Lời giải đề thi học sinh giỏi Hóa học lớp 9 phần 1

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DE SO DE THI HOC S1I\IH GiOl HOA HOC LOfP 9, TP HQ CHI MINH NAM HOC 1998 Cdu I Cdu I I Viet phuang Cdu Viet phuang trlnli khdc trinh phdn > Fe(0H)2 FeCh > Fe(0H)3 mang (NHJ.SO,; dugc NaoCOs; de dicu die muoi ling de bieu dlen FeCls III Co ong nghiem NaOH; chiiSi ZnCl2 bien ^ FeSO, PbiNO^)^; so ndo dUng chat ndo? Viet phdn CaCl icng minh Fe(N0s)2 > Fe so tii den chica Ba(N03)2; hoa sau: > > FesOs ddnh - 1999 cdc dung dicli: Hay cho bict 6'ng hga Biet rdng: a) Dung dich (2) cho ket tila trdng vai cdc dung dich (1), (3), (4) ''• b) Dung dich (5) cho ket tua trdng vai cdc dung dich (1), (3), (4) c) Dung dich (2) khong tgo ket tila vai dung d) Dung dich (1) khong tgo ket tua vai cdc dung dich e) Dung dich (G) khong phdn (5) f) Dung dich (5) bi trung i(ng vai dung hoa bdi dung dich (5) dich (3), (4) dich HCl g) Dung dich (3) tgo ket tua trdng vai HCl, dun nong ket tua se tan » CduIV a) Nong Cau • dung Tinh dich bdo hoa KCl d 40°C la tan cua dung ) Xdc dinh AgNOs lugng dich KCl a ciing AgNOs tdch 28,57% nhiet Idm Ignh bdo hoa a 60°C xuong 10°C Cho biet 2500 gam dung dich tan cua AgNOs o 60°C Id 525 gam, a 10°C Id 170 gam V (A) la dung Trgn Lay them dich H2SO4; (B) la dung 0,3 lit (B) vai 0,2 lit (A) dugc 20 nd (C), them tic tit dung dich dich NaOH 0,5 lit (C) mot it quy tim vdo thdy HCl 0,05M c6 mdu xanli tai quy tim ddi thdnh Sau mdu tim thdy het 40 ??il axit Trgn 0,2 lit (B) vai 0,3 lit (A) dugc Lay 20 ml dung Sau them dich (D), them tii tii: dung dich Tim dich mol/l dung I r(i n i A i n c T U I u n r C I M M n i f i i H H A H f i n Q mot it quy tim vdo thdy c6 mdu dich NaOH tim thdy het 80 ml dung nong 0,5 lit (D) 0,1M tai quy ddi thdnh NaOH (A) vd (B) mdu Cdu VI Xdc dinh cong thiJtc ciia hai oxit sdt A vd B, biet • 23,2 gam (A) tan vica dil • 32 gam (B) k/iii bdng Ho tqo thdnh r&iig: ' ^ ) ^ 60"C, t r o n g 525 + 100 b a n g 625 g a m d u n g d i c h c6 525 g a m AgNOg v a 100 g a m H2O T r o n g 2500 g a m d u n g d i c h c6 x g a m AgNOa v a 0,8 lit HCl IM sdt vd 10,8 gam IhO y g a m nxidc: L6\I Cdu I Cdu 11 * Zn + CI2 ZnCla > Z n O + 2HC1 • Z n S + BaCl2 X Fe + 2IIC1 + BaS04i > FeCl2 + H t FeCl2 + N a O H > Fe(0H)2^ + N a C l (trdng Fe(0H)2 + H2SO4 2Fe + 3CI2 > B a S ^ + Fe(N03)2 FeCla + 3NaOH > FeCOIDai + 3NaCl (ndu do) • = 2100 (gam) 400 g a m ni/dfc CJ 10°C, CLf 100 g a m ntfcJc h a t a n 170 g a m AgNOa 400 g a m niJcJc h o a t a n z g a m AgNOa z = Do khoi 400x170 100 liTgng AgNOa = 680 (gam) AgNOa ket t i n h k h i lam lanh: 2100 - 680 = 1420 (gam) Cdu V Phan ufng: — > Fe203 + 3H2O Fe203 + C 625 & 60°C t r o n g 2500 g a m d u n g d i c h c6 0 g a m A g N O s v a xanh) > 2FeCl3 2Fe(OH)3 Vay > FeS04 + H O FeS04 + Ba(N03)2 2500 X 525 y = 2500 - 2100 = 0 g (nxidc) > ZnCl2 + H2O > ZnCh = H2SO4 + N a O H > Na2S04 + H O L a n thuf n h a t q u y t i m c6 m a u x a n h chufng t o diT N a O H T h e m a x i t — > F e + 3CO2T Cdu I I I Theo cac duf k i e n de b a i neu r a , cac lo d i i n g cac h o a c h a t sau: H C l de t r u n g h o a N a O H dif: HCl + NaOH f ) => L o so d i i n g : Na2C03 > N a C l + H2O (2) g) L o so d i i n g : Pb(N03)2 L a n l a m quy t i m c6 m a u chufng t o H2SO4 dir d) L o so chufa: Ba(N03)2, lo chuTa CaCl2 T h e m N a O H de t r u n g h a H2SO4 d i / N a O H + H2SO4 a ) => L o chijfa: (NH4)2S04 (NH4)2S04 + Ba(N03)2 > B a S i + 2NH4NO3 (NH4)2S04 + Pb(N03)2 > P b S i + 2NH4NO3 (NH4)2S04 + CaCl2 (3) Theo cac p h a n ufng (1), (2), (3) t a c6 phiiofng t r i n h : 0,3y - > C a S i + 2NH4CI ' 0,3x - Cac p h a n i l n g c o n l a i hoc s i n h tii vie't Cdu rv X 0,2x = 0,05 0,2y 0,1 x 0,08 X 0,04 x 500 20 500 20 = 0,05 = 0,1 (a) (b) G i a i h e p h u a n g t r i n h (a), (b) => x = , M v a y = 1,1M a ) G o i S l a t a n cua K C l d 40°C Cdu VI K h o i lUcfng d u n g d i c h t h u dUgfc: (S + 100) g a m N o n g p h a n t r a m cua c h a t t a n t r o n g d u n g d i c h b a o h o a : S + 100 S + 100 - 3,5S » > Na2S04 + 2H2O G o i X , y l a n l u g t l a n o n g cua H2SO4 v a N a O H e) =^ L o chijfa: N a O H P h a n irng: (1) S + 100 G o i c o n g thufc o x i t s a t l a FexOy Fe^Oy + y H C l (mol) 0,4 0,8 S = 40 (gam) I fi\l n £ T u i u n r s l u u Rini Hni Hnr a ^^IfilAlflg THI HOC SINH GIOI GIO HOA HOC > x FeClayx + y H (1) Ta c6: I I H C I = 0,8 X = 0,8 (mol) PHAN B Bai toan Theo de: Bai 23,2 = — (56x + 16y) y 56x = 42y X _ fx = y [y = FexOy + yH2 vd K2SO4 dugc trgn Idn theo tl Ic Hoa tan hon hgp vdo 102 gam nUac thi thu dugc dung dich A Cho 1664 gam dung dich BaCL 10% vdo dung dich A Lgc ket tua, them H2SO4 du vdo tiudc vUa Igc tin thdy tgo 46,6 gam ket tua Xdc dinh nong phdn tram cua NosSOj > xFe + y H 0,6 (mol) Co mot hon hgp gom Na2S04 I : ve so mol (1 mol Na2S04 yd.2 mol K2SO4) Vay oxit sSt (A): Fe304 (sdt tii oxit) • 1- (2) vd K2SO4 dung dich ddu • Bai 0,6 Tinh CM cua dung dich H2SO4 vd NaOH, biet rdng 10 nd dung dich H2SO4 tdc dung vUa du vdi 30 ml dung dich NaOH Neu lay 20 ud 10,8 Ta c6: 18 dung dich H2SO4 = 0,6 (mol) lugng du ndy tdc dung vita du vdi 10 nd NaOH ' TheodI: ^ ( x + 16y] = « y ^ x = LdlGIAI - = -=>• y [y = PHAN => Cong thufc oxit (B): FeaOs- Cdu DE SO NaOH + N H C I N a O H + FeCl2 PHAN A Li thuyet Chi dugc diing thuoc thii de nhgn biet cdc muoi sau: NH4CI, MgCl2, NaCl, AICI3 (Gidi thich vd viet phuang trinh phdn ling 2Fe(OH)2 + - O2 + H2O > Fe2(S04)3 b) MgCOs > Fe(0H)3 > MgS04 > MgCOs > FcsOa > FeCls > MgCl2 > CO2 > Mg(0H)2 > Ca(HC03)2 III Chi dugc dung guy tini vd dung dich AgN03 IV Sdt nguyen chdt khong thi khong CO tap chdt de Idu ngdy khong hien tugng a N a O H + FeCls >Fe >FeCl2 phdn biet cdc dung dich: NaOH, NaCl, HCl, H2S, Cdu > 2Fe(OH)3 Mau CO ket tija nau la FeCls: trinh phdn ling bieu dien cdc bien hoa sau: > FeClz Cdu , II Viet phuang a) Fe > Fe(0H)2i + 2NaCl trang xanh FeClo, neu c6) Cdu — > NaCl + N H g t + H^O Mau tao ket tiia trSng xanh, hoa nau khong k h i la FeC]2: TP HO CHJ MINH NAM HOC 1998 - 1S99 FeCls, Cho dung dich NaOH dii Ian liicft vao cac mau thif t r e n va dun nhe; Mau thuf nao c6 k h i m i i i khai bay la N H C I : DE THI CHOIV flOl TUYEN KQC SINH GIOIHOA HOC 9, qUAAl TAN BINH I A Li thuyet I - T r i c h m6i lo mot i t lam mt\ thuf - Cdu cho tdc dung vdi 2,5 gam CaCOs thi axit du vd ^ > Fe(0H)3i + 3NaCl nau > AgCl Mau tao ket tua keo trSng sau d6 tan NaOH dif la AICI3: > MgO N a O H + AICI3 > CaCOs > A l ( H ) t + 3NaCl keo trdng c6 sdn, neu each A ( H ) + NaOH H2SO4 > NaAlOa + 2H2O Mau tao ket tua trSng la MgCl2: bi han gl, nhUng sdl Igi bi han gl Hay gidi ' thich N a O H + MgCl2 ^ Mg(0H)2 + 2NaCl Mau khong c6 hien tJong gi la NaCl Lfll GIAI Oi THI HOC SINH GIOI HOA HOC Lfll GIAI T H I unr emu ni/\i un» unr a ^ Cdu Cciu II a) 2Fe + 6H2SO4 damdac 2Fe(OH)3 - K h i t r o n g s a t b i I a n t a p c h a t , de l a u n g a y b i b a n g i x a y r a s u a n m o n k i m l o a i tufc l a bie'n sat t h a n h h o p c h a t cua sat — > FeaOs + 3H2O Gidi — > 2Fe + SCOat l u o n g o x i n e n c h u y e n Fe -> Fe^\ F e + 2HC1 > FeCla + H a t V a o x i hoa t a n t r o n g niidc theo qua t r i n h : O2 + 2H2O 2FeCl2 + CI2 > 2FeCl3 Sau Fe^' k e t h o p vdri O H " 2FeCl3 + Fe > SFeC^ M o t p h a n F e ( H ) b i o x i hoa t a o F e ( H ) (nau MgCOg + H2SO4 loang Mg(0H)2 > MgCl2 + B a S i > MgCOIDai + N a C l —> M g O + CO2 — MgCOa + CO^^t + HgO > MgSOi ^ PHAN Bai B Bdi rr - T a co: > C a C O a i + NaaCOg + 2H2O hoac Ca(HC03)2 + C a ( H ) 2 C a C i + 2H2O HI M a u k h o n g c6 h i e n t i i o n g l a N a C l • Nhufng m a u l ^ m quy t i m h d a do: I I C l , H2S, H2SO4 Sau cho dung dich AgNOa Ian li/gt vao cac m a u l a m quy t i m hoa M a u nao tao k e t t u a t r S n g la H C l : > A g C l i + HNO3 • 10 BaS04i + 2NaCl X (mol) 2x -> (1) X > B a S i + 2KC1 2x ^ (2) 2x V i k h i t h e m d u n g d i c h H2SO4 vao n L f d c loc l a i t a o k