Boi duong hoc sinh gioi hoa hoc 9 p1

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ClIC I0fi\ CtitJ ChiiOng I DICuc luc Vd Cflf A L I T H U Y E T C d B A N VA NANG CAO GAG HOP CHAT V GO Chrfrfng I C A C L O A I H P C H A T V C d A L I THUYET CCJ B A N V A NANG CAO 3 B B A I TAP THEO CHU DE 12 C B A I TAP L U Y E N T H I 24 ChtfrfnglL K I M L O A I A L I THUYET C d B A N V A N A N G CAO B B A I TAP THEO CHU DE C B A I TAP L U Y E N T H I Chtfcfng I I I PHI K I M - B A N G T U A N HOAN 83 83 86 98 155 B B A I TAP THEO CHU DE 158 C B A I TAP L U Y E N T H I 172 194 A L i THUYET CO B A N V A N A N G CAO 194 B B A I TAP THEO CHU DE 196 C B A I TAP L U Y E N T H I 210 Chrfrfng V D A N X U A T C U A HIDROCACBON - P O L I M E 244 A Lf THUYET C d B A N V A N A N G CAO 244 B B A I TAP THEO CHU DE 246 C B A I TAP L U Y E N T H I 256 ChrfoTng V I C A C B A I T O A N K H O Oxit Oxit Oxit Axit Axit Bazo Bazd Mud'i Mue'i CO oxi kheng tan I 2NaA102 + H2O AI2O3 + 6HC1 > 2AICI3 + 3H2O "77 Oxit trung tinh (hay oxit khong tao m u o i ) : L a nhUng oxit khong tac dung v d i axit, bazd, nUdc Vidu: NO, CO * Lu^i y : H2O cung la mot oxit I I I T m h chsYt hoa hoc *> T i n h chaft cua oxit bazcf: a) C n g thtfc chung la MjOn hay M,Oy DHI i ) ; MlkM hifu chung cho c^c kim loai (c6 ho^ tri n) : Li K Ba Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Hg Ag Pt Au CaO CaO CaO b) * Hoa tinh MjOn + niidc (M tir Li -) Na) M(OH)„ bazcf tan (kiem) M2(C03)„ + oxit axit (M tir Li -> Na) M20„ ^^"^^ Cac phan iJng +SO2 M2(S03)„ MC1„ +HC1 + axit +H2SO4 + nHjO SddoCaO + nudc Ca(0H)2 / Ca CaC03 - +O2 HCl goooc +axit CaO CaCl2 CaS04 H2SO4 Ca(0H)2 * Cac phan iJng : D i e u c h e : 2Ca + O2 CaC03 — a) C6ng thtfc chung RxOy hoac R20n N,P ) CaCOs CaSOs (R la ki hi?u chung cho cic phi kim C, S, b) Hoa tinh cua oxit axit | " i 71 - Tac dung vdi n\idc tao axit - Tdc dung vdi bazd tao mudi va nirdc - Tac dung vdi oxit bazcJ tao muo'i * (Khisunfurd) Sdd6S02 S +02/t° — +H2O +02/t° FeSz H2SO3 900" C > CaO + C02t H2SO4 +H2SO4 dac SO2 Na2S03 +Ca(0H)2 CaSO, +K2O ^ H2S04d,c iCli^ +oxit baza •* K2S03 +BaO , +CO2 , CaCOj 4FeS2(r) +SO2 CaS03 (dung san xuat voi song) CaS04 Na2S03 +dd bazcJ Cac phtfdng trinh phan tfng : D i e u c h e : S(r, + 02(K) — dot > SO2 (K) (dilng cong nghiep) 2CaO + + H2O H2O ; c) Hoa tinh va dilu ch§' SO2 Ca(N03)2 Ca(0H)2 — ^ CaO + H , O t H2O Ca(0H)2 2HC1 CaCl2 Hoa tinh: CaO + CaO + CaO + SO2 — + H2O HNOa^ i+oxit axit ' CO2 — > Ca(N03)2 Ca(N03)2 +NaOH c) Hoa tinh va dilu chd'CaO * 2HNO3 Tinh cha't cua oxit axit M2(S04)„ 2M(0H)„ MzOn + nC02 -> M2(C03)„ MzOn + nS02 -> M2(S03)„ MzOn + 2nHCl > 2MCI„ + nH20 M20„ + nH2S04 > M2(S04)„ + nHzO MzOn + + + d6't ^ 2Fe203 (r) + 8SO2 (K) (dung cong nghiep) Na2S03w + H2S04(dd) — Na2S04(dd) + HjOd) + S02(K) Cuw + 2H2S04(dd) - CuS04(dd) + S02(K) + 2H2O0) + 1102(K) Hod tinh : La oxit axit cd ten goi anhidrit sunfurd SO2 (K) + H20(i) > H2SO3 (dd) Axit sunfurd S02(K) + 2NaOH(dd) > Na2S03 (dd) + H2O Natri sunfit S02(K) + Ca(0H)2(dd) > CaSOjtr) + H2O (,) Canxi sunfit S02(K) + K20(r) > K2SO3 (r) Kali sunfit S02(K) + BaO(r) > BaS03(r) Bari sunfit aoi amng hoc sihh gfoi ttoa HOC a, if - uao Htfu VirtFT II TINH CHAT HOA HQC CUA AXIT I Tinh chS't hoa hpc chung cue axit +Quy tim Hoa V , - D i ^ u che": +Kim loai (trU'(tc H) M u o i + H2 Cu(OH)2(r, + 2HCl(dj) AgN03,dd) + HCl(dd) CaC03(r) + 2HCl(dd) H2 (k) + CI2 (k) 2NaCl(o + H2S04dac —> —> CuCl2(dd) + 2H20(|) AgCl„ + HN03(dd) CaCl2(dd) + H ( , ) + C02(k) ^"''"^"g > 2HCl(k) —> Na2S04(r) + 2HC1• (k) III Tinh cha't hoa hpc va dieu ch§' axit sunfuric +Oxit ba/.() M u o i + H2O + G i a y quy t i m +Baz(( M u o i + H2O +Mum f K i m loai M u o i (mdi) + Axit (mdi) FeS / / +Oxit bazo +012/1° +A1 +CuO +NaOH Baza +Cu(0H)2 tAgNOa \i +CaC03 H S O (loang) +Cu(OH)2 CUSO4 FeClj AICI3 +BaCl2 o c / ^ ^ BaS04 +Muoi +CaCO CuCU NaCl CuClj AgCl CaCU * C a c phu'dng t r i n h p h a n tfng : - Hoa tinh: Fe(r) + 2HCl(jj) -> FeCl2 (dd) + H2 (k) 2A1(,, + 6HCl„,o -> 2AlCl3,dd) + 3H2(k, CuO(o + 2HC1,,,, -> CuCl2(dd, + H20(|) Fe203( r) + 6HCl(Ud) -> 2FeCi3(dd) + 3H20(|) NaOH(dd, + HCl(jj) -> NaCl(dd) + HzOd) Na2S04 XCaS04 H2S04 (dd) + Fe F e S (dd) 3H2S04(dd) + 2A1 Al2(S04)3(dd) + H ( k , H2S04(dd) + ZnO(r) ZnS04(dd) 3H2S04(dd) + Fe203(r) Fe2(S04)3(dd) + H ( i ) H2S04(dd) + Cu(0H)2,rt C U S O (dd) + 2H20(„ H2S04(dd) + 2NaOH(dd) Na2S04(dd) • 2H20(„ H2S04(jd) + BaCl2,dd, BaS04i H2S04(dd) + CaC03(r) CaS04 (r) 4FeS2(r, Cac phu'dng I r i n h p h a n uTng : Fe2(S04)3 +O2 cm +H2O S O S O +NaOH HCl Hoa tinh : > ZnS04 ^ +Bazcf +Fe203 M^P, + H S , d a c / + O x i t bazdi ^ +02/t» +02/t» Hoa +Fe / + K i m loai / (trirdc H ) FeS04+H2t i^Al2(S04)3+H2T II Tinh chat hoa hpc va dieu chS'axit clohidric +Quy t i m Hoa +Fe + 02 (k) 2S02(k, + 02(k) S03(„ + H20,,, + H2O (!) iriing + 2HCl(dd) + H20(,) + C02(k) 2Fe203 (r) 1102(k) S(r) + H2 (k) + 8S02(k) S02(k) t",x, 2S03,i, H2S04(i) Chu y : T i n h cha't o x i hoa manh cua H2SO4 dac, tac dung v d i hau he't cac k i m l o a i , Ichong g i a i phdng H2 (ma giai phong k h i SO2 ) a) V i du : T i n h chfi't hoa hpc v a d i ^ u c h d ' N a O H + Gia'y q u y t u n Hoa H2SO4 dJc + Cu CUSO4 + N h i l u k i m loai (trtr Au, Pt) + Ag ^ + Cha't hQu CO (C12H22O11) * Cha't chi thi mau Na 2H2S04dac + CU(,) Hod than (C) NaCl — ^ CuS04(dd) + S02 (k) + 2H20 (1) Ag2S04(r) + SO2 (k) +2H2O (,) H2SO4.IIH2O +12C 2H2S04dac + 2Ag(„ H2S04dac + Ci2H220n > III TfNH CHAT HOA HQC CUA BAZd I T i n h cha't hoa h p c chung c u a bazcf T i n h cha't hoa h p c chung c u a bazoT tan nvtdc (chS't k i l m ) LiOH, K O H , B a ( H ) , Ca(0H)2, NaOH Quy tim Cha't chi thi mku Phenolphtalein Xanh •Hong + Oxit axit Muo'i + NxJdc M(OH)„ dd + dd mu6'i +CO2 Na20 dpdd Oxit axit / / +NiAJc +SO2 NaOH Cu(OH)2 +FeCl3 Fe(0H)3 +HC1 +Nir 2NaOH(dd) +H2(k) + Cl2(k) b) V i du 2: T m h cha't hoa hpc v a dl4u chS' Ca(0H)2 Cha't chi th) mau t" (M: Ba, Ca) +CUSO4 + dd mu6'i BazcJ mdi + M u o i mdi + Axit Xanh Hong NazCOs NazSOj Phenolphtalein j Ag2S04 + SO2T Cic phiTcJng t r m h p h a n tfng : /1 + SOat Quj- tim / / M u o i + Ni/dc / MzOn + Nirdc Ca +NUiJc Ca(0H)2 Qu}' tim Phenolphtalein ^ +CO2 Oxit axit +SO2 + dd mu6'i +Na2C03 +FeCl3 CaO +HC1 + Axit +H2SO4 Xanh Hong CaC03 CaSOj CaC03 Fe(0H)3 CaCl2 CaS04 CaO * Cic phUdng trlnh phSn tfng hod hoc : - Hoatinh: ^ j Fe2(S04)3(dd) Ca(OH)2(dd) +C02(k) -> CaC03(r) + H2O,,) Ca(OH)2(dd) + S02(k) -^CaSO3(0 + H2O„) Ca(OH)2(dd) +Na2C03,dd) AgN03(dd) CaCOj, r) + 2NaOH(dd, + + 6KOH(dd) -> NaCl(dd, 2KC103(r, 2KC1(.) CaO(,) 3Ca(OH)2(dd) +2FeCl3(dd) -> 2Fe(OH)3(r) +3CaCl2(dd) CaC03(r) Ca(0H)2,dd, +2HCl,dd, -> CaCl2(dd) + 2H20(i) +H2S04(ddf -> CaS04(r, + 2H20(„ 2Cu(N03)2(r) 2CuO(o Ca(0H)2(dd) 2KN03(o 2KN02(r, \(0H)2,.) — ^ CaO(,) +H20(k, 2AgN03(r, c) Thang p H : D i l n g de bieu thi axit hoSc bazd cua dung dich : 2Ag(,) — 2Fe(OH)3(,) +3K2S04(dd) A g C l (r) + NaN03(dd) + 302(k) I + C02t(k) + N ( k , + 02(k) : t / + 02(k) + N ( k ) + 02(k) - Ne'u p H = 7: dung dich trung tinh I I Phan uTng trao dSi - N e u p H > 7: dung dich c6 tinh bazd; p H cang Idn bazd dung dich cang Idn D}nh nghia: Phan tfng trao d o i la phan tfng hoa hoc hai hdp cha't - Ne'u p H < 7: dung dich c6 tinh a x i t ; p H cang nho axit dung dich cang Idn Tinh chat hoa HQC chung cua bazd khong tan ntfdc: Mg(0H)2, A1(0H)3, Zn(0H)2, Fe(0H)2, Fe(0H)3, M(OH)„ (ran) * Cu(0H)2 M20„ + AB + j ' CD > A D + CB D i l u k i $ n xay r a phan tfng trao doi Phan iJng trao d d i giffa dung dich cac chS't chi x a y neu san pham tao cd m o t chat de bay h d i hoac chat khong tan - I I I PhSn b o n hoa hoc H2O v d i dung dich m u o i PhSn b o n ddn: Chtfa mot nguyen to dinh duTdng chinh la d a m ( N ) , Ian (P),kali ( K ) a) PhSn dam: IV TINH CHAT HOA HQC CUA MUOI, PHAN HOA HQC MOI QUAN H E GlQfA UfNG TRAO DOI - PHAN BON CAC HCJP CHAT VO Cd Tmh cha't hoa HQC cua muo'i + K i m loai M ' (M' dufng trU(5c M) +.Axit + dd baza dd m u o i kirn loai M + dd muoi Co can n h i ? t phan 10 V i d u : +axit •> muoi + nur6c K h o n g tac dung l e n chat chi t h i , khong tac dung v d i oxit axit, k h o n g tac dung Vi du: tham gia phan uTng trao d d i v d i nhffng phan cau tao ciia chung de tao nhCfng hdp chat m d i Muoi + M - ' Ure (NH2)2CO tan niTdc, chuTa 46% nitd (NH2)2CO + 2H2O ^ ' '^ ' (NH4)2C03 - A m o n i nitrat NH4NO3 tan nu'dc ; chiJa 35%N - A m o n i sunfat (NH4)2S04 tan n i ^ c ; chtfa % N ; " ' b) PhSn ISn: ^ Muoi m d i i + Axit m d i t M u o i m d i + Bazd m d i i - Photphat tif nhien Ca3(P04)3: khong tan nifdc, tan cham daft chua - Supe