X U A T BAIN D A I NHA HOC Q U d c QIA 16 H a n g C h u o i - Hai Ba Trimg - Ha HA Pi6l Npi PHLfCJNG PHAP AP DgNG DjNH LUAT BAO TOAN KHOI L U O N G Di^n t h o q i : Bien tap - C h e b a n : (04) 39 714896 VA BAO TOAN NGUYEN TO H a n h chinh: (04) 39714899; Tong bi§n tap: (04) 39714897 A N O I D U N G PHlJCfNG P H A P Fax: (04) 39 714899 Chiu Gidm doc Tong bien Bien Che track nhi^m Trinh tham gia phdn iing bang tong khoi lU0ig cdc chat tao thdnh " D i e u giup ta giai bai toan h6a hoc mot each ddn gian, nhanh chong X6i phan tfng: A + B -> C + D PHAM THI TRAM tap : T a l u o n c o : niA + me = mc + moCl) T H U HtTClNG * LUu y: D i e u quan nhat ap dung phu'dng phap la viec phai xac djnh dung lifdng chat (kho'i liTdng) tham gia phan u'ng va tao (c6 KHANG V I E T : bay A p dung djnh luat bao toan khoi lifdng ( B T K L ) : " Tong khoi lUctng cdc chat ban: PHUNG QUOC BAO : tap: ban xuat 1, NOi d u n g chii y den cac chat ket tua, bay hdi, dac biet la khoi li/dng dung djch) Cac d a n g bai toan thifoTng gap KHANG V I E T bia: Doi tdc lien ket xuat H$ qua 1: B i e t tong khoi lifdng chat ban dau «-> khoi lu'dng chat san pham Phifdng phap g i a i : m(dau) = m(sau) (khong phu thuoc hieu sua't phan u'ng) ban: qua C o n g ty T N H H MTV D W H K H A N G V I E T Tong phdt 2: Trong phan u'ng c6 n chat tham gia, ne'u biet k h o i liTdng cua (n - 1) chat thi ta de dang tinh khoi lu'dng cua chat l a i //e ^Ma i ; B a i toan: K i m loai + axit -> muoi + hdnh: '^mue'i ~ "^Icimloai C O N G T Y TNHHtVlTV D!CH Vy VAN HOA KHANG V I E T - Bie't k h o i luTdng k i m l o a i , khoi lu'dng anion tao muoi (tinh qua san pham khi) -> khoi lUdng m u o i - K h o i lu'dng anion tao muoi thudng dUdc tinh theo so mol thoat ra: • V d i axit H C l va H S O loang Email: khangvietbookstore®yahoo.com.vn + 2HC1 ^ Website: www.nhasa + H nen S04^' " K J C O , (irong dung dich) = 0,1 + 0,02 = 0,12 Hifdng dSi^giai mol Taco: m „ , u - i = m K L + m ^ n; cho = 0,06 mol (Irong K2CO3) ^ "c (trong B a C O j ) + "c (trong K H C O ) 0,02 = 0,06 + a (a la so mol KHCO3) KHCO3 0,06 0,06 0,06 CO2 + K H ^ 0,04 1,5 "cr(OH)3 K2CO3 + H2O Dap an B Hon hdp X gom Fe(N03)2, Cu(N03)2 va AgN03 Thanh phan % khoi li/dng cua nitd X la 11,864% C6 the dieu che diTdc toi da bao nhieu gam hon hdp ba kirn loai tiT 14,16 gam X? 0,5 B 7,68 gam C 3,36 gam Hifdng d§n giai Ta c6: % N = 11,864% ^m^ = 0,12 mol => n = , ^ - ^ ^ = 1,68 (g) I vJL) =0,12 mol NO3 =>DapanD OH ''^i • ' • = 0'5 - 0,3 = 0,2 mol ^ ' " ' ' \ mc,(OH)3 = 0,2.103 = 20,6g ' D 6,72 gam lit dung dich X Lay lit dung djch X tac dung vdi dung dich BaCl2 (diT) thu diTdc 11,82 gam ket tua Mat khac, cho lit dung djch X vao dung djch CaCl2 (dir) roi dun nong, sau ket thuc cac phan iJng thu difdc 7,0 gam ket tua Gia trj cua a, m ti/dng vlng la A 0,04 va 4,8 _ = 14,16-0,12.62 = 6,72 (g) NO3 B 0,07 va 3,2 C 0,08 va 4,8 D 0,14 va 2,4 (Trich de thi tuyen sinh Dai hoc khoi A nam 2010) Trich de thi tuyen sink Dai hoc khoi B nam 2011} Ta c6: mki = mx - m Cl Cfiu 7: Cho m gam NaOH vao lit dung djch NaHC03 nong a mol/1, thu du^dc = 1,4M ^ 0,1 nN = n _ = 1,8 mol = n => D a p a n B 0,08 A 10,56 gam ^ = — - — = 0,9 mol Cr(0H)3 + OH- ^ Cr(OH); (tan) 0,3 0,3 iiKOH = 0,14 mol => [KOH] = — Cfiu 5: n(K.Na) Cr^^+ H - > C r ( H ) 0,5 =>a = 0,06 CO2 + KOH Trong m gam hon hdp Na, K c6 n(K, N-D = n Trong 2m gam hon hdp Na, K c6 Ap dung djnh luat bao toan nguyen to C ta c6: z:>0,l+ =31,95 gam cr 31 95 Vay phan ifng CO2 vdi KOH ngoai muoi K2CO3 c6 muoi KHCO3 "c =:> m cr 0,12 Ta thay; n; = 0,12 "C (trongCOj) _ (m + 31,95) = m + m ci BaCOji + 2KC1 0,12 D 51,5 gam (Trich de thi tuyen sinh Cao dang nam 2011) 0,1 BaCl2 + KiCOs C 30,9 gam ^=0: "BaC03 = Hil"(Jng d§n giai mol, n(,^,co3 = 0'07 mol PhiTdng trinh phan u-ng; NaOH + NaHCOj t> nw! r •: ^ > Na2C03 + H2O (1) ' Ap dung DLBT nguyen to C, so' mol C hai ket tua phai b^ng Ma ta thay 0,06 mol ;^tO,07 mol => Sau phan ifng (1) NaHCOjCon dir Vay dung djch X CO HCO3" va C03^' * Khi cho dung djch BaCl2 (dir) vao X: NaOH + NaHCOj , 0,06 mol 0,06 mol > Na2C03 + H2O 0,06 mol - V-'v-j:,t^/; u:0.y' > vt'v' Phuang phip Ba^* + ^m * ky thugt giSi nhanh BTTN H6a dgi cuong - v6 co - Dg XuSn Hung COj^" BaCOji 0,06 0,06 mol ^/ (2) x= — = 0,06.2.40 = 4,8g ) Na2C03 + COst + H2O 0,2mol > 'I I, (3) 0,1 mol CaCl2 + Na2C03 100% = 25% • Cachll Chpn so' mol cua hon hdp la Gpi so' mol cua N2 la x, thi ciia H2 la - x, so' mol N2 phan lirng la a Khi cho dung djch CaCl2 (dif) vao X roi dun nong: 2NaHC03 = 0,25 mol => H = — I N2 + 3H2 >-CaC03 + 2NaCl (4) 2NH3 , |j Ban dau: a - a PhaniJng: x 3x 2x Ta c6: so mol Na2C03 (4) = so mol Na2C03 (1) + so mol NazCOs (3) Sau phan ufng: a-x l-a-3x 2x => so mol NazCOj (3) = 0,7 - 0,6 = 0,1 mol HonhdpX: => so mol NaHCOj lit dd = so' mol NaHC03 (1) + so mol NaHC03 (3) Hon hdp Y c6 so' mol la: a - x + - a - 3x + 2x = I - 2x 0,7 mol 0,7 mol : ;; 28a + 2(1 - a) = 1,8.4 =^ a = 0,2 my = ( - x ) m a mx = m Y ( D L B T K L ) = 0,6 + 0,2 = 0,8 mol => a = 0,8.2/2 = 0,8 mol/1 => Dap an C ( - x ) = 1,8.4 Cfiu 8: Hon hdp X gom N2 va H2 c6 ti kho'i so vdi He hlng 1,8 Dun nong Hieusua'tphanu'ng= ^ ^ x l O O = : % 0,2 Y CO ti khoi so vdi He bkng Hieu suat cua phan iJng tong hdp NH3 la =:>DapanD B 36% C 40% D 25% Htfdng d§n giai 0i nong den phan iJng hoan toan, thu difdc 8,3 gam chat r^n Khoi liTdng A 0,8 gam ^ C 2,0 gam ' D 4,0 gam (Trich de thi tuyen sinh Dai hoc khoi A nam 2009) Siir dung sd dudng cheo ta c6: H2: ' B 8,3 gam Taco: M x = 1,8.4 = 7,2; M Y = 2.4 = HvCdng dSn giai Ap dung dinh luat bao toan khoi liTdng: ^5,2 ^7,2 ; ^ CuO CO hon hdp ban dau la Cach 1; 28 ^ , Cfiu 9: Cho luong CO (dU") di qua 9,1 gam hon hdp gom CuO va AI2O3 nung (Trich de thi tuyen sink Dai hoc khoi A nam 2010) N2: x = 0,05 X mot thdi gian binh kin (c6 bpt Fe lam xiic tac), thu du'dc hon hdp A 50% ,;fi;d.> _ 5,2 20,8 ]_ n i h : : mo= 9,1-8,3 = 0,8 (g) => n o = - V = 0.05 (mol) 16 20,8 ^ , V => ncuo=no= 0,05 (mol) => mcuo= 0,05.80 = (g) => Gia sur ban dau c6 mol N2 v^ mol H2 => mhh X = + m,^^ = 1.28 + 4.2 = 36 (g) =^DapanD Cfiu 10: Cho 3,68 gam hon hdp gom A l va Zn tac dung vdi mot liTdng vifa du Ap dung D L bao toan khoi lifdng ta c6: dung djch H2SO4 10% thu diTdc 2,24 lit H2 (d dktc) Kho'i liTdng dung djch mx = my => 36 = Hy M Y => 36 = nY.8 => ny = 4,5 mol thu difdc sau phan iJng la A 101,48 gam B 101,68 gam Pu' : N2 + 3H2 2NH, D 88,20 gam (Trich de thi tuyen sinh Dai hoc khoi A nam 2009) Di/a vao puf ta c6 : Hifdng d§n giai mol N2 phan drng thi sau phan iJng so' mol h6n hdp giam - = mol Vay X mol N2 phan uTng thi sau phan tifng so' mol hon hdp giam - 4,5 = 0,5 mol C 97,80 gam Ta c6: n „ „r, = n „ H2SO4 H2 =^^=0,lmol 22,4 ''~m l>'.' r • PhLiong phap va ky thugt g\i\h BTTN H6a dgi cuang - vO co - D5 Xuan HLfng _ 9,8x100 mddH2S04 = = 98 gam 10 Ap dung dinh luat bao toan khdi liTdng: =^ nb = 0,01n-0,01 (2) ^ Matkhdc: Ma+(2M+]6n)b = 2,9 => Ma + 2Mb + 16nb = 2,9 nih6nh(?pKL+ Tl^jj j^^SO^ - mddsaiiphantfng + T l p , ^ => m j d sau phan iJng=nih6n hdp K L + m^jj^^^Q^ - i; Va: a + 2b = 0,5.0,04 = 0,02 => na + 2nb = 0,02n m H2SO4 = 0,1x98 = 9,8gam The (2) vao (3) ta diTdc: ' Ma + M b + 16(0,01 n - , ! ) = 2,9 ^ m,j^ = 3,68 + - , x = 101,48 gam ' (3) => M(a + 2b) + 0,16n = 3,06 , Ma + 2Mb + 016n = 3,06 0,02M + 0,16n = 3,06 ^ M + n = 153 => Dap an B Cfiu 11: Nung nong m gam PbS ngoai khong sau mot thdi gian, thu dU'cJc h5n hdp rdn (c6 chtfa mot oxit) nSng 0,95 m gam Phan trSm khdi lUdng PbS da bj do't chay la M 145 137 129 =:> M la B a n (Ba) ^ A 74,69% B 95,00% C 25,31 % D 64,68 % (Trich de thi tuyen sinh Dai hoc khoi A nam 2009) HuT^ng dSn giai m(g) n Cfiu 13: Khur hoan toan mot oxit s^t X cT nhiet cao can vijfa du V lit CO (d dktc), sau phan fj^ng thu du"dc 0,84 gam F e va 0,02 mol CO2 Cong thtfc Ap dung D L B T K L ta c6: A F e O v a 0,224 B F e j O j va 0,448 C Fe304 va 0,448 D Fe304 va 0,224 m o = m - 0,95m = 0,05m (g) => no = 3,125.10"^m (mol) Ta c6: npbsphiinifng= npbo = n o => %PbS (da bi ddt chay) = (Trich de thi tuyen sinh Cao ddn^ khdi A,B nam 2009) = 3,125.10"^m (mol) 3,125.10"'m.239.100% Hifdng dSn giai Ta c6: = 74,69% m Cfiu 12: Hoa tan hoan toan 2,9 gam hon hdp gom kim loai M va oxit cua no vao nirdc, thu di/dc 500ml dung dich chiJa mot chat tan c6 nong dp 0,04M va 0,224 lit H2 (d dktc) Kim loai M la B.Ba C K = 0,02 mol => V^.^)^ = 0,448 lit ^ Goi cong thiJc tdng quat cua oxit la FcxOy: = 0,02 = =^ X la Fe,04 => Dapanc' Cfiu 14: Nung nong 16,8 gam hon hdp Au, Ag, Cu, F e , Zn vdi mot liTdng du" O2, den cac phan lirng xay hoan to^n, thu diTdc 23,2 gam cha't rdn X HuTdng d§n giai M : amol The tich dung dich HCl 2M vuTa du de phan iJng vdi cha't rdn X la M „ : b mol A 600ml 2M + 2nH20 ^ B 200ml 2M(0H)„ + nHz na C 800ml D 400ml (Trich de thi tuyen sinh Cao ddn^ khdi A.B nam 2009) n Hi^dng dSn giai Y Theo D L bao toan khdi liTdng : M n + n H ^ 2M(0H)n T a c o : n„^ = n^Q^ V a : nFe= 0,84:56 = 0,015(mol) THe: ^ = ^ y D Na (Trich de thi tuyen sinh Dai hoc khoi B nam 2009) b = ncophanung Mat khac: no(,rongoxi.)= ncophantfng = n^^o^ = 0,02 mol => Dap an A A.Ca ^j,; cua X va gia tri V Ian lu'dt la 0,95m (g) hh (PbO va PbS di/) + SO2 ) D a p an B niQ^ = 23,2-16,8 = 6,4 (g) 2b ^'' no = 6,4/16 = 0,4 mol Phan urng cua HCl vdi chat riln X CO the diTdc bieu dien vdi sd do: = 0,01 na = 0,02 (1) O'0,4 + 2H^ 0,8 H2O 1,' PhUdng phap va ky thujt giii nhanh BTTN H6a dgi cuong - v6 co - D5 Xuan Hung (lit) = 0 (ml) => Dap an D VHCI = 0,8:2 = 0,4 , Q^U C S u 15: Cho 11,36 gam hSn hdp g6m Fe, FeO, Fe^O,, va Fe304 phin iJng he't vdi dung djch HNO3 loang du", thu diTcfc 1,344 lit NO (san pham khi^ nha't) d (dktc) va dung dich X Co can dung djch X thu diTcfc m gam muo'i khan Gia trj cua m la: A 38,72 B 35,50 • • i Hrftfng d i n giai: 1,344 H N O viTa du, thu NO- Gii trj cua A 0,04 dung djch X (chi chilfa muo'i mol