e t t u a n e n t r o n g niidc loc dtf BaCla (mol) > B a S i + 2HC1 0,2 «o F e ( H ) (trdng F e ( H ) + O2 + 2H2O > Fe(N03)2 + A g C U MgCl2 + N a O H - T r e n be m a t k i m l o a i c6 I d p nUdrc a m , d a h o a t a n m o t thidi: FeaOg + SCO M g S + BaCla Cau b e n t r o n g k h o n g k h i a n h i e t t h u n g > F e ( O I I ) i + 3Na2S04 FeCla + A g N b) 1^02(804)3 + 3SO2T + 6II2O > Fe2(S04)3 + N a O H IV - s a t n g u y e n c h a t k h o n g b i b a n g i v i n6 dugc bao ve bori I d p Fe203 ^ ^ ^ ° " > kgiSi ^ ^ ^ 74 + 2HNO3 M d u k h o n g c6 h i e n t u o n g g i l a H2SO4 i f i l R I I f i F TH] H f i r SlUH f;inr u n A u n r n = X 142 Ml GIAI DE THI HOC SINH GIOI HOA HOC ^ ^ 200 X + 0,4 X x 174 + 102 = 200 (gam) 100 = 14,2% 04 11 Bai T a c6: 2.5 = 0,025 (mol) 100 nC a C O , Tafco them P h i i n ufng: H2SO4 + CaCOa (mol) 0,025 (mol) a 2NaOH -> Ong 3: dung dich bac Na2S04 + 2H2O 20 Mot hon hap NaCl vd MgCl2 them nuac vao hon hap ta c6 dung dich A = 0,0125 (mol) Them dung dich AgNOs vao dung dich A, phdn iing ket thuc 30 m l d u n g d i c h N a O H chufa y m o l N a O H 2a 30 X 10 Chia B lam phdn: a vd b Phdn a: Sau c6 can tiep tuc dun nong thl diigc mot hon Jigp = 6a (mol) C Cho hon hap qua binh dUng KOH H2SO4 + N a O H (mol) Na2S04 + 2H2O Phdn b: Cho vdo lugng dU dung dich HCl thi thu dugc ket tua trdng D 0,0125 - > 0,025 Viet cdc phuang V i p h a n ufng t r u n g hoa n e n 6a = ^' "2SO4 " 0,01 logi bo chat ket tua trdng, phdn dung dich Igi la dung dich D T r o n g 10 m l d u n g d i c h N a O H chijfa 2a m o l N a O H y = Em hay cho biet dng ndo chat ket tua Gidi thich vd viet phiXang trinh phdn icng 10 m l d u n g d i c h H2SO4 chtfa x m o l H2SO4 = nitrat Sau cho them axit nitric vao ca ba ong nghiem T r o n g 20 m l d u n g d i c h H2SO4 chufa 0,025 m o l H0SO4 X vao: Ong 2: dung dicli natri cacbonat 2a 10 X 0,025 ngUdi ta cho diig l:~dung dich kali cacbonat -> CaS04 + C O a t + H2O < - 0,025 H2SO4 + 6'ng nghi&m deu dung dung dich bari clorua, (mol) trinh phdn dng Lam the ndo de phdn biet cdc Ig hoa chat diidi day md khong dung theni hoa chat ndo khdc: MgCl2, H2SO4, NaCl, CuSO^, = 1,25M dugc NaOH Mot hon hap gom: CuO, FeO, AI2O3 Lam cdch ndo de tdch chiing 0,025 ^6 2,5 0,03 khoi M Cdu II Bai todn Cho 26,Ig MnOo tdc dung vdi dung dich HCl c6 20 gam HCl Cho hct OE SO qua mot lit dung dich NaOH lodng dU DE THI HOC SINH GiOl HOA HOC QUAN TP HQ CHJ MINH NAM HOC 1998 - 1999 a) Lugng HCl c6 du de phdn dng het vdi Mn02 Cdu Li thuyet b) Tinh nong mol 11 cua muoi thu dUgc phdn I'ing giOa vd NaOH Tit Ig hoa chat sau, em c6 the dieu chc nhUng chat nuo? Axit sunfuric; natri hidroxit; amoni sunfit; sdt sunfua vd kim logi keni Ta H2SO4 CO may cdch de dieu che nitrat; canxi cacbonat; c) Nung qugng pyrit sdt 'de tgo SO2 Cho SO2 sue vdo dung natri chda muoi khoi lugng pyrit dung dich dich tren Sau them vdo mot lugng du Ba(N03)2 CaSOJ Viet cong thUc vd ten ggi muoi diing nong nghiep (phdn dam, phdn Idn vd plidn kali) Hay gidi thich tgi ngUdi ta khong trgn tro bep vdi phdn dam de ban rugng? 19 khong? can dung Tim khoi lugng ket tua va Biet rdng lugng SO2 tdc dung vUa du muoi (Cho Mr, = 55; O = 16; H = 1; CI = 35,5; S = 32; Fe = 56; Ba = 137) ^^F: , 13 L d l GIAI Cdu I Li dng thuyet 1000°C Zn + -> CaO + C O a t FeS + H2SO4 —> — NH4NO3 > Na2S04 + S O s t + H2O MgCl2 + 2AgN03 Phdn 2AgN03 —> > CaS04 + 2H2O Phdn M g O + N t + O-^t —> b: K h i cho H C l d u vao d u n g d i c h B t h u diJOc k e t t u a D i e u H C l + AgNOs > A g C l i + HNO3 (D) > CaS04 + C O a t + H2O Chu y: Con nliieu cdch klidc, xin nJiuang ban doc! C o n g thufc ba m u o i d u n g t r o n g n o n g n g h i e p : 6, T a n h a n t h a y : t r o n g t a t ca cac d u n g d i c h t r e n t h i c h i c6 m o t d u n g d i c h CO m a u x a n h l a : CUSO4, cac d u n g d i c h l a i l a k h o n g m a u • CO(NH2)2 ( % n i t a ) : U r e ( p h a n d a m ) T r i c h m o i lo m o t i t l a m m a u thijf • Ca(H2P04)2: Supe p h o t p h a t k e p ( p h a n I a n ) - • KRO'i [Ca(H2P04)2 • m a n g t i n h k i e m n e n n g i / d i t a k h o n g t r o n p h a n d a m vcfi t r o bep Ong 2: BaCl2 + K2CO3 M a u thCf cho ke't t i i a m a u x a n h la N a O H : CUSO4 + N a O H kali nitrat (phan kali) hoac K C l : k a h clorua Ong 1: Cho d u n g d i c h CUSO4 I a n lifot vao cac m a u thuf t r e n : • 2CaS0, P h a n d a m de b i p h a n h i i y t r o n g m o i t r t f d n g k i e m , v i t r o bep - > B a ( N ) + C t + H2O BaCIa + Na^CO^ > BaCOgi + 2NaCl H N O + BaCOa -> Ba(N03)2 + C02t + H2O > C u ( H ) i + Na2S04 Cac m l u l a i k h o n g c6 h i e n tu'ong D u n g N a O H l a m thuoc thiif, cho I a n liiot vao cac m a u ihxi l a i : • Mau t h L f CO k e t tua mau t r S n g la MgClg: N a O H + MgCla > B a C O g i + 2KC1 H N O + BaCOg K N O + K N O + H2O n a y cho t a t h a y r a n g t r o n g d u n g d i c h A g N O s dif Cdch 4: Tac d u n g vcri muo'i cija c a n x i : hoac supe pho't p h a t dcfn -02t — > 2Ag + N t + t 2NO2 + K H CaS04 + II2O Cdch 3: Tac d u n g v6i bazcJ C a ( H ) : H2SO4 + CaC03 NaNOa + > CaS04 + H a t H2SO4 + CaO H2SO4 + Ca(0H)2 —> 2Mg(N03)2 D i e u che CaS04 t i f H2SO4: Cdch 2: Tac d u n g vdri CaO: Mg(N03)2 a: K h i d u n n o n g d u n g d i c h B : NaNOs ) N t - H ^ t + 2H20 Cdch 1: Tac d u n g vdri c a n x i : H2SO4 + Ca > A g C U + Mg(N03)2 v a A g N O s dif > N a O t + 2H2O >^°°°^ > A g C U + NaNOg Sau k h i Ipc bo k e t t u a t h i d u n g d i c h (B) g o m : N a N O s , > N a N O s + N H a t + H2O Na2S03 + H2SO4 hoac P h a n ufng: N a C l + A g N ZnS04 + S t + 2H2O -> FeS04 + H S t N a O H + NH4NO3 NH4NO3 mac dti l a a x i t H N O v i t h e t r o n g o n g c h a t k e t t u a > ZnS04 + H t 2H2S04dac > Ba(N03)2 + A g C l i Trong ong nghiem t h i ta thay ket tija AgCl k h o n g t a n axit Zn + H2SO4 loang hoac BaCla + A g N A g C l + HNO3 T a d i e u che duac cac c h a t k h i sau: CaCOa 3: • M a u t h i i tao d u n g d i c h t r o n g suot va toa n h i e t m a n h l a H2SO4 N a O H + H2SO4 • > Mg(0H)2 + 2NaCl > Na2S04 + 2H2O M a u k h o n g c6 h i e n t u g n g l a N a C l b ) T h e tich dung dich thu dUcfc chinh 1^ the tich cua N a O H tufc l a : T a c h C u O , F e O , AI2O3 r a khoi nhau: CuO AI2O3 +C0, ^NaAlOg N a O H da CuO_ FeO -> A l ( O H ) , + HC1 du FeO^ Fe(OH), cliaii kliong So' mol cua N a C l p h a n ufng (2): 0,1375 (mol) + N H O H dil FeCl,, +HC1 [Cu(NH3)J(OH)2 CuCl^ Vduiigdich = (lit) ->A1,03 ->CuCL +NaOH du C^ -^Cu(OH), -»CuO AI2O3 + N a O H (mol) > A l ( H ) i + NaHCOg CuO + 2HC1 — > CUCI2 + H2O F e O + 2HC1 -> FeCl2 + H2O > Na2S04 + N a C l + H2O (2) - > 0,1375 Ba(N03)2 + Na2S04 Txi (3) ^ ^ ne^so, = ^i^.,so, > [Cu(NH3)4](OH)2 (3) + 2NaN03 > Ba^O^i 0,1375 (mol) (i) 0,1375 (mol) 0,1375 < - 0,1375 > Cu(0H)2i + N H C I Cu(0H)2 + N H ^ SO2 + N a C l O + N a O H -> AI2O3 + 3H2O CuClz + 2NH4OH — > 2Fe203 + 8S02t 4FeS2 + I I O > 2NaA102 + H2O - = 0,1375M -^FeO N a A l O z + CO2 + 2H2O 2A1(0H)3 = c) P h a n ufng: P h a n ufng: • =C„ 0,1375 = 0.1375 (mol) mg^so^ = 0,1375 x 233 = 32,04 (gam) (xanh t i m , tan) [Cu(NH3)4](OH)2 + 2HC1 CUCI2 + N a O H Cu(0H)2 Cdu TO (1), (2) + N H T + H2O > CuCOIDal + N a C l np^3 ^so, =^ _ 0,1375 = (mol) " (gam) -> F e ( H ) i + N a C l chankhong DE SO FeO + H2O I I Bai todn T a c6: :^ 0,1375 => mpes, = - ^ - r — X 120 = 8,25 > CuO + H2O FeCl2 + N a O H Fe(0H)2 > CuCh OE THI HOC SiNH GIOI HOA HOC , CAP TP HO CHI M I N H N A M HOC 1999 - ZOOO n MnO., 26,1 87 = 0,3 (mol): n^j^j = M n O a + 4HC1 CI2 + N a O H 36,5 0,55 (mol) Cdu (1) 0,1375 -> N a C l + N a C l O + H2O (mol) 0,1375 I Khi cho kirn loai vdo dung Cdu I I Viet cdc phuang (2) Fe 0,1375 trinh > FesO^ TCr p h a n ufng (1) t a c6 t i l e : = 0,3> Cdu n HCl 0,1375 V a y l i r a n g H C l n a y k h o n g d i i de p h a n ufng h e t liicfng M n da cho h a y I I C l p h a n ijfng h e t v a M n d i i phdn mud'i c6 the xdy nhitng phan I I I Mot