photphat Ca(H2P04)2: c) P h a n k a l i : tan dU'dc nu'dc K C l va K2SO4 deu de tan nu'dc PhSn b o n kep : ChiJa hoac nguyen to dinh dtfdng la N , P, K muoi mdi - H o n hdp san pham CU(r, + 2AgN03 (dd) Cu(N03)2(dd) + 2Ag,r) BaCl2(dd) + H2S04(jd) BaS04(r, + 2HCl,dd) V i du : Phan N P K la hon hdp cac m u o i : NH4NO3, (NH4)2HP04, K C l deu de tan cung cap cho cay dong thdi d a m , Ian, k a l i - V i du : K N O ( d a m va kali) ; (NH4)2HP04 (dam va Ian) PhSn b o n v i Irfc/ng: ChiJa mot liTdng rat it cic nguyen to hoa hoc di/di dang hdp chat can thiet cho sU" phat trien cua cay, nhiT hdp chat cua B , Z n , M n v.v rV M6'i quan M gi^a cdc h^p chfi't v6 ccf OXIT AXIT OXIT B A Z d CO2 + Ca(OH)2 SO2 + Ca(OH)2 > CaC03(r) + > CaS03 (r) + H2O B ^ n g phtfdng phap hoa hoc hay nhan b i e t : a) Dung dich H B r va dung dich K2CO3 + Axit H2O ,, b) Dung dich K C l va dung dich H2SO4 + Bazcf O x i t bazcJ " •'"' V i e t cdc phtfdng trinh phan iJng xay ' ' ' " '^^^ * ' ' HuTong dan giai + H2O a) Diing dung dich B a C l de nhan dung djch K2CO3, phan uTng tao ket tua tr^ng: MUOI BaCl2 + K2CO3 Chat khong phan urng v d i BaCh BaCl2 BAZd + 2KC1 la dung dich H B r + > BaS04(r) + 2HC1 H2SO4 Cha't khong phan iJng v d i dung dich B a C b la K C l + K i m l o a i bazd + Muoi BaCOjfr) b) Dung dung dich B a C l de nhan dung dich H2SO4, phan uTng tao ket tua tr^ng: + Axit oxit axit > Hoac dung quy t i m de phan biet + O x i t bazd, m u o i B^ng phuTdng phap hoa hoc hay nhan biet chat ran m a u t r ^ n g CaS04, AXIT CaC03, CaO, Ca(0H)2 V i e t cac phiTdng trinh hoa hoc, neu c6: B B A I T A P T H E O C H U D E HuTdng d i n giai • T r i c h cdc m a u t h ^ cho hoa tan vao niTdc Dgng 1: Nh$n bi^t - T^ch h6n hdp - Tinh ch^ c^c ch^t I B d i tap c6 lofi gi^i Co hon hdp gom BaO va Fe203, neu phuTdng phap hoi hoc tich rieng Fe203 V i e t phiTdng trinh phan iJng - M a u thuf nko hoa tan va toa nhieu nhiet la CaO - M a u thuf nao hoa tan it la Ca(0H)2 - Hai mau thuf khong tan la CaS04, CaCOs ta cho chung Ian lifdt tac dung v d i dung dich H C l , m a u thuT nao phan iJng sui bot k h i la CaC03 UvCdng d i n giai C a C O j + 2HC1 Hoa tan h5n hdp nifdc lay du", Fe203 khong tan diTdc tach BaO bi hoa tan phan tfng: BaO + H2O > M o t dung djch bao hoa k h i C O nirdc c6 p H = H a y g i a i thich va viet Ba(0H)2 phiTdng trinh hoa hoc cua C O v d i H2O K h i cho quy t i m vao dung dich tren Co hon hdp gom AI2O3 va Fe203, neu phi/dng phap hoa hoc de tdch rieng quy t i m CO mau gi? Fe203 V i e t phtfdng trinh phan tfng Htftfng d§n giai Hi^dng d i n giai K h i C O la mot oxit axit c6 kha nang tac dung v d i ntfdc tao axit H2CO3 Hoa tan hon hdp dung dich N a O H lay diT, Fe203 khong tan diTdc tach AI2O3 bi hoa tan phan uTng: AI2O3 + 2NaOH > 2NaA102 nen dung dich c6 tinh axit p H = + H2O K h i CO duTdc dung lam chaft dot cong nghiep, c6 Ian tap cha't la cic C O va S O L ^ m the nao c6 the loai bo diTdc nhffng tap chat khoi CO b^ng hod cha't r^ tien nha't ? V i e t cdc phiTdng trinh phan ufng hod hoc xay Hifdng dan giai D a n hon hdp C O , SO2, CO I p i tir tiT qua dung dich C a ( H ) dM, C O va SO2 b i hap thu phan tfng : 12 > C a C l + H2O + C t C02 BSi + H;0 > H2CO3 tap lis gi^j Nhffng k h i nao sau day c6 the lam kho bang H2SO4 dSc: CO, H2, C O , SO2, O2, NH3 ( k h i n6 c6 Ian h d i niTdc) B^ng phuTdng phap hoa hoc nhan biet chat long : dung dich N a O H , dung dich H2SO4, dung dich N a C l , nirdc , , •i 13 Cty TNHH MTV DVVH Khang Vi&t D^ng 2: X^c djnh ch^t phSn Qng I Bdi tSp CO Idri giSi CaC03 Co nhffng oxit sau: NazO, BaO, MgO, FeaOj, Fe304, N2O5, NO2, SiOj, ZnO, CaO AI2O3 Hay cho biet nhffng oxit nao tac dung dffcJc vdi : a) Nffdc b) Axit sunfuric loang b) c) Dung djch NaOH Hifdng d i n giai a) Cac oxit tac dung vdi nffdc gom : Na20, BaO, N2O5, NO2 + H2O BaO + H2O N2O5 + H2O 3NO2 + H2O — > 2HNO3 H2O —)• CaO + CaClz + H2O + 2NaOH CO2T CaO + CO2 - CaCOj CO2 + Na20 > Na2C03 CaO + H2O > Ca(0H)2 Ca(OH)2 -—> CaCOji nao tac dung dffdc v d i : a) Nffdc 2HNO3 > + Co nhffng oxit sau: SO2, CuO, Na20, CO, CaO, CO2 Hay cho biet nhffng oxit Ba(0H)2 > 2HC1 + CaCOj CaO + CO2T - ^ ^ ^ Na2C03 + 2NaOH > — CaCOj Vid't cAc phffdng trinh phan ffng Na20 CO2 Ca(OH)2 + Ho^n th^nh phaong trinh phdn Cfng Di^u c h ^ + NO b) Cac oxit tac dung vdi H2SO4 loang gom: Na20, BaO, MgO, Fe203, Fe304, ZnO, AI2O3 b) Axit HCl c) Dung dich NaOH Viet cac phffdng trinh hoa hoc xay Hif^ng d i n giai a) SO2, Na20, CaO, CO2 tac dung dffdc vdi nffdc : S02 + H2O > H2SO3 Na20 + H2O > 2NaOH CaO + H2O > Ca(0H)2 CO2 + H2O > H2CO3 NazO + H2SO4 ——> Na2S04 + H2O BaO + H2SO4 ——> BaS04 + H2O MgO + H2SO4 — MgS04 + H2O Fe203 + 3H2SO4 ——> Fe2(S04)3 + 3H2O Fe304 + 4H2SO4 Fe2(S04)3 + FeS04 + 4H2O CuO + > CUCI2 + H2O + H2SO4 ZnS04 + H2O 2HC1 ZnO Na20 + 2HC1 > 2NaCl + H2O AI2O3 + 3H2SO4 ——> Al2(S04)3 + 3H2O CaO + 2HC1 > CaCl2 + H2O ——> Luu y : N2O5, NO2 chi tac dung vdi nffdc dung dich H2SO4 loang b) CuO, NazO, CaO tac dung dffdc vdi dung dich HCl: c) SO2, CO2 tac dung dffdc vdi dung dich NaOH : c) Ci.c oxit tac dung vdi NaOH gom : N2O5, NO2, Si02, ZnO, AI2O3 SO2 + 2NaOH > Na2S03 + H2O Bo tiic cac phan ffng theo sd sau : CO2 + 2NaOH > n&jCOi + H2O a) Ca > CaO > Ca(OH)2 *C02 -> > C&CO^ > CaO > CaCh , Nhffng oxit nao c6 the dieu che bkng : a) Phan ffng hoa hdp ? Viet phffdng trinh hoa hoc Na2C03 b) Phan ffng hoa hdp va phan ffng phan huy ? Viet phffdng trinh phan ffng hoa hoc >• CaC03 CaCOs CaCOj A)H20 B)CuO C)Na20 D) CO2 E) P2O5 Hff^ng d i n giai CaO ->> Ca(0H)2 a) H2O, CuO, Na20, CO2, P2O5 deu c6 the dieu che bang phan ffng hoa hdp : Hufdng d i n giai a) 14 2Ca-*-'^^' O2 > 2CaO CaO H2O > Ca(OH)2 + 2H2 + O2 2H20 2Cu + O2 2Cu0 4Na + O2 2Na20 > 4P Hxidng d i n giai o CO, 5O2 2P2O5 900'* c CaCOj b) CuO, CO2 CO the dieu che bang phan ufng hoa hdp va phan dng phan huy + C02t CaO + H2O > Ca(OH)2 Ca(OH)2 + CO2 > CaC03(r) + H2O Cu(0H)2 CuO + H2O CaC03 + 2HC1 > CaCl2 + CaCO., CaO + C02t CaCl2 + NajCOj > CaC03(r) + H20 + C t 2NaCl CaC03 + 2HNO3 > Ca(N03)2 + H20 + C t Can phai dieu che' mot lU'dng muo'i dong sunfat, phu'dng phap nao sau day Trung hoa dung dich Ca(0H)2 bang dung djch H3PO4 xay phan uTng trung tiet k i e m du'rtc H2SO4: a) H2SO4 tac dung vdi CuO b) H2SO4 dac tac dung v d i Cu hoa Hay vie't cac phU"dng trinh hoa hoc theo t i le khac ve so m o l cac chat tham gia phan uTng G i a i thich cho cau tra I d i Hrfdng d i n giai HtfcJng d i n giai PhiCcfng phap a, tiet k i e m dtfdc H2SO4 : CuO H2SO4 + > CUSO4 mol + H2O mol SO2 FeS2 -iiU SO2 — ^ H2SO3 -i^Na2S03->S02 (9) Na2S03 SO2 + I/2O2 SO3 + H2O 2H2SO4 + SO2 + H2SO3 IIO2 + 8SO2 > CUSO4 H2O > H2SO3 + 2NaOH > Na2S03 + 2H2O Na2S03 + 2HC1 >• 2NaCl + SO2 SO2 + 2NaOH > NajSOj + H2O H2SO4 + 2NaOH - > Na2S04 + 2H2O (1:1) CaHP04 + 2H2O 2H3PO4 (3:2) Ca3(P04)2 + 6H2O a) Nirdc b) A x i t H C l c) Dung dich N a O H ^ B C A + + 2H2O SO2 H • + HC1 ^ D +Na2C03 'Jfs ^ A Y X Z ^ + H2O c) FeS2 — >>SSO2 02- ->> BaSOj BaS03- > Ba(HS03)2 d) H2SO4 - > K2SO3 -> K2SO4 > SO2 Y Z jijjt > SO2 > BaSO, ->S03 > BaS04 Dang 3: Tinh theo c6ng thQc phUdng trinh ph^n Lfng, higu su^t ph^n Qng, n6ng dQ dung dich >• Baitap c6 Idfl gidi 1- Cho mot liTdng dung dich H2SO4 10% viTa du tac dung het v d i 16 g CuO (3) > Ca(OH)2 H3PO4 X, Y, Z la hdp cha't cua canxi hoSc bari Vie't cac phu'cfng trinh hoa hoc theo sd (2) + nhi^ng oxit nao tac dung du'dc v d i : b) X H2SO4 (1) Ca(0H)2 A, B, C, D la hdp chat cua canxi, goi ten ? SO3 > CaC03^CaO- 2H3PO4 Co nhffng oxit sau: SO3, FeO, K2O, CO, BaO, NO, NO2, CO2 Hay cho biet a) A 2Fe203 Cu Ca(H2P04)2+ 2H2O + V i e t cac phu'dng trinh phan i^ng theo sd : Hi/d^ng d i n giai + ' V i e t cac phu'dng trinh hoa hoc xay Na2S04 4FeS2 ^ II Bdl tap \\s gidi SO3 — ^ H2SO4 («) (l;2) Ca(0H)2 3Ca(OH)2 + V i e t cac phu'cfng trinh hoa hoc theo sd (5) CaC03 - i ^ ,0 16 CaO CaCl2 CaCOj (6) T i n h nong phan tram cua dung djch muo'i thu difdc ^^^JM^gi^lnSJai >Ca(N03)2 ncuo= : 80 = , m o l ^ ' f l f f i / wjPf^ TiNHBiNH THUAN ^1./ / 7/9/*^/ JO ' ' ' 17 Boi dii&ng hoc stnh gtot Hoa hoc sr^- Hvldng d S n g i a i tJao Him Wnft Theo phiTcfng trinh phan uTng: CuO + 0,2 > CUSO4 H2SO4 ^ 0,2 + 0,2 H2O a) Tinh nong % dung dich B: Oca = : 40 = 0,1 mol Theo cac phu'dng U-inh phan iJng: (mol) m H S = 0.2.98 = 19,6g =^ m^^^^so, = dOO : 10).19,6 = 196 g 2Ca + O2 CaO + H2O j ,^ > 2CaO , >Ca(OH)2 m c u S = 0.2.160 = 32 g Ta thay: nca(OH)2 = "cao = nca = 0,1 mol => mca(OH)2 = 0,1-74 = 7.4 (g) i " d d C u S = m c u o + m j j H S = 16 + 196 = 212 g Mat khac: C%(CuS04) = (32 : 212).100% « 15,1% Vay Hoa tan 4,7 g K O vao 195,3 g niTdc Tinh nong % cua dung djch thu di/cJc C% = - ^ 1000 100%= CO2 5) Hxidng d§n g i a i = mc,o + m H = 0,1-56 + 994,4 = 1000 g + 0,1 " K = 4,7 : 94 = 0,05mol 0,74% , > Ca(0H)2 CaC03(r, ' M + H2O 0,1 (mol) V a y V c o = 0,1.22,4 = 2,24 lit Theo phufdng trinh phan tfng: K2O + H2O 0,05mol niKOH Cho 11,2 g CaO t^c dung viifa du vdi V lit dung dich hon hdp HCl 0,2M va > 2K0H H N O 0,2M Tinh V va tinh khoi lifdng cac muoi thu diTdc 0,1 mol = 0,1.56 = 5,6g; mjdKOH = " ^ + ninifcic Hifdng d i n giai = 4,7 + 195,3 = 200 g ncao = 11,2 : 56 = 0,2 mol; C%(KOH) = (5,6 : 200) 100% = 2,8% Mot loai da voi chtfa 80% CaCOj Nung mot tan da voi loai c6 the thu duTdc bao nhieu tafn voi song Biet hieu suaft phan iJng la 90% CaO 80 = 0,8 tan 100 Theo phiTdng tnnh phan ufng va hieu sua't phan uTng 90% (tuTc 0,9): Cdch]: CaCOj 100 (g) "^'^"'^ > ^ Cdch 2: + CO2T 0,9.56 (g) 0,8 (tan) Ta CO : CaO + 2HC1 0,1V < - 0,2V CaO + (mol) -> 0,1V < - 0,2V -> + COjt ^ 0,1V > Ca(N03)2 ^ + H2O 0,1V mcaci2 = 0,1.111 = 11,1 g ; mca(N03)2 = 0,1-164= 16,4 g ' •'••^ 10x200 nNaOH= het A vao 994,4 g niTdc thu difdc dung dich B 100x40 0,5 mol ,ufOw:!