B a i giai PhSn tfng dien phan n6ng chay K C l : 2KC1 dpnc 2K + 2H2O > 2K + cut (A) (B) K O H + H2T (D) (1) FeO + CO2 -> A1(N03)3 + N t +H2O — > ZnS04 + H2ST +H2O Z n + H2S04djc > 1r » B a i giai (C) + 8SO2 4FeS2+1102 IFejOi 2 y A l + 3Fe,0y 3xFe + y A l 3 Fe^Oy + (y - x ) C O xFeO + (y - x)C02 8A1 + 3OHNO3 V i theo cdc phan iJng ( , 2, 3, 4) so mol H C l = so m o l K O H nen k h i cho het 4Zn + 5H2S04d5c D vao E dung dich trd trung tinh va quy chuyen t r d l a i mau t i m Bai Hon hdp A g o m AI2O3, M g O , Fe304, CuO Cho luong H2 dif qua A nung 2HC1 > (3) dd H C l (E) K h i cho quy t i m vao E, quy t i m chuyen m ^ u K h i them D vao E xay phan uTng: KOH + HCl > K C l + H2O (4) -> khong phan uTng M g + CUSO4 -> MgS04 + Cui M g + 2AgN03 -> Mg(N03)2 + A g i 2A1 + 6HC1 2A1 + 2NaOH + 2H2O 2A1 + 3CUSO4 A l + 3AgN03 > 2AICI3 + 3H2T > 2NaA102 + S H j t 8A1(N03)3 + N t + I5H2O -> 4ZnS04 + H2ST +4H2O - B a i giai CAc phan uTng: Mg + NaOH ;3nt! = i i.nq -; ^ HNO3 loang (dir C O k h i nhat N O bay ra) V i e t cdc PTPU xay B a i giai MgCl2 + H2t Cdc phan tfug: 'UO 3Fe + 4H2O Fe304 + 4H2 CuO + H2 -> Cu + H2O Chat ran B g o m Fe, Cu, AI2O3, M g O : T^c dung cua B v d i dung dich N a O H va vcli CO2 - AI2O3 + 2NaOH Al2(S04)3 + C u i > 2NaA102 + H2O CO2 + 2H2O + NaA102 -> Al(N03)3 + 3Ag>l Tic Al(0H)3i +NaHC03 dung v d i dung dich HNO3: FeCl2 + H t M g O + 2HNO3 Mg(N03)2 + H2O Fe + N a O H -> khong phan ufng Fe + 4HNO3 Fe(N03)3 + 2H20 + N O t Fe + CUSO4 -> FeS04 + Cui 3Cu + 8HNO3 3Cu(N03)2 + 4H2O + N Fe(N03)2 + A g i N e u d i r A g N thi: Fe(N03)2 + A g N 1- Chosddobienhod: -> A , -> Fe(N03)3 + A g ^ B a i Can blng cac phuTcfng tnnh phan iJng: FeS2 + O2 •ad: Tdc dung v d i H2: Fe + 2HC1 Fe + A g N aii^r v^ chat rin D Sue k h i CO2 dU vao dung dich C va hoa tan D bhng dung dich N a O H , C U S O A g N O j V i e t cac phiTdng trinh phan ufng xay Mg + 2HC1 j^nti ^ht>^ n6ng diTdc chat r ^ n B Hoa tan B vao dung dich N a O H diT dtfdc dung dich C B a i Cho cac k i m loai M g , A l , Fe Ian liTdt tdc dung v d i cac dung dich HCl, 102 Fe^Oy + C O H2 + CI2 Fe + AI2O3 A l + HNO3 (2) A l + Fe.Oy -> FezOj + SO2 +x -> A2 +Y -> A : -> Fe(OH)3 |Fe(0H)3 B, +z ^ B2 +T B 103 Cty TNHH MTV DWH Khang Viet Tim cong thuTc cua cac cha't vCng vdi cac chat A,, B, Vie't cdc phifdng trin}, SO3 + H O phan tfng theo sd do Cho sd b i e n hoa: ^ A, +Y +x ^ A2 CaCO., CaC03 +z B, +T > B phan ufng theo sd H H2 + CI2 2HC1 2Fe + 3CI2 > 2FeCl3 (8) > 2FeCl3 + 3H20 (2) A g C U +Fe(N03)3 (3) 2H2O + Ba Ba(OH)2 + H t (4) Fe203 + CO B a ( H ) + Na2C03 BaCOji (5) X2 + X (6) X2 + X 4 X5 + X , - X7 + Xs - X y + X|() Fe(N03)3 + N a O H -> F e ( H ) ; + N a N G h i c h u : Co the su" dung cac phan ufng khac, axit khac -> CaO + C02T CaCOj (1) CaO + H2O -> C a ( H ) (2) C a ( H ) + 2HCI -> CaCl2 + 2H2O (3) 2CO2 + B a ( H ) -> Ba(HC03)2 N a O H + Ba(HC03)2 > B a C O j ^ + Na2C03 + 2H2O C a ( H ) + Ba(HC03)2 > C a C i + B a C i + 2H2O C a C l j + Na2C03 B a i Tuf cac nguyen lieu ban dau la khoang vat pirit, m u o i an, nU'dc, cac chai xuc tac va thiet b i can thiet, vie't cac phU'dng trinh phan tfng dieu che' Fe B a i giai 10 4FeS, + 1 , 2SO2 + O2 2Fc203 + S t V2O5 -> 2SO3 (1) (2) (10) Bai T i m cac chat X , , X2, X ihich hdp va hoan cac PTPlJ sau: Co the silr dung cac phan iJng khac, chat khac Cac phan iJng can thiet NaHS04 + H C l t (9) Co the siir dung cac sd khac r -> Fe,0> + X i > —>• > B a S i +Na2S04 + C t + H2O B a S i + Na2S04 + CO2T + H2O A g i + K N O + H2O Ca(H2P04)2 Fe2(S04)3 + S02T + H O Xy + X|() > Fe2(S04)3 + S t +H2O Xy + X|0 > Fe2(S04)3 + S t +H2O Xy + Xlo > Fe2(S04)3 + S t + H2O Xy + XlO —— > Fe2(S04)3 + S t +H2O Xy + X||) — > Fe2(S04)3 + S t +H2O CaC03^ +2NaCl FeClj, F e ( H ) , NaHS04 IPeCh) > F e ( H ) i + 3NaCl NaCl,,i„, + H2S04dac — ^ FczO, + 6HC1 +2NaOH (6) (7) (1) - ^ > FeClj + H21 Fe203 + H t + 3AgN03 • (4) Fe + H C I 2Fe(OH)3 FCCI3 + CI21 + N a O H > 2Fe + H O '" FeCl3 + N a O H Cac phan ufng: (3) (hoac 2FeCl2 + CI2 B a i giai 104 N a C l + 2H2O — Fe203 + 3H2 T i m cong ihtfc cua cac cha't uTng v d i cac chaft A i , B, Vie't cac phiTdng trinh > H2SO4 X|ii + X | Ag2S04 + S t +H2O BaC03i+H20 X,2 — > X3 + X,3 -— > BaCOsi + C a C i + HjO X,; + X , -— > FC(N03)2 + X , X3 + Bai giai Hoan chinh cac phU'dng trinh phan iJng: XFC2O3 + ( x - y ) C > 2Fc,0y + ( x - 2y)C02 2NaHS04 + Ba(HC03)2 B a S i + Na2S04 + C t + 2H2O 2NaHS04 + BaCO, —> B a S i +Na2S04 + C t + H2O 2AgN03 + 2KC(ti —> AgzOJ' + 2KNO3 + H2O Ca3(P04)2 + 4H3PO4 2Fe + 6H2S04dac 2FeO + 4H2S04dac 4H2S04dac CUSO4 + K H -> Fe2(S04)3 + S02T +6H2O 4H2S04dac Fe2(S04)3 + S t +4H2O Ba(HC03)2 + Ba(0H)2 Ba(HC03)2 + Ca(0H)2 10 Fe + Cu(N03)2 Chu de 3Fe2(S04)3 + S t + IOH2O 2FeO + NaHS04 + BaCl2 ^ B a S i + NaCl + HCl K O H + BaCl2 -> khong NaHS04 + KOH NaCl khong c6 phan iJng vdi chat c6n lai Tren cd sd cac phan tfng ta c6 the nhan biet: - Cha't tao ket tua trang la BaCl2 -> 2BaC03i +2H2O - Chat tao I ket tua trang la NaHS04 BaCOai +CaC03i + 2H2O -> Fe(N03)2 + Cu4' B a i giai Tri/dc het cho vao moi dung dich mot mieng kem (hoSc s^t, magie), ndi nao CO thoat la HCl: > ZnCl2 + H t Sau trpn dung djch HCl vdi dung dich NaCl dtfdc dung dich A, va trgn dung dich HCl vdi dung dich NaN03 dUdc dung djch B Cho hai mieng Cu vao dung dich A va B, ndi nao c6 khong mau thoai ra, ho^ nau khong la hon hdp HCl va NaNOs Cac phan iJng xay ra: 3Cu + 8HC1 + 8NaN03 > 3Cu(N03)2 + 2N0 T + 8NaCl + 4H2O 3Cu + 8HC1 + 2NaN03 > SCuCls + N T + 2NaCl + 4H2O > 2NO2 - Chat tao ket tua trang, ket tua xanh la CUSO4 - Cha't tao ket tua xanh la KOH Bai Cho mau kim loai A, B, C, D mau sang bac giong Ian Itfdt tac dung vdi cac dung dich HNO3 dac, ngupi, dung dich HCl va dung dich NaOH Ke't qua thu diTdc nhi/ sau: axit A HNO3 - HCl + NaOH - B C D + + + + - + + - - Hoi A, B, C, D la nhffng kim loai nao so ci.c kim loai sau: nhom, sat, kem, magie, bac? B a i giai Cho cac kim loai Ian liTdt tac dung vdi HCl, HNO3, NaOH ta c6 cac ket qua sau: A l + HNO3 dac nguoi 2A1 + 6HC1 > khong phan iJng > 2AICI3 + 3H21 2A1 + 2NaOH + 2H2O > 2NaA102 + 3H21 V$y A l khong ihuoc A, B, C, D B a i Co dung dich CUSO4, BaClz, NaHS04, KOH, NaCl Cho cac dung djcli Fe + HNO3 d6 tac dung vdi tirng doi mot Viet cac PTPLf xay Tren cd sd ca^ Fe + 2HC1 106 > NaKS04 + H2O khong c6 hifn tUdng dSc triftig -> Ag2S04 + SO2T + 2H2O khong: NaCl, HCl, NaN03 2NO + O2 C u ( H ) i +K2SO4 Chat khong c6 phan iJng gi la NaCl B a i Chi diTdc dung kim loai c6 the phan biet diTdc cac dung dich sau day hay Hoac -> khong N h S n bi^'t - T a c h hon hFe2(S04)3 + S t +4H2O Fe2(S04)3 + S T + IOH2O + 2H2S04dac Cic phanuTng: CUSO4 + NaHS04 10H2SO4dac 2Ag B a i giai -> Fe2(S04)3 + S t +6H2O 2FeS + piat nhan 3Ca(H2P04)2 2Fe304+ 10H2SO4 3ac 2Fe(OH)2 + ph^n ph5 i?ng d6 c6 nh$n biS't drfdc cdc dung dich khong, ne'u cdc dung dich bi dac, nguoi > khong phan uTng > FcCh + H2 T 107 Fe + N a O H -)• khong phan tfng V a y F e la kim loai A Zn + 4HNO3 ^ Zn(N03)2 + 2NO2T + 2H2O Zn + 2HC1 ZnCl2 + H t Zn + 2NaOH NazZnOj + H z t V a y Zn la kim loai B Mg + 4HNO3 -> Mg(N03)2 + 2NO2T + 2H2O Mg + 2HC1 Mg + 2NaOH -> khong phan ufng A g N + N t +H2O Ag + H C l khong phan uTng L a y mot phan dung dich B, them mot liTdng d\i HNO3 va sau 66 A g N tha'y j{6't tua trang: nhan bie't dtfdc NaCl AgN03 + NaCl >AgCU+NaN03 