc 11,2 lit khi (dktc) va 53,0g muoi. Tim khoi luWng hon hdp X. Hitdng dan: Gpi x, y, z Ian lutrt: la so mol cua Mg, Al, Fe c6 trong hon hdp )Q PTHH: t, Vdi 3 an, c6 2 phu'dng trinh. Tim khoi lu'dng 3 kim loai tiTc la tong Tach (2) ta du'dc: 24x + 27y + 56z 24x + 27y + 56z = 53,0 - 0,5 x Vi du 2: Cho 3,8 gam hon hdp A gom hoan toan vcfi oxi du' thu du'dc h 5,24gam. Tinh the tich dung djch H hoan toan B. HWdng dan: each 1: Dung phu'dng phap ghep an 0 672 nco2 = x + —y + z - mp = 5,24 - 3,8 = l, w 1 44 Theo (1,2,3,4) : - 0,5a + 0,75b + 0,5c + 0,5d = ^ - T h e o PTHH: n^^^^^ = n^^^ = 0 = 0,045mol - Nhu' vay con 5 g a m CaCOs n o m : CaCOj d i / , AI2O3, Fe203 va C Theo (5,6,7,8) : n^^,, = 2a + 3b + 2c + 2d = An^^ = 4 x 0,045 = 0,18nnol => \ ^ r , - - ^ = — = 0,181 = 180mI f v . . . Cach 2: Sau khi t i m ra so mol O2. ' ' Nhan xet: Trong cac cap chat phan (fng la : 1 va 5 ; 2 va 6 ; 3 va 7 ; 4 va 8 thay so mol axit luon gap 4 Ian so mol O2. %ALO, = 15,2 .100% 67 % F e , 0 3 -- — . 1 0 0 % = 1 4 , 6 67 %CaC03 - ; ^ . 1 0 0 % 67 Do d o : So mol HCI = 4 x 0,045 = 0,18 mol. 7,4% %CaO = 62,76% Tim ra the ti'ch dung dich la 180ml. Vi du 2: Cho m g a m hon hpp Na 7. PhUdng phap tu" chon lu'dng chat Nguyen tac cua phu'dng phap: Phan tram lydng chat trong dung dich hoac thu du'dc cho tac dung vdi Ba( trong hon hop nhat dinh la mot dai lu'dng khong d o i . Khi giai ngu'di giai tu' trong khdng khi den lu'dng kho chon lu'dng thi'ch hdp de giai bai toan 2.2.7.2. Xay di/ng va sir dung bai tap hoa hoc theo phWdng phap giai bai lu'dng moi kim loai ban d a u . Hi/dng dan: - PTHH xay ra khi cho m gam ho tap til chon iWdng chat trong day hoc d trWdng THCS Cac e m thu'dng lung tung va khong xac dinh hu'dng giai khi gap dang nay vi 2Na + 2HCI -> 2NaCI + H2 de bai chi cho lu'dng chat du'di dang tong quat hoac khong noi den lu'dng chat Fe + 2HCI -> FeCb + H2 nhu'ng biet du'dc ti le giiJa cac chat - PTHH xay ra khi cho dung dich Khi gap dang nay cac e m c6 the chpn lu'dng chat c6 m o t gia tri nhat dinh de tien viec giai. Co the chpn lu'dng chat la mpt mdl hay m p t so mol theo he so ty lu'dng trong phu'dng trinh phan iTng; hoac lu'dng chat la lOOg,... Vi du 1: Hon hdp g o m CaCOs Ian AI2O3 va Fe203 trong d o AI2O3 chiem 1 0 , 2 % , Fe203 chiem 9 , 8 % . Nung hon hdp nay 6 nhiet d p cad t h u du'dc chat ran c6 lu'dng bang 6 7 % lu'dng hon hdp ban dau. Ti'nh % lu'dng chat ran tao ra. Hitdng dan: FeCl2 + Ba(0H)2 -> Fe(0H)21 + - PTHH xay ra khi nung ket tua t 4Fe(OH)2 + O2 - Gpi m = mpe + mNa = 100 gam =^"^Fe203 - Gpi khoi lu'dng hon hdp ban dau la 100 g t h i : m^, ^ 2 3 = 10,2g - - ^ 2Fe203 + -100gam=^n,^^o^ - T h e o P T H H ( 4 ) : n,^,„,,^ = 2.n,^ - Theo PTHH ( 3 ) : n , ^ „ ^ m^co. - 8 0 g - PTHH xay ra khi nung hon hdp: CaC03 - T h e o PTHH ( 2 ) : n^^ := n^^j,,^ = 1 — > CaO + CO2T - Thed bai ra, lu'dng chat ran thu du'dc sau khi nung chi bang 6 7 % lu'dng hon hdp ban dau. Nhu vay d p giam khoi la do CO2 sinh ra bay d i . 