nhd hoa hoc dieu dang ben ngoai vCng theo chuSi bien hoa sau: FeCh > Fe(0H)3 > FezOs FeCl2 > Fe(0H)2 > FeO M n C l a + Cl2t + 2H2O 0,55 (mol) 20 che dugc mdu (mdu sdc) va da tim dugc phuang chong dng lay cdc mdu NaOH, ket qua dugc ghi tron^^M^:^,^^"^ L f l l R l A i n £ T u i u n n S I U H R i n i HflA H t l R kim log.i cho tdc dung — kirn loai gidng phdp phdn ve biet vdi^axitjm_d^g nhanh dich 6SNH THUAN • _ J O 17 Thuoc Axit HCl Axit HNO3 Dung dich thii Kim loai I - Kim loai II Kim loai III K i m loai I : A g (Bac) + + K i m loai I I : A l (Nhom) + - + K i m loai I I I : Zn (Kem) - + + NaOH Trong ddu (+) de chi trudng hap kim loai hoa tan, ddu (-) chi Cdu I I I T h e o de b a i t h i : • triiang ciiu, viet phuang HCl, HNO3 ddc, AgNOa, KCl, Vii't cdc phuang trinh phdn ling vd gidi • - dung dich A Cho 1664 gam dung dich BaClz 10% vao dung dich A, xudt hien ket tua Loc bo ket tua, them H2SO4 dU vdo nilac Igc thi tlidy tgo ket tua Xdc dinh nong phdn dung dich A ban l a HNO3 Cu + 4HNO3 dac - +) M a u CO k e t t u a m a u x a n h l a K O H Cu(N03)2 + K H - N e u k i m l o a i cho vao k h a c k i m l o a i t r o n g m u o i t h i x a y r a p h a n Fe + CUSO4 nhiet la H C l N e u k i m loai t r u n g v i k i m loai t r o n g muoi t h i xay r a p h a n tfng tiT oxi hoa > SFeClg Cdu I I P h a n ufng: F e + 202 ^° Fe304 + 8HC1 > Fe304 > K C l + H2O KOH + HCl M a u l a i l a K C l Cdu V G o i k i m l o a i hoa t r i I I l a A va c6 a m o l => o x i t l a : A O A O + H2SO4 (mol) Theode: a ^ 2Fe(OH)3 a « , , „ , ^ = ^ " ^ d d Hi,S04 a ( A + 96)100 C%ddASO, > FeCOIDgi + N a C l —> FeCla + N a O H FezOs + SHsO > Fe(0H)2i + 2NaCl TT-^rt chan khong > FeO + H2O > ASO4 + H2O a > FeCla + 2FeCl3 + 4H2O FeCl3 + N a O H • Cho d u n g d i c h K O H vao m a u l a i , m a u nao c6 p h a n uTng > FeS04 + Cu Fe + 2FeCl3 > 2KNO3 + C u ( H ) i +) Cac m a u l a i k h o n g c6 h i e n t u g n g K h i cho k i m l o a i vao d u n g d i c h m u o i c6 t h e x a y r a : Fe(0H)2 > Cu(N03)2 + N t + 2II2O Sau cho d u n g d i c h viia t h u di/oc a t r e n I a n liicft vao cac m a u l a i ddu t h e - o x i h o a khijf - > Cu(N03)2 + A g +) M a u thuf nao vifa t a o d u n g d i c h m a u x a n h va c6 k h i n a u bay r u tram cua A^a^SOj vd K2SO4 Ld\I - trUcfng Cho b o t k i m l o a i Cu d t i I a n l u a t vao cac m a u t h t f t r e n : Cu + A g N (Cho Na = 23; S = 32; K = 39; Ba = 137; CI = 35,5) Cdu I A l k h o n g tac d u n g v d i HNO3 v i A l b i t h u d o n g t r o n g m o i +) M a u t h i i nao d u n g d i c h id k h o n g m a u chuyen sang x a n h la AgNOs vd K2SO4 dilgc trgn Idn iheo ti le I : ve so mol Hoa tan hon hap vdo 102 gam nUac thi thu dilgc 46,6 gam v d i NaOPI v i k h o n g p h a i l a k i m ufng Cdu IV T r i c h m o i lo m o t I t l a m m a u thijf KOH trinh phdn dug xdy qua trinh nhgn biet VI Co mot hon hap gom Na2S04 hoat HNO3 dac n g u o i day: Cdu V Hoa tan oxit cua kim logi hoa tri II mot lilgng vita dil dung dich H2SO4 20% thi thu dicgc dung dich mudi c6 nSng 22,69c Xdc dinh kim logi Cdu v d i H C l v i A g dufng sau h i d r o t r o n g day l o a i iLfdng t i n h thich vi kim loai khong tdc dung vai cdc chat dd cho Cdu TV Chi diing kim loai, hay nhgn biet cdc dung dich sau uTng dong k i m loai A g k h o n g phan hap kim loai khong tdc dung vai dung dich kicm hay axit Hay xdc dinh kim loai nghien Ag khong phan = — ™ d d AS04 Ma: => a(A + ) x l 0 a x 98 x 100 • = a(A + 16) + 22,6 20 A = 24: M a g i e (Mg) ufng toa VI T a c6: Cdu „, Cdu = 10 x 1664 ^ ^ 100 X 208 ^^^'2 '^NajSO,, Gidi Cdu > BaS04i + 2NaCl X < - X - > 2x 2x + 2KC1 (2) 2x K l i i t h e m dung dich H2SO4 vao lo niidfc loc thi tao ket tua nCfa nen BaCl2 + H2SO4 > BaS04i + Ca{OH), lAl 'Cdu ->CaCl, III P h a n ufng: (Bl N a + 1120 CO,, + NH011 +CO.,*U.,0 -^Na^CO, N a O H + Al2(S04)3 ->NaHCO., CaC03 iooo"c CaO + CO2T (A) CaO + I I O (X) Ca(0H)2 — CO2 + N a O H > Na2C03 + > 2NaHC03 (Z) — -> C a C O y i + N a N + CO.T + II2O Cho k i r n l o a i B a vao cac m a u ihii t r e n , dau t i e a c6 p h a n ufng: B a ( H ) + H2T M a u nao cho k e t t i i a t r f i n g x a n h la FeCl2: Ba(01i)2 + FeCL — -> CuO + H2O ^ — > Cu + H2O AI2O3 + H C - > 2AICI3 + 3H2O CuO + 2HC1 - -> CuCla + H2O CuO + H2SO4 (mol) 0,25 -> — > CUSO4 + H2O 0,25 0,25 K h o i iLfoag d u n g d i c h H2SO4 % : II T r l c h m o i d u n g d i c h m o t i t l a m m a u thuf B a + 2H2O — H2O (Y) N a I i C + Ca(N03)2 Cu(0H)2 - > AI2O3 + 3H2O V i E t a n p h a n n e n E chufa: CuO, Cu, AI2O3 > CaCO^i + 2NaCl NaaCOs + CO2 + I I O - CuO + H2 - > CaCl2 + 2H2O CaCl2 + NaaCOa 2A1(0H)3 -> C u ( H ) i + Na2S04 AI2O3 + H2 (B) C a ( H ) + 2HC1 > A l ( O H ) i + 3Na2S04 N a O H + CUSO4 — P h a n ijfng: 22 -> N a O H + ^ H a t CaCO;, CaCOg Cdu > BaCl2 + 2H2O C o n l a i l a BaCla ThiTc h i e n chuoi p h a n ufng: CaO > BaCl2 + N H t + 2H2O M a u cho d u n g d i c h t r o n g suo't v a toa n h i e t l a H C l : Cu = 64, Ag = 108, Ba = 137 Cdu I > A l ( O H ) i + SBaCla M a u c h i CO k h i m t i i k h a i bay r a l a NH4CI: 14, O = 16, Na = 23, S ^ 32; CI = 35,5, Co = 40, Fe = 56, LCfl > B a S ^ + 2NH3T + 2H2O M S U cho k e t t i i a keo t r a n g la AICI3: ddu C de trung > F e ( O H ) i + 3BaCl2 > FeCOIDai + 2BaCl2 I rii p i A i n c T u i u n r Qtwu n n i unA u n r Q 0, 25 X 98 X 100% 20 = 122,5 K h o i l u g n g CUSO4: 0,25 x 160 = 40 (gam) (gam) K h o i lu'gng d u n g d i c h sau p h a n ufng: ' ^ d u i i g d i c h s a u p h a n iMig — niCuO + = 0,25 X '^^ddn^SO^ 80 + 122,5 = 142,5 (gam) d ) nc = Tii bang t r e n t a thay: = 1,66.10^^ (mol) D u n g dich n a o cho v a o t a o r a I a n ket t u a l a d u n g dich Na2C03 va So n g u y e n tuf C = 6,02.10^^ n g u y e n ti:f/mol.l,66.10^ m o l • AgNOa (cap d u n g dich 1) D u n g dich n a o cho v a o t a o r a I a n k e t t u a = 9,993.10^° n g u y e n tuf g) n la dung dich Na2S04 v a BaCla (cap dung dich 2) D u n g dich n a o cho 1,1 = 0,025 (mol) 44 CO2 vao t a o r a I a n k e t t i i a l a dung dich M g C l s v a Ba(N03)2 (cap dung dich 3) So p h a n td CO2 = 0,025 x , ' ' = 1,505.10^^ p h a n tit h ) MNa = 23 g/mol 2, niNa.- MgQ = 64,06 g/mol => a) Ta nco ( b a n ddu) = = mol => 64,06 -,23 6,02.10' mco +) L a y m o t h a i chat a c&p dung dich I a n lirgt cho v a o dung = 3,82.10"^^ g/nguyen t t f 6,02.10'23 i) c6: 23 dich cap 2, n e u c6 t a o r a k e t t i i a : thi chat cho v a o l a Ba(N03)2, l a i l a MgCl2 C h a t t a o r a ket t u a a cap l a Na2S04, c o n l a i l a BaCl2 = , ' " ' g/nguyen tuf +) L a y Ba(N03)2 d a tim diTOc a cap cho vao h a i d u n g dich d cap 1, ( b a n d s u j = 28 neu CO ket t u a t h i : C h a t tao r a ket tiia vcfi Ba(N03)2 l a Na2C03 gam Do t a n g k h d i liiOng: 36 - 28 = gam = mo => n^^ = 0,5 mol lai l a AgNOa V a y CO 0,5 mol C O ket h o p v d i 0,5 mol O cho r a 0,5 mol CO2 Cdu T h ^ n h p h a n h o n hap k h i l a : 0,5 mol C O v a 0,5 mol CO2 Bat so mol Na2C03 v a NaHCOa 25 m l dung dich A I a n liTot l a x, y b ) T a c6: n^^ (c6 t r o n g FeO v a Fe203) = n^^ l a y r a ( d t r e n ) = 0,5 m o l Doi v(5i t h i nghiem 1, t a c6: Na2C0a + 2HC1 G o i a = npeo v a b = np^^p^ V a y t r o n g X c6 (a + b ) m o l O D o do: a + b = 0,5 (1) Va: 72a + 160b = 30,4 (2) G i a i (1) v a (2) dirge: a = 0,2 m o l FeO v a b = , m o l FeaOg Vay: mpeo = 0,2 x 72 = 14,4 g a m "^Fe.,03 = 0,1 X 160 = 16 (mol) 1^ ( t h u dLToc) nipe ( t h u duoc) x k > N a C l + H2O (3) , x = , mol So mol H C l da tac dung vdri dung dich A l a : (4) Boi vdfi t h i nghiem 2: Cdu BaCl2 + NaaCOa L a y m o t d u n g d i c h b a t k i cho v a o d u n g d i c h c o n l a i , t a c6 b a n g sau: 50 y 2x + y = 0,1 - 0,028 = 0,072 56 = 22,4 g a m Na2S04 Na2C03 BaCl2 Ba(N03)2 AgNOa MgCl2 Na2S04 - - i i i Na2C03 - - i i i - BaClz i i - - i - Ba(N03)2 i i - i i - - AgNOa - MgCl2 - i - i - - (2) So mol H C l dir sau p h a n ufng (1) v a (2) l a : 0,014 x = 0,028 mol = npe ( t r o n g FeO) + npe ( t r o n g Fe203) = 0,4 > N a C l + CO.T + H2O So mol H C l 100 m l dung dich l a : = a + 2b = 0,4 mol Vay: y H C l (dii) + N a O H H|| T h e o d i n h l u a t bao t o a n n g u y e n t o t h i : npe (mol) (1) 2x NaHCOa + H C l I gam X > N a C l + COgt + H2O i LOI GIAI at THI HOC SINH GIQI HOA HOC > BaCOai + 2NaCl (5) Sau k h i loc bo ket tua, lay niidfc loc, ni/dc riia chufa N a H C O a cho tac dung vefi dung dich H C l ||^|^' ^ • NaHCOa + H C l (mol) y > N a C l + CO2 + H2O y ^ i a i r a t a c6: y = 0,026 x 1,0 = 0,026 mol x = 0,023 mol nong mol cua NaHCOa l a : 0,023 : 0,025 = , M nong mol cua N a H C O g l a : 0,026 : 0,025 = 1,04M i J ^ i i l i l n f (6) Cdu IL 300 g a m C H C O O H ^ 300 m l d u n g d i c h N a O H M chufa nNaOH P h a n l i n g : CH3COOH + N a O H 0,5 = 0,3 x icng dieu che CaCOs vdi Mg thi dot chdy hap kim do, oxlt tqo nen c6 khoi = 0,6 m o l doi khoi lugng > C H a C O O N a + H2O 0,5 tri/ih phan Cho kim loai sau: Be, Na, Al, Ca Hoi kim loai ndo tqo hap kim \ So m o l C H C O O H = 0,5mol (mol) phuang a) Viet V i % chOra 30 g a m C H C O O H Cho hgp kim ban ddu Kim T i le p h a n lifng l a : loai Hda V a y sau p h a n ufng N a O H = 0,6 - 0,5 = , ( m o l ) tri Khoi lugng nguyen ti'i Be Na Al Ca H I HI H 23 27 40 d u n g d i c h A c6 t i n h bazcf c) Dot chdy phdm m ( d u n g d i c h A ) = m (dung d i c h CH3COOH) + m ( d u n g d i c h N a O l I = 300 C% ( C H g C O O N a ) = + Vd,.>gdkh 0,5 X 82 X X d = 300 100% 660 X 1,2 = 660 du gam Tinh = 6,21% Cdu 660 0,6% todn tqo thdnh khoi lugng HI Dot chdy dich 5,6 lit (dktc) dich NaOH inol trung 2M de tqo thdnh cua mpt kim loqi chi cd hai hda' tri vd 3) tqo nen c6 nong Idng Lam Iqnh xud'ng vd dung muoi Id 23% Xdc dinh cong thUc cua tinh the DE THI HOC SINH GIQI HOA HOC 9, CAP TP HO CHJ MINH NAM HOC 2002 - 2003 Cdu muoi trung hda MeS (kim loqi rV Cho m gam hon hgp CaCOs Me oxi du Hda tan chdt viia du dung dich H2SO4 29,4% dp 34,5% the hidrat tdc dung viCa binh ciia hon hgp ban ddu ilng bdng mpt lugng 2,9 gam tinh hdn hgp gom H2 vd H2S sdn la nuac vd mpt chdt Lugng 1,76 gam sunfua cdc hgp chat rdn sau phan DE SO 12 Cdu hodn vdi 100 ml dung = i i L f ^ , «(NaOH) + 300 gap biet: 0,5 T i n h k h o i lufOng d u n g d i c h A : lugng dung dich dich Dung thi thu dugc Iqi cd nong dp hidrat vd FeS tdc dung vUa du vdi V ml dung dich HCl (D ^ 1,1 g/ml) thi thu dugc a lit hon hgp klii X (dktc) c6 I a) Cho biet dp tan mot loai muoi neu lam Iqnh thi phqm nhiet dung c6 nong dp % ciia muoi dp theo bang sau khodni^ ndo? dp Dp tan (trong b) Co dung lOOg dich nicac) dugc ddnh mpt chdt gSm: BaCh, n,3g 10° C 20°C 30°C 40°C 15,2g 18,9g 24,8g 36,7g so ngdu nhien H2SO4, luat thuc hien cdc thi nghiem NaOH, tic den 6, moi dung c/?' HCl, MgCl., NazCOa i nghiem 1: Dung dich (2) cho ket tua vai cdc dung dich (3) vd (4'i' Thi nghiem 2: Dung dich (6) cho ket tua vai cdc dung dich (1) vd ^4' Thi nghiem dung dich (4) cho bay len tdc dung dich (3) vd (5) Hay xdc dinh s6 cua cdc dung dich vai mol trung binh Id 40,67 (gam/mol) vd dung dich Y cd khoi ugng la b gam - ud dugc ket qua sau: lugng c) Neu chi sii dung Thi 3: Dung khoi Id 22% tie 50°Cl vi de bdt ddu xudt Men ket tinh cua muoi Nhiet chiia dich bii'n doi theo nhiet Cdu khoi hay tinh % theo khoi lugng V Mpt hon hgp A gom Na2C03 (iich HCl du thu dugc 56 (gam/mol) ^"(0H)2du Tinh vd NaoSOs hon hgp X cd klidi C/io 0,224 tit (dktc) •Sau thi nghiem lugng phdi diing gam Tinh mol trung dich ciia X la cua hdn hgp ban ddu cho tdc dung lugng vdi mol trung X di qua lit dung 50 ml dung binh a dich dung binh Id Ba(0H)2- HCl 0,2M de trung hda ' % so mol mdi X vd % khoi lugng moi chdt hon hgp A Tinh nong dp mol/lit ''^'10 sd lieu: H= 1,0= VTHI HOC S I N H Rini HOA HOC cua dung dich Ba(0H)2 ban ddu 16, Na = 23 S = 32, CI = 35,5, Ca = 40, Fe = 56, Ba = 137 53 L d l GIAI 2H2 + O2 Cdu a ) Dap b) H S + so: - K h o a n g n h i e t x u a t h i e n k e t t i n h tit 30°C -> 40°C - Kho'i l i j g n g muo'i k e t t i n h k h o a n g [24,8 < Thi nghiem m„,u6i (mol) < 36,7] (4) => d u n g d i c h (3) l a H2SO4 va d u n g d i c h (4) l a Na2C03 hoac ngagc l a i Thi nghiem |02 2: D u n g d i c h (6) cho k e t t u a v i d u n g d i c h (1) v a = 0,1 (mol) 0,1 ^- Cdu d u n g d i c h (1) l a N a O H ; d u n g d i c h (2) l a BaCl2 d u n g d i c h (3) l a H2SO4, d u n g d i c h (4) l a NaaCOg d u n g d i c h (5) l a H C l ; d u n g d i c h (6) l a M g C l y: Cdc em c6 the ke bang 22,4 Vay: de nitgn CO2 biet cdc (mol) ™ Mhh = 0,1 X X = 14,8 (g/mol) MeaOs + S t a a Me203 + 3H2SO4 3a — a ^ (2Me + 8 ) x l 0 > CaCOsi + 2NaCl T h e o de: %dd = 34,5 Mea (304)3 ( M e + 48)+^-^ x 98x^-0°^ ' 29,4 > C a C O a i + C t + H2O > C a C i + Na2C03 + 2H2O Ca(HC03)2 + N a O H N a K C + Ca(N03)2 CaC03i + NaN03 + h) IWO Oxit ^^^liap k i l l ! J^^hdp k i n i k l u dot Be BeO 33 65 Na NaaO 47 102 Al AI2O3 51 142 Ca CaO 64 96 100Me +14400 M e + 524 Me = 3678 65,5 = 56 (Fe) K h o i iLCgng d u n g d i c h b a n dau: n i d u i i g dich ban dau = (Be) = 34,5 lOOMe + 14400 = , M e + 18078 Vay k i m loai can t i m la B e r i Me2(S04)3 + 3H2O G o i c o n g thtfc cua t i n h t h e h i d r a t : M e ( S ) x H > CaCOs^ + H2O CaCl2 + Na2C03 hoac 34 + 0,15 0,25 (mol) lo -> CaCOa Ca(0H)2 + C O - Ca(HC03)2 (mol) 2MeS + - O 2 (mol) CaO + = 0,25 nj^^ = 0,25 - 0,1 = 0,15 II a) (mol) /// G o i a l a so m o l ctia MeS dem do't chay l a i d u n g d i c h (2) l a BaCla Luu 0,2 > Na2S03 + H2O 5,6 I^hon hap — 3: D u n g d i c h (4) cho k h i bay l e n v d i d u n g d i c h (3) v a (5), (2) 0,2 (5) p h a i l a H C l , d u n g d i c h (1) l a N a O H , d u n g d i c h (6) l a M g C l a , coii Vay: = H9O 0,1 SO2 + 2NaOH (4) Do d u n g d i c h (4) l a NaaCOs, d u n g d i c h (3) l a H2SO4 => d u n g dich X (1) SO2T + d u n g d i c h (1) l a N a O H v a d u n g d i c h (4) l a Na2C03 hoac ngu'cfc l a i Thi nghiem — > 0,1 HNaOH 1: D u n g d i c h (2) cho k e t t i i a v d i d u n g d i c h (3) v a -> 2H2O — "^Fe^Oy "^HgSO^ a = |(56 x dd 3a 98x100 ) - x - ^ - = M ^ x ( l ) + 0,03 98 x 0 = 11,6 X 29,4 , (gam) b ) K h i m = 1,44 gam; V = 400 m l ; b = 440,83 gam Khoi lUctng dung dich luc sau: mdung dkh sau = 11,6 - 2,9 = 8,7 (gam) Trong (400 + 18x) gam Fe2(S04)3.xH20 t h i chufa 400 gam Fe2(S04)3 => Vay 2,9 gam Fe2(S04)3.xH20 t h i chura a gam Fe2(S04)3 Vay khoi lifong Fe2(S04)3 b i t^ch ra: 2,9 400 X 72 x +y 1160 = 2,001 400 + 18X c:> 1600 + 72x - 1160 = 800,4 + 36,018x o 35,982x = 360,4 => => %mFeS = 40,67 X = 2y lOOx xlOO = — X100 = 69,44% loOx + 88y 0 y + 88y %m ^^^°3 x = 10 Vay cong thufc cua mudi t i n h the hidrat: 88y = lOOx + 88y xlOO = Na2C03 + 2HC1 CaCOg + 2HC1 (mol) X -> > CaCla + C02t + H2O 2x X FeS + 2HC1 (mol) y y (mol) (li NaaSOs + 2HCI (2i y c:> 34(x + y) + lOx = m + 1,1V - b „, 34 22,4 (mol) a X 22,4 (mol) 44x + 34y X + y lOOx + 88y - m 44x + 34y = 40,67x + 40,67y X m lOOx + 8 x - = m o x = —— 144 The' (**) vao (*): — m- ^ + 1,1V - b 144 22,4 34a 22,4 a = > BaSOgi + H2O (4) b > BaCh + 2H2O (5) 0,005 CUSO4 + H2O (1) C u + 2H2SO4 > CUSO4 + S O a t + 2H2O (2) Dung dich B l a CUSO4; k h i C l a SO2; dung dich D l a K H S O : SO2 + K O H > KHSO3 K2SO3 + B a S O s + 2IIC1 K H S O + BaCl2 V lit COo a diSu boo nhieu vd BafOHJs vd H2SO4 hgp A gom 0,2M rdn C Xdc chdt 0,02M the tich dung dugc iCng sau: logi KOH B C kirn HCl dich dugc A B vd klii C Khi hon dich cdng thi'cc phdn rdn CuCls mdu CuSO^ trinh dugc > CuS + •? > tU tU tUng thai gian dugc vai NaOH trinh phdn + ? sau mot mol binh 1,344 nong KOH cdc phuang a) Cu(N03)2 Cho dich vila tdc dung Viet cdc phuang Hodn khi, H2SO4 dgc, vai dung vai BaCl2 khong l,18g nong 4,56 gam DE THICHON HOC SINH GIOIHOA HOC 9, TJNH HA NAM NAM HOC ZOOZ - 2003 hgp rdn OE SO 13 nong tan vd dung b) Nung 2i 1 Nung Hda a) Xdc dinh V a y so' mol cua Ba(0H)2 ban dau: 0,004 + 0,006 + 0,005 = 0,015 (mol) Cdu chudn So jnol B a ( H ) dH = ]- so' mol cua H C l = 0,005 (mol) Vay: hgp dich B can de trung ling, cho rdng 375 ml dung (II) a + b«0,01 = a + b nx = han hay nhe han khong hdn the tich dung phdn Theo de: A chUa ' dich B chUa hon hgp HCl 0,05M 126.1 a 106a + 126 don dinli 100 = , % X i o a + 126b NajSOa Dan luong Ho di qua dug tliiiy tinli chiia 28 gam bgt oxit nung nong Sau mot thai gian thu dugc 24 gam chat rdn Xdc khoi lugng hai nitac tgo thdnh? 