pai M u o i N a C l b i Ian cac tap chat Na2S04, MgCl2, CaCl2, C aS04, NaBr Trinh bay phu'dng phap hoa hoc de thu dtfdc N a C l nguyen cha't Bai giai Hoa tan hoan toan N a C l vao niTdc (CaS04 v i liTdng i t nen tan het) sau Na2S04 + BaCl2 > B a S i + 2NaCl Na2C03 + MgCl2 > M g C i + 2NaCl Na2C03 + CaCl2 > C a C O j i + 2NaCl ' -> khong phan uTng V a y Ag la kim loai C B a i Trinh bay phi/dng phap hoa hoc ngan gpn nhat de nhan bie't tiifng muoi natri mot dung dich gom: Na2C03, NaHC03, Na2S03, Na2S04, NaCl, NaNOj B a i giai Tru-dc het cho lu-png du" dung dich Ba(CH3COO)2 vao dung dich ban dau thu diTdc ke't tua A va dung djch B Ket tua A gom BaS04, BaS03, B a C O j > 3CuCl2 + N O T + N a C l + 4H2O th6m luTdng du" BaCl2 de ke't tua het muo i sunfat: Ag + 2HNO3 Ag + N a O H 3Cu + 8HC1 + 2NaN03 MgCl2 + H t V a y Mg la kim loai D Sau them tie'p Cu vao thay dung dich c6 m a u xanh, c6 k h i khong mau thoat hoa nau khong k h i : nhan bie't dtfdc NaNOs: Na2S04 + Ba(CH3COO)2 > BaS04 i + 2CH3COONa NasCOj + Ba(CH3COO)2 > B a C i + 2CH3COONa Na2S03 + Ba(CH3COO)2 )• BaS03 i + 2CH3COONa Phan nirdc loc l a i (gom NaCl, NaBr, Na2C03) cho tac dung v d i dung dich HCl d i / , l u c d : Na2C03 + 2HC1 > 2NaCl + H2O + CO2 T ' Trong dung dich c h i lai NaCl, NaBr Sue k h i CI2 t d i du" va c6 can dung dich ta se c6 N a C l nguyen cha't 2NaBr + Cl2 )• 2NaCl + Br2 Bai Co hon hdp M chiJa cdc chat CaC03, Fe203, AI2O3, Si02 H a y t n n h bay phiTdng phap hoa hoc de lay tiifng cha't rieng le nguyen cha't, k h o i liTdng khong d d i B a i giai ' Hoa tan ke't tua A bkng dung dich H C l diT, mot phan khong tan la BaS04: Cho hon hdp vao niTdc r o i sue k h i CO2 drf vao L u c x a y phan iJng hoa nhan bie't diTdc Na2S04 tan C a C O j , l a i chat r ^ n R Na2C03 + 2HC1 > 2NaCI + H2O + CO2 T Na2S03 + 2HC1 > 2NaCl + H2O + S O Nhan bie't hon hdp CO2, SO2 nhiT cac bai tren Nhif vay ta bie't dufdc Na2C03, Na2S03 Phan dung djch B c6 NaHC03, N a C l , NaNOj, C H C 0 N a , Ba(CH3COO)2 CO2 + H2O + CaC03 > Ca(HC03)2 Lay dung dich thu du'dc dem dun nong ta c6 C a C O s Ca(HC03)2 — ^ CaC03 + H + C02T Si02 + H C l > khong phan u-ng ra: nhan bie't N a H C O j Fe203 + 6HC1 > 2FeCl3 + 3H2O AI2O3 + 6HC1 > A I C I + 3H2O 108 > N a C l + H2O + C O (2) Hoa tan R bkng dung dich H C l diT, thu dUdc S iO j La'y mot phan dung dich B cho tac dung vdi dung dich H C l diT, thay bay NaHC03 + H C l (1) I ^ '' ' (3) J (4) 109 Lay dung djch thu dUcJc (gom FeCh, A I C I , HCl du) cho tac dung vdi duiig dich NaOH dU' Cac phtfcJng trmh phan i?ng xay ra: (Iri) tC HCl + NaOH > NaCl + H2O FeCl3 + N a O H > AlCl3 + N a O H > A1(0H)3 Al(OH)3 + NaOHdu > NaAlOj + 2H2O Fe(OH)3 + N a C l i + 3NaCl (5) Fe203 + H t ^ siu dong thi vat bing nhom rat ben khong bi htfhong, trai lai cdc vat b^ng s^t, dong thi bi hoen ri (6) (7) (8) Lay ket tua nung nhiet cao ta c6 Fe203 2Fe(OH)3 Tai nh6m hoat dong hdn s^t, dong nhuhg de cic d6 vat hkng nhom, Bai giai £)5 v$t b^ng bac de khong van giiir duTdc anh kim vi Ag khong tdc dung vdi O2 cua khong A g ; nhiftig khong nhiem ban H2S fhi bi den xam phan tfng tao Ag2S (den) sau: 2Ag + 0,502 + H2S (9) > Ag2S + H2O Khi bi rdi vai thuy ngan ra't doc (do bay hdi), khong the thu gom, v l Hg bi Sue C O vao phan ni/dc loc (gom N a A va NaOH) sau nung d nhiet phan tin nhffng hat ra't nho, 66 ngiTdi ta phai r^c liTu huynh, luc 66 cao ta diTcfc AI2O3 t^o H2S khong bay hdi, ta c6 the thu gom de dang: > > C O + NaOH > NaHC03 C O + H O + NaAIO2 2A1(0H)3 > A1(0H)3 i + NaHC03 ^ Al203 + H t (10) (11) (12) Bai7 Hg + S > HgS Nhom Ih kim loai hoat dong hdn s^t, dong, nhtfng cac vat hkng nhom de ISu khong khong bi hoen r i nhom tic dung vdi O2 (cua khong khi) tao mot Idp mang raft mong bao ve cho A l phia khong phan tfng vdi O2 Co the tach s^t kim loai khoi hon hdp kim loai sat, dong, nhom, tach s^t kim loai khoi s^t sunfua hlng each diing nam cham hay khong? Mot loai dirdng kinh bi Ian mot it cat Lam the nao de c6 dufcJng nguyen chat Co the lay diTdng tiT dung dich di/dng tan riTdu etylic dtfdc khong? Bai giai Co the dung nam cham de tach sat kim loai khoi hon hdp sat, nhom, dong vl Bai -A C6 hai dung dich loang FeCl2 v^ FeCl3 (gan nhiT khong mau) Ta c6 the diJng dung djch NaOH ho|c nxXdc brom, hoac dong kim loai de phSn biet dung dich dd Hay giai thich bling cic phiTdng trinh phan tfng Co ong nghiem diTdc d^nh so thi? tiT 1, 2, 3, 4, M o i ong diTng dung dich sau day: Na2C03, BaCh, HCl, H2SO4, NaCl Neu lay ong 66 vko nhom, dong kim loai khong bj nam cham hut, nhiftig khong the tach Fe khoi ong thay c6 ket tua; lay ong vao ong tha'y c6 tho^t ra, lay ong FeS VI day la hdp chaft, nam cham khong the pha v9 hen ket giiJa Fe va S vao ong thay cd ket tua Hoi ong n^o difng dung dich gi? TriTdc het hoa tan diTdng kinh vao nu'dc, loc (gan) bo phan cat khong tan, sau d6 c6 can can than ta thu dufdc dtfdng kinh Co the lay diTcJng tif dung dich dirdng riTdu bang each chifng ca't de lay rifcJu, diTcJng khong bay hdi do ta thu diTdc nhffng tinh the dirdng tr^ng Bai Tai vat b^ng bac de khong van giff difdc anh kim, nhi^ng neu khong bi nhiem ban H2S thi vat bang bac bi nhanh chong doi maU den Khi v6 tinh lam v3 nhiet ke thuy ngan, thuy ngan kim loai luc 66 rdi va' kh^p nha nhQ-ng hat nho l i ti khong thu gom het V i thuy ngan rat dpc nen ngu'di ta dung bien phap r^c bot lu\ huynh vao nhffng cho c6 thuy ng'"' rdi vai Tai sao? Bai giai 1- C6 the dilng NaOH de phan biet FeCh thknh Fe(0H)2 mau tr^ng dnh luc v^ Fe(0H)3 mau nau C6 the dilng nurdc brom (mau nau do), vi FeCh lam ma't mau niTdc brom, c6n PeCl3 khong phan iJng: F e C l + 3Br2 > 4FeCl3 + 2FeBr3 Cd the dfing dong kim loai vi luc dd FeCh khSng phan tfng vdi Cu, nhiftig f'eClj hoa tan di/dc Cu CuCb mau xanh 2FeCl3 + Cu > CUCI2 + 2FeCl2 • "^rong dung dich: Na2C03, BaCb, HCl, H2SO4, NaCl ta nhan thay chi cd ^aClz tao ket tua vdi Na2C03 va H2SO4: BaCl2 + Na2C03 m FeCU (rat loang) v i Itic d6 tao • BaC03 >l + 2NaCI iO) 111 > B a S i + 2HC1 BaCl2 + H2SO4 N h i f v3y ong phSi 1^ B a C h ; ong phii Ik NazCOs \i cho v ^ o ong cjj k h i bay ra, va ong p M i Ih H C l NazCOj + H C I > 2NaCl + H2O + CO21 > B a S i +2FeCl3 Fe2(S04)3 + 3BaCl2 ' ^- O n g m H2SO4 va ong m N a C l Bai 10 B p t dong oxit b i Ian bpt than (hon hdp A ) 2FeCl3 + 3Ag2S04 6AgCl>l +Fe2(S04)3 Fe(OH)3 + 3HC1 FeCb + 3H2O FeCl3 + N a O H -> Fe(OH)3^ + N a C l ^ Cu + CI2 -> CUCI2 FeCl2 + Cu i T r i n h b ^ y phiTdng phap vat l i de lay rieng CuO Fe + CUCI2 L a y m o t i t hon hdp A nung n6ng chan khong (khong c6 mat cua oxit) (C6 the thay Fe h\ng Z n , Mg, v.v C6 the d i e n phan t d i k h i cac phan uTng xay hoan toan G i a i thich sif bie'n d o i m ^ u ciia hon * hdp b^ng cac phiTdng trinh phan tfng N e u nung n6ng hon hdp A khong Y dp CuCb k h i t h i hien tifdng xay nhiT the nao? (C6the V i bpt CuO ra't nang, bpt than ra't nhe nen ta c6 the dilng phiTdng phap l ^ n g gan de tach lay CuO: cho hon hdp A vao coc, t h e m ni^dc vao, khuay ^ Cu + C l s t ) -> CU + 2H2S04da Bai giai CuS04 + S t +2H2O Cu + HgS04 CUSO4 + H g ) a CUSO4 + Fe ^ FeS04 + C u i ''"^ (Co the thay Fe bang Zn, Mg, v.v ) deu r o i l i n g gan, bpt than nhe se troi theo niTdc ngoai, lap d i lap l a i vai ba Ian ta c6 CuO sach K h i nung chan khong t h l xay phan tfng i - i J j B + eOiS^ CUCI2 + Ag2S04 > 2AgCl i CUSO4 + BaCl2 > B a S i + CUCI2 + CUSO4 (latO >JU X>? W^' } t N e u it than: 2CuO + C Na'u nhieu than: CuO + C > Cu + CO I C6 the dieu chc Fe b^ng each khur s i t oxit theo cic phan ufng sau: (hoac: CO2 + C > 2C0) a) Fe.Oa + CO -> Fe + m ^ u v ^ n g cua C u ; neu nhif diT C hoac dif CuO t h i h6n hdp c6 b) FczOj + H2 ^ m a u Ian den (tuy t i le Cu va CuO hoac C) c) FejOj + A l d) Fc.Oy + A l e) FC2O3 + C > 2Cu + CO2 N e u t i I9 so m o l C: C u O tuT : den : t h i mku den ciia hon hdp bien N e u nung hon hdp U-ong khong k h i thi c6 the coi than chay he't CO2 l a i C u O m a u den ( v i Cu b i oxi CuO) , ^ Chu de B d tuc p h a n tifng - D i e u ch6' u2 A Fe + %Fe$^!:^o Fe + Fe + Hoan cac PTPLf tren Theo em phan ufng nao duTdc dung de san xua't gang tiir cac quang oxit s i t Bai V i e t cdc phiTdng trinh phan uTng theo cac sd bien hoa sau: Fe2(S04)3 < > Fe(0H)3 Cu < 2- Vie't cac P T P U theo sd bien hoa > CuCh + x FeCb to 1.1a 112 CUSO4 Bai giai Fe2(S04)3 + N a O H > 2Fe(OH)3 i + 3Na2S04 2Fe(OH)3 + 3H2SO4 > Fe2(S04)3 + 6H2O A Fe203 +Y FeCl2 B •+ T Trong A, B, X, Y, Z, T la cac cha't khac 113 Bai giai Hoan cac phan uTng: a) , FejOa + 3C0 — ^ 2Fe + SCOj • ' - • • b) c) d) e) If, FeaOj + 3H2 — ^ 2Fe + 3H2O FezOa + 2A1 2Fe + AI2O3 3FexOy + 2yAl — ^ 3xFe + yAlzOj FezOa + 3C — ^ 2Fe + 3C0 phan iJng a) difcJc dung de san xua't gang Cic phiTdng trinh phan tfng: a) FezOj + 3C0 b) Fe + CUCI2 c) FezOj + 6HCI d) 2FeCl3 + Fe Bai Cho sd bien hoa +x t" — ^ 2Fe + 3CO2 (X c6 the' la H2, Al, C ) > FeCh + Cui > 2FeCl3 + 3H2O > BFeClj (T c6 the \h Cu, KI) ' 3Fe304 + 8A1 > Zn + CUCI2 > ZnCl2 + Zn > ZnCl2 + H + 2HC1 D — C — ^ 3Fe + 4H2O 9Fe + 4AI2O3 Fe + 2HC1 > FeCl2 + (B) 2FeCl2 + Cl2 (E) Fe304 + 8HC1 (D) > 2FeCl3 (C) > FeCl2 + 2FeCl3 + 4H2O Hat CuCh ^-i-lil tkm I/ Cu4' 'C ' Cic phan uTng trifc tie'p tao kim loai tfif muo'i: Zn + 2AgN03 > Zn(N03)2 + A g i 'i AgNOj Ag + N O + I / O ' • CUCI2 + Cl2t ^ ^ C u C^c phan tfng tao NaOH + 2H2O > 2NaOH + H2 T Na20 + H20 Biet ring A + HCl > D + C + H2O Tim cac chat tiTdng tfng vdi cac chuT cai A, B, va viet cac PTPl/ Bai giai Nhin sd ta thay A phai Ih oxit sat, va vi A + HCl tao hai loai muoi ncn A phai Ih Fe304: Fe304 + C — ^ 3Fe + 4CO2 J ^ Fe304 + 4H2 CU + CI2 2Na > Fe pjay vie't loai phan uTng tri/c tiep tao muoi tit kim loai va loai phan ^jig tT\ic tiep tao kim loai ttf muoi < , j^ay vie't loai phan ufng tao NaOH ^ ^ Bai giai : , •, J Cac phan tfng trifc tiep tao muoi tiT kim loai: Ca(0H)2 '^^ > 2NaOH + Na2C03 NaHC03 + Ba(0H)2 > CaC03i + 2NaOH > v^adu hoacdu BaCOs i + NaOH + H2O v.v Bai Cho mau Na vao dung dich sau: ZnCb, FeCl2, KCl, MgS04 Viet cac PTPU xay Bai giai Trirdc het Na tac dung vdi niTdc ' 2Na + 2H2O > 2NaOH + H j t ' ' ••^ ' ;v / - Sau do: 2NaOH + ZnClj > Zn(OH)2 i + 2NaCl , N a O H du- + Zn(0H)2 > Na2Zn02 + 2H2O 2NaOH + FeCl2 > Fe(0H)2 i + 2NaCl , ,, Neu de khong khi: 4Fe(OH)2i +2H2O + O2 KCl + NaOH 2NaOH + MgCl2 > 4Fe(OH)3i > khong phan urng > Mg(0H)2 i + 2NaCl 115 B a i Tic cac chat A, B, C thich hdp B > C viS't cdc PTPLf theo sd b i e n ho^ FeS04 + Cu>l Fe + C U S O (H) > D B ^ H -> F e S i + N a S FeS04 + NazS M (M) -> G B i c t M la hdp cha't cua k i m loai p h i kirn B , C, D, H 1^ cac hdp chf, f chtfa liTu huynh, E, F, G, H la hdp chat cua s^t hoSc s^t k i m loai p a ' ^' J V i ^ t l?i '^^"S ^^^^ N2H«C03, H2P20sCa, C4H,„06Ba M phai la FeS Cac PTPLf each la: > FciOi (A) > thi chinh l a i cho dung (B) 2HBr + — ^ + 6H2S04dac (F) FeCl2 Fe2(S04)., + S t + 6H2O H.T (G) FeS04 + NazS (H) FcSi +Na2S04 (M) CO 2FeO + CO2 > Ca(H2P04)2 canxi dihidrophotphat BaC4H,„06 > Ba(CH3C00)2.2H20 tinh the bari axetat ngam 4H2S04dic + 2FcO (F) FcO + H2 -> Fe2(S04)3 + S02T + H O (B) CuO CUCO3) > FeCh 2FeCl2 + CI2 > 2FeCl3 FeCl3 + N a H + Cui (2) > Fe(0H)3 i + 3NaCl (3) (CO the thay N a O H bang K O H , Ba(0H)2 v v ) Fe203 + H t Fe203 + 3H2SO4 Chu de ,i if! (4) (5) > Fe2(S04)3 + 3H2O (6) > B a S i + 2Fe(N03)3 T i n h theo phi/cftig trinh p h a n ufng ^ De m i e n g A I nSng 5,4g khong k h i mot thdi gian thu di^dc cha't r ^ n (G) -> C U S O + Fe + C u C b (1) H i ^ u suS't phan ufng - Nong dp dung dich - > Fc + H2O (hoSc CO) >FeCl2 + H2t Fe2(S04)3 + 3Ba(N03)2 (F) 116 H2P20xCa 2Fe(OH)3 Cach21a: (hoac Cu, amoni cacbonat Fe + 2HC1 Hoac A g C l i + FCSO4 H2SO4 + > (NH4)2C03 hai phan tur H2O Ag2S04 + F c C l : (C) N2HSCO3 (D) FeCb + FC2O3 + ;:i.n Viet lai cong thuTc dung va goi ten: > Ag2S04 + H2O Fc + 2HC1 Fe(N03)3 B a i giai 2Fe + 3H2O (B) + Ag20 FeO Fe2(S04)3 (C) H2SO4 Fe(0H)3 — ^ FeCh (C) (F) 2Fe Fe H2SO4 (hoac CI2, O , H2O2 ) Fe203 + 3H2 Viet cac phiTdng trinh phan iJng triTc tiep theo sd d6 b i e n hoa, neu ndi n^o sai + 2SO2 (E) SO2 + 2H2O + Br2 phan cho dvldi day v£k goi ten c'liing, ^^^^ nd'u c6ng thi?c nao sai dUdc phep chinh l a i chi so' cua m p t nguyen to': B a i giai 2FeS + 7/2O2 A ^- Hoh tan hoan toan A b^ng dung dich H C l tha'y bay 6,5856 l i t H2 (dktc) H2O Tinh kho'i lu^dng ran A va %A1 b i o x i hoa oxit - (D) 117 B a i giai Khi 3/ a < b tiJc du3/ khong Al bi oxi hoi mot phan th^nh oxit: > AI2O3 2A1+ - O (1) Cic phan ufng hoa tan: 2A1 + 6HC1 2AlCl3 + H t D t Dung dich X c6 a mol FeS04 va (b - a) mol C U S O ; chat r^n Y c6 a m o l Cu pai 3- Nhung mot mieng nhom kim loai nSng 10 gam vao 500 ml dung dich C U S O 0,4M Sau mot I h d i gian lay mieng nhom ra, ruTa sach, say kho, can nang 11,38 gam t * ! >,;: • Tinh khoi l i / d n g dong tho^t bam vao micng nhom (gia suT ta't ca dong thoat bam vao mieng nhom) Tinh nong dp mol cua cac chat sau phan vlng (gia suT the tich dung dich van 500 ml) ^ B S ( ) Cich 1: AI2O3 + 6HC1 Khoi lifdng Al con: (3) A I C I + 3H2O nAiciin= —^H-, - — X 3 Ttfc 0,196 X 27 = 5,292 g Khoi liTdng AI2O3 lao thanh: _1 CUSO4 ^'^^^^ -0,196mol 22,4 " " ' ^ ' Al p h a n iJng la 2x, l i f d n g C U S O p h a n iJng la "5y ^ •J-, • B a i giai Ta can xet cac triTdng > FeSO^ + Cui (1) hdp: CuO + CO — ^ C u + C02 C02 + Ba(OH)2 - 2- Tinh so mol cac chat: _ 2/ Trurdng hdp 1: n, = n2 hieu suat h% = s^t Dung dich X c6 b mol FeS04; cha't r^n Y c6 b mol Cu va 118 (a - b) mol Fe ' V °2 ="1 a "CuO ="2 (1) ^BaCOji+HjO 1/ a = b ttfc cac cha't tdc dung vdi vijfa du Dung dich X chi c6 a mol FeS04; chat r^n Y chi c6 b mol Cu a > b ttfc du- ' ^- Cac phu'dng trinh phan uTng: B a i giai Fe + C U S O 3x Dya treh nguyen tic l a Al phan tfng thi kho'i liTdng mieng Al bi giam, Cu tao bam vao mie'ng Al nen khoi l i f d n g tang len, ta c6 phu'dng trinh: - 2x X 27 + 3x X = l l , g a m Giai ta di/dc: x = 0,01 mol Khoi l i / d n g Cu thoat bkng 3x x 64 = x 0,01 x 64 = l,92g Tinh nong mol n 0,01 ^ 0,5x0,4-3x0,01 ^ ^ ' CA12(SO4)3 = = " ' ^ ^ ' ^ '^CuS04 = 0,34M Bai Cho V lit CO (dktc) di qua ong stf diTng a gam CuO nung nong Sau ket thiic thi nghiem cho di khoi oag si? hap thu vao dung dich Ba(0H)2 tha'y tao m gam ket tua 1- Viet cac PTPl/xayra Tinh hieu suat ciia phan i?ng khur CuO theo V, a, m tfng: gnrdv: Phan tfng: 2A1 + 3CuS04 > Al2(S04)3 + 3Cu>l' "f'^^ "f ' Goi X la so mol Al2(S04)3 tao t h a n h , nhu* vay l i f d n g Cu tao t h a n h la 3x, luTdng 5,4-5,292 , h„^ = • X - = 0,002mol "AI2O3 27 Turc 0,002 X 102 = 0,204g Vay khoi liTdng A = 5,292 + 0,204 = 5,496g 5,4 Cach 2: 54 Ne'u khong bi oxi hoa thi 5,4 gam Al tao ra: —-—x — x22,4 = 6,72 l i t H Theo (1, 2) so mol nguyen tuT oxi phan iJng bkng so mol H2 • = 0,006mol 6,72-6,5856 22,4 Vay khoi liTcJng oxi da tham gia phan iJng b^ng: 0,006 x 16 = 0,096 gam Do 66 khoi liTdng A = 5,4 + 0,096 = 5,496g Bai Cho a mol bot Fe vao dung dich chtfa b mol C U S O Sau ket iho"^ phan iJng ta thu diTdc dung dich X va chat r^n Y Hoi X, Y c6 nhiWr chat gi? Bao nhieu p i o l ? Phan , B a i giai (2) m i B a C O , = "3 = ^^^^% hoac n2 f ' f M ^^^^% n, 119 .