^- > NaCl + H2O 0,5 0,5 Ta c6: mHci = 0,5 36,5 = 18,25 (g) m,uHa = ^ ^ ' f ^ ' j ^ " = 500 (g) 3,65 mwaci = 0,5 58,5 = 29,25 ( g ) ; mjjNaci = 500 + 200 = 700 (g) h-j ;T J; b) Tinh the tich C O (do d dktc) tac dung vilfa du vdi dung dich B de tao muoi CaCOj , , = 0,5 -> X (mol) , Hi^dng d a n g i a i Cho g canxi tac dung he't vdi O2 (khong khi) thu diTdc cha't ran A, hoa tan 18 H2O Ta c6: 0,1V + 0,1V = 0,2 : ^ V = (lit) NaOH + HCl => m c a o = 0,72.10''.56 = 40,32.10' g = 403,2 kg a) Tinh nong % dung dich B + dich NaOH 10% Tinh C% dung dich muo'i thu diTdc 0,9 (mol) 0,8.10*^ , „ (mol) -> Ta c6: X = 0,72.10^ (mol) CaO = 0,2.V (mol) Can diing bao nhieu gam dung dich HCl 3,65% de trung hoa 200 gam dung X (tan) "^^"'^ ) HHNOJ > CaCh 2HNO3 x = " ' ^ " ' ^ - ^ ^ = 0,4032 tan = 403,2 kg 100 ^ CaCO, = 0,2.V (mol); , Theo cac phu'dng trinh phan tfng : Hvldng d§n g i a i LMng CaC03 nguyen chat mot tan da voi la : UHCI C%(ddNaCl) = 100% = 4,18% C6 200 ml dung dich HCl 0,2M (c6 khoi liTdng rieng D = g/ml) Dung djch chiJa 20 g NaOH da hap thu hoan toan 11,2 lit C O (dktc) Hay a) De trung hoa dung dich axit can bao nhicu ml dung dich NaOH 0,1M ? Tinh nong mol/1 ciia dung dich muo'i sinh chobie't: a) Muo'i nao diTdc tao b) Neu trung hoa dung dich axit tren bang dung dich Ca(OH)2 5% thi can bao nhieu gam dung dich Ca(OH)2 Tinh nong % cua dung dich muoi sau phan uTng b) Khoi lu^dng muoi la bao nhieu Hxidng dan giai HHCI a) ' HCl mol: 0,04 + NaOH > NaCl 0,04 ,; r mol: + 0,04 -> mol: "NaOH = 0,5 : 0,5 = : - > NaOH 0,5 Phan uTng tao muoi axit CO2 > NaHCOi 0,5 -> 0,5 + > Ca(0H)2 0,02 CaClj + mjjca(OH)2 = (1,48.100) : = 29,6(g) CaO mcaci2 = 0'(^2.111 = 2,22 ( g ) ; mjdcaci2 = 29,6 + 200 = 229,6 (g) C%caci2= (2,22: 229,6) 100% =0,97% .p, v t i 0,5 84 = 42 g Ucoj = 1,68 : 22,4 = 0,075 mol > Ca(0H)2 H2O + 0,05 mol 0,05 mol Theo ciic phU'dng trinh phan iJng : + CO2 Mot lit nirdc tinh khiet d 25"C c6 the hoa tan toi da 0,027 mol Ca(0H)2 a) Hay xac dinh tan cua Ca(0H)2 niTdc d 25"C So mol ban dau: 0,05 0,075 b) Neu tron g C a ( H ) vao niTdc cat, ngi/di ta thu diTcJc mot the tich la 250cm' Somolphaniyng: 0,05 0,05 d 25"C Hay cho biet hien tiTdng cua hon hrtp thu diTdc va giai thich c) Sau loc hon hcJp thu dUdc c! tren, ta du'dc mot dung djch suo't la nirdc voi Dan C O di vao niidc voi cho tdi thu difdc liTdng ke't tua toi da Hay cho biet pH cua nUdc voi thay ddi the' nao qua trinh phan tfng ? d) Tinh khoi li/dng cha't r^n (neu c6) lai tren gia'y loc Htfdng d i n giai mca(OH)2 = 0,027.74 = 1,998 g ; C02 + Ca(0H)2 -o; 0,05 0,025 0,05 Sau phan uTng tren so' mol C O du' se hoa tan mot phan CaCOv CO2 + H O + CaCO, — ^ So mol ban dau: 0,025 So molphan rfng : 0,025 Ca(HC03)2 0,05 -> 0,025 0,025 0,025 0,025 ' (*) pH cua dung djch giam dan tao moi IniUng trung tinh ( H O ) d) Thco phu"ctng trlnh phan tfng (*) "caco., = nca(0H)2 = 0-027 m o l ^ mcaco3 = 0,027.100 , j, 11 Cho 15,5 g NaiO tac dung vdi nxidc diTdc 0,5 lit dung djch A b) Tinh the tich dung dich ^ > CaC03(r, + H O , , fifufjn,, a) Tinh nong M cua dung dich A 0,2 g 1000 So mol sau phan iJng : +H2O > CaC03(rt V$y khoi liTdng ke't tua tao la : 0,025.100 = 2,5 g 1,998.100 w Ca(0H)2 So mol sau phan ufng: lit H O nSng 1000 g a) Do tan la so gam chat tan tan difdc 100 g ni/t'lc tao dung dich bao hoa 20 ' ' '•'•^ Hi^ng d i n giai '^'^^ ncao = 2,8 : 56 = 0,05 m o l ; c) M ^**^''" toan 1,68 lit C O (d dktc) Hay cho biet c6 bao nhieu gam ke't tua tao 2H2O 0,02 Ta c6: mca(OH)2 = 0'02.74 = l,48(g) ^ _ Do tan S = ^'V; 10 Cho 2,8 g CaO vao nU'dc dU'dc dung djch A Dung dich A da ha'p thu hoan b) mddHci = 200.1 =200 g 2HC1 '• b) Khoi liTdng muoi NaHCO,, la : = 0,04 : (0,2 + 0,4) « 0,067 moI/1 - = 20 : 40 = 0,5 mol ; nco2 = 11,2 : 22,4 = 0,5 mol 0,04 VayrVauNaOH = 0,04 : 0,1 = 0,4 lit = 400 ml C M (Naci) a) nco2 + HjO '< : Hvldng d i n giai nNaOH = 0,2.0,2 = 0,04 mol •>i- -iv' *•-»'.< H2SO4 20% (d = 1,14) can de trung hoa dung dich A c) Tinh nong mol/1 cua cha't c6 dung dich sau phan dng Hi/dng din giai nNa20= 1-'^,^'' : 62 = 0,25 m o l ; = 2,7 g Thco phU'dng trinh phan ufng : y ^ ^ , > 21 Cty TNHH MTV P W H Khang Viet hoc 8, - Bclo HUk Vim B6i dtfffng hoc sitih gtoiHoa B a i giai C a c phan iJng ke't tua: AgNOj + HCl A g C l i +HNO3 (1) BaCl2 + H2SO4 > B a S ^ + 2HC1 (2) 87 r iChi them tiep H C l vao thi k e t tua A g P i b i tan ra, tao ket tua A g C l mautr^ng Ag3P04 + 3HC1 -> A g C U + H3PO4 M a u trang C a c phan urng: Theo phan iJng (1): nHci = nAgci = —— = 0,02 mol 143,5 Theo phan urng (2): nH2S04 = " " f ^ " ^ ' ^ ^ ™ " ^ Vay n o n g d o m o l C H Q =CH2S04 " ^ " ^ ' " ^ " " ^ ^ ^ • 'I,, 2P + 5/2O2 -> P2O5 P2O5 + 3H2O 2H3PO4 (2) H3PO4 + N a O H NaH2P04 + H2O (3) -> Na2HP04 + H2O (4) NaH2P04 + N a O H Na3P04 + H2O Na2HP04 + N a O H Cac phan u'ng trung hoa: > N a C l + H2O HCl + NaOH H2SO4 + N a O H > NajSO^ + 2H2O (3) (1) (5) M u o i hay muo'i khac dU'dc tao tiiy thuoc ty le so m o l H3PO4 va NaOH D o trufdtc het can tinh so'mol cac chat: (4) Theo cac phan u'ng (3, 4) tong so mol N a O H can de trung hoa flNaOH = 0,02 + X 0,02 = 0,06 G o i V la so mol N a O H can dung: mol V = Q'^^^^Q^O ^ 0 m l 0,2 Theo phan urng ( , ) : U p - n H j P o ^ = ^ So m o l N a O H : DNaOH = 0,2mol = 0,8 x 0,6 = 0,48 m o l i B a i 30 2nH3P04 < '^NaOH < 3nH3P04 K h i tron dung dich A g N O j v d i dung djch H3PO4 khong thay tao ket X 0,2 < 0,48 < X 0,2 tua N e u them N a O H vao thi thay ket tua mau vang, neu them tiep dung Do ta difdc hon hdp m u o i Na2HP04 va Na3P04 Sau phan u'ng (4) thu dich H C l vao thi thay ket tua mau vang chuyen ke't tua mau trang dircfc 0,2 m o l Na2HP04 va liTclng N a O H duT: 0,48 - 0,40 = 0,08 m o l , luTdng G i a i thich cac hien tiTcJng xay bang cac phu'dng trinh phan u'ng NaOH tac dung v(3i Na2HP04 de tao 0,08 m o l Na3P04 va lifdng m u o i D o t chay hoan toan 6,2 gam photpho thu du'cfc chat A Cho chat A tac dung vdti 800 ml dung dich N a O H , M thi thu dufdc m u o i g i , bao nhieu gam? B a i giai K h i tron dung dich A g N O j v d i axit H3PO4 khong the tao ket tua v i axit HNO3 tao manh hdn axit H3PO4 do hoa tan ke't tua Nghla la: H3PO4 + A g N > A g P i + 3HNO3 khong xay Phan u'ng x a y theo chieu ngU'dc lai: Na2HP04 0,2 - 0,08 = , m o l K h o i liTdng m u o i Na2HP04 = 0,12 x 142 = 17,04g ^ - K h o i lirdng m u o i Na3P04 = 0,08 x 164 = 13,12g Bai 31 Trong coc diTng 19,88 gam hon hdp M g O , AI2O3 Cho 200ml dung dich HCl vao coc, khuay deu Sau k h i k e t thuc phan u'ng, cho bay hdi dung djch thay c6n l a i coc 47,38 gam chat ran khan Cho tiep vao coc 200 m l dung dich H C l ( d tren) khuay deu Sau k h i ket thuc phan u'ng, l a m bay hdi dung djch thay l a i coc 50,68 gam chat r^n khan A g P ; + 3HNO3 > H3PO4 + A g N K h i them N a O H vao, v i H3PO4 b i trung hoa tao Na3P04: : 3NaOH + H3PO4 > A g P i + 3NaN03 Mau 72 2- Tinh % k h o i liTdng m o i oxit hon hdp dau > Na3P04 + 3H2O Va ket tua vang du^dc tao theo phan u'ng A g N + Na3P04 1- Tinh C M cua dung dich H C l B a i giai Cac phan urng xay ra: M g O + 2HC1 AI2O3 + 6HC1 " ' ^ * ' tj • ' > MgClz + H2O > 2AICI3 + 3H2O (1) (2) ' vang 73 Vi CO can dung dich sau Ian iM hai khoi Itfdng chaft r^n khan tang len chtfng to sau Ian thi? nha't cac oxit chUa tan het noi cdch khac axit HCl thieu Theo cac phan tfng (1, 2) ciJ mol HCl tham gia phan iJng khoi luTdng muoj clorua tang len so vdi oxit mot khoi li/cfng