CujS vao axit sunfat) va nha't la: B 0,075 C 0,12 ^ D 0,06 Hrfdilg d§n giai > Fe2(S04)3 Ta CO sddo: 2FeS2 • ^ , 0,06 0,12 CujS - +HNO3 a 2CUSO4 2a Sd phan iJng: Ap dung djnh luat bao toan nguyen to S: Fe, FeO, Fe203, Fe304 + HNO3 ^ Fe(N03)3 + NO + H2O 0,12.2 + a = 0,06.3 + 2a -> a = 0,06 mol =^ Dap an D C S u 18: Nung hon hdp bot gom 15,2g Cr203 va m(g) A l d nhiet cao Sau Gpi X la so mol Fe(N03)3 phan iJng hoan toan, thu du^dc 23,3g hon hdp ran X Cho toan bp h6n hdp X Ap dung djnh luat bao toan nguyen to N, ta c6: "NCtrongHNOj) = " N (trong F e ( N ) ) + "N(trongNO) = + ^,06) phan uTng vdi axit HCl diT thoat V(/) H2 (dktc) Gia trj cua V la: mol A 4,48 lit Diravkosc(d6tathay: n a (Trich de thi tuyen sinh Dai hoc khoi A nam 2007) = 0,06 mol 22,4 a diTdc D 34,36 (Trich de thi tuyen sink Dai hoc khoi A nam 2008) -''^•:v*^''''^ Ta c6: n NO C 49,09 17: Hoa tan hoan toan hon hcJp gom , mol FeSa va =ln B 7,84 lit C 10,08 lit D 3,36 lit (Trich de thi tuyen sinh Dai hoc khoi A nam 2007) = ^(3x + 0,06) = (l,5x + 0,03) mol Hi^dng d i n giai Ap dung dinh luat bao toan khoi iiTdng, ta c6: Ta c6: n Cr203 mhh + m ^ ^ o ^ = mpg(^o^j^ + " ^ + m ^ ^ o => 11,36 + (3x + 0,06).63 = 242x + 0,06.30 + (1,5x + 0,03) 18 => X = 0,16 mol '"feCNO^jj 15,2 = 0,1 mol 152 1«' k;- Ap dung djnh luat bao toan kho'i liTdng: = 0.16.242 = 38,72 (g) =^ Dap an A 81 mAi= nihh- mcr203 = 3 - 15,2 = 8,1 (g) HAI = C S u 16: Hoa tan het 7,74 g hon hdp bpt Mg, AI blng 500ml dung djch hon hdp HCl I M va H2SO4 0,28M thu diTdc dung djch X va 8,736 lit H2 d (dktc) Co Phanu-ng: 2A1 + Cr203 0,2 can dung dich X thu du^dc lu'dng muoi khan la: A 38,93 gam B 103,85 gam C 25,95 gam 0,1 AI2O3: 0,1 , ,, 0,15 0,2 mj^^5o^ = mmuoi + ;| Cr + 2HC1 - > CrCl2 + H2t = 0'28.0,5 = 0,14 mol + mHci + mbl 2A1 + 6HC1 - > 2AICI3 + 3H2t 0.1 =0,5.1 =0,5 mol Ap dung dinh luat bao toan khoi Iifdng: mhh 0,1 mol Hon hdp X + dung djch HCl: Taco: n^ - ^ - ^ =0,39 mol "2 22.4 "H2SO4 0,2 03 Vay hon hdp X gom: Cr: 0,2 mol Hifitng d i n giai nHci + AI2O3 = fAl dir: 0,3-0,2 = 0,1 mol D 77,86 gam (Trich de thi tuyen sinh Cao dang khoi A,B nam 2008) 2Cr 27 m^^^ => m , w i = 7,74 + 0,5.36,5 + 0,14.98 - 0,39.2 = 38,93 gam=> Dap an A => n „ 0,2 = 0,15+ 0,2 = 0,35 mol ^ V H2 =7,84 lit => Ddp an B 13 PhuOng ph&p va ky thujt g i i i ntianh BTTIM H6a dgi cuong - vO c o - D5 X u a n Hung Cau 4: Sue het mot liTdng clo vao dung dich hon hdp NaBr va Nal, dun nong thu Cau 1: Hoa tan hoan to^n 2,8Ig hon hdp gom FcjOj, MgO, ZnO 500ni| diTdc 2,34g NaCl So mol hon hdp NaBr va Nal da phan iirng la: B 0,15 mol C 0,02 mol D 0,04 mol dung djch H S O 0,1M viTa du Co can dung dich sau phan iJng thl thu diTOc A.O.lmol HUdngdSngiai bao nhieu gam muoi khan: Ap dung djnh luat bao toan nguyen to Na ta c6: A 6,81g B.4,81g C 3,81g D 5,81g 2,34 (Trich de thi tuyen sink Dai hoc khoi A nam 2007j UNaBr + " N a l - i N a C l = 0,04 mol => Dap an D 58,5 Hif(}ng dan giai cau 5: Cho 16,3g hon hdp kim loai Na va X tac dung het vdi HCl loang, dtf Ta c6: n H , ~ " H S O = 0,1.0,5 = 0,05 mol thu di/dc 34,05g hon hdp muoi khan A The tich H2 thu du'dc la bao nhieu lit? Ap dung djnh luat bao toan khoi lU'dng: mhh + ITIH2SO4 ~ "^H^o A 3,36 B.5,6 c! 8,4 D 11,2 =^ m = 2,81 +0,05.98-(0,05.18) = 6,81 (g) ^ Dap an A HifcJng din giai c1* Cau 2: Cho 24,4g hon hdp NajCOj va K2CO3 tac dung vCfa du vdi dung dicli Ta c6: m„,u6'i = m k i m i o a i + m^^_ •1 BaCl2 Sau phan iJng thu du'dc 39,4g ke't tua Loc tach ket tua, c6 can dun^ =:>m = , - , = 17,75 =^ n _ = 0,5 mol djch thu di/dc m gam muoi clorua Gia trj cua m la: cr ci A 2,66 B 22,6 C 26,6 D 6,26 Phu'dng trinh phan iJng: '"' ' Hifdng din giai 2Na + 2HC1 > 2Na^ + C r + H2 39,4 2X + 2nHCl > 2X"^ + 2nCr + nH2 Ta c6: n'BaCl2 n.^, ~ = "n^B a C O j ~ = 0,2 mol C BAI TAP AP DgNG Theo phu'dng trinh phan iJng ta c6: n^,^ = = "O'S = 0,25 mol Ap dung dinh luat bao toan khoi lU'dng: mhh + 'TiBaCi2 ~ " ^ k e t t i i a + m = 5,6 (lit) Dap an B => m = 24,4 + 0,2.208 - 39,4 = 26,6 gam => Dap an C Cau 3: Cho 0,52 gam hon hdp kim loai Mg va Fe tan hoan toan dung Cau 6: Hoa tan 10,14g hdp kim Cu, Mg, Al bing mot liTdng vHa du dung dich djch H S O loang, dU' thay c6 0,336 lit thoat (dktc) Khoi lifdng hon hdp HCl thu diTdc 7,84 lit A (dktc) va l,54g chat ran B va dung djch C Co can muoi sunfat khan thu dufdc la: dung djch C thu dufdc m gam muoK Gia tri cua m la: A gam B 2,4 gam C 3,92 gam D 1,96 gam A 33,45 B 33,25 C 32,99 D 35,58 Hifdng dan giai Ta CO muo'i thu du'dc gom MgS04 va FeS04 Theo dinh luat bao toan kho'i liTdng: m„u,^i = m k i , „ i o a i + m , (1) Tac6 :n„ = so| =0,015 mol 22,4 PhiTdngtrinhphanurng: Mg + H2SO4 > Mg^* + 804^" + H z t Fe + H2SO4 > Fe^* + "2 804^" Theo phUdng trinh phan tfng ta c6: " ^ ^ - ~ " H ~ mol Tfif (1) = > m n , u ^ i = 0,52 + 0,015.96 = 1,96 gam => Dap an D + H2t Hifdng d§n giai 7,84 = 0,35 mol ^ n =2n, =2.0,35 = 0,7 mol Taco n„ = — "2 22,4 cr "2 Theo djnh luat