100 hdp thu hodn IM vd Ca(OHh 0,75M todn vdo 200 //^ thu dUgc 12 ga»^ 2raiS03 + 2NaOH CUSO4 + K H — Na2S03 + K2SO3 + 2H2O -> C u ( H ) i + K2SO4 , H o ^ n t h a n h cac p h u a n g t r i n h p h a n ufng: a) C u ( N ) + NazS b) Cu + CI2 > CuSi + 2NaN03 > m a u n a t r i k i m l o a i d e n dii vko d u n g d i c h AICI3 t l i i K h i cho tijt tii tiing t h a y CO k l i i bay l e n , r o i c6 k e t t i i a , sau c u n g k e t t u a t a n (Hoc s i n h t u v i e t p h i i o n g t r i n h p h a n lifng) C n k h i cho v^o d u n g d i c h CuSO; t h a y CO k h i bay l e n , r o i c6 k e t t i j a , k e t t i i a k h o n g t a n (Hoc C2H4 + H2 '° = 0,2 n c , ( O H ) , = 0,2 mol; x 0,75 = 0,15 mol C a ( H ) + CO2 thi > C a C O g i + H2O 0,12 ~ (1) = 0,12 V (CO2) = 0,12 X 22,4 = 2,688 l i t 0,12 b ) P h a n ufng: mol + (0,15 - 0,12) CuO ' (3) 0,12 2CO2 > Ca(HC03)2 (4) 0,06 CuO + H2 ban dau: ncuo = > Cu + H2O 28 — 80 = 0,35 (1) = 0,35 m o l hay 0,35 x 64 = 22,4 g a m CuO i hon => T o n g so m o l H ' ' t r o n g l i t d u n g d i c h B : 0,05 + 0,05 X = 0,15 mol H " + OH- — Ba^^ + SO^- mol da p h a n ufng vori H2 l a x, t h i s o m o l H2 tham gia p h a n ufng v a so' m o l Cu, H2O tao t h a n h sau p h a n ufng c u n g l a x; 2-^ g a m c h a t r S n t h u di/gc sau p h a n ufng g o m CuO diJ v a Cu t a o t h a n h T a CO p h u o n g t r i n h : 64x + 80(0,35 - x) = 24,0 (1) > H2O (2) > BaS04i T a c6: t r o n g 0,2 l i t d u n g d i c h B c6 n T h e o (2) => n v6 l i , CO n g h i a l a p h a n ufng xay r a chaa h o a n t o a n , duf CuO so' m o l u nhe a ) T r o n g l i t dung dich A t a c6 tong so mol O H ' = 0,02 + 0,005 x = 0,03 22,4 g a m n ^ y l a i n h o h o n 24 g a m c h a t r a n t a o t h a n h sau p h a n ufng la Goi ^^ci = 0,65 Ian b) T h e o b a i r a , t a c6 cac p h i i o n g t r i n h p h a n ufng sau: N e u p h a n ufng x a y r a h o a n t o a n t h i so' m o l d o n g dago g i a i p h o n g l a : ncu , + 1,5.26 + T h e t i c h d u n g d i c h B can l a y Vdungdich(B) = ? ^ = , l i t 0,15 CO2 = 0,2 + 0,12 + 0,06 = 0,38 m o l V (CO2) = 0,38 X 22,4 = 8,512 l i t thi Do t i le kho'i l i j g n g h o n h o p A so v d i k h o n g k h i b a n g : T r o n g l i t d u n g d i c h B co nnci == , m o l ; n^^^o^ = , m o l > C a C O s i + H2O 0,12 Ca(0H)2 (mol) (2-) 0,2 T o n g so m o l b a n d a u = - 1,5 - = , l i t D u n g d i c h A chufa h'on h o p K O H , M v a B a ( H ) , 0 M ; d u n g d i c h > KHCO3 C a ( H ) + CO2 (mol) =1,5 lit B chtia h o n h o p H C l , M v a H2SO4 , M K O H + CO2 0,2 (2) b a n dau = ,rV„ dh6n h?p AJkk = +) N e u t a o m u o i : (mol) > C2H6 ^ n , H2 v a C2H2 d i u he't V„ „ +) N e u c h i t a o m o t m u o i CaCOs p h i i o n g t r i n h : (mol) (1) n h a t l a i l a C2H6 Cac p h a n ufng (1) v a (2) x a y r a h o a n „ n K O H = X 0,2 C2H4 h o n h o p (A) g i a m l a V^j^ p h a n ufng = - = l i t Cdu So ^ ^ - ^ sinh ti; v i e t p h i i o n g t r i n h p h a n tfng) a) C2H2 + H2 CuCh „_ = 0,05 x 0,2 = 0,01 m o l 2, = n^^.,_ = 0,005 m o l so' => So m o l i o n SO^" c5n diT = 0,01 - 0,005 = 0,005 m o l V i t h e t i c h d u n g d i c h k h o n g t h a y d o i n g h i a l a t h e t i c h t o n g cong 1,2 l i t m a n , = 0,02 m o l ; n^,_ = 0,01 m o l v a K n „_ = 0,005 mo CI N e n n o n g mol/1 cua m u o i clorua: G i a i r a : x = 0,25 m o l , suy r a so g a m h o i H2O t a o t h a n h \k: 0,25 X 18 = 4,5 g a m [KCl] = Ml 1,2 = , 0 M ; [K2SO4] = 1,2 = 0,0042M Cdu a) Hoa tan hon hgp A vao dung dich H C l chi c6 A l tan: 2A1 + 6HC1 Ta c6: va Khoi lugng 1,344 l i t CH4 2AICI3 + SHat K h o i lugng cua 2,688 l i t A: (1) nAi=-n„ =x 0,03 "2 = 0,02 So mol H C l tham gia phan ufng (1) = 2nj^ mol CH4 = 0,03 x = 0,06 mol 3,6.22,4 2,688 « 30 gam + (1) H (2) + ^ - I I O2 Dung dich Ba(0H)2 hap thu CO2 c6 the xay ra: Ba(0H)2 + Theo (4) CO > 2CO2 Ba(0H)2 + CO2 (3) Ba(HC03)2 - > BaC03i + (4) IhO = 0,18 mol BaCOgi difgc tao 0,18 mol CO2; 197 f so mol Ba(0H)2 dii la 0,24 - 0,18 = 0,04M = 0,06 mol, dieu chufng to phan ufng (3) khong xay 0,18 mol CO2 la (1) va (2) tao ra; (1) Do n^j^,, = n A i = 0,02 mol nen [AICI3] = 0,02 0,375 = 0,053M chi 1,344 22,4 = 0,06 0,18 - 0,06 = 0,12 la (2) > AI2S3 tao mol CH4 chay; So' mol CO2 lai 688 b) K h i nung 3,54 gam hon hcfp A binh khong c6 oxi, xay phan ufng: 2A1 + 3S 4,56 - 0,96 = 3,6 gam C02T + 2O2 CHyO, + Coi V dung dich B thay doi khong dang ke, gan bang the tich dung dich B ban dau = 0,375 ht 0,375 " Goi so nguyen tuf C, H hgp chat A la x, y ta c6: iiHci ban diu = 0,375 x 0,2 = 0,075 mol Sau phan ufng dtr HCl = 0,075 - 0,06 = 0,015 mol; vay A l taii het, dung dich B binh phan ufng gom HCl dii 0,015 mol va A1C1;J, (S khong tan) NhU vay [HCl] = ^^22 4^^^ Vay khoi lugng mol ciia A: 0,672 = 0,03 mol 22,4 H, (dktc): = 0,12 mol C J i y O , chay tao Trong 1,18 gam hon hgp A c6 0,02 mol A l hay 0,54 gam A l : Do vay, ta c6 the ket luan phan ttf C^HyO^ chi c6 nguyen tuf C So g S lai la: 1,18 - 0,54 = 0,64 gam hay ng = 0,02 mol Mat Uiac, phan tuf khoi ciia C J I y O , = (4,56 - N h i i vay, t i le so mol A l va S la ^' = 1-: 1; ma 3,54 gam h6n hgp A 0,02 54 V dung gap — — = Ian, nen c6 so mol A l = so mol S - 0,06 mol 1,18 0,96) 22,4 2,688 = 30 (gam) Nen cong thufc phan ttf CxHyO^ la CH2O, c6 cong thufc cau tao la: H C H O O E SO Theo phan ufng (2): t i le so' mol A l vk S phan ufng la —; 14 T H I C H Q N HOC SINH GIOI H O A HOC 9, Q O A N , TP H C M N A M HOC 0 - 0 3 N h u vay S phan (ing het, A l d\J 0,02 mol; phan ufng tao t h a n h0,54 0,02 mol AI2S3 X 100 = 15,25% ; %Al2S3 = 84,75% %A1 = 3,54 Nong (%) cac chat dung dich D %A1 = 275 X 100 = 0,29^ ; %Al203 = i ^ M i O ^ ^ 100 = 1,09% 275 o>) Hay chgii chat rdn khdc iihau de clio moi chat tdc dung dung dich HCl ta thu duac chat trinh phan ling xay khdc Viet cdc uoi pliUang ^) Co dng nghiem, moi dug chiia mot dung dich mudi (khong triing kini loai cung nhu gdc axit) clorua, sunfat, nitrat, cacbonat cua cdc kini loai Ba, Mg, K, Pb ^oi " dung dich mudi ndo da chiia dng nghiem phuang phdp de phan biet dng nghiem 'HI H f i r Q i u u n i n i u n A u n r « trcn «3 Cdu II a) Viet pliUang trliih phdn iciig de bieu dien chudi phdn I'cng sau: ^ SO3 > H2SO4 > SO,, b) Neu hien tiigng xdy vd uie't phuang trinh - Sue CO2 vdo dung dich nUac voi - Cho til tit dung dich HCl vdo dung dich Cdu HI Trgn Ba(0H)2 C&u I- \ > SO2 ^ FeSs LCfl GIAI 100 ml dung dich Fe2(S04)3 phdn Na2C03 1,5M vai 150 ml dung die], trinh • phdn ling Tinh khoi lugng D vd E - iCng ket thuc khong thodt nUa, thi Igi dung dich 17% muoi sunfat tan Xdc dink — > CaCl2 + S02t + H2O CaCl2 + C2H2t 14,7% Sau phdn - Mau thtf nao c6 hien tLTomg sui bpt k h i la K2CO3: K2CO3 + 2HCI - them vdo 10 gam muoi FeCl2 khan Dun nong de hba tan het Khi de ngugi den nhiet ban 24,3 gam tinh thi muoi hidrat thiic cua tinh the hidrat, > PbClsi + 2HNO3 > 2KC1 + CO2T + H2O f Cho dung dich NaOH vao hai mSu l a i : b) Lay 40 gam dung dich bdo hoa cua FeCl2 ddu thi Idng xuong Mku thijf nao cho ket tiia trSng la Pb(N03)2: Pb(N03)2 + 2HC1 khoi lugng nguyen tii cua kim logi Xdc dinh biet rdng dung dich bdo hoa chiia 38.'' Mau iiao cho ket tua trSng la MgS04: MgS04 + 2NaOH cdiig khan > MgCOIDai + Na2S04 C n l a i la BaClz Can II a) Vie't p h a n ufng: V Cho 100 gam dung dich Na2C03 16,96%, tdc dung vai 200 gam dui'S dich BaCh CaSOs + H C N h a n bie't ong nghiem tren: a) Hoa tan mot lugng muoi cacbonat cua mot kim logi hoa tri II bdng Cdu — - > M n C l + CI2 + 2H2O Cho dung dich H C l Ian li/cft