b^ng 71 16 =55 gam i; Khoi liWng ciia CI Khoi liWng cua O NhU" vay tdng so mol HCl da phan iJng bang Gpi X la nong dp mol cua A I C I ta c6 phUdng trinh 0,lx ( , - X 0,lx) = f^g't tua ciTc dai ket tua bi tan NaOH di/ 0,14 ket tua c6n lai Giairax=l,6M pai 33- ^ '^""^ '^^^^ A I C I , B la dung dich NaOH IM Them 240 ml dung dich B vao coc dirng 100 ml dung dich A, khuay deu tdi phan iJng hoan toan thay c6 6,24 gam ket tua, them tiep 100 ml dung dich B vao coc, khuay deu tdi phan tfng hoan toan thay c6 4,68 gam ke't tua Tinh nong dp mol cua dung dich A Bai giai C^c phan uTng c6 the xay ra: (1) A I C I + 3NaOH > A1(0H)3 i + 3NaCl (2) A1(0H)3 + NaOH > NaAlOz + 2H2O ImoK Vay CHC, = — = 5M 55 0) !Z Sau Ian them thuT hai dung dich HCl, cac oxit phai tan het vi neu chUa tan het tuTc HCl thieu hoSc viTa du thi khoi lu'dng muoi tang 55 gam so vdi oxit (vi HHCI = X 0,4 = mol), thiTc te chat r^n chi tang , - 19,88 = 30,8 gam Gpi X, y la so mol MgO va AI2O3 ta c6 he phifdng trinh Ta c6: nwaOH Ian = 0,24 x = 0,24 mol 40x + 102y= 19,88 HNaOH Ian = 0,10 X = 0,1 mol 95x + 133,5 x 2y = 50,68 Giai ta CO x = y = 0,14 mol nAi(OH)3 1= = O'O^ '""^^ "Ai(OH)3 ^ = ^ = 0,06 mol 40 X 14 X 100 Vay %MgO = -j— = 28,17%; %Al203 = 100 - 28,17 = 71,83% Vi them tiep 0,1 mol NaOH lifpng ket tua giam 0,08 - 0,06 = 0,02 mol 19,88 chiJng to sau Ian thtf nha't them NaOH, lu'png A I C I van vi neu A1(0H)3 Bai 32 X la dung dich A I C I , Y la dung dich NaOH 2M Them 150 ml dung dich da Idn nha't thi ket tua phai tan hoan toan, sau Ian thiJ hai them NaOH thi Y v^o coc chu-a 100 ml dung dich X, khuay deu tdi phan tfng hoan toan thay NaOH d\i, da xay phan ifng hai: c6 7,8 gam ket tua Them tiep 100 ml dung dich Y vao coc, khuay deu tdi Gpi X la nong dp mol cua A I C I ta c6: :yp phan tfng hoan toan thay c6 10,92 gam ket tua Tinh nong dp mol ciJa dung i| 0,lx (0,34-3.0,lx) = 0,06 dich X Ket tua ciTc dai ket tiia bi tan NaOH diT ket tua c6n lai Bai giai Giai rata c6 x = IM Cdc phan iJng c6 the xay ra: Bai 34 A la dung dich hon hpp HCl 0,1M va H2SO4 0,2M B la dung dich hon A I C I +3NaOH > A l ( H )3 i +3NaCl (1) hdp NaOH 0,05M va Ba(0H)2 0,1M Lay 50 ml dung dich A, them it quy tim Al(OH)3 + NaOHd^ > NaAlOj + 2H2O (2) vao, quy tim c6 mau Them V ml dung dich B vao dung dich A thay quy Ta c6: UNaOH Ian = 0,15 X = 0,3 mol tim trd lai mau tim Tinh the tich V nNaOH Ian = 0,10 x = 0,2 mol • ^ Bai giai nAi(0H)3 Ian = ^ = 0,lmol; nAi(0H)3 Ian = ^ ^ = 0,14mol Cac phan ufng trung hoa: -> NaCl + H2O (1) HCl + NaOH Vi li/dng ket tua A1(0H)3 tang len d Ian 2, chiJng to sau Ian liTdng A I C I -> BaCl2 + 2H2O (2) van con; vi them tiep 0,2 mol NaOH liTdng ket tua chi tang 0,14 - 0,1 = 2HC1 + Ba(0H)2 0,04 mol chiJng to sau Ian lu-png NaOH diT, vi neu khong diT thi ket tua phai H2SO4 + N a O H (3) Na2S04 + 2H2O H2SO4 + Ba(0H)2 tang — = 0,0667mol (4) BaS04 + H O ^ ^ ^ ^ ^ - - ^ X2= 75 Tinh so' mol cac chal: nHci = 0,05 0,1 =0,005 mol; nH2S04 = 0,05 x 0,2 = O.Olmol V 0,05 = 0,00005V mol; n^^^Qy^^^ = Vj ^ x 0,1 = 0,0001V mol nNaOH= ,: Theo cac phan iJng ta nhan tha'y mol H S O tiTdng di/dng mol HCl Con mol Ba(0H)2 tiTOng diTdng mol NaOH, do ta c6 phiTcfng trinh: i 0,005 + X 0,01 = (0,00005 + x 0,0001) V Riitra V = 100ml 3x56x100 Fe304thi %Fe = — = 72,41% f>0^ Vay la Fe203 |,, , Goi M la kim loai hoa tri n Theo cong thiJc phan tuf M2(C03)„: ^^""^"^ = 40; rut M = 20n * 2M + 60n ^ , 3MxlOO„ E)6'i vdi muo'i photphat: M3(P04)„ ta co: %M = ^ ^ ^ ^ ^ ^ % ,; «w 3x20nxlOO -„_,o, Thay M = 20n vao cong thtfc ta co: %M = 3^20n + 95n " ' ; Bai Hay ke cac giai doan chinh qua trinh san xuat axit sunfuric tif khoang vatpirit TH mot tan pirit chtJa 90% F e S co the dieu che dtfdc bao nhieu lit H S O dac 98% (d = 1,84 g/ml) biet hieu suat dieu che la 80% Bai giai Cac giai doan chinh san xuat axit sunfuric tiiT pirit: a) Dot chay pirit de dieu che SO2: ;• ( F e S + 1102 — ^ 2Fe203 + S O :r '' ^ ' b) Oxi hoa SO2 SO3: X Chu de X a c dinh phan h o n hgfp Bai Tinh % khoi lufdng cua cdc nguyen to cac hdp cha't sau: H2O; H2SO4; C H O H ; C H - COOH; C„H2„ ^ ; Bai giai Fe.Oy H20:%H = ^ ^ = U % 18 %0 = ^ ^ ^ ^ ^ = 88,9% hoac= 100- 11,1 =88,9% 18 H2SO4 %S = ^ ^ ^ ^ = 32,65%; %0 = 2%S = 32,65 x = 65,30% : Va %H = 100 - 32,65 - 65,3 = 2,05% CH3COOH: 60 c n c/t7 56xxxl00 ^ FCxOy: % F e = CnH2n + 2: Bai 60 „^ %; 60 %0 = 56xx + 16xy 56xx + 16xy S^t tao du^dc oxit: FeO, Fe203 va Fe304 Neu ham liTdng cua F e oxit la 70% (khoi liTdng) thi la oxit nao cua s^t? Neu ham Itfcfng phan tram cua mot kim loai muo'i cacbonat la 40% thl ham liTcfng phan tram cua kim loai muo'i photphat la bao nhieu? Bai giai F e O thi %Fe = = 77 % 72 + O2 '"-"^"^ > 2SO3 c) Cho S O hap thu vao H S O dac de thu di/dc oleum va tijT oleum dieu che dung dich H S O co nong mong muon Cdc phan ufng: F e S + 1IO2 > 2Fe203 + S O (1) ' ' % ^ ^ ^ J n ^ ) x l 0 ^ ^ j n ^ l ) x l 0 ^ ^ ^ %C = l ^ ^ % = i l i l ^ % 14n + 7n + l 14n + 7n + l Trong 2SO2 16xyxl00 „ ^ S O + O2 > 2SO3 SO3 + H2O > H2SO4 (2) ' (3) ^ 1x10*^x90 ,^nnno.^i Theocdcphanu'ng(l,2,3): nH2S04 -^^F.S2 = X ^QQ^^^Q =l5000mol Nhirng VI hieu suat dieu che chi co 80% ntn so mol H S O Ihi/c te thu diTdc on bkng 15000 x — = 12000 mol ^ 100 M a t khac bie't I lit H S O dac chtfa: ^ " " " ' ^ ' ^ ^ ^ ^ ^ = IS-^mol H2SO4 11 2000 Do 66 the tich axit dac blng di = 652,2 lit o» ^ Bai Trpn 500 gam dung dich C U S O 4% vdi 300 gam dung dich BaCb 5,2^, thu ke't tua A va dung dich B Tinh kho'i li/dng ke't tua A va n6ng % cu; cac chaft dung dich B Bai giai 500x4 , 300x5,2 _ , Tmh np,,co = 0,125mol; nn.,,->, = = 0,075mol 100x160 '^'"''2 100x208 Trirdc C U S O + BaCh > BaS04 + CuClz So molphan u-ng: Trirdc phan iJng: 0,125 0,075 Phan u-ng: 0,075 0,075 caccha't ' Sau phan iJng: 0,05 0,075 0,075 Khoi lifdng ket tua A (BaS04) = 0,075 x 233 = 17,475g Nong phan tram B: * 0,05x160x100 , %CuS04 = —• 1,02%; 500 + 300-17,475 %CuCl2^ 0.075x135x100 500 + 300-17,475 Bai Tron 50 ml dung dich NazCOj 0,2M vdi 100 ml dung djch CaCb 0,15M thi thu 6mc mot lu^dng ket tua dung bang liTdng ket tua thu diTdc trpn 50 ml NajCOj cho tren vdi 100 ml BaClj nong dp amol/1 Tinh nong dp a Bai giai Cac phan iJng xay ra: NazCOj + CaClz NajCOj + BaCl2 > CaCOj + 2NaCl > BaCOj + 2NaCl Vi so mol Na2C03 = 0,05 x 0,2 = 0,01 mol it hdn so mol CaCl2 = 0,1 X 0,15 = 0,015 mol (1) (2) Do so mol ket tua CaCOs = so mol Na2C03 = 0,01 mol Khoi liTPng ket tua bkng 0,01 x 100 = Ig Theo phan uTng (2) so mol ket tua cua BaC03 = Nong dp cua dung dich BaC^: a = 78 = 0,005 mol = 0,05mol/1 0,1 PhSn bon Railpjnh nghla phan bon hoa hpc PhSn dam, phan Ian, phan kali 1^ gl? Phan v: ^ lifdng la gi? Hay ke mot v^i loai phan vi liTdng mk em biet Vig't cong thiifc va gpi ten mot so loai phan dam, phan kali, phan Ian thong thifdng Bai giai phftn bon hoa hpc la nhi^ng hod chaft lam phi nhieu dat dai, chtfa cAc nguyen te'dinh dirdng can thid't cho cay phan dam la loai phan chtfa nguyen to nitd (N), phan Mn chiJa nguyen to photpho (P), phan kali chtfa nguyen to kali (K) Phan vi lufdng la loai phan bon chuTa cac nguyen to ma cay chi can mot lifdng raft nho cdc nguyen to dd nhifng lai ra't can thiet cho sif phdt trien cua cay (khoang 10 gam den 100 gam tren hecta da't trong) Cic phan vi Wdng quan trpng nhat la: molipden, dong, mangan, bo, kern, coban, iot Cac phan dam: NH4CI amoni clorua, NH4NO3 amoni nitrat, (NH4)2S04 amoni sunfat, CO(NH2)2 ure - Cac phan Ian: hon hdp Ca(H2P04)2 va CaS04 supephotphat ddn, Ca(H2P04)2 supephotphat kep - CAc phan kali: KCl kali clorua, K N O kali nitrat, K S O kah sunfat Ngoai c6 amophot vifa chtfa ca dam va Ian; NH4H2PO4 amoni dihidro photphat; (NH4)2HP04 amoni hidrophotphat Bail 1- Tinh ham lifdng % cua nitd cdc loai phan dam NH4NO3, NH4CI, (NH4)2S04, CO(NH2)2 Can bao nhieu m^ dung dich H N O 63% (d = 1,38 g/cm') va bao nhieu dung dich NH3 25% (d = 0,9 g/cm^) de san xuat 10 tan phan N H N O Bai giai ^-Tinh ham liTdng % cua N cac phan dam: NH4NO3: 2x14x100 ^3^^^ 53,580 c (NH4)2S04: %N= ^ ^ =26,17% =21,2% NH4CI: %N= ^ ^132 CO(NH2)2: %N= ^^111^=46,67% 60 79 Cty TNHH MTVTJWfr-Kmng Phan tfng tao NH4NO3 H N O + NH3 Supephotphat kep la Ca(H2P04)2- Danh tijf kep d day c6 nghIa la qua trinh (1) > NH4NO3 i l u che g o m hai giai doan: 10x10^ T h e o p h S n u - n g C l ) : nHNOj =nNH3 -nNH4N03 = — — = 1,25xlO^mol oU Ca l i t dung dich H N O CO ^'38x63 ^ ^ ^^^^ 100x63 V a y t h ^ ach dung dich H N O c i n b^ng ^ ^ i ^ ^ ^ i ^ - 9,06 X ^ l i t = 9.06m\ ' (3) Theo phan iJng (1) d phan ta thay cu" mol H S O tao m o l Ca(H2P04)2 va m o l CaS04, trJc tdng