bao toan kho'i liTdng: m = m(Ai + M g ) + m^^_ = (10,14 - 1,54) + 0,7.35,5 = 33,45 gam => Dap an A cau 7: Hoa tan 28,4g hon hdp muo'i cacbonat cua kim loai thuoc nhom lA bang axit HCl thu diTdc 6,72 lit (dktc) va dung djch A Tong kho'i li^dng muoi clorua dung dich thu diTdc la: A 3,17 B 31,7 C 1,37 D 7,13 15 Phuong phap va ky thujt g'A't nhanh B T T N H6a d^i cuong - vfl cO - D XuSn Hmg Hrfdng dan giai Hifdng din giai Goi cong thiJc chung cua muoi cacbonat kim loai nhom lA la R2CO3 f 2RCI + CO2 + H2O R2CO3 +2HC1 Ta c6: n H ~ " c O i '2 6,72 22,4 = 0,3 mol; DHCI = n(.Q = 0,6 T a c : ng^ci^-Hg^so^)j mol ri2> = ^8,2 + 0,3.208 - 69,9 = 30,7 (g) , f | iJ's./Jtv.;;! CSu 11: Cho 6,2g hon hdp gom mot s6' kim loai kiem vao dung djch HCl du" thu =>m = 28,4 + 0,6.36,5 - (0,3.44 + 0,3.18) = 31,7 (g) => Dap an B diTdc 2,24 lit H2 (dktc) Co can dung djch thu diTdc sau phan ilng se thu diTdc CSu 8: Tron 5,4 gam nhom vdi 6,0 gam FcaOs roi nung nong de tht/c hien phan bao nhieu gam chat ran: A 1,33 B.3,13 iJng nhiet nhom Sau phan ufng ta thu difdc m gam hon hdp chat ran Gia tri cua m la: B 9,40 gam Ap dung djnh luat bao to^n khoiJurdng: mhSnhdp + mBaCi2 ~ =>DapanB, nihh + niHci = m + m , , ^ + m^^ Q A 2,24 gam = 0,3 mol => m = mnan H^P + ^n.a, ' n^-' Ap dung djnh luat bao loan khoi liTcJng ta c6: J2 69,9 C 10,20 gam C 13,3 D.3,31 Hvldng d§n giai: D 11,40 gam )1 Htf^ng d i n giai 2,24 Taco nH^=22^^ = 0,1 mol Theo djnh luat bao toan khoi lufdng Ta c6: mnn sa., = mnn tnrdc = 5,4 + 6,0 = 11,4 Ma: n gam CSu 9: Thoi mot luong CO dU' qua o'ng siJ di/ng m gam hon hdp gom CuO, =2nH = 2.0,1 = 0,2 mol => m„.5-i = m k i „ „ o , i + ni^^_ = 6,2 + 0,2.35,5 = 13,3 (g) =^ Dap an C Fe203, FeO, AI2O3 nung nong thu diTdc 2,5g chat ran Toan bp thoat sue vao nu'dc voi diT tha'y c6 15 gam ket tua trang Khoi liTdng cua hon hdp oxit kim loai ban dau la: A 7,4 gam B 4,9 gam C 9,8 gam dung djch H2SO4 loang thu diTdc V lit d (dktc) va 7,48g muoi sunfat khan Gia trj cua V la: D 23 gam B 1,008 + yCO — - — > x M + yC02 Ca(0H)2 + C O - > CaCOj + H2O Hu ^C02~ "CaCO, "77^" ^''^ lUU ma: moxit= mki,„ioai + m o x i moxit= ; > • 2,5+ 0,15.16 = 4,9 gam C&u 10: Mot dung dich chiJa 38,2g hon hdp muoi sunfat ciia kim loai kiem A va kim loai kiem tho B tac dung viTa du vdi dung dich BaCl2 thu duTdc 69,9g ket tua Lpc b6 ke't tua va c6 can dung djch sau phan tfng thu du'dc bao nhieu gam muoi khan: : ^ ©: 30X- '••''''^ ; C 7,03 ' =n H2 + SO^ 5,76 ,2J4 so 96 = 0,06 mol ' V = 1,344 (lit) H2 => Dap an A Cfiu 13: Cho 2,81 gam h§n hdiJ A g6m oxit Fe203, MgO, ZnO tan viTa du 300ml dung djch H2SO4 0,1M Co can dung djch sau phan ijTng, kho'i liTdng hon => Dap an B • A 3,07.; f , D 3.36 ,2- Ta c6 sd do: H2SO4 Theo djnh luat bao toan nguyen to ta c6: =nco = C 1,12 Htfdng dSn giai: m ,_ = , - 1,72 = 5,76 (g) ; Ta c6: mn,u6'i= mki,„ioai + m so so,2- Cac phiTdng tnnh h6a hoc: no(trongoxit) ^, A 1,344 Hxidng d§n giai MxOy Cfiu 12: Hoa tan het 1,72 gam hon hdp kim loai gom Mg, A l , Zn va Fe b^ng D 70,3 hdp cac muoi sunfat khan tao la: A 3,81 gam B 4,81 gam C 5,21 gam D 4,8 gam Hi^dng dSn giai dung djnh luat bao tohn kho'i lifdng: moxii+ ni^^so^ ="^H20 + mn„,fi-i = moxii + mH S O 'muoi = "^HjO THl/ VIEN TINH B!NH THUAN Phiiang ph^p va ky thugt giSi nhanh BTTN H6a dgi CLiong - vO CO - B5 Xuan Hung Cfiu 17: Sue k h i clo vao dung dich NaBr va N a l den phan tfng hoan toan ta thu^ Trong do: nj.,^o = n^^^Q^ = 0,3.0,1 = 0,03 m o l Vay: mn,u6-i= (Ji/dc l , g N a C l X a c djnh so m o l hon hdp NaBr va N a l c6 dung djch 2,81 + , - , = 5,21 gam =:> D a p a n C ban dau? Cfiu 14: Hoa tan het lOg hon hcJp muoi cacbonat M g C O j , CaCOj, NaaCOs, K C O siA.O.lmol b^ng dung djch H C l duf thu dugc 2,24 lit (dktc) va dung dich Y Co can dung B 11,1 C 11,8 24 n N a B r + " N a l = HNaci = »'v«> ''ifv T a c : n H c i = n c o = ^ = 0,2mol; n^^Q = n^o^ = , m o l A p dung dinh luat bao toan khoi lu'dng: mhh + mHci = m + m^^Q^ + m^^^ m = + , , - ( , 4 + 0,1.18)= 11,1 (g) =:> D a p a n B C 13,2g ' G i a t r i c u a V i a : ^ '»•'(! B 0,112 l i t C 5,6 l i t D 0,224 l i t B 15,8 gam C 2,54 gam D 25,4 gam Gia trj cua m l a : A 1,58 gam Htfdng dan giai Ta nhan thay, k i m loai tac dung vdi oxi va H2SO4, so mol n„ = n Trong do: mo = Sau phan ^ng k h o i lu'dng ong suT giam 5,6g Co can dung djch A thu difdc m(g) muo'i Gia trj cua m l a : "H, = n D 39,65 ^ Hvtdng d§n giai Kh6'i liTdng dng si? giam chinh la k h o i liTcJng ciia nguyen to o x i ~ " H O O ~ " o (trong oxit) = ^ = 0,35 mol = SO]' hay =nr, T dung dich A L i / d n g k h i H2 tao dan vao o'ng stj" difng CuO dU' nung n6ng " H , H2 (dktc) Co can dung djch thu diTdc m gam muoi khan A 2,24 l i t Cfiu 16: H o a tan 14,8g hon hdp A l , Fe, Z n bang dung dich H C l viTa du thu diTdc Ap dung djnh luat bao to^n nguyen to' ta c6: D Phan 2: T a n hoan toan dung djch H2SO4 loang thu diTcJc V l i t i::>m = 1 , + , , - ( , 4 + 0,1.18) = 12,6 (g) => D a p a n B C 27,575 an - "H2O ~ " c O j ~ B.28,8 Dap Phan 1: B j o x i hoa hoan toan thu diTdc 0,78gam hon h d p oxit " ~ D 12,3g A p dung djnh luat bao toan khoi lufdng: mnn + mHci = m + m^^^^ + m^^^Q A 20,6 => -J - Hifdng d i n giai Ta c6: nHci = ^^co2 ~ ^'^ mol , b i n g nhau: thu diTdc 2,24 lit CO2 (dktc) Khoi li/cJng muoi clorua tao la: B 12,6g = 0,02 56,5 , Cfiu 18: Chia 1,24 gam hon hcJp hai k i m loai c6 hoa trj khong do'i phan Cfiu 15: Cho l , g hon hcfp gom ACO3, B2CO3, R2CO3 tan het dung djch HCl A 16,2g D 0,02 m o l ' ' i ' A p dung djnh luat bao toan nguyen to' natri, ta c6: D 14,2 Hifdng dSn giai ur; C 0,015 m o l JHrfdng dSn giai djch Y thu diTdc x gam muoi khan Gia trj cua x la: A 12 B 0,15 mol moxit - m i c i n , i o a i 2- = " o = ^ SO4 24 = 0,78 - ^ = 0,16gam g.^l.l - = 0,01 m o l => V = 0,01.22,4 = 0,224 l i t 16 => D a p a n D ^- nimuoi = mi(in, loai + m , so| 24 = - — + 0,01.96 = 1,58 gam => D a p a n A , Cfiu 19: H o a tan hoan toan 10 gam hon hdp M g v a Fe dung djch H C l dir thay tao 2,24 l i t k h i hidro (dkTc) Co can dung djch sau phan ilng thu diTdc => n = n„ = 2.0,35 = 0,7 mol muoi khan K h o i liTcJng m u o i khan thu difdc l a : A 1.71 gam B 17,1 gam D 34,2 gam Hifdng dSn giai => m„„ D a p &n D C 3,42 gam Tac6:n = n „ =2.( — cr "2 ) = 0,2mol '22,4 , Phuong ph^p va ky thugt giSi nhanh BTTN H6a dji cuong - vO co - D XuSn Hung Ma: 16 + 28.0,3 = mpe + 0,3.44 ^ mpe = 1 , gam + m^^_ = 10 + 0,2.35,5 =17,1 gam=> Dap an B, m,„ua-i = mki,„ioai Dap an D CSu 20: Hoa tan hoan toan 20 gam hon hcJp gom Mg va Fe vao dung dich axit HCl dirthafy c6 11,2 lit (dktc) thoat va dung djch X Co can dung djch X B 45,5 gam C 55,5 gam 'r«,KB - :u A1(0H)3 + NaHCOj AI2O3 -H 3H20 2Ai(OH)3 Do do: UAJ(band^u)= n AI2O3 sau phan iJug qua dung djch Ca(0H)2 diT, thay tao 30 gam ket tua Kho'i lUUng s^t thu duUc la: ^ ? , « ~ C 9,6 gam rt M r,nco= nAi(bandiiu)=2n DK.* xFe + yCOz (1) =0,4 mol " -''^ ~' ^ (trong F e , y = — = 0,3 mol ^ = 0,3 mol mol => m^ = 0,1 27 = 2,7 ' gam =0,1 27 = 2,7 gam mAi 102 n A i ( i ) = n A i ( b a n d u ) - n A i ( ) = 0,1 -0,02 = 0,08 mol Theo djnh luat bao toan khoi liTdng nguyen to' oxi, ta c6: ^ ' V - ' - CO2 + Ca(OH)2 ^ CaCOa + H2O CaC03 = — = 0,1 102 = — = , mol "Ai(band5u)= n A i ( i ) + nAi(2) => 22,4 n "^W,;"^^ ' •"Fe o = 9.66 - 2,7 = 6,96 gam => Dap an A AI2O3 ^ (3) ,m D 11,2 gam Hufdng d i n giai yCO + FexOy — Nhan xet: Ta't ca liTdng Al ban dau chuyen het ve AI2O3 ( ) CSu 22: Thdi 8,96 lit CO (dktc) qua 16 gam FcxOy nung nong Dan toan bo liTdng B 6,4 gam v = 1,3 mol =^m = 38,6 + 1,3.35,5 = 84,75 gam => Ddp an B A 9,2 gam 0,03 i ^nl A „ 6,96-0,12.16 , npe = =0,09 mol Theo dinh luai bao toan Idio'i Ii/dng c6: mp^ Q + mco = m p e + ^002 1,5.0,08 = 0,12 mol s- ' ^^ n c-(,./«/V ^ t • 56 100 => nco > n^Q^ -> CO diT va Fe^Oy het A l ) = "Fe: no = 0,09 : 0,12 = : => CTPT la Fe304 => Dap an C ' ' Cfiu 24: Khur hoan toan 32g hon hdp CuO va Fe203 bang H2dirthay tao 9g H"20 Kho'i liTdng hon hdp kim loai thu diTdc la: A 12 gam B 16 gam C 24 gam D 26 gam P h i / a n g p h S p v k y I h u j t g i i i n h a n h B T T N H a d^i c u o n g - vfl CO - Fe, "•, D5 X u a n H u n g > 3Fe(NO.,)2 + 2Fe(N03)3 Qud trinh nhan electron: 0,05 0,1 ' ••'d LiTdng sat d ca phan ilng la n^e = 0,1 + 0,05 = 0,15 mol , „ , , 4H^ m = 0,15.56+ 2,4= 10,8 (g) lOjr + =>DdpanC Cfiu 10: Hoa tan hoan to^n hon hdp gom a mol FeS^; 0,5a mol FeS va 0,06 mol CU2S vao axit H N O (vifa du), thu du'dc dung dich X (chi chiifa muoi sunfat) A 0,24 mol B 0,20 mol C 0,12 mol D 0,06 mol Hil'dng dfin giai I => -> 0,06 - > 2CUSO4 dung dich H N O I M thu diTdc dung djch B va hon hdp C gom 0,05 mol N2O; 0,1 mol NO va lai 2,8(g) kim loai Gia tn cua V la: A 1,15 B 1,22 C 0,9 Sau phan i^ng lai 2,8g kim loai, la khoi lu'dng cua Fe diT npeprf = 0,35 - ^ = 0,3 mol ' ' r'.*,"'''" ft: ^ ^ ;,j V i Fe diTsau phan iJng nen muoi tao la muoi Fe^* In, " ' > • „i,;„, = 0,7 mol (v6 1i) • + NO, + 8e NII4' +3H:0 0.2 I n , = 0,4 + 0,5 + 0.25 =1,15 mol 1,15 = 1,15 lit => Dap an A > Mg 0,3 +2 > Fe " Ine nhuang = 0,3 + 0,6 = 0,9 mol C 0,075 D 0,08 , Goi a, b Ian lu-dt la so' mol Fe phan i^ng va so mol oxi phan uTng v 5a.p +3 Fe 3e C ; i • ;:^::;,:c,.sx ^ - • -> Fe 3a' )ua trinh nhan electron: 1"; -2 O2 +3 +2 B.0,07 IIiMng dan giai b 0,6 >f mol = 0.9 mol > A 0,035 Qua trinh nhu'dng electron: 0,3 = 0,7 la: Ap dung dinh luat bao toan electron, ta c6: 0,05 Qua tnnh nhu'dng electron: , D 1,1 Hrfdng dfin giai Fe - 2e ,3 hoan toan oxit sAt dung dich I I N O thu du'dc 0,07 mol N O Gia tri cua a Cfiu 11: Cho hon hdp A gom 0,15 mol Mg va 0,35 mol Fe phan iJng vdi V lit 2e + 0,4 , Cau 12: a mol sat bj oxi hoa khong difdc 5,04 gam s;tt oxit, hoa tan 0,12 , + Be -> N O + H O In,, m,,,,,,, V,U N O , ;v*"f>J < 2a + 0,5a + 0,06 = 0,5a.3 + 0,25a.3 + 0,12 => a = 0,24 =^ Dap an A 0,15 „h.nn = 0,3 + 2H2O •=> Con mot qua trinh nhan electron niTa, qua