vao mau thi^ t r e n : IV muoi Mn02 + H C Trich moi dung dich mot i t lam mau thuf ke) dung dich H2SO4 -> BaCla + H2ST ta c6: BaCl2, MgS04, Pb(N03)2, K2CO3 b) Tinh khoi lugng mol chat tan dung dich B (coi the tich thay dd'l khong dang BaS + H C — b) Cac dung dich muoi chufa ong nghiem phai la muoi tan nen du vdo dung dich B thi tdch ket tua E a) Viet phuang > N a C l + COat + H2O CaC2 + 2HC1 - 2M thu duge ket tila A vd dung dich B Nung ket tua A troniy BaCl, ZnCl2 + Hat NaaCOa + 2HC1 ling khong den khoi lugng khong doi thu dUgc chat rdn D Them Cdu Zn + 2HC1 a) 10,4% Sau phdn ling, Igc bo ket tua dugc dung dich ^ Tinh nong % cac chat tan dung dich A Tii 9,8 gam H2SO4 c6 the dieu che 4FeS2 + I I O -> 2Fe203 + 8S02t 502 + - O ^ SO3 ^ 503 + H20 — dUgc: a) 1,12 lit SO2 (dktc) cho tdc dung vai kim logi (2H2SO4 + Cu b) 2,24 lit SO2 (dktc) cho tdc dung vai muoi SO2 + NaOH - Viet vd can bdng cdc phUang trinh phdn iCng xdy Cho so lieu: H ^ 1, C = 12, O = 16, Na = 23 S = 32, CI = 35,5, Mg = 24, ' 2NaHS03 • Na2S03 + H2SO4 Fe = 56, Ba = 137 I rti m i l T H I HflK SIKIH m i l H Q A HO'' ' k , ' "^'^l OE THI HOC SINH GIOI HOA HOC • H2SO4 -> C U S O + SO2T + 2H2O) -> NaHSOg Na2S03 + S02t + H2O > Na2S04 + S02t + H2O 65 b ) H i e n tiJcrng K h i sue k h i CO2 vao dung dich nUdc voi t h i lam due nUdc voi troiijj CO2 + Ca(OH)2 (mol) 0,15 Neu sue tiep CO2 t h i ket tua tan dan v^ dung dich suot: NaHCOg + NaCl Theo de: "Ba(oH) x 1,5 = 0,15 X = 0,3 0,3 ^ (mol) 0,2 > 3BaS04i + 2Fe(OH)3i 0,3 (1) 0,05 So mol cua Fe2(S04)3 0,15M + 102,4 = — M + 17 17 2,55M + 1740,8 = 15M + 1440 o 38,5 (2) -> Ci' 0,15 = 0,15 - 0,1 = 0,05 (mol) T i n h k h o i lifcfng D: Theo (1) va (2), ta c6: X 40 25,4 - niE = me^so^ = 0,15 x 233 = 34,95 (gam) b ) T i n h nong mol/1 cua Fe2(S04)3 2540 ifli m i l O F THI Hnn SINH GIQI HOA HOC + 15,4 = 25,4 (gam) m^.o = 50 - 25,4 = 24,6 (gam) 100 = 0,2M = 15,4 (gam) 15,4 25,4 Do do: 60,63 gam FeClg + 100 gam H2O ^ bao hoa T i n h k h o i liXcfng E: Theo (3), ta c6: The tich dung dich thu dugc la: 0,1 + 0,15 = 0,25 (lit) 10 Do t a n ciia FeCl2 of n h i e t dp b a n dau: mo = mg^so^ + mp^^o^ = 0,3 x 233 + 0,1 x 160 = 85,9 (gam' " M = 24: Magie (Mg) Sau k h i them lOg FeCl2 vao dung dich bao hoa t h i : mdungdich= > 2FeCl3 + B a S i Nong mol cua Fe2(S04)3 d^r: ~ J x 1007C! = 17'7r K h o i liJOng FeCl2 k h a n dung dich bao hoa: 0,1 Fe2(S04)3 + SBaCla x44 = 17 100 Theo de: dung dich (B) chufa Fe2(S04)3 di/ (mol) 0,15 ( M + 60) + 0 - , 0,2 ^^—^ Fe203 + 3H2O -> = 14,7 (gam) b ) Goi cong thufc ciia muo'i t i n h the hidrat: F e C l x H 0,1 2Fe(OH)3 (mol) (mol) 3Ba(OH)2 + Fe2(S04)3 (mol) = 0,15 0,15 0,15(M + 96) , ( M + 60) + 93,4 ^ ^ ^ ( ^ , ^ = C M - V = 0,1 H2O = 0,15 (mol) 98 0,15(M + 96)X100 III '^^ 14,7 > NaCl + C t + H2O tugng sui bgt tiic khdc c6: 100 C% MSG, Liiu y: Neu cdc em cho ngUgc lai, ticc Id cho Na^COs vao HCl tin c6 hi 1, a ) Ta 0,15 14,7 X100 n, K h i het Na2C03 ma vSn nho tiep H C l t h i c6 hien tLfcfng siii hot k h i \ Cdu ^ CO h i ^ n tUcmg gi v i : H C l + NaHCOg MSO4 + CO2T + Ta xet lOOg dung dich H2SO4 14,7% > Ca(HC03)2 H C l + NaaCOa Cong thile muo'i caebonat: MCO3 Phan ufng: MCO3 + H2SO4 > CaCOs^ + H2O CO2 + H2O + CaCOa IV - a ) Goi k i m loai hoa t r i la: M 24,3 127 + 25,4- 18X 100 = 60,63 (gam) 160,63 gam dung dich , — ^ x 127 + X 127 3086,1 437,4x - 60,63 , 127 + 18x 127 + X 127 + 1491,498 18X - 26519,562x 127 + X 133159,754 + 18873,036x = -26519,562x + 308610 45392,598x = 175450,246 => x * Vay cong thufc t i n h the hidrat: FeCl2.4H20 I BE THI HOC SINN GIOI HOA HOC 67 Cdu V DE SO 15 T a c6: m^^^,^^ = ^ 0 => n^Na,C03 m„ ^ p, = x — 200 O A O , = 20,8 P h a n ufng: NaaCOg + BaCla CO Vi _ J _ J L NAM HOC 2002 - 2003 C&u I- Hodii ^ (gam) > BaCOs^ + N a C l 0,1 = 0,16 > (1, 0,2 (1) => n N a C i = 0,2 ( m o l ) => TCr (1) ^ HN^^COJ = - 0,1 a) Co lo mat nhdn NaCl, = 0,06 NaoCOa, hay nhdn b) Phan khi, T a c6: n„^,„^ = b) Diet oleum H = ^^' ^"^^^ X ndy trung => K i m loai l a Cu CUSO4 + SO2T + 2H2O dung dich 0,05 22,4 = 1,12 ( l i t ) (mol) ^ 0,1 the sdt sunfat md khong dich NaOH dung NaOH, quy tim, diing them hoa nUac FeS04.7H20 cho den diC, roi dun nong rvCa sgch, can ngng nggm say kho vd nung gam Hoi muoi sunfat vdo a nhiet khong cao sdt c6 tinli > Na2S04 + S O g t + H2O -> c6 cong thilc la H2S04.nS03 0,1 Vyo,= 0,1 X 22,4 = 2,24 ( l i t ) Luu y: Cdc em c6 the dung chat klidc nliU cacboii (C) Ldl GIAI BE THI HOC SINH GIOI HOA HnC 200 ml dung dich lodng H2SO4 Biet nho han ba Idn so vai nguyen ten cdc kim loai VI Nu?ig nong Aj nil gam Cu ms gam dung vd A3 Todn 0,15M tH khdi nguyen rdng 10 ml dung dich (diing trUac H ddy vai dung tH khdi cua kim loai dich H2SO4 cila kim loai dan sau Ti le so mol hon hgp Id : ^dc dinh Cdu rdng ling Id 85% 0,5M Xdc dinh n? c6 hoa tri II vd III tdc dung 0,45 mol Biet tieu thu bao nhieu Hoa tan 6,76 gam oleum ndy hoa vica het 16 ml dung dich NaOH Idn lugt sinh ta phdi Biet hieu sudt phan Cdu V Cho 5,4 gam hdn hap gom hai kim loai b ) D e t h u di/gc 22,4 l i t SO2 t h i ngg^ = n^^g^^ => muoi 1&: NaaSOg Na2S03 + H2SO4 Chi dugc tan voi chiia 85% CaO, nguai vdo nude thdnh 0,1 Vyo, = 0,05 sau: BaClz, khong'? Beketop) ^ dich vd AySOJs- kg dd voi clii'ia 94% canxi cacbonat a) De t h u diicfc 1,12 l i t SO2 t h i n^.^^ = ^ n „ (mol) thu dugc a) De sdn xudt 280,3 Cu + 2H2SO4 , Fc Cdu IV 280,3 Na.COsdu Fe(0H)2 > FC3O4 biet tiing dung dich KOH vd Al2(S04)3 Igc ket tila tgo thdnh, khiet 100 > biet tiing Ig biet hai dung San pJidm = 100 + 200 - 19,7 = 280,3 (gam) X chi'ia rieng Cdu I I I Hoa tan 15 gam tinh (mol) m k e t tua = , x 197 = 19,7 ( g a m ) 11,7 > FeCNOsJs > FesOs H2SO4, NH4CI nUac, them dan dung => nidung d,ch > FeS04 chat ndo khdc m N a C i = 0,2 x , = 11,7 ( g a m ) => nif^^ C O , , , , = 0,06 X 106 = 6,36 ( g a m ) Ma: bien hoa sau day: Cdu 11 = 0,1 =i> sau p h a n ufng NaaCOa di^ Do d o , d u n g d i c h A g o m : N a C l tao t h a n h v a Na2C03 du Ttf chuoi > Fe(0H)3 0,1 < - 0,1 ^ R a f ! thien Fe < = ± Fe2(S04)3 208 BaCl, ^Ma OE THI HOC SINH GIOI HOA HOC LifP (VONG ) , QUAN THU flUfC, TP HCM 20 = — — = , (mol) m„„, (mol) ^^^^^^ = - ^ = ' ( m o l ) 10,4 va ^ m2 gam O2 thu dugc sdn phdm dich H2SO4 98% tan h^t dugc bg A3 dugc tgo 2,3 gam muoi USO4.5H2O hdn hgp hap thu bdi 200 ml dung Dem c6 can dung Aj Dun dung dich dich Ao thu dugc dich NaOH 30 gam Neu cho Ao tdc dung NaOH vai NaOH, IM mai tao dugc dung thu dugc, dich lugng phdi dung sau phdn ling thdy khoi lugng mj, m2, ni3? b) Tinh khoi lugng sdt dd tan vdo dung c) Tinh khoi lugng muoi c6 VII Trgn sau trgn Mat khdc, ndy thi Xdc dinh dung 40 ml dung A1(0H)3 + N a O H 0,8 gam dich sau nhung dich H2SO4 sau trgn moll I cua hai dung dich NaOH c6 nong vai 60 ml dung dU NaOH dich H2SO4 > NaAlOa + 2H2O Dung H2SO4 l a m thuoc thur va Ian liigt cho vao cac mSu thuf cbn l a i M a u thijf CO k e t tiia t r a n g khong t a n l a BaCl2: H2SO4 + B a C l a vai 40 ml dung dich > A l ( O H ) i + 3Na2S04 Al2(S04)3 + N a O H dich dich? dung dich H2SO4 Mau thCf cho ket tua keo trSng, sau d6 bi tan dan den het Ik Al2(S04)3: ket tua tan sdt vdo dung sdt tang chUa mot muoi axit vd du H2SO4 neu trgn nong - dicli - 120 ml dung dich 300 ml dung ket tua toi da Cho lugng HCl viCa du Sau nhung a) Tinh Cdu it nhdt vd NaOH - M a u thtf CO h i e n tLfcfng siii bot k h i l a Na2C03: Dung H2SO4 + NaaCOg 0,1M dicli c6 nong sdt? > B a S i + 2HC1 - > Na2S04 + C t + H2O M a u thtf k h o n g c6 h i e n tiXOng l a N a C l NaOll 0,16M Cdu III- T a c6: nFeS04.7H20 ban ddu 15 ~ "^FeSO^ 278 FeS04 + 2NaOH — Biet: Fe = 56, S = 32, O = 16, H = 1, Na = 23, C = 12, Ca = 40, Mg = 24, -> F e ( H ) i + Na2S04 278 LCil GIAI 278 ^ 2Fe(0H)34 2Fe(OH)2 + - O + H2O Cdu F e + 6H2SO4 dac — > Fe2(S04)3 + S S O g t + 6H2O Fe2(S04)3 + 2A1 > Al2(S04)3 + F e Fe2(S04)3 + F e (mol) > 3FeS04 F e S + Ba(N03)2 Fe(N03)2 + N a O H (mol) > F e ( H ) i + 2NaN03 2Fe(OH)3 Cdu 15 