Tuyển tập các dạng điển hình và phương pháp giải nhanh bài tập trắc nghiệm hóa học 12: Phần 2

B a i 13: Cho 1,35 gam hon hop gom Cu, Mg, A l tac dung het vdi dung dich HNO3 thu duac h5n hdp k h i gom 0,01 mol NO va 0,04 Tinh khoi laang muoi tao dung dich A 10,08 gam B 6,59 gam C 5,69 gam Hit&ng dan mol NO2 B a i 14: Cho 1,35 gam hon hap A gom Cu, Mg, A l tac dung vdi HNO3 du dirac 1,12 l i t hon hop X gom NO va NO2 (dktc) c6 khoi lucfng mol trung binh la 42,8 Tong khoi lUcJng muoi nitrat sinh la: A 9,65 gam D 5,96 gam Cdch 1: n^= Dat X, y, z Ian lifcft la so mol Cu, Mg, A l Cu'^ + 2e Mg ^ x ^ x - ^ x Thu e: N + 3e ^ Mg'^ + 2e y - > y - > y N (NO) N + le ^ 0,03 x\^^Q - m = mcu(No,), + Ap dung dinh luat bao toan khoi luang: => ni„u6i = 5,69 y z > 2y > 3z Khoi liidng muoi nitrat sinh la: 0,06 mol • m„,u6i 2x A l ^ Al^^ + 3e => 2x + 2y + 3z = 0,07 = 2.0,04 + 4.0,01 = 0,12 mol 1,35 + 0,12.63 = > Mg ^> Mg2^ + 2e mMg(NO,), + = 1.35 + 62.(2x + 2y + 3z) ^Aum,), = 1,35 + 62.0,07 = 5,69 gam Chon dap an D Cdch 2: Ap dung cong thiJc: + 0,01.30 + 0,04.46 + 0,06.18 m gam muoi nitrat = " kim loai + molgpk.so nhan.62 Chon dap an C I Cdch 3: Dung cho trSc nghiem fiai 15: Nung m gam bot sat oxi, thu dugc gam h6n hcJp chat rSn X Hoa tan het hon hdp X dung dich HNO3 (dtf), thoat 0,56 lit id dktc) NO (la san pham khuf nhat) Gia t r i cua m la nimudi = mu + Hspk.so e nhan.62 = 1,35 + (0,04.1+ 0,01.3).62 = 5,69 gam Chon dap an C = 1,35 + (0,01.3+ 0,04.1).62 = 5,69 gam A 2,52 gam B 2,22 gam C 2,62 gam D 2,32 gam Hit&ng dan Hitctng dan gidi Dat hai kim loai A, B la M Cdch 1: m gam Fe + O2 gam h6n hop chat r ^ n X HNO, du > 0,56 Ht NO ThUc chat cac qua t r i n h oxi hod - khuf tren la: Cho e: Nhan e: Fe Fe'^ + 3e m 56 3m 56 + 4e O2 3-m 32 3m 56 20'-; N +5 mol e 2H^ + 3e 0,075 mol m = 2,52 gam + (mol khu-so 3x-2y-0,025.3 0,15 mol Chon dap an A B a i 16: H5n hop X gom hai kirn loai A va B dufng trUdc H day dien hoa va c6 hoa t r i khong doi cac hop chat Chia m gam X hai phan hhng nhau: - Phdn 1: Hoa tan hoan toan dung dich chiJa axit HCl vk H2SO4 loang tao 3,36 l i t H - Phdn 2: Tac dung hoan toan vdi dung dich HNO3 thu duoc V l i t NO (san pham khuf nhat).Biet cac the tich k h i cf dieu kien tieu chuan Gia t r i cua V la A 2,24 l i t B 3,36 l i t C 4,48 l i t D 6,72 l i t -> 0,3 N^' 0,1 mol C 1,7M D 1,2M gidi 7Q9 n„ = i ^ ^ = 0,08mol ^ 22,4 N2 28 ^ ^ 37 NO2 = 0,03 mpe = 0,045.56 = 2,52 gam Hiidng dan • X = 0,045 " ^ V N O = 0,1.22,4 = 2,24 l i t = 0,7.3 + (0,025.3.5,6) = 2,52 gam 56x + 2e 0,3 M^* + Theo (1): So mol e cua M cho bfing so mol e cua 2W nhan; -> 4(3 - m) - Ph^n 1: M - + n H * - Phan 2: 3M + 4nH^ + nNOa'-^ 3M"^ + nNO + 2nH20 Chon dap an A mpe gidi 46 (MN Ta c6: +MNO = 9,25 x - 37 = ^ - ^ X ) ^ la trung binh cong khoi ItfOng phan tuf cua hai N2 va NO2 nen: HN, =nNo, - ^ va 2NO3- + lOe mol NO3" + le N2 0,08 m x = 0,2 ( + 160) = , a) T a c6: F e O + H2 ^ X 145,2 "Fe(No,), = giai Q u y h o n h a p X ve h o n h a p h a i c h a t F e O v a Fe203 v d i so m o l l a x , y , > Fe(N03)3 + N O + H O 0,2 m o l < dan D 112 m l 6: H n h a p X g o m ( F e , F e , Fe304, F e O ) v d i so m o l m i c h a t la 0,1 m o l , h o a t a n h e t v a o d u n g d i c h Y g o m ( H C l v a H2SO4 l o a n g ) 6xi t h u diiac d u n g d i c h Z N h d tCr tCf d u n g d i c h C u ( N ) I M v a o dung d i c h Z cho t d i k h i n g U n g t h o a t k h i N O T h e t i c h d u n g d i c h C u ( N ) can d u n g va the t i c h k h i t h o d t r a (d dktc)? A 25 m l ; 1,12 l i t B 0,5 l i t ; 2 , l i t C m l ; 2,24 l i t D m l ; 1,12 l i t HUdng Fe203 + H N O dan gidi y/2 - > 3y Quy h n h o p 0,1 m o l FegOs va 0,1 m o l F e O t h a n h 0,1 m o l Fe304 H o n h o p X gom: Fe304 0,2 m o l ; F e 0,1 m o l + dung dich Y ' Fe304 + 8H^ 0,2 Fe + H " He phuong t r i n h : lOx 0,06 ^NO dung d i c h HNO3 d i / thu duoc 2,24 l i t k h i NO2 (dktc) l a s a n p h a m khiJ 0,1 m o l n h a t G i a t r i c u a m l a =^ VNO = 0,1.22,4 = 2,24 l i t A 11,2 g a m B 10,2 g a m n e „ , K , = | n ^ = - mol =— H o a t a n h n h o p X v a o dung d i c h HNO3 dU t a c6 F e + 6HNO3 B a i 7: N u n g 8,96 gam F e t r o n g k h o n g k h i difoc h n hop A g o m F e O , Fe304, Fe203 A hoa t a n hoan toan t r o n g dung dich chufa 0,5 mol HNO3 t h i vCfa du, t h a y t h o a t r a k h i N O l a san p h a m khuf n h a t So C 0,03 D 0,02 Hiicfng dan gidi > F e ( N ) + 3NO2 + 3H2O M < 0,1 m o l => So m o l cua n g u y e n tuT F e t a o oxit Fe203 l a 8,4 56 mol N O thoat r a l a 0,1 V a y : mx = m^^ + m^^^o^ ^ 0,35 , _ >n F e , , = 0,16mol O D ta CO phaong t r i n h : 2Fe + O2 X F e O + 4HNO3 0,1 -> > F e ( N ) + N O + 5H2O -> 0,1 0,1 m o l x/3 m , C h o n dap a n A -> F e + 3O2 0,05 y/2 3FeO + IOHNO3 X - > lOx/3 Ta c6: 0,15 mol > 2Fe203 ^ > F e ( N ) + NO2 + 2H2O F e + O2 X Fe203: < > 2FeO F e + 3O2 y ' • Quy hdn hap X vi hat chdt FeO vd Quy h o n hop A gom ( F e O , Fe304, F e j O g ) t h a n h h o n h o p ( F e O , Fe203) 0,35 3.2 mx = - ^ + - ^ = 11,2 g a m C h o n dap a n A = ^ 56 D 6,9 g a m • Quy hdn hop X vi hat chdt Fe vd FegO^: = 0,05 l i t (hay 50 m l ) B 0,04 C 7,2 g a m HtCdng dan gidi Chon dap a n C A 0,01 = 0,02 mol c h a t r S n X g o m F e , Fe203, Fe304, F e O H o a t a n m g a m h n h o p X v a o > 3Fe'^ + N O t + 2H2O 0,1 V = B a i 8: N u n g 8,4 g a m F e t r o n g k h o n g k h i , s a u p h a n ufng t h u diiOc m g a m Dung dich Z: (Fe^^: 0,3 m o l ; Fe^^: 0,4 mol) + Cu(N03)2: ^ y = 0,1 m o l C h o n d a p a n