k h o i Itfdng supephotphat ddn b ^ n g : 234 + 136 = 506 gam X c , ; Bai NgircJi ta thiTdng tinh ham Itfdng dam theo N (nitd), h ^ m liTdng Ian theo P2O5 va ham IiTdng kali theo K2O Hay tinh ham lifdng cua N c6 kg NH4NO3, ham lifdng K O kg K2SO4 va ham liTdng P2O5 kg Ca(H2P04)2 B a i giai T i n h the tich C O va NH3 (d dkte) de san xuat 1,5 tan ure biet hieu suaft dieu che'm 60% B a i gial Tinh ham lufdng N , P O , K O : nnn x: x Trong k g NH4NO3 c6: ' " ^ ^ 80 Ui ~ 14 ; = 350 gam N J,^ '"-P^"') > CO(NH2)2 + H2O m o l NH3 V a y de d i e u che 1,5 tan ure tiJc 1,5 x 10^60 = 2,5 x lO'' m o l can X lO'' m o l NH3 Nhirtig vi hieu suat , Trong k g K S O c6: (chu y cu" m o l K S O c6 m o l K O ) Theo phan uTng, m u o n dieu che' m o l ure tuTc 60 gam ure can m o l C O va 2,5 X 10^ m o l C O va X 2,5 X 10" = > 3Ca(H2P04)2 PTPLT: > C O ( N H ) + H2O Phtfcfng trinh d i e u che ure: C O + 2NH3 Ca3(P04)2 + 4H3PO4 > Do k h o i liTdng supephotphat ddn thu diTdc bkng: 506 x l O ' g = 50,6 tan B a i Phan dam ure diTdc d i e u che bang each cho k h i C O t^e dung v d i NH3 a C O + 2NH3 > Ca(H2P04)2 + 2CaS04 (it tan) (2) 10 tafn H S O 98% cd ^ ^ ^ ^ ^ ^ ^ - ^ = 10^ m o l H S O V a y the tich dung dich NH3 can b ^ n g i ^ ^ ^ ^ ^ = 9,44 x 10^ l i t = 9,44 m ' 13} 24 n h i ^ t do, ap suaft cao (c6 mSt chat xue tdc) theo Ca3(P04)2 + H S O in V 10^x98 13,8 ^OOQxO'9x25^^ 100x17 CiJ H i t dung dich N H C : vtex , , ' 1000x94 ^ _ -=540gK2O 174 Trong kg Ca(H2P04)2 c6: (chu y cur m o l Ca(H2P04)2 c6 m o l P O ) 1000x142 I : ; = 606,8g P2O5 234 chi 60% nen the tich C O va NH3 thuTc te bang: - f,? ! Bai Li/dng dam ( N ) , Ian (P2O5) va kali ( K O ) c6 tan phan xanh tiTdng Bai Vco2 ^ 2.5 X10^ X 0 , 22.4 = 9,33 X ^ l i t = 933 m^ diTdng v d i lufdng dam, Ian, k a l i c6 10 kg ure, 20 kg supephotphat kep V N H = V C O = 3 x = 1866m^ v^ kg K C l H a y tinh lifdng dam, Ian, k a l i c6 tan phan xanh B a i giai LiTdng dam c6 10 kg ure (CO(NH2)2) b i n g : The nao la supephotphat ddn, supephotphat kep? Cho 10 tan dung dich H2SO4 98% tac dung het v i mot liTcJng viTa du Ca3(P04)2 thi thu dtfdc bao nhieu tan supephotphat ddn, biet hieu suat di^u che la 80% LiTdng Ian c6 20 kg supephotphat kep (Ca(H2P04)2) bkng 2 Ca(H2P04)2 + 2CaS04 (it tan) = 4,67 kg N 60 X 94 L u d n g k a l i c6 kg k a l i clorua ( K C l ) b^ng: =3,15 kg K O 2, X /4, J (1) Vaiy Itfdng d a m , Ian, k a l i c6 tan phan xanh la: 4,67 X = 23,35 kg N ; 12,14 3,15 X = , k g K • ; ,, x = 60,7 kg P O ' ^7 • 81 T^on rapr y - uao ntm vmn Can lay bao nhieu gam C U S O 5H2O de dieu ch6' 50 kg dung dich C U S O 2%? daft mg d6ng (dU'di dang C U S O ) Hoi d j , Ngtfcfi ta can bon tten moi bao nhieu lit dung dich CUSO4 2% (d = 1,0 g/ml) de bon cho hecta dat ^ 1.1 THUYET C d BAN VA NANG CAO J Tinh chS't vat I i : Hau het cac kim loai deu c6 tinh deo, tinh dan dien, tinh din nhiet, va c6 anh kim Bai giai Khoi liTdng C U S O can thiet n ^^"^ Bo hoat dong hoa hoc cua cac kim loai giam dan tiT trai qua phai theo day: = kg 250 gam tinh the Vay khoi liTdng tinh the C U S O 5H2O can lay b^ng: ^-• ^ fnghTa: a) Kim loai dufng dau day (tir L i -> Na) tac dung vdi nifdc d dieu kien thufdng tao dung dich kiem va giai phong H2 Ctf moi ra^ dat can mg Cu ttfc can = 12,5 mg C U S O Vay the tich dung dich C U S O 2% can thiet b^ng = ^ b) Kim loai diJng triTdc H day diTdc H khoi dung dich axit (HCl, H2SO4 ) c) Kim loai dtfng triTdc day dufdc kim loai diJng sau khoi muoi cua no (trif cac 64 Vay hecta (10.000 m^) can: 12,5 x lO" mg = 125 g Cu = 6,25 lit Bai Mot loai phan amophot chiJa hon hdp NH4H2PO4 va (NH4)2HP04 vdi ti le kim loai tan nufdc tijf L i Na) I I I Tinh chSt hoa HQC )> Tac dung \6i oxi - Cac kim loai tiT L i Na de dang tac dung vdi O2 dieu kien thifdng 4Na(r) so moi tu-dng u-ng la : Tinh so m^ NH3 (dktc) va so tan dung djch H P O - 49% de san xua't 100 tan amophot noi tren Bai giai H3PO4 + N H X2 H3PO4 + 2NH3 3Fe(r) > NH4H2PO4 (1) > (NH4)2HP04 (2) Theo phan tfng 1, ta thay de c6 diTdc hon hcJp gom moi NH4H2PO4 va moi (NH4)2HP04 tiJc 115 + X 132 = 379 g amophot can moi H3PO4 va moi N H Vay de c6 100 tan = 100 x 10^ gam amophot can: iNTTT 5x100x10^ 379 5x100x10^22,4 , Va the tich NH3 = — ^ = 29,55.10 379x10^ C-^' i T i r ^ 3x100x10^ „ w ^ u - ' ' U M T T - SomolH3P04= — , 82 > 2Na20(r) (natri oxit) + 302(k) + 202(k) '" ) — ^ 2Al203(r) Fe304 ' (uhom oxit) (s^ttuToxit) Ca(r) + H2(k) — — > CaH2(r) (canxi hidrua) Hau het cac kim loai deu tac dung vdi CI2, Br2, S tao muoi 3Cl2(k) — ^ 2AlCl3(r) (nhomclorua) 2Al(r) + 2Fe(„ + 3Cl2(k) 2A1 + 3S '" > Fe + S '" > FeS 2FeCl3(r) (sat I I I clorua) AI2S3 (nhom sunfua) ' (sat I I sunfua) ^' Tac dung vdi ntfdc : = 0,79xl0^mol Goi X la so tan dung dich H P O can diing, ta c6 ^"^^^^^^"^ 100x98 R u t r a x = 158tan 02(k) - Tdngquat:2xM + yOz '" > 2M,0y •> Tac dung vdi phi kim : H2, Clj, B r j , S - Cac kim loai dau day tac dung difdc vdi H2 tao muoi hidrua kim loai - So moi NH3 = + Cac kim loai tiif Mg -> Hg tac dung manh vdi O2 d nhiet cao 4Al(r) Phan tfng dieu che' amophot: X, D^y Li K Ba Ca Na Mg A l Zn Fe Ni Sn Pb H Cu Hg Ag Pt Au Theo cong thtfc tinh the CUSO4 SHaO ta thay muon c6 160 gam C U S O can - KIMLOttl Chiidngll Bai = 0,79 x 10^ ~ Tij* L i -> Na tac dung vdi nu-dc d t" thiTcfng 2Na(,, + 2H20(,) ' > 2NaOH(jd) + H2(k) 83 - A l tdc d u n g v d i ntfdc d Idp b e n ngo^i sau d6 dilTng l a i khSng tdc d u n g tidp , A l ( O H ) k h o n g tan mSdc b a o ve cho A l b e n k h o n g tdc d u n g tiep, 2A1(,, - + 6H20(,) >2Al(OH)3(„ + 3Fe + 4H2O CaO y I^guy^n iic H2O + — ^ FeO + H••2 Fe304 + 4H2 + > Si02 CO2 CaSiOj s a n xuS't t h e p y l An m o n k i m loai ^ J p i n h nghTa: Sir pha huy kim loai, hdp kim moi triTdng tiT nhien diTdc ggi la sir an mon kim loai > Na2Zn02(dd) + Hjd,) 2NaOH(jd, N g u y e n n h S n : Do kim loai hoSc hdp kim tac dung vdi moi trirdng (nirdc natri zincat 2A1 + khong giau O2 de oxi hoi cac nguyen to can loai fc; V ^ Z n , A l t^c dung v d i dung dich k i e m nhiet thi/dng + + CaO Loai khoi gang p h i n Idn c^c nguyen to' C , S i , Mn, S , P b i n g cdch dilng T a c d u n g v 2NaA102(dd) + B H z f k ) ' ^ * " " natri aluminat T a c d u n g v d i d u n g d j c h a x i t a) Cac k i m loai diJng tru'dc H day difdc H k h o i dung dich axit thong thiTdng (HCl, H S a , ) khong khi, dat ) Nhffng yd'u t6' a n h hrfdng d n srf a n m o n k i m l o g i ^ * Thanh phan moi tru"dng u; .' * Thanh phan kim loai \h ; * Nhiet C a c h c h S n g a n m o n k i m l o a i 2A1(„ + 3H2S04(dd) > Al2(S04)3(dd, + Fe(r) + H2S04(dd) >FeS04(dd) + 3H2(k, H2(k) b) H2SO4 dac, H N O dac t i e dung vdi hau het cac k i m loai (trijf Au.Pt) khong g i a i phong H2 2A1(„ + 6H2S04d.c 2Fe(„ + 6H2S04(dac) — ^ * Cich li kim loai vdi moi trirdng : Sdn, ma, boi d i u md * Che tao hdp kim It bi an mon AI2O3 Fe2(S04)3,dd, + 3S02(k, + 6H20(i, c) Al, F e khong tac dung v d i H2SO4 dac nguoi, HNO3 dac nguoi - VII.Hoatinhvadi^uche'Al >Al2(S04)3,dd, + SSOjck, + 6H20(,) T a c d u n g v d i d u n g djch mu6'i ; H MT +s AI2S3 +CI2 AICK + dd N a O H AI2O3 NaAlOj + H j K i m loai dufng tru'dc day du'dc k i m loai dtfng sau k h o i dung dich m u o i cua no (trilf cac k i m l o a i L i -> + H2SO4 d a c Na) 2A1 + 3CuS04,dd, Fcirt + CuS04(dd) + CUSO4 > Al2(S04)3(dd) + 3CU(„ > FeS04(dd) + Cu,,, V I I I H o a t i n h v a di4u c h e F e I V N g u y e n t ^ c s a n xuS't g a n g * DOng C O khijf oxit sat d nhiet cao * Cac phan uTng xay 16 l u y e n gang : - O x i hoa than c o c : - K h i C O khuf oxit silt quang sat: 3CO + O2 '" > CO2 C + CO2 '" > + FC2O3 — ^ 84 FcjOj 2C0 Fe304 - — > FeO FeS FeCl, + H2SO4 loang Fe + C , t° > Fe / \ \ : +O2 +CI2 t° + C O , t" 3C02 + 2Fe Qua trinh khu" theo biTdc: Fe203 — > Al2(S04)3 + C u -t-S C Al2(S04)3 + SO2 + H j O H S O dac, -t- C u S O t" _^ FeS04 + H2 Fe2(S04) + SO2 + H j O F e S + Cu 85 B B A I TAP T H E O CHU D g n g 2: ot D g n g 1: Nhan bi^t - T a c h h6n hpp X ^ c djnh ch^t ph^n H o a n th^nh phifdng trinh ph^n i/ng - Tinh c h ^ c ^ c chdt BSI tap CO lofi giSi I B^i t§p CO lori giij Co ba k i m l o a i la nhom, bac, s^t Hay neu phrfdng phap hoa hoc de nhj^ J Ong - Di^u c h ^ ^ , V i e t phiTcfng trinh hoa hoc b i e u dien b i e n d o i sau day : AI2O3 bie't tifng k i m l o a i Cac dung cu hoa chat coi nhtf c6 d u V i e t cac phuo^ AICI3 AI2O3 A1(0H)3 AICI3 _(5L^ Al trinh hoa hoc de nhan bie't Hxidng d§n giai Fe FeS04 — ^ Fe(0H)2 - J ^ FeCh FeCh T r i c h cac m a u thijT cho tac dung vdti dung dich N a O H , mau ihvt nao b i hoa tujl — ^ FeCb va l a m sui