278 278 15 15 278 556 K h o i liiong Fe203 thu diicfc: — > Fe203 + 3H2O SFeaOg + C O — > ^Fe^Oi + C O a t Fe304 + C —> F e + C t V i theo gia thuyet 556 15 —(mol) 556 160 = 4,32 (gam) < nip^ p h a n tog, n e n muoi sunfat khong tinh khiet Cdu IV a) II Hip^^Q^ = 15 CaCOs 1000°C > C a O + CO2T (kg) 100 56 C h o quy t i m I a n liiot vao cac mau thuf t r e n : (kg) X 850 - M a u thijf l a m quy tim hoa x a n h l a : N a O H - M a u thtf l a m quy t i m hoa l a : H2SO4 85 K h o i Itrang C a O thu dt/gc: 1000 x - — = 850 (kg) a ) T r i c h m6i lo mot it l a m mau thtf D u n g N a O H l ^ m thuoc thuf cho I a n lifot v^o cac m a u thif Kho'i lugng C a C O s : x = 850_x_100 ^ 1517,86 (kg) 56 l a i , dun nong nhe - (2) (3) Fe203 + 3H2O TU (1), (2), (3): So mol F e t h u di/oc: —> 2Fe(OH)3i 2Fe(OH)2 + - O + H2O 15 2Fe(OH)3 > Fe(N03)2 + B a S i (1) 15 15 (mol) Cu = 64, CI = 35,5 (mol) K h o i lugng da voi can dung: 1517,86 x M a u thijf c6 k h i mui k h a i bay r a \k: N H C I NH4CI + NaOH — > N a C l + N H T + H2O I GlAl n£ TLJI i m n m i n i e^in* uni unp n 100 94 = 1614,74 (kg) 71 b) Ta c6: n n a O H = 0,5 2NaOH (mol) 16 = 0,008 (mol) 1000 x + H2SO4 - 0,008 -> (mol) > Na2S04 + 2H2O 6,76 Theo de bai, ta c6: » 98 + 80n Goi b la so' mol SO2 tham gia phan Ta 4,5n = 13,5 =^ n = Vay cong thufc cua oleum: H2SO4.3SO3 c6 so' mol Ian I j o t la 3a va a; K i m loai X c6 hoa t r i I I < K i m loai Y c6 hoa t r i I I I Phan ufng: (mol) X + H2SO4 - 3a XSO4 + Hat -! 3a (mol) 3a a Ta c6: n M a t khac: mk.m H, -2- 3a = 3a + — loa = 0,45 o a = 0,1 (mol) (1) CO he: (2) (2) CuO + H2SO4 dac - > CUSO4 + H2O (3) — 2a b -> b 2NaOH (mol) 0,24 > NaHSOa 0,12 > C u ( H ) i + Na2S04 ["^„u,o, =126a + 104b = 2,3 =^ a = b - 0,01 0,02 (mol) 0,12 - 0,02 = 0,1 (mol) => So mol CuO tao t h a n h phan ufng (1): 0,1 (mol) => So mol dong dif tham gia phan ufng (2): 0,02 (mol) => So mol dong tham gia phan ufng (1): 0,1 (mol) Vay so mol Cu ban dau: 0,1 + 0,02 = 0,12 (mol) m i = mcu = 0,12 X 64 = 7,68 (gam) So' mol O2 tham gia phan ufng (1): nig = m^^ = 0,05 x 32 = 1,6 (gam) So mol H2SO4 tham gia phan ufng (2) v^ (3): 0,02 X + 0,1 = 0,14 (mol) Theo phan ufng (6), (7) nNaOH " (5) (6) : , = 2c + 0,24 = 0,3 x =i> c = 0,03 (mol) Vay so mol H2S04ban dau la: 0,14 + 0,03 = 0,17 (mol) K h o i l i / g n g H2SO4: 0,17 x 98 - 16,66 b + CUSO4 CUSO4 + SOot + 2H2O SO2 + 2NaOH 0,03 => So mol CUSO4 tao t h a n h phan ufng (2): Vay: Cudu + 2H2SO4 dac (mol) a + b = " , = ^ n ^ u = ^ x 0.1 = 0.05 (mol) a) Phan ufng: -> NaaSOa + H2O = 2a (5) 0,02 (mol) VI > 2CuO fnv^nH tfng So mol SO2 tao t h a n h phan Ofng (2): Tir (1) va (2) =^ X = 9: Beri (Be) va Y = 27: Nhom (Al) 2Cu +O2 (gj 30 So mol ciia CUSO4.5H2O t h u d\igc: — = 0,12 (mol) 250 ^ ncuso, = " c u S O , H , o = ' ( n i o l ) => = 3aX + aY CuCh + 21120 So' mol cua NaOH tham gia phan ufng (4), (5): 200 m l d u n g dich H2SO4 chufa x m o l H2SO4 10 (7) 2c Fe + CuCl2 T r o n g 10 m l d u n g dich H2SO4 chuTa 0,004 m o l H2SO4 0,004 X 200 c -> C u ( H ) + 2HC1 0,004 X = > Na2S04 + 2H2O H2SO4 + N a O H A p dung: ^ C% = m,, ddHjSO^ (gam) x 100 = m„ = m^jXlOO C% 16,66 =— x100 98 , , = 17 (gam) b ) T i n h kho'i l u o n g s S t t h a m g i a p h a n uTng: DE SO 16 Tii p h a n tTng ( ) : Fe + CuClz (mol) (mol) X DE THI TUYEN HOC SINH GIOI HQA HOC (VONG 1), QUAN 9, TP HCM > FeCla + Cu 1 NAM HOC 0 - 2003 X Theo de t h a n h sSt t a n g 0,8 gam n e n t a c6 phiicrng t r i n h dai so s a i ; 64x - 56x = 0,8 o 8x = 0,8 => x = 0,1 So mol sat t h a m gia p h a n ijfng: 0,1 mol mpe = 0,1 X 56 = 5,6 (gam) c) ' K n h khoi liicfng cac muoi c6 dung dich sau k h i nhiing s&t T a c6: = "cuso, = 0.12 (mol) n^^^^H),^,,,^, So mol CuCl2 t h a m gia p h a n ijfng (9): 0,1 (mol) V a y so mol CuCl2 diT sau p h a n ufng (9): 0,12 - 0,1 = 0,02 (mol) m^^ci, = 0.02 X 135 = 2,7 (gam) ==> mp^ci, = 127 X 0,1 = 12,7 ( g a m ) ra: m„u,6, = 12,7 + 2,7 = 15,4 C i l a nong cua H2SO4 (mol) H2SO4 + N a O H > N a H S + H2O (' 0,04C2 • Na2S04 + 2H2O (^ 0,08Ci H2 ndi tren Biet rdng tap chat khong tham gia phan I'Cng Cdu V KhvC hodn todn 4,64 gam mot oxit kim logi thi can 1,792 lit CO (dktc) Neu lay todn bg lUgng kim logi thu diigc d tren cho vdo dung dich HCl du thi thu dugc 1,344 lit H2 (dktc) Xdc dinh cong thi'Cc hoa hoc cua oxit ndi tren - " a> Tinh nong phan trdm cua dung dich H2SO4 6,95M (D = 1,39 g/ml) phuang trinh sau: 0,16 0,06C2 - 0,08Ci = 0,016 G i a i (a) v a (b) ta di/oc: C i = 0,4M v a • ^) Trong mot binh kin, ngUdi ta thUc Men mot phan Theo de t a c6 phiiong t r i n h : 0,1 b) Tinh khoi lugng hon hgp E vUa du de phan Ung hodn todn vdi V lit Cdu VI + 0,04Ci ^ 0,06C^ - , C j Cdu III a) Can lay bao nhieu gam CUSO4 hoa tan vdo 400 ml dung dich CuSO^ 10% (D - 1,1 g/ml) de tao thdnh dung dich C cd nong Id 20,8% a) Tinh V C2 l a nong ciia NaOH T r o n I a n 1: Cdu II- L^y "^9^ ^ ' ^ n hap gom 6,9 gam natri vd 6,2 gam natrioxit vdo 500 ml niCac tao dung dich A Can lay bao nhieu gam NaOlI (c6 lun 20% tap chat) cho vdo dung dich A de dugc dung dich B cd nong Id 2M Biet the tich thay doi khong dang ke qua trinh thuc hien tren Cdu IV Cho 7,73 gam hon hap gom kem vd sdt cd ti Ic nzn • np^ -5:8 vdo dung dich HCl du ta thu dUgc V lit H2 (dktc) Dan todn bg lugng H2 ndy qua hon hgp E (gom Fe203 chiem 48%; CuO chiem 32%, tap chat chiem 20%) c6 nung nong (gam) Cdu VII Goi phuang trinh phan jJCng dieu die O-? ma em da hoc a chuaiig lap 8, ghi du dieu kien phan ling (neu cd) phuang phdp hoa hoc, em hay trinh bay each tdch rieng Fe203 hon hap c6 chiia Fe203 vd CaO b) Khi nhiet dung dich C xudng 12°C thi thdy c6 60 gam muoi CUSO4.5H2O ket tinh, tdch khoi dung dich Tinh tan cua CuSOj d 12°C (Dugc phep sai so nhd han 0,1%) So mol F e C l tao t h a n h p h a n ufng (9): 0,1 (mol) Suy Cdula) Viet trinh b) Bdng _ (b) = 0,8M 3A + 2B — vCng hoa hoc theo > C + 2D Trong A, B, C, D Id cdc hgp chat hoa hoc Tong so mol cac chat ban ddu la 1,5 mol ThUc hien phan Ung dUgc phut ngUdi ta dUng Igi thi luc tong so mol cdc chat binh la mol Hoi luc diing phan iing thi so phan tii cua moi logi hgp chat C vd D thu dugc Id bao nhieu? Cho: Na = 23; O = 16; H = 1; Cu = 64; S = 32; Ca = 40; Fe = 56; Zn = 65; CI = 35,5 b ) K h i h a n h i i i d u n g d i c h C x u o n g 12°C t h i : L d l GIAI Cdu 60 x 160 I '^CiiSO I ^SicU r a khoi dung dich — a ) p h a n ufng d i e u che O2 m -> K C l + KCIO, = 38,4 250 (gam) 60^90 H^O tach khoi dung dich 250" = " ^ ^ ^ ' ^ ^ S ^ ™ ^ T r o n g 0 g d u n g d i c h (C) c6 104 g a m CUSO4 v a 396 g a m H2O 2KMn04 HgO -> H g + m ^Oat m dien phan H2O K l i i h a n h i e t x u o n g 12°C t h i : '—> K2Mn04 + t + M n Hat + ^ Oat ^ ^^0^ = 0^ = 396 - 21,6 = 374,4 (gam) 65,6x100 ,0^,, H2O, FeaOs k h o n g t a n t a d u n g p h e u de loc FeaOs CaO + H2O Cau 11 T a c6: 6,9 n 23 ^dung = 0,3 ( m o l ) ; n , 62 Z n + 2HC1 (mol) (1) (mol) > 2NaOH 0,1 (2) TO p h a n ufng (1), (2): = 0,3 + 0,2 = 0,5 ( m o l ) Cau 40 X 1^ = 80 (mol) 25 (gam) FeCl2 + (mol) FeaOs - + 0,009m CuO • + 0,004m a ) Goi a l a kho'i l i i g n g C U S O c a n t i m 400 x 1,1 = 440 (gam) ^ a + 100 100 (gam) va => (3) -> Cu + HaO (4) 0,004m a = 60 (gam) = ^"2°3 %mcuO = n^ 100a + 4400 = 20,8a + 9152 79,2a = 4752 %m„„ „ n K h i p h a t r o n t a c6 p h i f o n g t r i n h ciia c h a t t a n : ( a + 440) 20,8 -> 2Fe + 3H2O 0,003m < m, 440 x 10 (2) = (0,05 + 0,08) x 22,4 = 2,912 ( l i t ) Ha T h e o de: K h o i li/grng d u n g d i c h C U S O 20,8%: a + 440 Hat G o i k h o i liJcfng h o n h o p E l a m (gam) III K h o i iLTgng d u n g d i c h C U S O 10%: (1) (dktc) 3Ha V a y k h o i liicfng N a O H (chila % t a p chat) can t h e m vao dung dich X np^ = , ( m o l ) b ) T i n h k h o i li/cfng h o n h o p E (FezOg v a CuO) - 0,5 = 0,5 ( m o l ) 0,5 " z n = 0' 05 ( m o l ) 0,08 Ti^ (1), (2): V „ Suy r a , so' m o l ciaa N a O H cho vao d u n g d i c h (A) l a : (A) l a : (gam) 0,05 0,08 "2 0,5 x = ( m o l ) na^ngdichA = 17,52 -> ZnCl2 + H t 0,05 0,2 So' m o l cua d u n g d i c h (B): =7,73 = : Fe + H C 0,3 NaaO + H2O (mol) ^zn : = 0,1 (mol) dich 0,3 374,4 mzn + ^ F e a ) T h e o de b a i , t a c6 h e : = 500 ( m l ) ^ 0,5 ( l i t ) N a + H2O — -> N a O H + i H a t (mol) la: IV > Ca(0H)2 6^ (gam) H ^ O CO t r o n g d u n g dich bao h a d " C V a y t a n cua C U S O cr 12°C b ) Cho t o a n bo h o n h o p vao niJdc d i i va k h u a y deu, CaO t a n h e t 104 - 38,4 = 65,6 C U S O CO t r o n g d u n g d i c h b a o h o a u n r C I M M nini unA u n r a X 100 m 48 X m 160 X 100 m X = 0,003m (mol) 100 m 32 X m 100 X 80 TCr (1), (2), (3), (4) ^ lui ^ = , 0 m (mol) , 0 m + , 0 m = 0,13 =i> m = 10 (gam) 77 Cdu V G o i o x i t c a n t i m l a MxOy c6 a m o l M^Oy + yCO (mol) a - » ay (I ax 2M + 2nHCl (mol) DE SO 17 > x M + yCOat (2 0,5anx CduJd) Trong 792 Tii (1): dung (3 nco = ay = (4, = 0,5anx = = 0,06 (mol) = a ( M x + 16y) = 4,64 (gam) o M a x + 16ay = 4,64 ZZ, 0,06 16 0,5n X CalNOs)?, M = 28n (6) M 28 56 84 K e t qua loai nhan loai Thi todn tinh dich dugc ddnh vd hai nude khiet? NaOH, so ngdu nhien BaCl2 Thue 2: Dung y 0,06 x (2) tdc dung vd bi hda tan nhd du dung nghiem 3: Dung Cdu hon Iigp gom SO2, Na2S0j, tie den gom: hien cdc thi ngliiem vd dugc dich X = sa bien b) Trong ly = vai dung dich trdng (1) cho ket tila dich (2) (4) tdc dung vai dung dich (5) khong cd ket ndo'? 11 a) Viet phuang 0,08 dich tua Hoi Ig ndo chiia chat diing C o n g thijfc o x i t : Fe304 trlnh phan iing de bieu dien hda sau: phong thi ngliiem ngudi ta NaOH thudng NaOII cdc hda clidt la H2SO4 dgc, CaO de ldm kho cdc chat Hoi plidi diing chat ndo de ldm kho cdc dm sau day: SO2, CO2, O) Hay gidi CM = n V(l) n.lOOO M'"^^^^ m.lOO.lOD C%xlOD m,dd D mdd-M M V(ml) Vdri C M = 6,95; D = 1,39 ^ Mat dung C% - C^.M 6,95x98 lOD 10x1,39 3A + 2B 3x ^ X ~ = 0,25 (mol) So p h a n tiif hcfp c h a t (C): 0,25 x 6,02.10^^ = 1,505.10^^ So p h a n tuf h g p c h a t (D ): 0,25 x x 6,02.10^^ = 3,01.10^^ I rti niki vila du dung todn rui unr CIMU dich dich vao dung HCl (du) 10,52% HCl thu dugc dich MgCOs A tdc B Hoi neu dicJi A thi cd ket tua khong? 1: De trung 0,4M Sau dd c6 can dung Phan unA unr 2: Cho tdc dung 51,66 gam ket tua Tinh khd'i lugng ••Ste*' OE hda dung dich, dich bdng thu dUgc m gam muoi du dung dich dung dung dich thi can 250 ml dung vdi mot lugiig cdc chat vd KCl tdc (d = 1,05 g/nd) thu dugc Chia Y ldm phan Phan ^> Tinh nni nUac thu dUgc dung gam hon hgp X gom K2CO3, KHCO3 Y vd 6,72 lit COo (dktc) ~ X ta lay 36,8 gam hSn hgp gom CaCOs, nguai vdi mot lugng ^oi V ml dung 2x x = 0,5 => todn Vi sao? T o n g so m o l c h a t t a o t h a n h sau p h a n ufng: « khde ^ a u / / / Cho 39,09 So m o l c h a t dii sau p h a n ijfng: 1,5 - x ( m o l ) = 1,5 - x + x sU lua cJign dd cho B hap thu hodn = 49% > C + 2D 2x thich c) Cho 45,9 gam BaO tan hodn b ) P h a n ufng: G o i so m o l c h a t A t h a m g i a p h a n ufng: x (mol) 7A each cho HCl tdc 1: Dung dich (4) tdc dung vai dung dich (3) cho ket tiia nghiem trdng n VL a) che CO2 bdng hda hoe, hay tdeh SO2 kiwi phdp Al(N03)3, Thi nghiem Thi ay 0,5anx dieu ket qua sau: 0,08 = 4,64 V a y M : Sat ( F e ) Cdu phuang c) Co lo dung thuang do CO2 bi Idn mot it hidroelorua SO3, O2 Bang-bien luan: Tii (3), (4): thi nghiem the ndo de ed CO2 hodn b) Bdng T h e (3) v a (4) v a o (5), t a c6: M phong vai CaCOs, Ldm = 0,08 (mol) TCr (2): V^: 20Q4 ^ pHAN BAT BUOC > 2MC1„ + nH2t ax OE THI HOC SINK GIQI HQA HOC , TP HO CHI M I N H M A M HOC Z 0 - dich NaOH khan AgNOs thu dugc hon hgp X V vd m THI Hnn siNH n i i i i HnA HOC 79 [...]... NaaCOa Cong thile muo'i caebonat: MCO3 Phan ufng: MCO3 + H2SO4 > CaCOs^ + H2O CO2 + H2O + CaCOa IV - a ) Goi k i m loai hoa t r i 2 la: M 24,3 12 7 + 25,4- 18 X 10 0 = 60,63 (gam) 16 0,63 gam dung dich 2 4 , 6 — ^ 1 8 x 12 7 + 1 8 X 12 7 3086 ,1 437,4x - 60,63 2 4 , 6 12 7 + 18 x 12 7 + 1 8 X 12 7 + 1 4 91 , 498 18 X - 265 19 , 562x 12 7 + 1 8 X 13 31 59, 754 + 18 873,036x = -265 19 , 562x + 308 610 45 392 , 598 x = 17 5450,246... ) l a : C2H6 13 6 = 544 " 250 -> Cu + C O 2 T p h a n ttjf k h o i cua ( Q ) : (mol) 10 ^ 1, 6 + 2H2O -> 2CO9T T a c6: (gam) 10 0 = 0,3539fc X1000 11 1 10 ' CaSO, %m Tir(l) 0, 392 .10 '^ Cu(OH), 2 (mol) X 98 = 0, 392 .10 ^ 250 n CaSO,, m 200 ,1 CO + CuO > C u ( 0 H ) 2 i + CaS04 _ 10 ^ axetat) III > Ca(0H)2 •^4 TCrd) Cdu (mol) K h o i iLTcfng d u n g d i c h Boocdo t h u dirge l a : 10 0 + 1 + 10 = 11 1 ( k g ) >... l ) (1) = 10 7 : 13 5 M + 17 y : 10 7 y M + 17 X 13 5 13 5xM + 2 295 xy = 1 0 7 M y + 18 19xy d 2Q°C t r o n g 1 l i t d u n g d i c h CaS04 bao h o a c6 0, 015 m o l CaS04 t r o n g 0,2 l i t d u n g d i c h CaS04 bao h o a c6 0,003 m o l CaS04 V i 0,6 .10 "^ < 3 .10 "^ n e n k h o n g c6 h i e n t i / g n g k e t t i i a Cdu M 2 0 y 16 x = 10 0,2 ( m l ) C^ = x — - — = 0, 015 M Mc,so, 13 6 0 ,10 02 0,0006 M = 39a C o... "Ba(oH) x 1, 5 = 0 ,15 X 2 = 0,3 0,3 ^ (mol) 0,2 > 3BaS04i + 2Fe(OH)3i 0,3 (1) 0,05 So mol cua Fe2(S04)3 0 ,15 M + 10 2,4 = — M + 17 17 2,55M + 17 40,8 = 15 M + 14 40 o 38,5 (2) trong do -> Ci' 0 ,15 = 0 ,15 - 0 ,1 = 0,05 (mol) T i n h k h o i lifcfng D: Theo (1) va (2), ta c6: X 40 25,4 - niE = me^so^ = 0 ,15 x 233 = 34 ,95 (gam) b ) T i n h nong do mol /1 cua Fe2(S04)3 2540 ifli m i l O F THI Hnn SINH GIQI HOA... 17 ,4 8 ,96 5% = 4 6 9 6 - 18 784a = 24 79, 5 - 4 3 5 0 a « 2 216 ,5 = 1 4 4 3 4 a Cong thuc oxit: 20°C: 10 0 g a m H2O + 0,2g CaS04 - > 10 0,2 g a m d u n g d i c h CaSO, CO d = I g / m l (mol) M2OX, V^^^^^ = 0,0006 "caci, = o 16 y = 20 : 27 2 M + 16 x • 2 M + 16 y x M + 8y 20 M + 8x y 27 2 7 M x + 216 xy = 2 0 M y + 16 0xy 2 7 M x + 56xy = 2 0 M y > CaS04 + 2 N a C l 17 y 17 x M a t khac: M + 17 x • M + 17 y X 0, 012 ... dkh sau = 11 ,6 - 2 ,9 = 8,7 (gam) Trong (400 + 18 x) gam Fe2(S04)3.xH20 t h i chufa 400 gam Fe2(S04)3 => Vay 2 ,9 gam Fe2(S04)3.xH20 t h i chura a gam Fe2(S04)3 Vay khoi lifong Fe2(S04)3 b i t^ch ra: 2 ,9 400 X 72 x +y 11 60 = 2,0 01 400 + 18 X c:> 16 00 + 72x - 11 60 = 800,4 + 36, 018 x o 35 ,98 2x = 360,4 => => %mFeS = 40,67 X = 2y lOOx xlOO = — X100 = 69, 44% loOx + 88y 2 0 0 y + 88y %m ^^^°3 x = 10 Vay cong... ufng (1) , (2): X xlOO = 30,56% 2NaCl + C 0 2 t + H2O b (mol) Khoi lifgng dung dich cua H C l : 1, 1.V (gam) m + 1, 1 200y + 88y a X > FeCh + HzSt 2y 88y V Fe2(SO4)3 .10 H2Q Cdu IV a) 22 4 X ^ =^a = 0,336 (ht) 34 X 1, 44 + 1, 1 X 400 - 440,83 44x + 34y (gam) 400 + iSx 11 60 4 - 400 + 18 x _ 23 _ ^ 8,7 10 0 Theo de: 67 c) Theo cSu a, 11 60 a = 400 + ISx a = %n so.2 a ^+ b X 10 0 = b a + b X100 = (I) —— 3 a+ -a X 10 0... khiet 10 0 > biet tiing Ig biet hai dung San pJidm = 10 0 + 200 - 19 , 7 = 280,3 (gam) X chi'ia rieng Cdu I I I Hoa tan 15 gam tinh (mol) m k e t tua = 0 , 1 x 19 7 = 19 , 7 ( g a m ) 11 ,7 > FeCNOsJs > FesOs H2SO4, NH4CI nUac, them dan dung => nidung d,ch > FeS04 chat ndo khdc m N a C i = 0,2 x 5 8 , 5 = 11 ,7 ( g a m ) => nif^^ C O , , , , = 0,06 X 10 6 = 6,36 ( g a m ) Ma: bien hoa sau day: Cdu 11 = 0 ,1 =i>... = 10 %) ^ dung - Trong A: - % X = 30,4% X _ 30,4 2Y 69, 6 % Y = 10 0% - 30,4% = 69, 6% _ bY TCrd), (2) X Y 60,8 69, 6 _ 74 ,1 60,8 25,9b 69, 6 74,1a X 25,9b Y 74,1a Dat 5 50 + 50 10 0 10 gam gam B + 2D + 4E a 2a (mol) = = 2a 4a Neu lay 2a mol A nhiet phan se tao t h a n h 7a mol k h i D i n h luat bao to^n khoi liicfng cho: M A X 2a = 22,86 x 7a a _ 2 b 5 => M A = 80 OE SO 9 BE THI HOC SINH GIOI HQA HOC 9, ... ~ J x 10 07C! = 17 '7r K h o i liJOng FeCl2 k h a n trong dung dich bao hoa: 0 ,1 Fe2(S04)3 + SBaCla x44 = 17 10 0 Theo de: dung dich (B) chufa Fe2(S04)3 di/ (mol) 0 ,15 ( M + 60) + 1 0 0 - 0 , 1 5 0,2 ^^—^ Fe203 + 3H2O -> = 14 ,7 (gam) b ) Goi cong thufc ciia muo'i t i n h the hidrat: F e C l 2 x H 2 0 0 ,1 2Fe(OH)3 (mol) (mol) 3Ba(OH)2 + Fe2(S04)3 (mol) = 0 ,15 0 ,15 0 ,15 (M + 96 ) 0 , 1 5 ( M + 60) + 93 ,4
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