Irlnh la qua trinh tao iniioi Vay: Ap dung djnh luat bao toan nguyen to vdi S, ta c6: - Ine Fe2(S04)3 CU2S Mg 2NO3 0,25 0,5a 0,25a 0,1 0,4 iOir 2FeS2 -> Fe2(S04)3 0,5a NO 0,3 N l I j N O , va n, „h„, = 0,9 - 0,7 = 0,2 mol Dung dich X chi chiJa muoi sunfat nen ta c6 sd do: 2FeS + 3e 0,5 Ta thay: va NO nha't Gia trj cua a la: a + NO3' 0,4 N + 4e -> 4b +4 + le - i( I'vs ^rA:i i H in ,C< -j ' N (NOj-) 0,07 0,07 Ap dung dinh luat bao toiin electron, ta c6: 3a = 4b + 0,07 (1) Mill khac, khoi liTdng oxit: 56a + 32b = 5,04 (2) TCr (1), (2) =^ a = 0,07 va b = 0,035 => Dap an B -fiu 13: Hoa tan vto du m(g) hon hdp FeO va r'e,04 dung dich chuTa 1,2 nioi HCl Co can dung dich du'dc 70,6(g) muoi khan Gia tri cua m la: A 37,6 B 32,8 C 30,4 D 26,8 Phuang phap va ky thujt giai nhanh BTTN Hba dgi cuong - vfl co - D3 Xuan Hung 16: Cho 14(g) bot Fe tac dung vdi lit dung dich FeClj 0,1M va CUCI2 HiftJng d§n giai K h i hoa tan hon hcJp oxit sat H C l , sif ket hdp giffa oxi oxit va fj+ 0,15M Ket thuc phan iJng thu du'dc cha't r^n A CO kho'i lu'dng: cua axit diTdc bieu dien nhiTsau: A 9,6g 2H* O^" + =>n„r,=-n H2O ^ , H2O »f' , nioxit =^ + moxi, mniudfi + m„^Q Phan uTng xay theo trinh tif: Fe = 70,6 + 0,6.18-1,2.36,5 = 37,6 (g) =^ D a p a n A B FeO C Fe304 15 100.36,5 2H^ + O'" 0,15 0,075 m p e (trongoxio -> = - 0,075.16 = 2,8 (g) => B 0,36 A 0,18 ni-e Fe — ^ Fe mpedtr + > - = 56.0,05 + 64.0,15 = 1?,4 (g) => D a p a n C mcu B.Ag,Cu C 0,54 > FeCh + 2AgN03 C Ag, Cu, Fe D Cu, Fe + Cu(N03)2 Vay: Fe(N03)2+ Cu , 0,01 3(g) => Sau phan ufng tren Fe diT => D a p a n C D 1,08 c a u 18: Cho 28 (g) Fe vao dung djch chiJa 1,1 mol AgN03, ket thiic phan ilng thu dufdc chat ran X va sau c6 can dung djch m u o i thu dufdc m(g) muoi khan Gia trj cua m la: ' B.96,2 ^ C 118,8 D 108 Hifdng dan giai 28 Ta c6: np^ 3x V a y : b = 18.4x=: 18.4.0,015 = 1,08 (g) 2Ag mx = m A g + mc„= 108.0,02 + 64.0,01 = 2,8 (g) < + Hj „ Theo phan ti^ng, t a c o : 3x = 0,045 => > 0,02 A 31,4 4x Fe(N03)2 + 0,01 3Fe + 4H2O 3x => D a p a n D = 0,02 Chi CO Fe304 bj H2 khur, goi X la so mol Fe304 '3X , Gia su' AgNOs va Cu(N03)2 phan iJng het, ta c6 trinh tiT phan i^ng: = 0,05 mol H i r i n g d§n giai 2HC1 - Hifdng d i n giai cua b la: + (2) 0,15 A.Ag,Fe H2O cha't ra'n A H o t a n het A dung dich HCl dU" du^dc 0,045 mol H2 Gia tri Fe + Cu Cu(N03)2 Phan lifng ket thuc diTdc chat ran X c6 khoi li/dng 3(g) Trong X c6: CSu 15: Nung a(g) hSn hdp A l j O * va Fe304 vdi H2 di/, thu du-dc b(g) H2O va c(g) X ' Cfiu 17: Cho bot Fe tac dung v6i dung djch chufa 0,02 mol AgN03 va 0,01 mol X _ np^ _ 0,05 Taco: - = = — => oxit Fe203 => D a p an A y no 0,075 Fe304 + 4H2 0,15 0,15 mA = 0,15 mol Sir ket hdp giffa H'" va oxi (trong oxit) diTdc bieu dien bang sd do: => (1) leo phan ufng (1), (2); chat rdn A gom Cu va Fe diT : 0,2 - 0,15 = 0,05 mol = d.V = 1,05.52,14 = 54,747 (g) 54,747.10 = 0,1 + CUCI2 -> FeCls Fe D FesOj va FeO Hufdng d i n giai "HCI 3FeCl2 2FeCl3 => So mol Fe de phan i^ng v6i C u C h la: 0,25 - 0,05 = 0,2 mol A FezOa ^ + 0,05 thifc cua oxit la: mddHC! np^^,, =0,1.1 = 0,1 mol = 0,15.1 = 0,15 mol 'CuCl CO: c a u 14: De hba tan gam Fe^Oy can 52,14 ml dung djch HCl 10% (d = l,05g/ml) Cong Taco: 14 np =— = 0,25 m o l ; 56 Taco: _^ = - , = 0,0 mol H"" ITIHCI = D l l , g C 12,4g Hif()ng d i n giai mo Vay: (trons o x i l l Vay: mpc =0,5.56 = m - , m => m = 112(g) =^ D a p a n D , i CO, ~ "caCO, ~ * ~' ^^^"^ Cachkhac: mi-e phsn Ifng = m - 0,75m = 0,25m (g) = 0,07.1 = ,1 (g) Goi X, y Ian lirdt la so' mol cua N O va N O m = n v , , - mo = 4,06 - 1,12 = 2,94 (g) => D a p an A Tacd: Cfiu 20: C h o m (g) h o n h d p FeO, Fe.O, va Fe^Oa tac d u n g v d i C O d u - d u n nong, 4H^ nifdc v o i t r o n g diTdiTdc (g) k e t l i i a G i a t r i c u a m l a : 4x r B.5,72 C 4,92 D 5,04 2H" Hir(?n{» dSn giai = > l o (trong oxii) + O -> = nco = H^.Q A^=» CO2 * ' NO 3x + NO3" + l e , + 2H2O X -> NO2 + H2O y y Jl ;J +2 Fe - 2e 0,25m 0,5m 56 'r C a u 21: C h o 0,24 m o l F e va 0,03 m o l Fe304 v a o d u n g d i c h HNO., l o a n g , k e t thuc > Fe A.48,6g , B 58,08g C 56,97g I A p dung dinh luat bao toan electron, ta c6: = 3x + 'y « ^56 =3.0,31+0,07=1 m = 1 ( g ) = > D a p an D Idn hdn 9,4 (g) Cong thufc cua oxit sat la: D o sau p h a n liYig sit c o n du-nen d u n g d i c h X cMa m u o i Fe(N03)2 A F e O A p d u n g sir b a o t o a n kho'i lu'dng (so' m o l ) Fe t a c6: => „^,„„^,^ = 0,27 mol ^ ' ^56 "^ 'fiu 23: 6,72 (g) F e tac dung vdi oxi tao oxit sdt nhat c6 khoi lirdng D 65,34g Hvldng d§n giai 0,24 + 0,03.3 = „ „ „ „ ^ , ^ ^ ' 56 p h a n i f n g t h u dUdc d u n g d j c h X va 3,36(g) k i m l o a i du" K h o ' i l U d n g muo'i c6 dung dich X la: 0.1 T i r ( l ) , (2) => x = 0,31 mol v a y = 0,07 mol Qua tnnh nhu'dng electron: m „ x i , = mp, + mo = 3,92 + , = 5,04 (g) => D a p a n D Ub n^^^ = x + 2y = 1,38 (2) joo => mo(,rong„xi,) = 0,07.16 = 1,12 (g) Vay: • % , = 0,07 m o l = u^.