D 0,1 m o l SFe'^ + NO3- + 4H^ X = 0,06 mol + 3y = 0,5 0,4 m o l > Fe^^ + H2T 0,1 0,3 X + y = 0,16 > Fe'^ + F e ' ^ + 4H2O 0,2 2Fe(N03)3 + SHaO -> 2FeO 0,1 mol SFeA 0,025 m o l = 0,1.72 + 0,025 160 = 11,2 gam n^ol _ _ 0,15 0,05 n CO Chii y: V a n c6 the quy h o n hcjp X ve h a i chat (FeO va F e ) hoSc (Fe va FeO), hoSc (Fe va Fe304) n h i m g viec giai t r d nen phuTc tap hcfn (cu the la t a p h a i dSt an so m o l m o i chat, lap he phtfong t r i n h , g i a i he phacfng t r i n h h a i a n so) • Quy hdn hop X ve mot chdt Id < 0,1.x = — 56 3x-2y ^ (trong oxit sit) CO2 => nco = no = 0,15 m o l -> mo = 0,15.16 = 2,4 gam =:> mpe = - 2,4 = 5,6 gam -> npe = 0,1 mol Theo phiicfng t r i n h p h a n ufng t a c6: - — n CO, Chon dap a n B Vay cong thufc quy doi la FeeOy ( M = 448) va teeO, 0,2 - 0,15 m o l -> ncodu = 0,05 mol CO + O X , > — = — mol y n,, n = — — — - 75% ThUc chat p h a n iJng khijf o x i t s^t la 0,1 mol 8,4 => n., = 75 100 > Fe(N03)3 + (3x-2y) NO2 + ( x - y ) H Fe,0y + ( x - y ) H N — m o l 3x-2y Fefiy %Vco B a i 2: (DH = 0,025 mol khoi B - 2009): Cho 61,2 gam hon hop X gom Cu va Fe304 tac dung v d i dung dich H N O loang, dun nong va khuay deu Sau k h i 6-2.7 cac p h a n ufng xay hoan toan, t h u diTcfc 3,36 l i t k h i N O (san p h a m khuf n h a t , d dktc), dung dich Y va l a i 2,4 gam k i m loai Co => mx = 0,025 448 = 11,2 gam can dung dich Y, t h u ducfc m gam muoi k h a n Gia t r i cua m la VI P H l / C l N G P H A P B A G * NOi d u n g : Phuang so mol electron T O phdp A N M O NGUYEN L bdo todn nguyen cho phep chung A 151,5 T C T ti2 vd phuang ta ggp nhieu phUang phdp trinh phdn ling C 137,1 Hiidng lai Idm mot, qui ggn viec tinh todn vd nhdm nhanh dap so B a i 1: (CD B 97,5 bdo todn nung nong diTng gam m o t oxit sat den k h i phan ufng xay r a hoan toan K h i t h u dirac sau p h a n iJng c6 t i k h o i so v d i hidro bang 20.C6ng khii Fe^") 3Fe304 + 28HNO3 9Fe(N03)3 D Fe304; 65% 3Cu + 8HNO3 ^ 3Cu(N03)2 K h i t h u dUdc gom k h i CO2 va CO du c6 M = 40 n^o, 0,2mol 44 28 — mol + 2N0 y 4H2O mol Cu + 2Fe(N03)3 -> Cu(N03)2 + 2Fe(N03)2 3x „ 3x 3x 3x mol 2 K h o i X p h a n ufng l a : 61,2 - 2,4 = 58,8 12 40- nco I4H2O o > xFe + yCOz S"co=|g = + 2y y Fe^Oy + yCO + NO 3x thufc cua oxit s^t va p h a n t r a m the t i c h cua k h i CO2 t r o n g h o n hdp C FeaOg; 65% dan giai giai K i m loai du la Cu va muoi s^t t h u diioc se la muoi s^t ( I I ) (Cu Khoi A - 2007): Cho 4,48 l i t CO (d dktc) tCf tvl d i qua ong su: k h i sau p h a n ufng la A FeO; 75% B FezOg; 75% Hiidng dan D 108,9 "NO = 232x + 64(y + 3x ) = 58,8 3,36 ^ ,^ X 2y = 0,15 = 0,15 m o l ^ 22,4 " (1) (2) Tii (1) (2) => X = 0,15, y = 0,15 M u o i t h u dtrcJc gom Cu(N03)2 va Fe(N03)2 c6 k h o i li/ong 