^ n, xlOO„ Trirdng hcJp 2: ni > 02 hi?u suat h% = — % n = 3r=>M = 97,5 loai E)6'i chieu vdi de bai, kim loai R 1^ Zn (kem) pai Cho 27,4 gam bari vao 400 gam dung dich Trirdng hdp 3: n, < nz hieu suat h% = 3,2% thu diTdc A, Icd't tija B va dung dich C "i Bai Cho 13,44 gam bot d6ng kim loai vao coc difng 500 ml dung dich AgNOs 0,3M, khuay deu dung dich mot thcJi gian sau d6 dem Ipc ta thu diroe 22,56 gam chat r^n A va dung dich B Tinh nong mol cua cha't tan dung dich B Gia suf the tich ciia dung J Tinh the tich A (dktc) Nung ke't tua B d nhiet cao den khoi liTdng khong doi thi thu diTdc bao nhieu gam cha't ran? Tinh nong % cua cha't tan dung dich C dich khong thay ddi Bai giai Nhiing mot kim loai R nSng 15 gam vao dung dich B, khuay deu de Cac phan iJng xay ra: c^c phan iJng xay hoan toan, sau lay kim loai R khoi dung Ba + 2H2O djch, can nang 17,205 gam (gia suT ta't ca kim loai thoat deu bam vko Ba(0H)2 + R) Hoi R la kim loai gi so cac kim loai cho sau: Na = 23, M g = 24, A l = 27, Fe = 56, N i = 59, Cu = 64, Zn = 65, Ag = 108, Pb = 207 Bai giai CUSO4 BaS04 ) Phdn tfng giffa Cu va AgNOj Cu + i A g N O j CUSO4 -> Ba(OH)2 + H t (1) -» BaS04i +Cu(OH)2 (2) ^ BaS04 Cu(OH)2 CuO + H O t (3) m 400x3,2 „ „ o , nga = = 0,2mol; np,, Cu(N03)2 + 2Ag ^ n 44 Tinh: ncu = —— = 0,21mol; 64 (1) n^gNOj = 0'5 x 0.3 = 0,15 mol Theo phan ufng(l): n^^ =nB,(0H)2 ""BU =0,2mol Gpi X la so mol Cu tham gia phan uTng (1) (nhd r^ng Cu chiTa phan tfng het) The tich H2: VA = 0,2 x 22,4 = 4,48 lit ta CO phufdng trinh ve khoi IiTdng cha't r^n A: Theo cac phan uTng (2, 3) chat ran gom BaS04 va CuO, vi Ba(0H)2 diT nen (0,21 - X ) X 64 + 2x X 108 = 22,56 Rut X = 0,06 mol i - 0,15-2x0,06 5^ = 0,06M; lifdng cac cha't ban dau truf lufdng H2 bay va li/dng ke't tiia 0,06 CCU(NO3)2 = " ^ = ^'^2M (0,2 - 0,08) X 171x100 Ta c6: Goi n va M la hoa tri va KLNT cua kim loai R ta c6 cac phan iJng: R + nAgNOj > R(N03)n + nAg i (2) 2R + nCu(N03)2 > 2R(N03)„ + nCu i (3) Kho'i lufdng R tang = 17,205 - 15,00 = 2,205 g Theo cac phan utag (2, 3) ta c6 phifdng trinh ve suf thay doi khoi liTdng R: M 2M (108 - — ) X 0,03 + (64 n ) X 0,06 = 2,205g n Riit M = 32,5n Khi C%B,(OH)2 = 5,12% (400 + , ) - , x - , x 233-0,08 x98 A la mot oxit kim loai chiJa 70% kim loai, can diing bao nhieu ml dung dich H2SO4 24,5% (d = 1,2 g/ml) de hoa tan vira du 8g oxit A ^ Bai giai Tim kim loai R Goi cong iMc cua oxit la RxOy, ta c6 ti le % khoi Irfdng: xR 70 16y 30 R= 112y 56 2y 3x X 56 -—n Trong n la hoa tri cua kim loai Khi n = => R = — loai; Khi n = : „ 56x2 , R = loai n = =^ M = 32,5 loai n = = > M = 65 d o l a Z n 120 i,v Trong dung dich C chi Ba(0H)2, khoi liTdng dung djch C b^ng tdng kho'i Vay nong mol cua: CAEN03 = ' nBaS04 =ncu(OH)2 = n c u O =0,08mol Kho'i lircfng cha't r^n = 0,08 x 233 + 0,08 x 80 = 25,04 gam 121 Bai Trong lo kin chiJa CO, bdm mot it NO2 (khi mau nau) vao binh E)fJ mot thdi gian thay mau nau lo bi mat di, nhutig sau md lo tiep x\i^ vdi khong lai thay xuat hien mau nau d mieng Ip Giai thich hien tiro^g bling cac PTPLf B a i giai J, Vi mau nau bi mat chuTng t6 r^ng NO2 da phan tfng vdi CO de tao NO, sau tiep xiic vdi khong NO bi tac dung vdi O2 NO2 man nau Cac phan i^ng: ,, v NO2 + CO > NO + C O Chat oxi hod chatkhuf 2N0 + O2 > 2NO2 Chat khuT chat oxi hoa B a i Hoan chinh cac PTPLf xay ra: ZnS + HCl > A + KCIO3 , ' > B ,1 + Mn02 ? NazSOj + HCl > C + Cho cac A, B, C tic dung vdi tilTng doi mot, viet cac PTPLf va ghi ro dieu kien Cho A va C tac dung vdi dung dich niTdc brom, viet cdc PTPU xay ra, bie't rkng ca deu cho cilng san pham , ,^ B a i giai t Cic phan ufng: ZnS ; + 2HCi > ZnCh + H2S t 2KCIO3 > 2KC1 + t (A) (B) Mn02 Na2S03 + 2HC1 Cac phan urng giffa A, B, C: a) ' 2H2S + 3O2 2H2S + b) > 2NaCl + H2O + — ^ SO21 2SO21 + 2H2O 02ihi(?u,• 2S + 2H2O ) 3S + 2H2O c) 2SO2 + O2 Cac phan iJng cua SO2, H2S vdi niTdc brom: (V2O5) SO2 + 2H2O + Br2 H2S 174 + 4H2O + 4Br2 ' > 2HBr + > 8HBr + H2SO4 H2SO4 (C) Chi Si N h a n bict - Tach hSn h CaCOa i + H O Lpc lay ket tua CaCOa dem nung len ta difdc CO2: iii.^; juu : CO2 * CaCOj — ^ CaO + C t Muon chuyen taft ca A CO2 thi oxi hoa het CO CO2 bang cac exit kim loai, chang han: CuO + CO ——> Cu + CO2 % Bai Co hon hdp CO va CO2 Lam the nao de lay rieng tiifng CO, CO2 nguyen cha't? Lam the nao de chuyen ca hon hcfp CO nguyen chat? Lam the nao de chuyen ca hon hdp CO2 nguyen chaft? '>^|j r ^ B a i giai Sue hon hdp qua dung dich kiem, vi du Ca(0H)2 diT CO2 + Ca(0H)2 > C a C O j i + H2O Khi lai la CO (de loai tap chat hdi niTdc cd the cho qua P O , CaO, ) Lay ket tua nung d nhiet cao: CaCOs — ^ C a O + COzt (c6 the cho tac dung vdi dung dich H S O CaCOj + H S O > CaS04 i + H O + CO2 T) 2' Cho hon hdp di qua cacbon diT nung do: ' # CO2 + C —^2C0 ^- Cho hon hdp qua ong difng CuO duf nung nong: CuO + CO '" ) Cu + CO2 ^ I ' ^- Cac di/di day bi Ian tap chat la hdi niTdc: HCl, CO2, NH3, H2, N2 Lam the lao de cd cac kho hoan toan ip!!;; ^- Trinh bay phu^dng phap hoa hoc loai H2S khoi HCl, SO2 khoi CO2, HCl khoi CO2, H2 khdi N2, O2 khoi N2 175 Bai giai I r ^ ^ ' y kS't tua hoa tan dung dich H C l V e nguyen t i c c6 the dilng cac chat hut a m tot nhiT CaO m d i nung, dac, CaClz khan, CUSO4 HJSQ khan, K O H r i n , P O , v.v ^ T u y nhien phai chpn chat hut am nao hoan toan khong t^c dung v d i k h i l ^ m kho Trong qui trinh hiit ^ m c6 the xay phan iJng hoa hoc nhiT: CaO + H2O > Ca(0H)2 P2O5 + H2O > H P O (CO the Viet H P O ) C U S O + 5H2O > CUSO4.5H2O H2SO4 + nH20 > H2S04.nH20, v v H2SO4 dac, ' ! ^ ,j P2O5 doi v d i C O CO the dung H2SO4 dac, P2O5 doi v d i NH3 CO the dung K O H rin, CaO doi v d i H2 CO the dilng K O H r ^ n , CaO, P2O5 C h u y: khong diing H2SO4 H2SO4 dac de l a m kho H2 v i H2 c6 the khuf mot phan d o i v d i N2 c6 the dung tat ca cac chat >CuSi+2HCl D u n g nirdc BTJ de tach loai SO2: > H B r + H2SO4 SO2 + 2H2O + Br2 > AgCU NaHCOj + HCl > N a C l + H2O + C O T +HNO3 D i i n g CuO (hoac Fe203 ) de loai H2: C u O + H2 — ^ Cu + H2O Sau tach l o a i niTdc nhiT tren > BaCh + H2O + CO2 T NH4CI NH31 + HCl T (NH4)2C03 — ^ 2NH31 + H20 + CO2 T _, , ~' ' m Cho hon hdp k h i qua dung dich Ba(0H)2 diT, k h i l a i la CH4, Ba(OH)2 + H2S > BaS + 2H2O CO2 + Ba(OH)2 > BaCOa i + H2O (•::: s ) ^ - vb; ( Lpc ket tua, nhiet phan diftfc CO2: ' *' — ^ B a O + C02T LSy dung dich (niTdc lpc) cho tac dung v d i dung dich H2SO4 ta c6 H2S H2SO4 + BaS > BaS04i + H2St Bai T r i n h bay phiTdng phap hoa hoc ngan gpn nhaft de nhan biet cac chat k h i mot hon hdp k h i g o m : H2, CO, CO2, SO2, SO3 v^ N2 Bai giai Cho hon hdp k h i d i qua dung dich BaCl2 duT c6 ke't tua t r ^ n g tao thanh: SO3 + H2O + BaCl2 > BaS04 i + 2HC1 Nhan biet diTdc SO3 H6n hdp k h i l a i cho qua ntfdc v o i trong, diTcd ket tua tr^ng g o m CaC03 va CaSOj > CaC03 i + H2O T r i n h bay phiTdng phap tach lay rieng tiJfng m u o i v d