^^^ = + N O ' + 3e y T r o n g qua t r i n h khCf o x i t sat bcfi C O , CO d a l a y o x i t r o n g o x i t tao CO2 Theo sddo: C O x + y = 0,38 (1) Qua tnnh nhan electron: Sau p h a n iJng t h u dUdc 3,92 (g) Fe S a n p h a m k h i t a o t h a n h d i q u a d u n g dich A 3,52 1,38 = 2.n,,^,^03)2 + ^ ' ^ ^ = ^ n F c ( N ) = 0,5 m o l = npeph,n„>yng B.Fe304 C FezOj HiMng d§n giai _ _ 6,72 T a c o : n, = - ^ - , mol 56 ' D F e O hoac Fe203 • Phuong u''.ap va ky thuat g\i\h BTTN H6a d?i cuong - vO co - D5 Xuan Hung Goi cong thtfc oxit s£t la FCxOy 9,4-6,72 =>^o> 16 Tile : ^ > n Fe '' Chon u 26: Hoa tan m (g) hon hdp Fe va Cu, Fe chig'm 40% kho'i liTdng bang dung djch HNO3 thu diTcJc dung djch X; 0,448 lit NO nhS't (dktc) va -0,1675 mol lai 0,65m (g) kim loai Kho'i liTdng muo'i dung djch X 1^: ^ ^ hay ^ > 1,396 => - < 0,716 , M W ? > i 0,12 - = - = 0,67 < 0,716 y A.5,4g fiorf: B.6,4g C 11,2g D 4,8g Hifdng dSn giai Ta c6: FezOa =^ Dap an C mpe = 40%.m = 0,4m (g) => mcu = m - 0,4m = 0,6m (g) Sau phan iJng c6n 0,65m (g) kim loai > mcu = 0,6m (g) Cau 24: Cho 2,11(g) hon hdp Fe, Cu, A l hoa tan het bdi dung djch HNO,, tao Vay: mpe A.9,62g Do Fe d\i sau phan iJng nen tao muoi Fe^"" C 7,86g D.5,18g + 3e N 0,7m 0,35m 56 56 56 0,06 N03 = mklmloai = 2,11 + 62.(0,06 + 0,04) = 8,31 (g) Taco: n, 'Fe(N03)2- Cfiu 25: Hoa tan m (g) hon hdp A gom FeO va Fe203 bang dung djch HNO., thu diTdc 0,01 mol NO Nung m (g) hon hdp A vdi a mol CO diTdc b (g) chat dn B roi hoa tan HNO3 thi du^dc 0,034 mol NO Gia trj ciia a la: B 0,036 C 0,03 0,448 22,4 = 0,02 ^""^"^ ) F e + 0,01molNO +3 >Fe +0,034 mol NO So mpl electron nhan bdi HNO3 qua trtnh (2): 0,034.3 =0,102 mol Taco: 2a+ (3.0,01) = 0,102 => a = 0,036 mol B 50,4 I n hm, IM '\;:»b • ^ C 50,2 D 50 ^j, , Hifdng d i n giai ' Ta c6: So mol electron cho bdi CO va nhan hd\3 tir(I): [2a + (3.0,01)] mol ^ 0,03 mol u 27: Hon hdp X gom Cu va Fe c6 ti le khoi liTdng tu-dng uTng la : Lay m (2) Dura vao sd ta tha'y: 55 "iFe(N03)2 = 0,03.180 = 5,4 (g) =^ Dap an A mla: A 40,5 (1) 0,35.4,8 5^ lai 0,75m gam chat r^n va c6 0,25 mol Y gom NO va NO2 Gia tri ciia Ta c6 sd chuyen h6a: A(Fe,Oy) 0,35m •gam X phan ilng hoin toan vdi dung dich chiJa 0,7 mol HNO3 Sau phan iJng D 0,04 Htidng dan giai =>DapanB N (NO) 0,7m lAp dung djnh luat bao toan electron, ta c6: '^^ = 0,06 => m = 4,8 + 62.nenhSn => Dap an B B > Fe •\bn Ta C6: mn„|o-i = midmioni + B< 2e 0,04 A 0,024 - , i' 0,35m > N (NO2) 0,04 - Qua trtnh nhan electron: N + 3e - + le , +5 0,02 + _ +2 Fe > N(NO) 0,06 + = 0,4m - 0,05m = 0,35m (g) Qua trtnh nhiTdng electron: Qud trtnh nhan electron: N phAn ?ng HvlSng d i n giai ^ => Kho'i lu-dng Fe diT : 0,65m - 0,6m = 0,05m (g) 0,02 mol NO va 0,04 mol NO2 Khoi lu-dng muoi khan thu diTdc la: B 8,31g ^^ mp^ = m =:0,3m(g); mcu = 0,7m (g) 3+7 Vi Fe phan iJng tri/dc Cu va sau phan tfng c6n 0,75m (g) chat ran [Cu chu'a phan iJng 0,7m ^ I Fe dir 0,75m - 0,7m = 0,05m => mpe p, = 0,3m - 0,05m = 0,25m (g) Fedir [chi tao muo'i Fe(N03)2 ( H N O J het (lim y chi H ^ het; NO3" mu6'i) PhifOng phap va ky thuat g&i nhanh BTTN Hoa dai CKOng - v6 ca ~ D6 XuSn Hung - Q u a t r i n h nhiTdng e l e c t r o n : Fe - 2e 0.25 m +2 • Qua trinh nhan electron: + H * + 3e ^ CI,',: 4a 28 NO3 + 2ir fa ^ "han = 3a => X e - 28 p h a n ufng l a k h o n g d a n g k e T i l e P1/P2 l a : m = 50,4(g) ^ Dap an B A 0,5 , ^ l i i t " ' - cvrr? crf-iV C , w + I + 2e O2 -2 -•12ghhX(Fe,FexOv) +H2S04d,t° — • D i / a v a o s d d o Ui thay F e la chat khu", O2 va H2SO4 dac la chat o x i hoa: , T a co: + -O2 > (' ) ;' , (1) ?.,{ F e z O j + 2CO2 _ , , , (2) ^ 2x = 0,75x mol k h i P h a n iJng (2) l a m tdng x P, _ ' ' " = 0,75x mol k h i T a thay lu'dng mol tang va giam b^ng P , = P2 h a y 12 - m , n^^ = — mol 32 ^ p i a n urng (1) l a m g i a m , +4 F e ' " ' + SO2 -3e rr HiMng dStt giai X' i +4e D 2,5 }m 2Fe203 + 8SO2 fceC03 r , , « m(g)Fe^ i ' , 1= '5., D 9,72 Turduy b a i toan theo sd d o : , C ' X Htfdng dSn giai: M ; , f.;, llx d a c n o n g d i f d c 2,24 l i t SO2 ( d k t c ) G i a t r i c u a m l a : B 9,62 B 4FeS2 + 1IO2 - c h a t ra'n X g o m F e , F e O , F e , FejO,, H o a t a n h e t X t r o n g d u n g d j c h H2SO4 V D a p an A b i n h l u c n a y l a P2 a t m B i e t r h n g t h e t i c h c h a t ra'n t r o n g b i n h tru'dc v a sau C S u : D e m ( g ) p h o i b a o F e n g o a i k h o n g k h i , sau m o t thcfi g i a n du'dc 12 gam A 9,52 + , = > m = 9,52 ( g ) p h a n uTng x a y r a h o a n t o a n r o i du'a b i n h v e n h i e t d o b a n d a u , d p suat k h i i r o n g + b = 0,45 ( m o l ) = 3a + b = 0,45 ^ ' b i n h k i n chiJa l i / d n g o x i du" A p suat t r o n g b i n h l a Pi a t m N u n g n o n g b i n h d e l b = 0,15 A p dung djnh luat b a o toan electron: 0,25m 0,1 Cfiu : C h o a g a m h o n h d p g o m FeS2 va F e C O i v d i so' m o l b a n g n h a u v a o m o t f a = 0,1 2b = 0,7 ^ l4a + 3m ^ 56 