1^: , (d\i), t h u dMc dung dich X Cho dung dich AgNOs (du) vao dung dich X, sau k h i p h a n ufng xay r a hoan toan s i n h r a m gam chat r a n Gia t r i cua m la A 68,2 g B 28,7 g C 10,8g HUdng d&n D 57,4g giai 0,24 + a = 0,18 + 2a o Jai 5: Cho h n hop A gom A l , Z n , M g D e m oxi hoa hoan toan 28,6 gam A bang oxi dtf t h u diigc 44,6 gam h o n hgfp oxit B Hoa t a n h e t B t r o n g dung dich H C l t h u dugc dung dich D Co can dung dich D difoc h6n hop muoi k h a n la A 99,6 gam c6 p h a n iJng A g * oxi hoa Fe^^ Vay chat rSn t h u diTOc gom A g C l va A g Goi so m o l cua FeCl2 la x t h i so mol cua N a C l la 2x B 49,8 gam C 74,7 gam Hii&ng dan (1) MaOn + n H C l np^ci, = 0,1 = 0,2 + 0,2 = 0,4 (2) Ap dung dinh luat bao toan khoi li/gng -> Tn.Q^ = 44,6 - 28,6 = 16 gam mol => no^ = 0,5 n^j Chon ddp an A Hoa t a n hoan toan h o n hcfp gom 0,12mol FeS2 va a mol CU2S vao dung dich H N O (vCfa du), t h u difcJc dung dich X (chi chufa h a i muoi sunfat) va k h i n h a t N O Gia t r i cua a la (cho ^ mol ->• n n c i = 4.0,5 - mol mol m^uoi = mhhki + m^,, = 28,6 + 2.35,5 = 99,6 gam Chon dap an A B a i 6: De khuf hoan toan 3,04 gam hon hop X gom FeO, Fe304, Fe203 Fe = 56 , Cu = 64 , S = 32): B 0,12 C 0,06 Hiidng dan D 0,04 gidi Cac nguyen to Fe, Cu, S se h i H N O oxi ho^ l e n so oxi hoa cao nhat V a y cong thufc cua h a i m u o i sunfat la Fe2(S04)3 va CUSO4 Ta CO so hop thutc: 2FeS2 2MCln + nH20 Theo phuong t r i n h (1) (2) -> n HCl ~ ' ^ • ^ O , mol m An = 0,4.143,5 + 0,1.108 = 68,2g B a i 4: (DH khoi A - 2007): gidi M + -n O2 = 0,1 np^ci, D 100,8 gam Goi M la k i m loai d a i dien cho ba k i m loai t r e n v d i hoa t r i la n 127x + 58,5.2x = 24,4 HAgci = n N a C i + a = 0,06 (mol) | T Chon dap a n C t o d n d day k h o n g chi don t h u a n c6 phan ufng trao doi ma A 0,075 2a (mol) A p dung d i n h luat bao toan nguyen to ta cd: FeCl2 va N a C l (c6 t i le so m o l tirong ufng la : 2) vao m o t lirong nyxdc HAg = a (mol) + T d n g so m o l S sau p h a n ufng: 0,06.3 + 2a = (0,18 + 2a) m o l B a i 3: (DH khoi B - 2009): Hoa t a n hoan toan 24,4 gam h o n hop gom X 2CuS + T o n g so m o l S trudc p h a n ufng: 0,12.2 + a = (0,24 + a) m o l (0,15 + M ^ ) i 8 + 3.0,15.180 = 151,5g Chon dap an A Bki CuzS ^ 0,12 (mol) Fe2(S04)3 0,06 (mol) can 0,05 mol H2 M a t khac hoa tan hoan toan 3,04 gam h6n hop X dung dich H2SO4 dac thu diioc the tich SO2 (san pham khuf nhat) d dieu kien tieu chuan la A 448 ml B 224 ml Hii&ng C 336 ml dan D 112 ml gidi Thuc chat phan iJng khuf cac oxit tren la H2 0,05 +0 -> H2O 0,05 m o l D a t so mol hon hop X gom F e O , Fe304, FezOg Ian lirot la x, y, z T a c6: no = x + 4y + 3z = 0,05 mol => n, , - (1) 3,04-0,05.16 , ' = 0,04 mol 56 => X + p a i 8: Thoi rat cham 2,24 lit (dktc) mot hon hop gom C O v a H2 qua mot ong sur dung hSn hop A I O , CuO, Fe304, Fe203 c6 khoi luTOng l a 24 gam d\i dang duoc dun nong Sau k h i ket thuc phan ufng khoi l i i o n g chat ran lai ong suT l a j 3y + 2z = 0,04 mol „ -> x+y => Z nso = _ = ^ = Khoi AI2O3 nung nong, phan ling hoan toan Sau phan uTng thu diioc m gam chat r S n va mot hon hop va hoi nSng hon khoi liTOng cua hon hop V la 0,32 gam Tinh V va m A 0,224 ht va 14,48 gam B 0,448 lit va 18,46 gam C 0,112 lit va 12,28 gam D 0,448 lit va 16,48 gam Hii&ng dan giai Thiic chat phan ufng khuf cac oxit tren la CO + H2 + O > CO2 > H2O Khoi lUtfng hon hop k h i tao nSng hon hon hgfp k h i ban dau chinh la khoi li/ong cua nguyen tuf oxi cac oxit tham gia phan iJng Do vay: mo = 0,32 gam 32 {n^Q ufng khuT cac oxit l a : C O + O + O > > CO2 H2O lUcJng chat rSn lai ong s i J la mrfn= 24 - 1,6 = 22,4 gam diing 16,8 gam hon hop oxit: CuO, Fe304, => 0,lmol mo = 1,6 gam B a i 7: Thoi tCf tCr V lit hon hop (dktc) gom C O va H2 di qua mot ong n,co D 16,8 gam no = npo + n„^ = O,lmol = 0,01 mol Chon dap an B ^ Thiic chat phan Vay: , Vay: VJ-Q^ = 224 ml = "o = + = 0,02 mol moxit = nichairiin + 0,32 16,8 ^ = m + 0,32 V^h(co.H,) Chon dap an A B a i 9: Cho m gam mot ancol (rugu) n o , dOn chuTc X qua binh dimg CuO (du), nung nong Sau k h i phan ufng hoan toan, khoi Itfcfng chat ran binh giam 0,32 gam H5n hop hoi thu diroc c6 ti khoi doi vdi hidro la 15,5 G i a tri cua m l a A 0,92 gam B 0,32 gam C 0,64 gam CnH2„^lCn20H + CuO ^ C n H „ l C H + C u i + H2O Khoi lirong chat r S n binh giam chinh la s o gam nguyen tijf O CuO phan iJng Do nhan duoc: 32 mo = 0,32 gam -> n,, = - — = 0,02 mol 16 C„H,„,,CHO : 0,02 mol H6n hop hoi gom: H,p : 0,02 mol Co M = 31 => => m = 16,48 gam = 0,02.22,4 = 0,448 lit D 0,46 gam HUdng dan giai Vay h6n hop hoi c6 tdng so mol la 0,04 mol n „ J = 0,02 mol Ap dung dinh luat bao toan khoi luong ta c6: Chon dap an D nhh(co.H,) H2 y/2 0,02 = —— > 3Fe2(S04)3 + SO2 + IOH2O C 20,8 gam Hiic/ng dan giai ^x/2 2Fe304 + IOH2SO4 y B 11,2 gam (2) Nhan hai ve cua (2) vdri roi trCr (1) ta c6: x + y = 0,02 mol Mat khac: 2FeO + 4H2SO4 > Fe2(S04)3 + SO2 + 4H2O x A 22,4 gam mhhhoi = 31.0,04 = 1,24 gam niancol + 0,32 = m h h hai mancol = 1,24 - 0,32 Chon dap a n A = 0,92 gam B a i 10: D o t chay h o a n t o a n 4,04 g a m m o t h o n h o p h o t k i m loai g o m A l , W Hii&ng F e , C u t r o n g k h o n g k h i t h u diTcfc 5,96 g a m h o n h o p o x i t H o a t a n FeO h e t h o n h o p oxit b k n g dung dich H C l M T i n h t h e t i c h dung dich FeCI2 IICl FegO/ H C l can dung A 0,5 l i t B 0,7 l i t C , l i t Hiiofng dan D lit dan gidi Fe(OH)2 (UiNn^ t^rongkk O 2^3 Fe(OH), FeCl, V i t r o n g h n hop