i iMng chat khong doi ttf S + Ca(OH)2 >CaS03i+H20 Lay ket tua dem hoa tan dung djch H C l diT dun nhe thu diTdc hon hdp T r i n h bay phiTdng phap lay rieng tuTng k h i tilf hon hdp k h i C O , H2S, CH4 v SO2 CUSO4 H2SO4 Vay the tich O2 can = ^ x 5,6 = 2,8 lit '' ' Bai Trong mot bmh k i n chufa SO2 va O2 theo t i 1$ s6' m o l : v^ mot i t bpt xuc tac K2SO3 V2O5 Nung nong binh mot thdi gian thu diTdc hon hdp k h i 66 san pham c h i e m 35,3% the tich T i n h hieu suat phan tfng tao SO3 B a i giai + O2 S -> B a i giai SOJT Phan tfng tao S O : 2SO2 + O2 j y^^^ ) 2SO3 ? ' " , ' SO2 + 2H2S 3S i + 2H2O SO2 + 2K0H K2SO3 + H2O gia phan tfng, h cung chinh la hieu sua't phan tfng Sau phan tfng ta c6 (1 - h) K2SO3 +2HC1 2KC1 + H + S t mol SO2; (1 - ^ ) SO2 + 2H2O + Br2 Cu CuO +2H2S04d,c + H2SO4 C U S O + H2S -> H B r + H2SO4 CuS04 + S t + H O ^ C U S O + H2O -> CuS i + H2SO4 Gia sur trirdc phan tfng c6 m o l SO2 v^ m o l O2 G o i h la so m o l SO2 tham m o l O2 va h m o l SO3 Theo dieu k i e n cho ta c6 bieu thtfc ve%S03 %S03= - i i ^ = (i_h)+(i-^)+h 2-^ • " Giai ta c6 h = 0,60 tiirc hieu suat phan tJng la % Cho mot dong k h i CO qua ong difng 20 g a m CuO nung n6ng v a cho k h i k h o i ong stf h a p thu hoan toan v a o nirdc v o i diT thay tao 16 gam k c t l u a 180 181 Tinh % CuOda bi khOr Neu hoa tan cha't rin lai ong stf b&ng dung dich bao nhi6u lit mau nau (duy nhat) bay (tinh theo dktc) Cic phan tfng khuT CuO , , J, tdc dung vdi nxidc voi t r o n g : C + Ca(OH)2 " ' "^'^' ncuobjkh* = n c u = (2) n^Q^ = nN02 100 0.05x500 250 0,02x500 ncu = x 0,16 = 0,32 mol thtfc tinh d theo a, b, c .> %)S03 = Khi hoa tan: 80n • x l 0 = : R u t r a n = 98 + 80n H2SO4.3SO3 + H O > 4H2SO4 4x 338g 8,6 gam ke't tua diT dung djch NaOH (dun nhe) thay bay r a 1,12 lit (dktc) ' Cach 1: Goi cong thuTc cua X la H2S04.nS03 Lay 200 ml dung dich A cho tac dung vdi lu'dng diT dung dich B a C l thay tao liTdng '' Bai giai / Lay 250 ml dung dich A cho tac dung vdi :,-.>>; X vao b gam dung dich H2SO4 c% du'dc dung dich Y nong d% Lap bieu Ta c6: lit (dktc) tCf X :?|? Bai giai *' Khi hieu chung muoi cacbonat va sunfat la M C O va M2SO4, M 98g bxc Tong khoi lufdng H2SO4 = — x x 98 + • ^ ' 338 \ Tinh tong khoi Itfdng muoi c6 500 ml dung dich\^ Too m clung d i c h Kho'i luTdng dung dich Y = a + b •If,;;* i ' 2MC1 + H O + C O (1) M2C03 + BaCl2 )-'BaC03i +2MC1 (2) ^ai Nung a gam bot nhom vdi b gam bot lu\ huynh mot thdi gian du'dc hon M2SO4 + B a C l > BaS04i (3) hdp X Hoa tan X b^ng dung dich HCl diT thay lai 0,04 gam chat r^n va + 2MC1 K i hieu (NH4)2Y la cic muoi amoni, Y la goc = C O hoSc = SO4 PhiTdng trinh phan iJng: 0/ > , (4) ^ Lay 100 ml dung dich A cho tac dung vdi dung dich HCLdu^ thay bay 0,224 MA ' 200 Bai Co mot loai oleum X SO3 chiem 71% khoi lufcfng Hoa tan a gam amoni(NH;) - ^ , = 0,05mol (3) Bai Dung djch A chtfa hon hdp cac muoi cacbonat va sunfat cua natri v,i - sot = = 0,lmol Vay the tich NO2 (theo dktc) = 0,32 x 22,4 = 7,168 lit - , (0,1 X 23) + (0,1 X 18) + (0,05 X 60) + (0,05x 96) = 11,9 gam > Cu(N03)2 + 2NO2 T + H O =2 x co _ , •=;0,05mol; n T6'ng kho'i lu'dng muoi c6 500 ml dung dich A b^ng: > Cu(N03)2 + H2O Cu + H N O 1,12 = 0,05 mol (trong 250 ml A) 22,4 = n NH3 vay so mol Na* = 2(0,05 + 0,05) - 0,1 = 0,1 mol Cac phan uTng hoa t a n chat r ^ n : ^ Theo phan u-ng ( ) : 0,01x500 2- ncaC03 " ^ ^ " ^ ' ^ ^ " l o l Vay % CuO da hi k h * = ^ ^ ^ x 0 ^ ^^^^ 20 CuO + H N O ^'^l^'^^^^^ = 0.02 mol (trong 200 ml A) 233 vay 500 ml dung dich A c6: (1) ^CaCOji+HjO =^'^ dac thi Theo phan iTng (4): n >Cu + C02 CuO + C O , _ , Theo cac phan iJng (1,2): leo phan urng (2, 3): n HNO3 (NH4)2Y + N a O H —!—> NaiY + 2H2O + 2NH31 0,224 Theo phan iJng (1): ncoj = n^^2- =-^:^= ^3" 22,4 a+b a+b CO 1,334 lit thoat (dktc) Cho di tCf tir qua dung dich Pd(N03)2 dif thay tao 7,17 gam ket tua den PbS Tinh cac gia tri a, b Bai giai (4) Cac phan iJng xay ra: ' 0,01 mol 2A1 + 3S — ^ AI2S3 ''''' (1) 183 AlzSj + 6HC1 S + HCl A I C I + 3H2S t (2) V $ y % k h o i liTdng cua Fe203 bkng Q'^^^^O^^OO _ 66,67% khong phan tfng - (0,04 gam S) 2A1 + 6HC1 > 2AICI3 + H2S + Pb(N03)2 3H21 Va cua CuO b i n g 100 - 66,67 = 33,33% Thanh phan % ciia m o i k i m loai chat r ^ n : (3) > PbS>l + H N O (4) Theo cAc phan tfng ( , 2, 3, 4) ta c6 scf do: 3S > AI2S3 > 3H2S > 3PbS Nhif vay so' m o l S da phan ifng: ns = npbs = 7,17 239 i I^eu thay C O b i n g H2 ta c6 cac phan iJng khiif: Fe203 + 3H2 = 0,03 m o l > 2Fe + 3H2O (3) > C u + H20 CUO + H2 K h o i lifdng S trirdc k h i nung = 0,04 + 0,03 x 32 = gam (4) So sanh cac phan urng 1, 2, 3, ta nhan thS'y so m o l H2 bang so' m o l CO, » ' ' 0,1x2x56x100 ^^^.^ %Fe = • = 63,64% ; , x x + 0,1x64 % C u = 100 - 63,64 = 36,63% d6 the tich H2 cung la 8,96 l i t So' m o l A l da phan tfng = — ns p h a n j n g = — x 0,03 = 0,02 m o l 3 Theo phan vCng (3), so m o l A l l a i Biki Cho hdi nU'dc qua than nong Gia suT luc 66 chi xay phan tfng: 2 '1,344 -0,03 = T " H = T (long so m o l k h i - nH2s) = r 22.4 = 0,02mol >-CO + H2 (1) C O + H2O > CO2 + H2 (2) Sau k h i phan iJug xong, l a m lanh hon hdp k h i de loai het niTdc va thu diTdc V a y k h o i liTdng A l tri/dc k h i nung = (0,02 + 0,02) x 27 = 1,08 gam (C6 the tinh tong so gam A l nhiTsau: theo c^c phdn iJng (2, 3), C + H2O hon hdp k h i kho A Cho 5,6 lit hon hdp k h i A d i qua nifdc v o i d\i tha'y l a i 4,48 lit hon hdp k h i B T i n h % the tich cua m o i k h i hon hdp A 2 "Al = 3nH2S + 3nH2 = ^ ( n H z S + n H ) = ' m o l TCf hon hdp k h i B m u o n c6 hon hdp k h i D v d i t i I the tich V H : V C Q =2:1 thi phai t h e m bao nhieu lit CO hoSc H2 vao hon hdp B Cho c^c the tich Va mAi = 0,04 x 27 = 1.08 gam d dktc X a c dinh p h i n hSn h „ , / , J Hoac 1881,45 = 1149,21 + 122,04x Rutra x = ^- Khi hoa tan cacnalit v l o nu'dc ta dUdc dung dich cic muoi KCl va MgCl2 ^Cac phan iJng them NaOH vao dung dich va nung ket tua: MgCl2 + 2NaOH Mg(0H)2 > Mg(0H)2 ^ + 2NaCl -> MgO + H t (1) , ^,., ; (2) 189 Theo cong thuTc cacnalit va cAc phan uTng (1,2): "cacnalit - nMgCl2 " "Mg(OH)2 - "MgO - , = 0,lmol Khoi Itfdng chat ran MgO b^ng 0,1 x 40 = gam B a i A la mot oxit cua luti huynh chuTa 50% oxi gam A chiem 0,3613 i;, (d dktc) Tim cong thiJc cua oxit A Baigiai Gpi cong thuTc cua cua A la S^Oy Theo dieu ki^n cho ta c6 he phiTdng irlnh: 32x = 16y tiJc y = 2x ' Khoi liTdng phan tuT cua A = 22,4 = 62 = 32x + 16y 0,3613 Giai he phu'dng trinh ta c6 x = va y = Cong thiJc cua oxit la SO2 B a i Hoa tan 12,8 gam oxit SO2 vao 300 ml dung dich NaOH 1,2M Hoi thu difdc muo'i gi? Bao nhieu gam? B a i giai Cac phan uTng c6 the c6 cho SO2 tac dung vdi dung dich NaOH: S02 + 2NaOH > NazSOs + H O (1) ) SO2 + H2O + NajSOj > 2NaHS03 (2) 12,8 = 0,2mol; nNaOH = 0,3 x 1,2 = 0,36 mol Tinh: nc Vi nso2 < nwaOH < x n^o^ nen thu dirpc hon hdp muoi Sau phan (Jng (1) thu diTpc ^ =0,18 mol Na2S03 va diT 0,2 - 0,18 = 0,02 mol SO2 va theo phan iJug (2) thu diTpc 0,2 x = 0,4 mol NaHS03 va nhif vay Na2S03 lai 0,18 - 0,02 = 0,16 mol G h i c h i i : Vi biet chac chin c6 hon hpp muoi nen c6 the viet cac phan ufng: SO2 + N a O H > Na2S03 + H O (3) SO2 + NaOH > NaHSOj (4) Goi X va y la so mol Na2S03 va NaHS03 ta c6 he phi/dng trinh: nso2 = X + y = 0,2 mol; UNaOH = 2x + y = 0,36 mol Giai he phiTcfng trinh c6 x = 0,16 mol va