bpt k h i la A l 2A1 + 2H2O + N a O H > 2NaA102 + SHzT > Fe(0H)3 c) FeClj FcjO, - i ^ Fe — ^ Fe304 — H a i mau thuf khong b i hoa tan dung dich N a O H la Fe, A g ta cho chunj tdc dung v d i dung dich H C l , mau thuT nao bi ho^ tan va l a m sui bpt k h i la Fc Fe + 2HC1 > FeCh a) Bac cdm (dang bot) c6 Ian tap chat dong, nhom L a m the' nao de thu difdc bac tinh khie't Cac dung cu, hoa chat coi nhiTco d u Hi^dng d i n giai Cho hon hcfp tac dung v d i dung dich A g N O j (lay dur), Cu va A l se bi hoa tan vao dung dich phan i?ng: A l + 3AgN03 > A1(N03)3 + 3Ag Cu + A g N > Cu(N03)2 + A g b) Loc la'y chat ran khong tan la Ag Trong phong t h i nghiem, ngtfcfi ta l a m kho cac k h i am b^ng each dan k h i na) di qua cac binh diTng cac chat hao nir^c nhiTng khong c6 phan tfng v d i can l a m kho Co cdc chat l a m kho sau : H2SO4 dSc, CaO D u n g hoa chat nao n i tren d l a m kho m o i k h i a m sau day : k h i SO2, k h i O2, k h i CO2 Hay giai thich siT U/a chon c) L a m kho SO2 va CO2 c6 the ddng H2SO4 d5e - L a m kho O2 c6 the dung H2SO4 dac hoac CaO v i chung khong tac dung vdfi II B^j tap ti/ gi§i Chi diTctc diing mot k i m l o a i , l a m the nao c6 the nhan biet diTdc ba dunS dich: HNO3, HgCl2, N a O H T r i n h bay phiTdng phap hoa hoc tach dong k h o i hon hdp ba k i m l o a i dong s^t, k e m (1) 2AI2O3 4A1 + AI2O3 + 6HC1 AICI3 + 3NaOH 2A1(0H)3 — ^ 2AI2O3 - i P = ^ 2A1 + 3CI2 Fe + H2SO4,,) - FeS04 + 2NaOH » F e ( O H ) ^ r + Na2S04 (2) Fe(OH)2 + 2HC1 > FeCb + 2H2O (3) 2FeCl2 +CI2 -—> 2FeCl3 (4) 3O2 — (2) 2AICI3 + 3H2O (3) A l ( H ) ( r , + 3NaCl - AI2O3 + 3H2O (4) 4A1 + 3O2 (5) (6) 2AICI3 (1) FCSO4 + H j t (5) 3FeCl2 2FeCl3 + Fe - FeClj + 3NaOH > F e ( O H ) , r ) + N a C l 2Fe(0H)3 - i ^ FezOj + 3C0 2Fe +3CO2 3Fe + 2O2 Fe304 Fe304 + 4H2S04(i, —- > Fe2(S04)3 + FeS04 + 4H2O Hrfdng dSn giai - Fe2(S04)3 Hifdng d i n giai + H2T M a u thur khong tan dung dich H C l la Ag ^ Fe20 + 3H2O (1) , , i^:-^n-; (4) (5) Cho cha't sau: A l , A l C U , A1(0H)3, AI2O3 Hay sa^p xe'p cha't day bie'n hoa ( m i day d6u g m cha't) va vie't phtfdng tfmh hoa hoc tiTdng tfng de thi/c h i e n day bie'n hoa > AI2O3 a) Al b) Al(OH)3 Hifdng dSn giai 0,01 mol FeS04 > AICI3 0,02625 - , = 0,01625 mol CUSO4 dff > AI2O3 > Al(OH)3 > AICI3 > Al j^hfi'i lifdng dung dich sau phan ffng : II Bai tap lii giii m,a = mpc + m , , c u s o " mcu = 0,01 56 + 25.1,12 - 0,01.64 = 27,92 g Viet phi/dng trinh phan tfng thiTc hien day bien hoa : jvj5ng % cac chat la : a) Fe — ^ FeS04 > Fe(0H)2 AI2O3—iiL>Al2(S04)3 Dien vao cho co dau hoi (?) va thay cac chCT cai A, B, C, D, E blng nhffng - , 0,01.152.100% C%FeS04= =5-44% 0,01625.160.100% C%cuS04du = = 9.31% cong thiJc hoa hoc phil hcfp roi can bang cac phan ffng sau : ^ Fe + ? > A + B A + NaOH > C + C + O2 D — ^ E + + H2O Cho 0,83 gam hon hdp gom nhom va s^t tac dung vdi dung djch H2SO4 loang NaCl a) Viet cac phffdng trinh hoa hoc xay > D b) Tinh phan phan tram cua moi kim loai hon hdp ban dau E + H2O B dir Sau phan ffng thu dffdc 0,56 lit d dktc Hifdng d§n giai — ^ Fe + ; nH2 = 0,56 : 22,4 = 0,025 mol ? Gpi X, y Ian lu^dt la so mol A l va Fe 0,83 g hon hdp Dang 3: Tinh theo c6ng thC/c phaong trinh phdn Qng, hieu su^t phdn Qng, n6ng dung djch Taco: 27x + 56y = 0,83 (I) Theo cac phffdng trinh hoa hoc : I Bai tap c6 Icri giii 2A1 + 3H2SO4 Ngam mot la sat co khoi lifdng 2,5 gam 25 ml dung dich CUSO4 15% cd X > Al2(S04)3 l,5x Fe + H2SO4 khoi Iffdng rieng la 1,12 g/ml Sau mot thdi gian phan iJng ngu'di ta lay la sat > FeS04 + y khoi dung dich, nJa nhc, lam kho thi can nang 2,58 g + 3H2t H2 y a) Hay vie't phffOng trinh hoa hoc b) Tinh nong phan tram cua dung dich sau phan ffng Giai he phffdng trinh (I,II) ta duTdc : x = 0,01 ; y = 0,01 Vay Hil'dng d i n giai 25.1,12.15 r.r.^^.c l,5x = 0,02625 mol 100.160 Khoi liTdng tang len cua la Fe la 2,58 - 2,5 = 0,08 g mot Iffdng Fe tan vao dung dich va mot liTdng dong dung dich bam vao la Fe Gpi X la so' mol Fe tham gia phan tJng : X ^ X Taco: 64x - 56x =0,08 > FeS04 X 88 0,025 ' (II) ^3 •'• ^ Cho 1,96 g bot sat vao 100 ml dung djch CUSO4 10% co khoi Iffdng rieng la 1,12 g/ml ^) Viet phu-dng trinh hoa hoc xay ''^^ ^) Xac dinh nong mol cua chaft dung dich sau phan ffng Gia thie't rang + Cu X %>Fe = (100-32,53)% = 67,47% (mol) the tich cua dung dich sau phan ffng thay doi khong dang ke Hxidng dSn giai => x = 0,01 mol b) Vay dung dich sau phan iJng gom co: = 0,83 , CUSO4 Fe + CUSO4 y 0.01-27.100% "cuso^ = a) + Taco: "FC = 1,96 : 56 = 0,035 mol; Ucuso = ^^"'^'^^'^^ = 0,07 mol 89 a) T h e o phiTdng irinh hoa hoc : Fe b) Somolbandau : 0,035 Somolphanu-ng: 0,035 CUSO4 + > FeS04 + Cu nH2S04 - O.^Sxmol; nwg = 3,6 : 24 = 0,15 mol Mg 0,07 - ^ 0,035 x/2 mol 0,035 Mg [0,035 mol FeS04 Vay dung dich thu diTdc gom : ,^ , ^ • [0,035 mol CuS04dur N6ng mol/1 cac cha't r^:C^^cnSO,) =CM(F.SO,) = + + 0,75x mol O-^^S : 0,1 = 0,35M 2HC1 >MgCl2 + H2T X mol H2SO4 > MgS04 + H2T 0,75x mol = x/2 + 0,75x = l,25x = 0,15 ^ x = 0,12 (1) ' " ^ Vay can dung 0,12 lit hay 120 ml dung dich HCl I M v^ H2SO4 0,75M Cho 1,41 gam hon hdp hai kim loai A l va Mg tdc dung vdi dung djch HjSO^ Ngam mOt la s^t dung djch CUSO4 Sau mot thdi gian, lay \i s^t khoi loang du", ngiTcJi ta thu diTcfc 1,568 1ft (dktc) Xac dinh phan % kho'i dung dich thay kho'i lufdng Id sat tang them 1,0 gam Hay tinh so gam s^t bi Itfdng moi kim loai hon hdp hoa tan va so' gam dong him tren la sat HiMngd§ngiai Htfdng d i n giai TCr the tich (d day la hidro - san pham cua phan i?ng giiJa kim loai vdi axit) suy so mol Tijf phu-dng trinh phan Ung ta lap diTcJc he phtfdng trinh : Tir so mol hidro se suy so mol hon hdp kim loai va khoi liTdng hon hcfp roi giai he phu'dng trinh an so Cachl: nH2 = 1,568:22,4 = 0,07 mol Mg + H2SO4 Hjt + > Al2(S04)3 3/2y = 0,07 ^ x + l,5y = 0,07 HhSn h p kim i„ai = H i M g + m.M = 24x + 27y = > FeS04 + 1,41 CUfr, 64 g tang them - = 8(g) 3H2t Tom tat: 56 g Fe tan 3/2y mol y mol nH2 = X + CUSO4 Cu" 56 g Fe bi hoa tan thi c6 64 g Cu sinh bam tren la sat lam khoi li/dng X mol 2A1 + H2SO4 Fe + 56 g > MgS04 + mol X Sat manh hdn dong nen Fe da day Cu khoi dung djch muo'i CUSO4 V i Cu sinh bam tat ca len s^t, va Mcu > Mpc nen kho'i lifdng la kim loai sau phan iJng tang them ; difa vao suT tang khoi lUdng va phiTdng trinh phan tfng se tim rake'tqua X (1) (2) Giai he phifdng trinh (1), (2) ta diTdc x = 0,025 ; y = 0,03 mMg = 24x = 24 0,025 = 0,6 g ; mA, = 27y = 27 0,03 = 0,8 I g Thanh phan phan tram cua hon hdp : g Fe tan > 64 g Cu > la sat tang g > > y g Cu 1g So gam sat tan la : x = (56 1): = g So gam dong bam len la sat la y = (64 1): = g Cach2: Fe + CUSO4 > FeS04 +CU(r) X mol X mol Khoi lUdng Fe phan iJng 56x g; Khoi lu'dng Cu sinh 64x g Bo tang kho'i lu'dng cua la sat bang hieu so' kho'i lu'dng dong sinh va kho'i "ft %Mg= Q'^-^^^ =42,55% ; %A1 = 100-42,55 = 57,45% 1,41 De hoa tan ho&n loan 3,6 gam Mg can phai diing bao nhieu ml dung dich ho" hdp HCl I M va H2SO4 0,75M Hxidng d§n giai Neu gia thiet phai dung x lit dung dich hon hdp axit, se suy so mol mo' axit phan uTng Tij" phu'dng trinh phan iJng va so mol Mg (suy tit khoi lu"dng' ta se tim x, suy so' mol dung dich So' mol moi axit c6 x lit dung dich axit la : n 90 = V.CM => nHci = x.l = x mol lufdng sat phan iJng (tan vao dung djch tao FeS04) la : 64x - 56x = 8x = 1=^ X = 1: = 0,125 (mol) Khoi lu'dng sat phan tfng bang 56x = 56 0,125 = g Khoi lu'dng dong sinh bam len la sat: 64x = 64 0,125 = g Ngam mot vat bang dong c6 kho'i liTdng gam 500 gam dung dich AgNO., 4% Chi sau mot luc ngiTdi ta lay vat va thay khoi li/dng AgNOs dung dich giam mat 85% Tinh khoi liTOng vat lay sau lam kho ' " Tinh nong phan tram cac chat dung dich sau lay vat khoi dung dich '-^ •» 91 Hvtdng d§n giai a) Cu la kim loai manh hdn Ag nen da day Ag khoi dung djch AgNOa, vl v^y lU'dng AgNOs giam di phan tJng, tir day ta tinh drfcJc lu'dng chat phai] iJng va san pham Khoi liTdng AgNOj ban dau: 500 (4 : 100) = 20 g V Khoi lu'dng AgNOs tham gia phan iJng (giam di) 20 (85 : 100) = 17 g (hay 17 : 170 = 0,1 mol) ;;^fj,,, Cu + 2AgN03 > Cu(N03)2 + 2Ag mol mol mol mol — mol 0,1 mol — mol 0,1 mol 2 Kho'i lu'dng Cu tham gia phan iJng: mcu = n.M = (0,1: 2).64 = 3,2 g Khoi lu'dng Ag sinh ra: = n.M = 0,1.108 = 10,8 g Khoi lufdng cua vat lay khoi dung dich : mCu(bandau) " nicu(phan i?ng) + mAg(sinhra) = ni(v5l) = - 3,2 + 10,8 = 12,6 g mdd(,,au phan ifng) = " I d d AgNOa ban daU ~ (vatlanglhcm) = 0 - (12,6- 5) = 492,4 g mcu(NO3)2 = (0,l:2).188 = 9,4g b) Nong cac chat dung djch : 9,4.100 , ^„ (20-17).100 C%Cu(N03)2 = = 