33 - V a y m^Q^ = 1 0 0 - 6 7 = 3 3 g => n^^^ = — = 0 , 7 5 m o l 14 n,^^„,, -Vay:%Fe = 70% % Na = 3 0 % . Vi d u 3: Hon hdp g d m NaCI, KC dich. Them AgN03 du" vao dung bang 2 2 9 , 6 % so vdi A. Tim % mo Hitdng dan: - PTHH xay ra: NaCI + AgNOa - > AgCI + NaNOa KCI + AgN03 - > AgCI + KNO3 Do chu'a biet X la kim loai hay p 2 3 1 X 2 16 8 Mx (1) (2) ' , : -nr 229 6 - Gpi rriA - lOOg m^g^, = 229,6gam n^g^i = 1 ^ = ^'^ - G o i PNaCI = X So mol AgCI sinh ra d phan uTng (1) la: x So mol AgCI sinh ra d phan iTng (2) la: 1,6 - x =>n^ci='"'6-x -Ta c6: MNaci-nNaci + MKCI-DKCI = 100 => 58,5x + 74,5(1,6 - x) = Vay: nNaci = 1,2 mol /O =i> x = 1,2. ^ |%KCI = 100% - 7 2 % = 29,8% 8. Phi/dng phap bien luan de tim cong thu'c phan tuT Nguyen tac: Khi tim cong thu'c phan tCr hoac xac dinh ten nguyen to thu'dng phai xac dinh chinh xac khoi lu'dng mol, nhu'ng nhiJng tru'dng hdp M chu'a c6 gia tri chfnh xac doi hoi phai bien luan. Pham vi uYig dung: Bien luan theo hoa tri, theo lu'dng chat, theo gidi han, theo phu'dng trinh v6 dinh hoac theo ket qua bai toan, theo kha nang phan uTng. Khi giai dang nay cac em thu'dng lung tung va giai den g\\jta chi/ng thi diTng lai vi luc do so an nhieu hdn so phu'dng trinh ma khong the ap dung cac phu'dng phap khac nhu' ghep an so, hay phu'dng phap bao toan khoi lu'dng. Luc nay cac em can tim each bien luan thfch hdp. Gia s(f mot phu'dng trinh c6 hai an so la khoi lu'dng mol (M) va hoa tri cua nguyen to. Ta c6 the bien luan hoa tri cua nguyen to theo khoi lu'dng mol VI du 1: Dot chay I g ddn chat X can dung lu'dng vCfa du 0,7 (I) O2 (dktc). Xac dinh X? Ht/dng dan: Goi x la hoa trj ciia X + |02 ^ 0,7 0,7 5,6x 22,4 1 M , = ^ = 8x 6,5x 16 vi du 2: Cho 3,06g oxit MxOy ta dung dich thay tao ra 5,22g m mot gia tri duy nhat. HWdfngdan: MxOy + 2yHN0 Bao toan nguyen to H: n „ „ = a Bao toan khoi lu'dng: 3,06 + 63 a = 0,02 mol - > n^^o, = 0^ m^^^,, = 1,2.58,5 = 70,2gam J%NaCI-70,2% 2X —> X2OX Bao toan nguyen to N: nMuoi = — Mmuoi = 5,22 .n = 130,5n ^ M 'HNOQ Bien luan M theo hoa tri n ta c6 n 1 M 68,5 Vay kim loai M la Ba. Vi dy 3 : Hoa tan 4,0g hon hdp g dich HCI thi thu du'dc 2,24 lit H2 cho vao dung dich HCI thi dung dinh kim loai hoa tri II? Hitdng dan: n^^ = n^^ = 0,1 mo A = -1^40^M<40 0,1 nHci = 0,5 mol M + 2HCI 0,25 < - 0,5 = MCI2 + H2 2.4 - = 0 , 2 5 - > M > 9,6 M 9,6 < M < 40 - > M la Mg. T