y = 0,04 mol - Khoi liJdng cac muoi: mNa2S03 = 0,16 X 126 = 20,16g; mNaHSOs = ^'^"^ ^^"^ = ^'^^S B a i Cho 4,9 gam kim loai kiem M vao niTdc Sau mot thc(i gian thay liTdng thoat da vuTdt qud 7,5 lit (d dktc) Hoi M la kim loai gi? 190 B a i giai phan tfng cua M vdi niTdc: (1) fheo phan tfng (1) so mol cua M phai vu'dt qua: x -^^^ = 0,67 mol 22,4 fjhif vay kho'i Itfdng nguyen tuT cua M phai nho hdn 4,9 = 7,3 0,67 V$y kim loai kiem la Li pai 7- Oxi hod hoan toan gam kim loai X can dilng mot liTdng vifa du 0,622 lit O2 (d dktc) Hoi X la kim loai gi? Oxit cua no c6 hoa tinh gi dSc biet M + 2H2O > M O H + H21 B a i giai phan uTng oxi hoa kim loai X hoa tri n 4X + n02 > 2X20„ (1) nx22,4 0,632 Theo phan urng (1) va ti 16 cho ta c6 ti le: 4X Rut X = 9n Nghiem thich hdp n = 3, X = 27 la Al AI2O3 CO tinh chat hoa hoc dSc biet la liTdng tinh, n6 tac dung difdc vdi axit lai tac dung ca vdi kiem Thi du: AI2O3 + 6HC1 AI2O3 + N a O H > 2AICI3 + 3H2O > 2NaA102 + H2O Bai 8, Co 29 gam dung dich Fe(N03)3 41,72% Lam lanh dung dich thay thoat 8,08 gam tinh the X Loc X, phan dung dich lai chiJa Fe(N03)3 34,7% Tim cong thiJc cua X B a i giai '^ Khi lam lanh dung dich tdi nong dp Fe(N03)3 viTdt qua dp tan d nhi$t dp d6 thi tinh the bit dau thoat diTdi dang tinh the ngam niTdc Fe(N03)3.nH20 2Qy 41 Cdch 1: So gam Fe(N03)3trong dung dich dau = ^ — ^ ^ = 12,lg 100 Khoi liTdng nirdc Ipc bang 29 - 8,08 = 20,92 g, c6 (12,1 - 242x) gam Fe(N03)3 (trong x la so mol Fe(N03)3 lai niTdc Ipc Ta CO he phiTdng trinh: 'x(242+ 18n) = 8,08 , •X100 = 34,7 20,92 (12,1-242x) _ ^ ;a, V: ;? i a " Giai c6: x = 0,02 va n = jVaiy cong thiJc cua tinh the la Fe(N03)3.9H20 Li 191 ZtH de Cach 2: So mol Fe(N03)3 luc dau = ^^'''^^''^^ = 0,05mol 100x242 So mol Fe(N03)3 lai niTcfc loc = (^9-8,08)x34,7 ^ 100x242 mol Fe(N03)3 c6 tinh the = 0,05 - 0,03 = 0,02 mol o no KLPT cua tinh the = = 404 dvC 0,02 404 — 242 Do n = Cong thtfc tinh the Fe(N03)3.9H20 18 I QQ^^^^ ^i: jl5i lit R ' nang hdn lit RH4 bao nhieu Ian (d ciing dieu kien Bai giai X 75 Baigiai Phan tfng xay ra: > R2(S04)„ + 2nC021 + 2nH20 (1) Trtfdc he't tinh khoi lU'cJng muo'i hidrocacbonat 500x6,478 ' i \1: NNguyen to' R tao hdp chat RH4 hidro chie'm 25% kho'i Itfdng va nguyen to R' tao hdp chat R'02 oxi chiem 50% kho'i lu'dng R va R' la cac nguyen to'gi? ' nhiet do, ap suat) ^ Me'u cf dktc, V, lit C H nang bang V j lit SO2 thi ti ie V,/V2 bang bao nhieu? Bai C6 500 gam dung dich muoi hidrocacbonat X nong 6,478% Them td tiJf dung dich H S O tdi v t o het thoat Sau dem CO can can than thi thy diTdc 27,06 gam muoi sunfat Tim CTPT cua X 2R(HC03)„ + nH2S04 B a n g t u a n h o a n c a c n g u y e n td' h o a h p c ffv = 32,39g, u 100 ^ • , 2(R + 61n) 2R + 96 Theophanifng (1) ta CO tl le: -^^ -= 32,9 27,06 RutraR=18n J Vi hidro chiem 25%, nen R chiem 75%, tiJc KLNT cua R bllng la cacbon (C), hdp chat la C H (metan) Vi oxi chiem 50% nen R' cung chiem 50% ttfc KLNT cua R' = 32, la liTu huynh (S), hdp cha't la SOj Vi cf C l i n g dieu 64 kien nhiet va ap suat nen khoi lu'dng ciJa lit SO2 n|ng gg'p lit C H = — = Ian 16 3, Vi khoi lu'dng cua hai bang nen ta c ti le sau m, = _^xl6 22,4 =-^x64: 22,4' V| 64 ^ — - — = Ian 16 t^'U,/'^: Bail Cho nguyen to: O, A l , Na, S Viet cong thitc phan tijT ciia cac hdp chat chiJa ho5c so' nguyen to' tren Cho n cac gia tri khac ta thay khong c6 cac kim loai phu hdp Vay n = 1, R = 18 i^ng vdi goc NH4 ' •/ * Nguyen to X c6 the tao \di A l hdp cha't kieu AlaXb, moi phan tuT gom nguyen tiJ, khoi lifdng phan tu" 150 Hoi X la nguyen to gi? ^, Vay X la muoi N H H C O (amoni hidrocacbonat) Bai giai Bai 10 Cho bie't phan cua mot loai quang apatit nhiT sau: photpho 18,459" 1- NajO, NazS, AI2S3, SO2, SO3, NaAlOs, Na2S03, Na2S04, Al2(S04)3 khong ke cac cha't Na202; SO; khong c6 chat Al2(S03)3 (khoi liTdng), oxi 38,1%, canxi 39,68% va flo 3,77% Hay bieu dien cong thuTc cua apatit duTdi dang muoi photpho = 12, muoi florua 2- Theo dieu kien cho ta c6 he phu'dng trinh: 'a + b = Cho apatit tac dung vdi H2SO4 duT, nong Viet PTPl/ "27a + X b = Bai giai D a t cong thtfc cua apatit la CaxPyO,.F, ^ , ^ 39,68 18,45 38,1 3,77 Ta co: X : y : z : t = — - — : — - — : ——: —— 40 31 16 19 (1) (2) The b = - a vao phiTdng trinh (2) ta c6: 27a + X(5 - a) = 150 rj ^ 150-27a Hay X = a 5-a ' Vay cong thiJc cua apatit la: CaF2.3Ca3(P04)2 Phan tfng cua H2SO4 vdi apatit: CaF2.3Ca3(P04)2 + IOH2SO4 > 10CaSO4>l + 6H3PO4 + H F t X 30,75 32 34,5 42 Loai Loai Ket luan Loai Vay X la lu'u huynh, hdp chat la 192 a S AI2S3 193 Chaong IV HWQOCaCSON - NHIEN U$U A L i T H U Y E T C d B A N V A N A N G C A O L Khai ni^m vl hflfp chat hi?u ccf va hoa hQC hffu cof ^y, Etilen C2H4 " Hoa hoc hi?u cd la nganh hoa hoc chuyen nghien cuTu ve cac hdp chat hi?u cd Hdp chat hiJu cd la nhifng hdp chat cua cacbon (truf CO, C O , H C O , muffj cacbonat kim loai ) Phanloai a) Hidrocacbon: CxHy ' b) Dan xuat cua hidrocacbon: C^HyO,, CHyO.N,, CxHyN, C2H5OH- xt, t° P, CH4 * Cac phan iJng: Al4C3(o + 12H20(„ CH4(k) + Cl2(k) xt H2SO4 (iac,ixo"c (phan iJng trung hdp) C2H4 + H O 2CO2 C H = C H + Br2 j>i; : +2H2O CH2Br-CH2Br" nCH2=CH2 ^^^^ (-CH2-CH2-)„ V Axetilen C2H2 CO2 (phan iJng chay) > CH = CH +Br2 (phan ufng cpng) —>• CO2 (phan iJng chay ) VI Benzen CaC2(r, + 2H20(„ C2H2(k) +Ca(0H)2(dd) 2C2H2 + 5O2 4CO2 + 2H2O CH=CH +Br2 CHBr=CHBr C H B r = C H B r + Br2 CHBr2-CHBr2 CfiHs -> 3CH4,k, + 4Al(OH)3(o C02,R, + 2H20(„ >C0 C , H ^ ^ + Br2/Fe CH3Cl(k, + HCl(k, Cac phUdng tfmh •jr^ zj J'-2n2(k) CfiHfid) 194 (phan iJng cong nifdc) (-CH2-CH2-)„ C2H4 + O +Cl2,as ^ > CH,C1 (phanu-ng the) '' t°, CH3-CH2OH * Cac phan iJng: > CH4(k) + 202(k) xt * Cac phu"dng trinh phan ufng : C2H,0H III Metan CH4 AUG, CH2 = CH2 +H2O, II Cfi'u tao phSn tv[ hgfp chS't hi?u cd Trong cac hdp cha't hffu cd hoa tri C b^ng IV, H bang I va O hkng II Cac nguyen tur lien ket vdi theo dung hoa tri cua chung Nhiyng nguyen tu" cacbon phan tu" hdp chat hifu cd c6 the lien ket trifc tiep vdi mach cacbon (thang, nhanh, vong) Moi hdp chat hffu cd c6 mot trat tif lien ket nhat dinh giiTa cac nguyen liJ phan tuf Cong thuTc cau tao la cong thiJc bieu dien day diJ lien ket giila cac nguyen tu' phan tuf Cong thu'c cau tao cho biet: - Thanh phan cua phan tuf va phan tuT khoi - Trat t\i lien ket giffa cic nguyen tur phan tuf CO2 (phan u-ng chay) ' C H B r - C H B r (phan iJng cpng Br2) +Br2 o; , 600"C, cacbon ^ ^ x j > '-6n6(l) + 7,502(k) ii' > 6C02(k) + 3H20(h) ^ - ' 195 CfiHfidi + Fct" Br2(i) -» Giai thich tai phan tur cung c6 lien ket d o i nhtfng benzen khong tac C f i H B r , , , + HBr(k) dung v d i dung dich midc brom V I I D ^ u mo va thien nhien Hi^dng dSn g i a i Dau mo la chat long sanh, mau nau den, khong tan nu'dc va nhc hdn nu-ae Benzen tac dung v d i h6i brom c6 mat bpt sit: phan uTng the' K h i thicn nhien phan chii yen la CH4, n ^ m du'di long dat CfiHs-H D a u mo va k h i thien nhien diTdc uTng dung l a m nhien l i e u va nguyen heu ch,, cong nghiep hoa hoc + Br-Br CsHj-Br C2H4 phan u-ng cong v d i brom: C2H4 + Br2 + HBr m W > CH2Br-CH2Br iM i ' ^" • V I I I Nhien lieu: V i benzen co ca'u tao v6ng dSc biet : c6 ba l i e n ket d o i xen ke ba L e n ket N h i e n heu la nhffng cha't chay difdc, k h i chay toa nhiet va phat sang ddn nen benzen khong tac dung vdti dung dich nU'dc b r o