1.91%; C%AgN03= ^^^'^ =^,61% Hai la kem co khoi lu'dng bang nhau, mot la diTdc ngam dung dich Cu(N03)2, mot la dtfdc ngam dung dich Pb(N03)2 Sau mot thdi gian phan tJng, khoi lu'dng la kem thu" nhat giam mat 0,15g Hoi khoi lU'dng la kem thiir hai tang hay giam bao nhieu gam Bie't rang ca hai trU'dng hdp kem bi hoa tan nhUnhau Hi/dng din giai Trong ca hai tru'dng hdp so' mol kem tham gia phan i^ng nhU' Zn manh hdn Pb nen da day chi khoi dung dich muoi VI Mpb = 207 g Idn hdn M^n = 65 g nen sau phan i?ng khoi lu'dng la kem tang len (Pb, Cu sinh deu bam len kem) Tiif c^c di? kien dau bai va phU'dng trinh phan iJng se tinh duTdc dp tang khoi liJdng cua la Zn thu" hai (diTa theo nz„ phSn ,ing = npb sinh ) • , Zn + Cu(N03)2 > Zn(N03)2 mol nizn (phan iJng) = 65x g ; mcu(sinh ra) = 64x g X + Cu(,) X mol Kh^i l i ' d n g la kem giam di Ik: 65x - 64x = x = 0,15 mol Zn t h a m gia p h a n iJng t r o n g ca triTdng h d p la 0,15 mol Zn thu" nha't + Pb(N03)2 0,15 m o l > Zn(N03)2 + Pb^^ 0,15 m o l '-A' •• mZn(phanJng) = 0,15 65 = 9,75 g ; mph(si„hra) = 0,15 207 = 31,05 g Khd'i lu'dng la kem tang them b^ng ' ; - ^ mpb - mzn(phJntfng) =31,05 -9,75 =21,3g ''un }0m:.:, QuSng oxit s^t tif chiJa 80% Fe304 Can dung bao nhieu ta'n kim loai quang de san xuat 100 tan gang c6 5% cac nguyen to' khong phai la s^t ? Bie't ring qua ttinh luyen gang li/dng sat bi hao hut la 4% Hi^dng din giai TriTdc h e t ta tim mpc c6 t r o n g 100 t a n va mp^ bi h a o h u t suy mpc can thie't, tif phU'dng t n n h p h a n iJng ta suy mpg^^ r o i suy m^usng 95 m,.,i.^-«: 95 + 3,8c6= 98.8 Trong 100 tan gang 95 ta'n(ta'n) s^t Kho'i liTdng sit hao hut: - — = 3,8 (tan) 100 Fe304 + C O — ^ 3Fe + C O T 232 g 3.56 g 232 tan 3.56 tan x tan 98,8 tan Khoi liTdng Fe304 can diing : x = ^ ^ ^ ^ ^ ^^^^^ 3.56 Khoi liTdng quang can dilng : 136,44 (100 : 80) = 170,55 (tan) II-B^i tap t i / g i i i 1- Hoa tan hoan toan 11 gam hSn hdp Fe va Al bing mot liTdng dung dich H2SO4 2M (vira du) ngi/di ta thu difdc 8,96 lit (dktc) 3) Tinh phan phan tram khoi lUdng moi kim loai hon hdp b) Tinh the tich dung dich H2SO4 da dung 2- Khu: 3,6 gam hon hdp hai oxit kim loai: Fe203 va CuO bing hidro d nhiet cao du-dc 2,64 gam hon hdp hai kim loai Hoa tan hon hdp hai kim loai tren dung dich HCl (dM) thi c6 V lit bay (do d dktc) Xac dinh kho'i Itfdng moi oxit hon hdp va tinh gia tri bang so' ciia V got dtmng HOC Stnrrgrm o; ? - r/tro nim tiotr vrnn- Goi k i m loai hoa t r i ( I I ) la R D ^ n g 4: L a p c n g thQc mQt c h d t ' 8,4 : 22,4 = 0,375 m o l nH2 I B ^ i tSp CO Idfi g i d i R Cho 9,2 gam mot k i m loai A phan iJng v d i k h i clo d\i tao 23,4 gam H2SO4 + > RSO4 0,375mol m u o i Hay xac dinh k i m loai A bie't rang A c6 hoa tri I m + 0,375mol H2T 0,375 m o l 21 Hifdng d i n giai Theo phtfcfng trinh phan tfng : 2A + CI2 j^guygn tur kho'i cua k i m loai R la 56 dvC d6 Id Fe > 2AC1 Fe 2A(g) (2A + ) g 9,2 (g) 23,4 (g) ^ , 2A Ta CO : 9,2 A + 71 = => + >FeS04 + H2SO4 Hat b) 0,375 m o l FeS04 ^ A = 23 (natri) 0,375 m o l + XH2O > 0,375 m o l 23,4 M,„h Cho 10 gam dung dich m u o i sat clorua 32,5% tac dung v d i dung dich bac =m:n nAgci * < ' = 278 • X 18 = 278 , , 278-(56.96) 18 Hvldng d a n giai So m o l A g C l ket tua ( t i n h the ) 0,375 m o l XH2O (56.96) s^tdadiang XH2O = 104,25: 0,375 = 278 g FeS04 nitrat diT thi tao 8,61 gam ke't tua Hay t i m cong thtfc hoa hoc ciia muoi FeS04 '*>.''" C6ng thu-c tinh the m u o i hidrat hod FeS04.7H20 = 8,61 : 143,5 = 0,06 m o l Hod tan 1,84 gam mot k i m loai k i e m vao niTotc De trung hoa dung dich thu LU'dng m u o i sat clorua da phan i^ng : ^ ^ ^ ^ ^ = 3,25 g diTdc phai diing 80 m l dung djch H C l I M Xdc dinh k i m loai k i e m dem hoa tan Hrfdng d i n g i a i D a t cong thtfc m u o i s^t clorua la FeClx Vi k i m l o a i k i e m tdc dung du'dc v d i midc tao bazd k i e m , nen difa vao FeCl, + xAgNOj > Fe(N03)x (56 + 35,5x)g , 56 + 35,5x liTdng axit da diing suy so' m o l bazO va suy so' m o l k i m l o a i k i e m cuoi cting suy nguyen tuf kho'i cua k i m loai, tra bang t i m ten k i m l o a i 0,06 m o l X = 3,25 0,06 Cong thiJc m u o i s^t la FeCU CO : xAgCl X mol 3,25 (g) „ Ta + _ => 80 ml = x = 0,08 (lit) ^ K h i hoa tan 21 gam m o t k i m loai hoa t r i ( I I ) dung dich = 0,08.1 = 0,08 m o l UHCI K i hieu k i m loai k i e m la R R + H O H2SO4 loang du', ROH ngtfcJi ta thu du'cJc 8,4 l i t hidro (dktc) va dung dich A K h i cho ke't tinh muoi '^R b) Xac dinh cong thtfc hod hoc cua tinh the m u o i hidrat hod d6 = "ROH = X f^jj-^ 0-3!;,: > R O H + H2T HCl 0,08 m o l dung dich A thi thu du'dc 104,25 gam tinh the hidrat hoa a) Cho biet ten k i m loai + > RCl + H2O 0,08 m o l = 0,08 m o l IR = m : n = 1,84:0,08 = 23 Nguyen tij* k h o i cua k i m loai la 23 dvC, Id natri Na K\i6ng d i n giai Tuf the tich H2 suy so' m o l H2, tiT phu'dng trinh phan iJng suy so m o l k i m loai TiJT so' m o l va kho'i lu'dng k i m loai se t i m dtfOc k h o i lu'dng m o l - > nguyen tur kho'i va ten k i m loai Tijf nH2 -> Hmuf-n - > nunhihd ^.^j; *' ^- Hai la k i m l o a i X hod tri ( I I ) c6 k h o i li/dng bkng nhau, mot Id diTdc ngdm dung dich Cu(N03)2, mot Id ngam dung dich Pb(N03)2- Sau mot thdi gian phan uTng, kho'i liTdng la thi? nhat g i a m mat 0,2%, la thiJ hai tang 28,4% so v d i ban dau Munhibd ••• T i m k i m loai X Bie't r^ng ca hai trtfdng hdp X deu hoa tan nhiT 95 94 Htfdng d§n giai RO Hod tri cua kim loai X bang hod u-i Cu, Pb muoi nitrat, nen chiing phan ifng vdi so'mol b^ng n Mx > Mcu va Mph > Mx- kim loai da giam di 0,002a (g) va tang them 0,284a (g) + > X(N03)2 Cu(N03)2 + ,5 Cu(,) n(mol) l i l i i ft^ty;!:;; n (mol) X + = >X(N03)2 + n mol • 22,4 lit H O (bay hcfi) + Rg V lit JfKo ,>«^;? •and'''- 7,82 g i l f i ^1 l^til lit' = R : 7,82 -> 7,82(R + 16) = 9,95R 9,95 7,82 16 = (9,95 - 7,82)R = 2,13R o R= = 58,74 2,13 ,(1), Pb(r, ?'•;•'•: n mol Khoi liTcJng la kim loai tang them • Mn\' • Chia phi/dng trinh (1) cho ( ) : n ( X - ) _ Q,002a 22 82 22,4 : V = R : 7,82 = 58,74 : 7,82 => V = ' ' = 2,982 (lit) 58,74 Khuf hoan loan 32g mot oxit sat bang CO ot nhi$t cao Sau phan iJng ket thiic khoi lifcfng cha't r^n lai 22,4g Lap cong thiJc oxit sat 207n - nX = n(207 - X) = 0,284a n(207-X)~ R Tim the tich hidro la : n ( X - 64) = 0,002a Pb(N03)2 > Nguyen tuf khoi cua kim loai R la 58,74 dvC la N i Khoi liTcJng la kim loai giam di nX - n (R + 16) g 9,95 g ' H ,>,.,,, Ne'u coi khoi li/dng ban dau ciia kim loai X la a (g) thi sau cic phan tfng 1^ X + (2) Hifdng din giai , Cha't ran lai sau phan ufng la Fe, tif khoi li/cJng oxit sat Fe^Oy va kho'i X-64 0,284a 207-X ~ 142X - 9088 = 207 - X Itfdng Fe ta suy khoi liTdng oxi c6 oxit Tijf cac dff kien tren ta tim 142 diTdc ti le so mol : Up^: no = x : y suy cong thtfc cua oxit X = 65 Fe.Oy Nguyen tuT khoi cua X = 65 dvC la kem Zn + xFe yCO , yC02 moxi = mp,,^Oy - mpc = 32 - 22,4 = 9,6 g Khi khuf 9,95 gam oxit cua mot kim loai hoa tri hai bkng hidro thu dtfdc 7,82 gam kim loai Xac dinh ten cua kim loai va the tich (do d dktc) hidrophai diing Cdch : no = vdi oxit) chinh la khoi lifcJng oxi So mol RO = so mol nguyen tuf oxi oxit va suy yO (56x + 16y) g 0,6 mol 56x + 16y _ 32 M R y mol 32 g Hidro khuf oxi oxit (ki hieu RO) nen giam khoi liTdng cua kim loai (so M R O > FexOy f'-^*^''''^ Hiidng d§n giai Tif suy 9,6 : 16 = 0,6 (mol nguyen tuT oxi) y 0,6 (56x + 16y) = 32y => 33.6x = 22,4y ~ (X6 • phiTdng trinh tren ta rut ti le no = (9,95 - 7,82): 16 = 0,133125 mol x : y = -'3 ong thufc oxit sit la Fe203 RO + 0,133125 H — ^ 0,133125 R + H O (bay hdi) 0,133125 (mol) ach : Tuf phu-dng trinh phan iJng ta c6 sd Fe^Oy > xFe + yO M R O = m : n = 9,95 : 0,133125 = 74,74 56x(g) 16y(g) RO = 74,74 -> R = 74,74 - 16 = 58,74 22.4(g) 9,6 (g) Nguyen tuT khoi cua R la 58,74 dvC la Ni 56x : 22,4 = 16y : 9,6 => 56x 9,6 = 22,4 16y The tich hidro phai dung 22,4 0,133125 = 2,982 (lit) 537,6x = 358,4y =^ 3x = 2y hay Bai toan ciSng c6 the giai theo each khdc sau day : Cong thu-c oxit sat Fe203 -i-v-;'!^'^'-^ x :y = :3 " ' ' *^ Cach : Tuf cong thiJc FcxOy ta suy ti le so nguyen tiJ 96 97 ^^>'=1^ ••^ MQ = ^••1^ 56 Cong thifc oxit s^t Fe203 II Bai tdp til giil 16 = : = : = : ^.