ban dau da c6 m o t lirgng Fe203 va coi Fe304 = FeO Fe203) n e n c h i x e t sir chuyen hoa qua t r i n h : F e O gidi Fe203 Do t a n g k h o i lirgng cua Fe203 so v d i h o n h g p dau l a k h o i lirgng o x i = m o x t - m k i = 5,96 mo no = 1,92 16 - 4,04 = 1,92 gam ma F e O l a y de tao Fe203 = 0,12 m o l no 1,2 can — H o a t a n he't h o n h o p b a o x i t b a n g d u n g d i c h H C l t a o t h ^ n h H2O n h u sau: H " + O^' 0,24 ^ Goi h h H2O no = 0,1 y = 0,05 C h o n d a p a n C B a i 1 : D o t c h a y h o a n t o a n , m o l m o t a x i t c a c b o n x y l i c dan chuTc c a - A 8,96 l i t B 11,2 l i t C 6,72 l i t Hit&ng dan D 4,48 l i t gidi 0,3 (mol) x + (1) 4y = (2) 0,3 mp,o = , % C h o n d a p a n C B a i : C h o m o t l u o n g k h i C O d i q u a o n g chufa ( h o n h g p A ) g o m , dirge , g a m c h a t r ^ n B g o m b o n c h a t H o a t a n c h a t r ^ n B b a n g d u n g d i c h H C l diT t h a y t h o a t r a , l i t H2 (6 d k t c ) T i n h so m o l F e t r o n g h o n h o p B B i e t r a n g t r o n g B so m o l F e b a n g A x i t c a c b o x y l i c d a n chufc c6 n g u y e n tuf o x i n e n c6 t h e d S t l a RO2 V a y : nQnjo^, + n g ^ p = noico^) + " o d i ^ o ) A 0,006 => n o (p.u) = 0,6 m o l B 0,008 C , Hii&ng n^^ = 0,3 m o l Hon hgp A = 6,72 l i t C h o n d a p a n C -^tong o so m o l F e O v a Fe203 0,1.2 + no,p„, = 0,3.2 + , FeO : 0,01 m o l Fe.Og : 0,03 m o l dan D 0,012 gidi + C O -> , g a m B ( F e , Fe203, F e O , F e ) tiTOng l i n g v d i so m o l l a : a, b , c, d ( m o l ) B a i : H o a t a n a g a m h o n h o p g o m F e O v a Fe304 h e t 0 m l d u n g d i c h H C l M dLTOc d u n g d i c h X C h o X t a c d u n g v d i m o t l u o n g d u n g d i c h NH3 d i i dirge k e t t u a N u n g k e t t u a t r o n g k h o n g k h i d e n k h o i liTOng k h o n g d o i diTOc a + , g c h a t r a n P h a n t r a m k h o i i J O n g c u a F e O hon hgptren la: A 28,47o oxit = x + y = 0,15 m o l F e O v a 0,03 m o l Fe203 d o t n o n g S a u k h i ke't t h u c t h i n g h i e m t h u cua V l a => npeo = 0,075 ( m o l ) B 24,6% C , % D 40,2% H o a t a n B b a n g d u n g d i c h H C l d u t h u dirge n H = ^ f f ^ ^ = , m o l 22,4 Fe + H C ^ FeCl2 + H2 a = 0,028 3Fe2(S04)3 + SO2 + IOH2O C 20 ,8 gam Hiic/ng dan giai ^x /2 2Fe304 + IOH2SO4 y B 11 ,2 gam (2) ... cua (2) vdri roi trCr (1) ta c6: x + y = 0, 02 mol Mat khac: 2FeO + 4H2SO4 > Fe2(S04)3 + SO2 + 4H2O x A 22 ,4 gam mhhhoi = 31.0,04 = 1 ,24 gam niancol + 0, 32 = m h h hai mancol = 1 ,24 - 0, 32 Chon... bang 9 ,25 Nong mol/lit HNO3 dung dich dau la A 0 ,28 M sinphdm ^ Chon dap an A Chon dap an A + 16y nhan => V N O = 0,1 .22 ,4 = 2, 24 l i t = 0,7.3 + (0, 025 .3.5,6) = 2, 52 gam 56x + 2e 0,3