m ; nhiTng c6 phan Phan loai: iJng the v d i hdi Br2 a) N h i e n lieu ran : Than mo, than gay, than m3, than non, than bun, go Bai tap \\i g i i i b) N h i e n lieu long : X a n g , dau, ru'du Co hai binh mat nhan, di/ng hai k h i khong mau metan va axetilen D i l n g c) N h i e n l i e u k h i : K h i thien nhien, k h i dau mo, k h i 16 coc, k h i 16 cao, k h i than B B A I T A P T H E O C H U phan iJng nao de phan biet hai k h i tren K h i metan c6 Ian tap chat la C2H4 va C2H2, l a m the nao de c6 difdc k h i DE metan tinh khie't Dang 1: Nhan bidt - Tach h6n hdp - Tinh ch6 cac ch^t Dang 2: Vi^t c6ng thC/c cdu tao Ho^n phaong trinh phdn Ong - O i l u ch^ I Bai tap c6 Idfi gi^j Co binh diing ba chat k h i la CH4, C2H4, C2H2 Chi dung dung dich broni co the phan biet diTdc ba cha't k h i tren khong ? N e u each tien hanh Bai tap co iGfi g i i i Hiidng dSn giai M o i quan he giffa cac loai hidrocacbon du'dc bieu dien bkng sd : Trich cac mau thuT C H , C2H4, C2H2 v d i nhufng the tich bang nhau, cho uic dung v d i nhCfng the tich dung dich Br2 nhu" nhau, mau thijT nao l a m dung dith Metan -> A x e t i l e n Br2 phai mau nhieu la C2H2 C2H2 + 2Br2 > CzHzBr^ T r i n h bay phu'dng phap hoa hoc dc phan biet ba cha't k h i khong mau : mcian Vie't c a c phUdng trinh p h a n tfng diTdi d a n g c o n g thUc ca'u tao rut gon de the hien c a c b i e n hoa sd tren I Htfdng d i n giai etilcn va cacbonic V i e t ph^dng trinh phan i?ng minh hoa 2CH4 Hrf(tng d § n giai - Benzen Eti e n C2H4Br2 M a u thu" khong lam phai miiu dung dich Br2 la CH4 -> Pd H2 q M a u thu' nao l a m dung djch Br^ phai mau it la C2H4 C2H4 + Br2 — > 600" C I5()0"C -> C H H C H + 3H2 Pel C H = C H + H2 > CH2=CH2 Dan tifng vao nu'ctc voi c6 d\i, nu'cfc voi van due la cacbonic CO2 + Ca(OH)2 > CaC03(r) + H2O 600" C 3CHsCH - Dan hai k h i l a i Ian lu^dt vao dung dich brom, triTdng hdp mau cua nii"'"^ iet cac phUdng trinh phan tfng du'di dang cong thtfc ca'u tao rut gon de the brom mat di hoac nhat di la etilen CH2=CH2 + Br2 > CH2Br-CH2Br V i c t phu'dng trinh phiin iJng (ghi ro dieu kien) xay k h i cho benzcn i''*; hien cac bien hoa sau: I CaC03 > CaCj > C2H2 > CfiH^ Hif^ng dan giai dung v6i hdi brom, etilen tac dung vc3i brom Ha i phan i?ng c6 ciing CaCOj hay khac loai? 196 > CaO Li > CaO + CO2 ' ' 197 CaO + C CaCz + H O 3CH = CH C2H2 b) CO ->CaC2+ + Ca(OH)2 II B^l tip tir giil Dira vao nhffng quy luat v e cafu tao phan tuT cha't ht?u cd, hay viet cong ca'u tao CO the c6 cua cac chat c6 cung cong thuTc phan tuf: a) C5H,2; b) C2H6O ; ^ ^ c) C H C I ; d) CjHsO thijj, Khoi liTdng ket tua thu dtfdc: m = 197a = 197 0,05 = 9,85 g Cdch 2: CO2 + Ba(0H)2 > BaCOaw + H O 0,15 mol 0,15 mol 0,15 mol C O lai 0,25 - 0,15 = 0,1 mol CO2 + BaCOs + H2O >Ba(HC03)2 0,1 mol 0,1 mol tan I B^i tap c6 icrl giSi: nQ^= 4,48 : 22,4 = 0,2 moi a) nh5„h CO2 + X 2O2 > CO2 + 2y = x + y = 0,25 n / o 2H2O y no => 0-2; y = o,o5 no2 =0,5x + 2y = 0,2j Thanh % C =phan phan tram so moi ciing la phan phan tram the tich hon hdp' = 80% ; % C H = 100 - 80 = 20% 0,2.100 0,25 mco = 0,2 28 = 5,6 g ; mcH4 = 0,05 16 = 0,8 g Phan tram khoi lu'dng cac cha't: %CO = 5.6.100 = 87,5%; %CH4 = 0 - 87,5 = 12,5% 5,6 + 0,8 198 mol; nBa(OH)2 = 0,2.0,75 = 0,15 moi TCf day c6 cac each giai nhu' sau : Cdch 1: C O + Ba(0H)2 > BaCOsfo + H O • ' ' a mol a mol a mol ^'^j':' 2CO2 + Ba(0H)2 > Ba(HC03)2 ' 2b mol bmol tan = a + 2b = 0,25 b = 0,l; a = 0,05 nBa(0H)2 =a + b = 0,15j Dang 3: Tinh theo c6ng thQc phuong trinh phdn C/ng, hi0u su^t ph^n Ong, n6ng dQ dung djch Dot chay hoan to^n 5,6 h't hon hdp hai CO va CH, can diing 4,48 lit oxi a) Tinh phan phan tram the tich va phan tr^m khoi Itfdng moi hon hdp Cac the tich d dktc b) Neu hap thu toan bo CO2 (sau phan tfng dot chay) vao binh chiJa 200 ml dung dich B a ( H ) 0,75M thay xua't hien m gam ket tua tr^ng Tinh m Hvidng, d§n giai Ne'u goi so moi moi hon hdp la x va y Tir phu'dng trinh phan u'ng do't chay se lap dtfdc he phifdng trinh lien quan den n^h va n o da dung Tic suy x va y Do'i vdi chat thi ti le so' moi cung la ti le the tich nco2»inhra = X + y = 0,25 Kho'i liTdng ket tua BaCOj thu diTdc: 197(0,15 - 0,1) = 9,85 g Cdch 3: nco2 : nBa(0H)2 = 0.25 : 0,15 = 25 : 15 ' '' ' = •• Phan u'ng CO2 vdi Ba(0H)2 tao hon hdp muo'i trung hoa va muoi axit 25CO2 + 15Ba(OH)2 > 5BaC03(„ + 10Ba(HCO3)2 + H O 25 mol 15 mol mol 0,25 mol 0,15 mol 0,05 mol ^ , , Khoi liTdng ket tiia cua BaC03 thu durdc: 197 0,05 = 9,85 (g) dktc 3,36 lit hon hdp gom metan va etilen c6 kho'i li^dng gam a) Tinh phan phan tram hon hdp theo the tich va theo khoi krdng b) Dan 1,68 lit hon hdp tren qua binh chura dung dich brom, nhan thay dung djch bj nhat m^u va khoi lifdng tang them m gam Tinh m, bie't rang phan tfng xay hoin toan -., Hrfdng d§n giai ^ ^j- Tirthe tich hon hdp suy -^jvs^i.* + ,|i Neu goi X va y la so mol moi h6n hdp x + y = Uhh Tijr kho'i liTdng mol cua moi va khoi liTdng hon hdp ta lap di/dc phiTdng trinh thu" McH4-X + Mc2H4-y = ^ff:5:.; , Giai cac phu'dng trinh tren se tim ke't qua " = 3,36 : 22,4 = 0,15 mol 199 I Uhh mhh ^) Uhh = I C2H4 ) = , ; y = 0,05 => X 0,3 16x + 28y = 3j ' = 66,7% ; = %C2H4= 3.0,3 + > 2CO2 mol + 2H2O 2.0,3 mol 5/2O2 > 0,25 mol 2CO2 ^ + H2O 0,2 mol , , M = 3.0,3 + 0,25 = 1,15 mol =^ VQ^ = 22,4.1,15 = 25,76 lit %Vc,H4 = 33,3% " ' ^ " ' ^ ' ^ ^ ^ = 53,3%; mol 0,1 mol nco2 = 2.0,3 + 0,2 = 0,8 mol => Vco^ = 22,4 0,8 = 17,92 lit Phan tram kho'i liTdng cdc chaft hon hdp la : %CH, 3O2 C2H2 Phan tram so mol cung la phan tram the tich cac chat hon hOp : %VcH4 = + *; ' 4, A la hon hdp hai metan va axetilen Tron 10 lil hidro vdi 10 lit hon hdp A 100- 53,3 = 46,7% roi cho tat ca di qua o'ng chiJa Ni nung nong Sau thi nghiem the tich b) Tim so' mol moi 1,68 lit hon hdp suy khoi liTdng C2H4 cung chinh lai 12 lil hon hdp B, cac the tich deu d cung dieu kien vc nhict va la kho'i lu'dng binh brom tang ap sua't Trong 3,36 lit hh c6 0,05 mol C2H4 Xac dinh the tich moi hon hdp A va B bie't rang axetilen ciing c6 1,68 lit hh CO 1,68.0,05 => a = a mol C2H4 phan ufng cong vdi hidro (c6 xl) nhUphan iJng cong vdi brom Phan iJng cong , = 0,025 mol xay hoan loan Htf(tng din giai 3,36 mc2H4= 0,025.28 =0,7 g C2H4 + Br2 Vi CH=CH phan i^ng cong vdi H2 theo hai each, theo lu'dng hidro nhieu hay it Sir giam the tich sau phan iVng chinh la the tich H2 lham gia phan i^ng cong, tif suy the tich C2H2 hon hdp — C H B r Kho'i lu'dng binh brom tang them m = 0,7 g CH^CH Dan 8,96 lit hon hdp etilen va axetilen vao binh dUng dung dich niTcIc brom du"; phan iJng xong, nhan thay kho'i lu'dng binh diTng dung djch brom IV tang them 11 gam b) Neu dot 8,96 lit hon hdp Iren thi can bao nhieu lit oxi vao tao bao nhieu lit CO2 — ^ IV iV CH2=CH2 (1) IV 2V CH3-CH3 (2) IV ^C2H2phanifng ~ ^sanphaniC2H4hoicC2H(s Cac the tich d dktc ' • Hifdng dSn giai = 8,96 : 22,4 = 0,4 = x + y C2H4 + Br2 > C2H4Br2 C2H2 + Br2 > C2H2Br2 C2H2 + 2Br2 -—> C2U2BT4 ' (1) Hon - hdp Hon hdp Khoi li/dng binh brom tang chinh la khoi lu'dng cua 8,96 lit (hay 0,4 mol) hon hdp 28x + 26y = 11 The tich hidro lham gia phan uTng = + - = lil '' Vay H2 thiTa - = lit Do do, chi c6 phan ufng (2) xay a) Goi X, y la so mol C2H4 va C2H2 hon hdp hoac H2 CH^CH + 2H2 — ^ a) Xac dinh phan phan tram the tich moi hon hdp Uhh + CH=CH + H CH3-CH3 4iit 4iit 