^ , Hoa tan 0,07 mol mot kim loai chtfa ro hoa tri dung dich HCl (lay du) ngtfcJi ta thu dmc 2,352 lit hidro (do d dktc) Xdc dinh ten kim loai de^j hoa tan Hoa tan hoan toan 9,2 g hon hdp gom mot kim loai hoa tri (II) va mot kim loai ho^ tri (III) dung dich HCl (diT), ngiTdi ta thu diTdc 5,6 lit Ha (do d dktc) a) Neu CO can dung dich sau phan tfng se thu diTdc bao nhieu gam hon hdp hai muoi khan b) Tinh the tich dung dich HCl 2M can dilng cho qua trinh hoa tan tren C BAI TAP LUYEN THI Chu de Cac khai ni#m - Clfu tao chS't Bai 1 Nhifng tinh cha't vat li dac triTng cua kim loai la gi? Khoi lUdng rieng c6 phai la tinh chat vat li dac tru"ng cua kim loai hay khong, tai sao? Hay chon cac kim loai va sap xep chiing theo thiJ tuf giam dan tinh boat dpng hod hoc so cac nguyen to cho diTdi day: C, Ba, Zn, Si, Li, Ca P, Na, Fe, Cu, Ag, Pb, Hg, Ni, Mg, CI, K Bai giai NhCfng tinh chat vat li dSc trU"ng cua kim loai la: tinh deo (de dat mong, kco dai); vang la kim loai deo nhat, ta c6 the dat mong vang tam suot, day khoang micromet lO'^m tiJc 0,001mm hoSc keo sdi nho tdi mtfc mat thiTcJng khong thay du"dc Tinh dan dien dan nhiet tot (bac dan dien, dan nhiet tot nha't); tinh anh kim (sang bong, moi kim loai c6 anh kim khac nhu" bac sang trang, vang sang vang v.v Tuy nhien can nh('< la anh kim phu thuoc vao cua be mat, thi du bac kim loai thoat cac phan tfng, thi du: Zn + 2AgN03 > A g i + Zn(N03)2 lai c6 mau den (nhifng hat rat nho mau den) Cac kim loai giam dan tinh hoat dong: Li, K, Ba, Ca, Na, Mg, Al Zn, Fe, Ni, Sn, Pb Cu, Hg (I), Ag Hg (II) 98 i Nh6m la mot nguyen to' hoa hoc tao drfdc cac oxit hidroxit luQng tinh Hay via't cdc PTPlJ ho^ tan oxit va hidroxit bang cac dung dich KOH va H2SO4 Bai giai Bai Al2(S04)3 AI2O3 + 3H2SO4 > AI2O3 + K O H >-2KA102 + H20 2Al(OH)3 + 3H2SO4 > Al(OH)3 + NaOH > NaA102 + 2H2O Al2(S04)3 + 3H2O ,ni-^ ^, mirpotki + 6H2O ' ' ^ fi^« - / /fci' Gang la gi, thep la gi? Viet cac phu'dng trinh phan tfug hoa hoc chu ye'u xay 16 cao (luyen gang), 16 Mactanh de luyen thep Bai giai Gang la hdp kim cua sat vdi cacbon ham li/dng cacbon tit 2% den 6% va mot so'nguyen to'khac Thep la hdp kim cua s^t vdi cacbon ham lifdng cacbon difdi 2% va mot so nguyen to'khac Cac phan ijfng chu ye'u xay 16 cao (luyen gang): KhuT oxit s^t: 3Fe203 + C >2Fe304 + C02 Fe304 + CO > 3FeO + CO2 FeO + CO > Fe + C O (Co the Viet phan ufng khur tri/c tiep Fe203, Fe304 Fe) CaCOj + Si02 > CaSi03 (xi) + CO2 Vi quSng si(t c6 Ian cac hdp chat cua Si, Mn, P, S nen c6 the xay cac phan drng nhif: Si02 + 2C > Si + 2C0 S^t nong chay hoa tan C, Si, Mn, P, S tao gang, dac bi$t gang c6 mot hdp chat quan la xementit Fe3C Cac phan iJng dot chay cacbon: ,j f^,; C + O2 — ^ CO2 CO2+C — ^ 2C0 5^-'' Cac phan iJng xay 16 luyen thep Mactanh: E>e luyen thep can phai loai khoi gang phan Idn cacbon, mangan, silic, Photpho, lull huynh theo cac phan iJng: C + O2 > CO2T Si + > Si02 S + O2 > SOat 4P+|o2 >• P2O5 99 2FeO + Si > 2Fe + SiOz FeO + Mn > Fe + MnO Cac oxit Si02, P2O5 FeO + C Bai giai >• Fe + CO tac dung v 2FeCl3 Neu sir an m6n c6 phat sinh dong dien (dong chuyen dcfi dien tuT, pin nguyen to) thi goi la an mon dien hoa Thi du: mieng Zn nguyen chat hoa tan rat cham dung dich HCl loang, nhiTng neu cho mot mau Cu tiep xuc vdi mieng Zn (hoac cho vai giot dung dich muo'i dong, thi du CUSO4 chang han), luc xay phan iJng: Zn + CUSO4 > ZnS04 + Cu i Thi qua trinh hoa tan xiy rat nhanh (khi H2 thoat rat nhanh, nhieu tif mieng Zn) Ivluo'n cho phan iJng xay nhanh, ngirdi ta khuay manh thuy ngan kim loai dung dich HgS04 bao hoa, luc thuy ngan la kim loai long bi phan tan nhi?ng hat rat nho, cac kim loai tao chat de dang tiep xuc phan iTng vdi dung dich HgS04 va tao cac muoi sunfat tan vao dung dich, loc tach Hg va rufa sach ta c6 thuy ngan tinh khiet Trirdc het hoa tan hon hdp kim loai b^ng dung dich HCl diT, luc xay cdc phan tfng: 2A1 + 6HC1 >2A1C13 + H T Fe + 2HC1 > FeCl2 + H21 2CU + O2 a Phu len be mat kim loai, hdp kim mot Idp bao ve nhir: boi dau, md, sdn, ma Ag + O2 > khong phan u?ng CuO + 2HC1 > CUCI2 + H O Ag + HCl > khong phan iJng b Cho vao moi trirdng mot chat lam cham an mon (chat vie che) Thi du cho chat urolropin (mot hdp chat hffu cd c6 tinh bazd) vao axit clohidric thi axil chi tac dung dirdc vdi silt oxit ma khong tac dung vdi sat kim loai Di/^' trcn tinh chat de lam sach be mat cac kim loai: hoen ri bi tan het va trcn be mat kim loai tao mang bao ve ''^ ' ' Phan lai la hon hdp Ag va Cu dem nung ndng khong tdi phan vlng hoan toan, luc chi c6 Cu bi oxi hoa CuO, hoa tan CuO bang dung dich HCl, lai Ag nguyen chat: Cac phirdng phap chinh de bao ve kim loai khoi bj an mon: kem, thiec, crom, vang bac ^ j.f — ^ 2CuO Ghi chu: Co the nung ca hon hdp kim loai, sau hoa tan cac oxit, lai Ag Tuy nhien tren thuTc te cac oxit Chii de AI2O3, Fe203 da bi nung thi raft kho tan G i a i thich hi^n tiTgrng - Vid't phi^oTng trinh phan tfng Bai De lam sach thuy ngan kim loai khoi cac kim loai tap cha't nhu" Zn, Al, M? Sn, ngu^di ta khuay thuy ngan kim loai can lam sach vdi dung dich HgSOj bao hoa, du" Giai thich qua trinh lam sach bang cac phiTdng trinh phan iJng Trinh bay phiTdng phap hoa hoc de lay du'dc Ag nguyen chli'l tiT hon hdp M' Al, Cu, Fc ^ai Dien phan ndng chay KCl dirdc cha't ran A va B Cho A tac dung vdi nirdc diTdc C va dung djch D Cho B tac dung vdi C roi lay san Pham hoa tan vao nU'dc du'dc dung dich E Cho mau giay quy tim dung dich E sau do tir tu" dung dich D vao dung dich E Viet cac PTPlJ va giai thich sir ddi mau cua quy tim, bie't rang cac phan iJng '^ay hoan toan 100 101 [...]... Bai giai 1 Trong cac oxit thi oxi luon luon hoa tri II va difa vao hoa tri cua oxi de tinh hoa tri cua cac nguyen to hoa hoc khac Hoa tri C, S, N, Mn trong cac oxit la IV, IV, VI, V, VII va cong thiJc cua cac axit tifdng u-ng la: H 2 C O 3 , H 2 S O 3 , H 2 S O 4 , H N O 3 , HMn04 2 Trong cac hdp chat hidro luon luon hoa tri I, oxi luon luon hoa tri II, con hoa tri cua nitd va clo trong cac hdp chat... nong 39, 5x5x10^ ri> 2 Khoi lu'dng dung dich bang V x d 2 Co bao nhieu gam muoi an trong 5 kg dung dich bao hoa muo'i an d 20"C, bie't dp tan cua muo'i an d nhiet dp do la 35 ,9 gam 120g 100 do mol , , 100 =t 795 g = 1. 795 kg Bai 7 1 Cho biet dp tan cua chat A trong nirdtc d 10"C la 15 gam, con d 90 °C la 50 gam Hoi khi lam lanh 600 gam dung dich bao hoa d 90 "C xuong 10"C thi c6 bao nhieu gam cha't A thoat... muoi cacbonat kim loai hoa tri I Them tijf tir dung dich H2SO4 10% vao coc cho tdi khi khi vilfa thoat het thu diTOc muo'i sunfat nong do 13,63% Hoi do la muoi cacbonat kim loai gi? Bai giai Gpi M la kim loai hoa tri I, ta c6 phan tfng hoa tan: M 2 C O 3 + H2SO4 > M2SO4 + H2O + C O 2 1 (1) Gia suf de hoa tan 1 mol muoi tu"c (2M + 60) gam can 98 g H2SO4 nguyen chal 98 X 100 hay — — = 98 0g dung dich, luc... trung hoa liTdng axit diT? B a i giai me = 100 X 1,137 + 400 - 23,3 = 490 ,4g V a y nong do % cua H2SO4 (dir) v^ H C l (tao thanh) b l n g : = 490 ,4 ,2.64%: %HC1 = Mii35:5>Na3P04 + 3H20 (1') ^''-''^ /.5:i^:'H 0,12 > 3Na2HP04 (2') ' 0, 09 • K e t qua cudi cung: 0, 09 m o l Na2HP04 va 0,12 - 2 x 0,03 Thidu: ; Ket qua cudi cung: 0,06 m o l Na3P04 va 0,15 - 0,06 = 0, 09 m o l Na2HP04 0,12 djch N a O H c6 the la m u d i axit hoac mot vai m u d i trung hoa HO»Hf B a i giai 0,15 Ca(HC03)2 +2HC1 B a i 19 b) ' Vie't cac PTPTJ xay ra va tinh sd m o l cac chat thu di/dc sau... hap thu he't khi A c) The tich dung dich sau khi trung hoa la: 0,5 + 0,1075 = 0,6075 (lit) Nong dp dung dich Na2S04 la: C M 2HC1 II B^i t i p W giSi 2H2O Khoi lircJng H2SO4 can dung la : 0,25 .98 = 24,5 g ^ + = 0,25 : 0,6075 = 0,41 M 12 .Hoa tan 6,2 g NazO vao 193 ,8 g nvCdc ta thu diTcfc dung dich A Cho A tdc 2* De trung hoa mot dung dich chiJa 1 09, 5 g HCl, dau tien ngiTcJi ta diing dung dich chiJa 112... (1) + yHiO (2) C'"^': ^ a i 9 Hoa tan hoan toan 1,44 gam k i m loai hoa t n I I bang 250 m l dung dich H2SO4 0 , 3 M De trung hoa li/dng axit diT can dung 60 m l dung djch N a O H 0.5M H o i do la k i m loai gi? 53 Ta cd nHC) = 0,3 x I = 0,6 mol Gpi M la K L N T cua R ta c6 ti le: Mx + 16y 2y l l , 2 y 56 2y 56 = => M = =— X— = —n 8 0,3 0,3x 3 x 3 Bai giai Gpi R la kim loai hoa tri I I CAc phiTcJng... 24n + 62m + (5m + n)16 , Khoi imng mo = (0,672 : 22,4) 32 = 0 ,96 Riit ra n = 2m Vay CTPT la Mg2P207 Khoi li/dng KCl = 2,45 - 0 ,96 = 1,49g Cach 2: V i Mg chiem 21,6% nen suy ra MgO chiem VI cha't r^n chi c6 kali clo (vi tdng % khoi liTdng bang 100%) NhiT vaj chat ran ufng vdi cong thtJc muoi duy nha't la KCl, vi khi thoat ra la oxi, ner 1 49 0 96 Ta CO ti 16 x : y = — : — = 0,02:0,06 = 1:3 ^ 74,5 16 Vay... qua trinh xiiy ra khi hoa tan mot chat vao nu'fJc'.' Nhiet hoa tan la gi? Tai sao khi hoa lan KOH, H2SO4 vao nifik' thi nu'dc bj nong len rat nhieu con khi hoa tan N H 4 C I , NH4NO3 vao nu'i'Jc thi lai l
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