8iit A c6 lit C2H2 va 10 - ' r = lit C H , >, B CO l i l C H , l i l C2H6 va lit H2 dir A la hon hdp hai metan va axetilen Dot chay 4,48 lil hon hdp A thu dirdc 7,68 lil C O (2) ^) Tinh thiinh phan phan tram the tich moi A, biet the tich cac Giai he phi/dng trinh (1), (2) ta dUdc x = 0,3 ; y = 0,1 Phan tram so' mol (ciing la phan tram the tich) cua hon hdp la : cf dktc ^) Neu dan lirdng CO vao binh dUng 200 ml dung dich Ba(OH)2 I M thi thu ^'^C2U, 200 = = 75% ; %Vc2H2 = 100 - 75 = 25% du'dc bao nhieu gam ket tiia Bie'l C O bi hap thu hoan toAn 201 E)p tang kho'i lu'dng cua dung dich brom chinh la HuTdng d i n giai a) CH4 +2O2 X > CO2 +2H2O (lit) C2H2 X + 5/2 O2 > 2CO2 (4,48-X) Tiy MA, V A (lit) a)^^ OA x = 1,28 (lit) = + 0,2 mol CO2 0,2 mol + BaCOj 0,143 mol = 15-6 g ^ T ; 0,7 = 3,36 : 22,4 = 0,15 m o l ; + Br2 > BaCOjcr, + H O ncH4 - X m o l v a nH2 = y m o l ta c6 : = = 0,15 - 0,03 - mc2H4 = 2,34 - 0,84 = 1,5 g = 0,09 mol va y = 0,03 mol X 16x + 2y = l,5 ' > Ba(HC03)2 tan Thanh p h a n p h a n tram (so'mol) the tich hon hdp l a : 009^00 neacoj lai = 0,2-0,143 = 0,057 mol ^^^^ , « 11,23 g Cdch 2: mol X X Ba(0H)2 y mol + > BaCOstr) + H O mol ;.,;8«rn OD 0,15 KhoiluTcJngkettuacuaBaCOsthudirdcla: 197.0,057 + CO2 = 0,12 m o l x + y = 0,12 0,143 mol Ba(0H)2 , > CH2Br-CH2Br mcH4 + H2O •• = 0,15 15,6 = 2,34 g mA nB,(OH)2 = 0,2 = 0,2 (mol) 0,2 mol + 22,4.0,4875 nc2H4 = 0,84 : 28 = 0,03 mol=> ncH4 + C c n t h i r a 0,343-0,2 = 0,143 mol CO2 ""^ " C H • mc2H4 = '50 tang kho'i liTdng dung dich brom = 0,84 g % V c H = - ^ ^ ^ ^ ^ = , % ; % V C H chiem 71,43% Cdch 1: Ba(0H)2 - va mA, tir tim n^j^^ va n^j • = V(„,) C2H4 Thanh phan phan tram the tich hon hdp A la : b) nco2 = 7,68 : 22,4 = 0,343 (mol); UA 22,4.mA + H2O 2(4,48-X) Vco2 = X + 2(4,48 - X ) = 7,68 suy ^02^^ nH2 - nc2H4 =^ % H = %C2H4 = ^"f^ = 20% b) Tim so mol moi 1,68 lit A Tim nca(0H)2 dung dich Tir x mol CO2 > Ba(HC03)2tan cac phan iJng dot chay ta tim du'dc UQQ^ Tit UQQ^ : n^^f^Q^^^^ biet du'dc phan tfng tao muoi nao ? So mol bao nhieu TiT tinh diTdc nong dung dich 2y mol sau phan tfug ^ nco2 = x + 2y = 0,343] ^ x = 0,057; y = 0,143 Khoi Itfdng ket tua BaCOj la 197 0,057 = 11,23 g A la hon hdp gom metan, etilen va hidro Cho 3,36 lit hon hdp A si^ qua dung dich brom nhan thay dung dich brom nhat mau va khoi lifcJng tanjthem 0,84 gam a) Xac djnh phan phan tram the tich cac hon hdp A, biet ran.^ 0,7 lit hon hdp c6 khoi liTdng 0,4875 gam Trong 3,36 lit A c6 0,09 mol C H , 0,03 mol C2H4 va 0,03 mol H Trong 1,68 lit A c6 0,045 mol CH4, 0,015 mol C H va 0,015 mol H CH4 + 2O2 > CO2 0,045 mol C2H4 + + 2H2O iiff: , 0,045 mol 3O2 > 0,015 mol H2 + 2CO2 >/ v; + 2H2O 5^"' -iMi' 0,03 mol 1/2021—> : H2O : , nco2= 0,045 + 0,03 = 0,075 mol; nca(0H)2 = 0,05.1 = 0,05 mol b) Dot chay hoan toan 1,68 lit A roi dan s a n pham vao lit dung di' - Ca(0H)2 0,05M (d = 1,025) Tinh nong phan tram cac chat duiL^ dich sau thi nghiem Cac the tich d dktc Hrfdng d i n g i a i DUa vao khoi liTdng cua 0,7 lit hon hdp A ta tim dufdc khoi li/dng cua Imol A (M 202 Cdch J: nco2 :nca(OH)2 = 0,075 : 0,05 = : Phan tfng tao hon hdp hai muo'i Viet phUdng trinh phan iJng theo diing ti Ic so' mol tren : 3CO2 + 0,075mol 2Ca(OH)2 > CaCOjio + Ca(HC03)2 + 0,05mol 0,05/2 mol 0,05/2 mol H2O 203 [...]... khong khi mot thcJi gian thu diTcJc chat ran /V CO kho'i lu'dng 2,802 gam Hoa tan chat ran A bang dung djch HCl di/ thay thoat ra 3,36 lit H2 , J Tinh % kho'i lu"dng ciaa Al va AI2O3 trong A 2 Tinh % Al bi oxi hoa thanh AI2O3 3 Neu hoa tan hoan toan cha't ran A bkng axit nitric dSc nong thi c6 bao nhieu lit khi mau nau duy nhat thoat ra Cho cac the tich khi do d dktc f Bai giai AI2O3 (1) MuA) 1 Cac... kim loai Cu - Al thanh hai phan bang nhau Phan thu" nha't nung nong trong khong khi tdi phan tfng hoan toan thu difdc 18,2 gam hon hcJp hai oxit Hoa tan hoan toan phan thu" hai b^ng dung dich H2SO4 dac nong thay bay ra 8,96 lit SO2 (d dktc) A; ', ^- Tinh so' mol moi kim loai trong hon hpp ^' Ne'u hoa tan hoan toan 14,93 gam kim loai X b^ng dung djch H2SO4 dac nong va thu diTdc mot Itfdng SO2 nhiT tren... loai vi lufdng Fe^Oy Idn hdn Fe203 d phan tfng (2) Bai 3 De hoa tan 3,9 gam kim loai X can diing V ml dung dich HCl va c6 LB^'' • lit H2 bay ra (d dktc) Mat khac de hoa tan 3,2 gam oxit cua kim loai Y cung can diing V ml dung dich HCl d Iren Hoi X, Y la cac kim loai gi? ' Baigiai Cac phan iJng hoa tan kim loai X hoa tri n va oxit kim loai Y hoa tri m >: X + nHCl > XC1„ + ^ H21 (1) 142 fjghiem thich... r^n b^ng: 0,1x40 + — x i x 160 = 8gam ^^TM^ 116_^ KL Fe203 Bai 7 Oxi hoa hoan toan p gam kim loai X thu dtfdc l,889p gam oxit, ho^ tan mu6'i cacbonat kim loai Y (hoa tri II) bang mot liJdng vijfa du dung dich H2SO4 9,8% thu diTdc dung dich muoi sunfat 14,18% !• Hoi X, Y la kim loai gi? ^ «,,1 • 2- A \n hdp kim loai X, Y ^) Hoa tan hoan toan 3,61 gam A bang dung dich HCI thu diTdc 2,128 lit H2 (dktc)... tua tr^ng Lay kim loai thu diTdc hoa tan hoan toin b^ng dung dich HCl thay bay ra 2,352 lit H2 (d dktc) Hay x i c djnh kim loai M v^ cong thtfc ciia oxit Bk\i MxOy + y C O > M2(S04)„ + nH20 c6ng thtfc ciia muoi 1^: Fe2(S04)3 XH2O, ta c6: (400 + 18x) X 0,014 = 7,868 Giii ra x = 9 ^ c6ng thu-c cua muoi la Fe2(S04)3 9 H 2 O ' " ' ' ^2 Hoa tan hoan toan a gam kim loai R c6 hoa tri khong ddi n v£lo b gam... tdng hoa trj cija nguyen to trong oxit cao nhat va trong hdp cha't vdi hidro b^ng V I I I (8) Trong do RO3 nguyen to R c6 hoa tri V I , nen trong hdp chat vdi hidro, no c6 hoa tri I I , cong thufc RH2 Difa vao thanh phan khoi liTdng nguyen to se tim diTdc nguyen tOr khoi ciia R va tra bang tuan hoan suy ra ten Nguyen lit khoi cua kim loai X b^ng 137 dvC, tra bang tuan ho^n ta biet do Ih bari: Ba 6 Hoa. .. HCl cho tcfi khi khi ngiTng thoat ra ihav con lai chat ran X Lay a gam chat r^n X nung trong khong khi tdi phan iJng Vay do la muoi canxi cacbonat CaC03 > 2 A I C I 3 + 3H21 > ZnClj + H21 Zn + 2HC1 tu'cR 8,585 gam ttfc R > 30 hoan loan thu dU'dc 1,36a gam oxit Hoi Al bi hoa tan hoan toan hay khong? 1 Cac phan... MgCOj Lay 100 gam X, nung mot thdi gian thay con lai 82 gam chat r^n Y De hoa tan hoan toan 10 gam Y can 173 ml dung dich HCl 2M Neu lay lOHg X nung tdi phan huy hoan toan muoi cacbonat thi con lai bao nhieu gam cha! r^nZ? B a i gial ^.f, Cac phan i^ng nung da X: - ' CaC03 — ^ C a O + C02t MgC03 — ^ M g O + C02t Cac phan tjTng hoa tan Y bkng dung dich HCl: + 2HC1 > CaCl2 + H2O (3) MgO + 2HC1 CaCOj +... cac P T P l / 3 Trinh bay phifdng phap hoa hoc de nhan biet cac k h i A , B trong hon hdp cua hoa tan la x Ta t i m difdc x = 0,062 m o l , a = 106,076 gam 0 0424 X 1 0 0 nHci = 0,0424 m o l va %AgN03 phan uTng bang — = 68,40% 0,062 chung 4 Cho biet 1 l i t hon hdp k h i A , B d dktc nang 2,1875 gam T i n h % k h o i liTdng moi chat trong hon hdp X > B a i 17 Hoa tan h e t 5,94 gam A l bang dung dich... 2 O + CO21 (Mv+96)xl00 ( M y + 6 0 ) + 1000-44 • ' - difdc mot dung dich chi chiJa mot muoi tan Hay xac dinh kim loai M va so mol muoi nitrat cua no trong dung dich 2 Hoa tan hoan toan 9,9 gam hon hpp kim loai A hoa trj n va kim loai B hoa tri m bang dung dich HNO3 loang thu diTpc dung dich X va 6,72 lit khi duy nhat NO (dktc) Tinh tong khoi lufPng muoi nitrat c6 trong dung dich X Bai giai 2A1 + 6HC1