b + b =0,25 -> S A p dung djnh luat b a o toan electron: a b +4 mm-, r,^^,,^im^jl.u^ + l e - > NO2 + H2O ••'•^ ^''v •:;,;2ls'^ 0: S + 2e NO + 2H:0 3a J, 32 +6 Qua trinh nhan electron: NO3 -2 O2 + e > 20 12-m 12-m Fe 0.25 m 56 - /-I so mol khong doi D a p an B rfiu 30: C h o 16,25 ( g ) F e C l j t a c d u n g v d i d u n g d i c h Na2S diT t h i t h u diTdc m ( g ) "Fe = ^ ( m o l ) ; n , o , = ^ Q u a trinh nhi/dng electron: = 0.1mol +3 F e - 3e m 3m 56 56 ket tua G i a t r i c u a m 1^: I A 10,4 B.3,2'''^""-"^- C C.1,6 D 12 ' Hrfdng d§n giai >F e T a c6: nFeCU 16,25 162,5 = 0,1 m o l 381 Ptiuang phap va ky thugt giai nhanh BTTN H6a dgi cUOng - vO CO - D5 Xuan Hung Phan urng: IFeCh 0,1 + NaaS -> 2FeCl2 + + 2NaCl 0,05 pe + 2HC1 ^ => m i = ms = 0,05.32 = 1,6 (g) => Dap an C 0,05 CSu 31: Hoa tan hoan toan liTdng bpt oxit F e vao liTdng dung djch H N Q vCra du thu diTpc 0,336 lit N^Oy (dktc) Co can dung dich sau phan iJng thu di/dc 32,67(g) muoi khan Cong thuTc cua oxit NxOy va khoi liTdng cua Fe,o lanliWtla: " '' * ^'^^ A NO2 va 5,52g ' B NO va 5,52g ,, FeCl2 + H2t 0,1 - rheo phan iJng (1), (3): nHci = 0,8 + 0,1 = 0,9 mol 0,9 = 0,9 (1) => Dap an A ,u 33: Hoa tan hoan toan 49,6 (g) hon hpp X gom Fe, FeO, Fe203, Fe304 b^ng VHCl kho'i liTcJng cii'a nguyen to'oxi hon hdp X la: Hifdng dan giai A 40,24% B 30,7% 32 67 • Ta c6: np^^j.^^,^ = 8,96 =0,135 mol =:> n^^3^ = 0,135 mol Ta c6: Qud trinh nhu'dng electron: Fe - le 0,135 0,045 - Qua trinh nhan electron: n^Q^ = ^"2 ^ 22,4 => 56x+16y = 49,6(g) - > 3Fe 0,135 +5 (1) X ' 0,015 Ap dung dinh luat bao toan electron, ta c6: iQud trmh nhan electron: S + 2e - 0,4 0,8 -2 >0 y 2y :::t:: Dap an C C 0,8 lit Ap dung djnh luat bao toan electron, ta c6: 3x = 0,8 + 2y h a y x - y = 0,8 (2) D 1,5 lit ; T i r ( l ) , (2) => X = 0,7 mol va y = 0,65 mol Vay: % (trong hhX) = 0,65.16 100% = 20,97% => Dap an C 49;6 Hifdng d§n giai = S O + 2e tich HCl toi thieu de h6a tan cdc chat r^n tren la: Hp +4 +6 I Cfiu 32: Cho Ian lifdt 23,2 gam Fe304 va 8,4 gam Fe v^o dung dich HCl I M The Ta c6: n^^ ^ = ^ = 0,1 m o l ; ^^304 232 3x B 0,015(5-n) = 0,045 => n = ( N ) B 1,1 lit +3 Fe jQua trmh nhiTdng electron: Fe - 3e >N 0,015(5-n) Mil +n N + (5-n)e = ^ ^ x = 10,44 gam ,.„„,,^„ = 0,4 mol Gpi: npe = x mol va no = y mol +3 - — = 0,15 mo! Cfiu 34: Nung 8,4g bpt s^t ngoai khong khi, sau mot thdi gian thu diTpc m(g) h6n '^"56 hPp X Hoa tan het X dung djch H2SO4 dac n6ng diT, thu difpc 2,8 lit SO2 PhiTdng trinh phan iJng: Fe304 + 8HC1 - > FeClz + 2FeCl3 + 4H2O 0,1 0,8 0,2 (1) 2FeCl3 + Fe - > SFeCU 0,2 0,1 (2) 382 D 37,5% ^ u y doi hon hdp X gom Fe va O (siir dung phiTcfng phap quy doi) +8/3 A 0,9 lit C 20,97% Hrfdng d§n giai n^, o = : ^ =0,015 mol ^xOy 22,4 Goi n la so oxi h6a cua nitd N^Oy =>"FeO (3) dung djch H2SO4 dac nong, thu di/pc dung dich Y va 8,96 lit SO2 (dktc) % D N2O va 10,44g C NOvalO,44g i jau phan ufng (2) con: 0,15 - 0,1 = 0,05 mol Fe, tiep tuc xay phan urng:;; (dktc) Gia tri cua m 1^: A 12 B I I C 10 HrfdngdSrtgiai Tac6: n so D 13 '•• ^'^ =0,125 mol; n^,=^ =0.15 mol 56 22,4 383 Phuang phAp v4 ky thuat giai nhanh BTTIM H6a dgi ciiong - vi3 cd - B Xuan Hung Fe la cha't khiJf, O: va H2SO4 dac la chat oxi hoa +3 Qua trinh nhu'dng electron: Fe - 3e > Fe (dktc) Cling lu-dng h6n hdp neu hoa tan het bang HNO3 dac nong du'dc 0,15 0,45 5,6 lit NO2 (dktc) Cong thu-c FcxOy la: -2 20 O2 + 4e — - Qua trinh nhan electron: m-8,4 A FeO B.Fe304 C Fe203 ni-8,4 < , +4 S +6 S + 2e Tac6:n,, 0,125 -i 0,25 i h A p dung djnh luat bao toan electron, ta c6: Fe = , bang nhau: - Hoa tan phan bhng V lit dung dich HCl M (viTa du) - Dan mot luong k h i CO du" qua phan nung nong dU'dc 33,6 gam sat V c6 gia D Khong xac djnh diCdc 0,05 mol FeCl2 ^^T::: 5:,0 ' ^ ' ' ^ ' y ^ ^ : : - ^ ^ , 0'> V ) ' 0,05 »£ai&!rfftil>l.nH,:.'' • 5,6 Ta c6: n.,„ = — — = 0,25 mol - 22,4 Fe - 3e Qua trinh nhu'dng electron: +3 > Fe 0,05 0,15 B.0,8 C 0,75 X Fe D 0,45 Hon hdp A gom FeO, F e j O j , Fe304 quy ddi Fe304 Phan du'dc dan qua CO dU', nung nong: -> 3Fe + 4CO2 + Fe304 T a c o : n.„ = ^ 'Fe 56 f O ' (3x-2y)e 7,2.(3x-2y) 56x + 16y 56x + l y > xFe •JiV IS! , h «n1AiH.'- N l e - ^ N ( N ( M = , mol "o(.rongF^e304)-0,2.4= - 7,2 Qua trinh nhan electron: 0,25 ifi.'^^^^.^^^.^,,: 0,25 A p dung dinh luat bao toan electron, ta co: Theo phan i^ng: npe^o4 = ^ " p e ^ ^-^-^ = ^'2 =^ ' H: H i f d n g dSn giai 4C0 o I'iul ' ' •' A 1,2 * ' Hi/dng d§n giai = - , = 7,2 (g) 11, C a u 35: A la hon hdp dong so mol gom FeO; Fe20i; F^iOa Chia A lam phan - 2HC|'Z 0,05 0,45 = -^^^^^ + 0,25 => m = 10 (g) => D a p a n C tnla: )' ' ^ 32 \ Fe^Oy b^ng H C l duTdc 1,12 l i t H2 •u 36: Hoa tan 10 gam hon hdp gom Fe 0,8 0,15 + " " ' ^ 56x + l y = 0,25 : o - = => FeO=> D a p a n A y mol lU 37: Dan luong k h i CO du" qua ong di/ng m gam hon hdp X gom Fe203 va * K h i cho phan tac dung vdi H C l , sir ke't hdp giffa H"" va oxi oxit di/dc bleu dien b^ng sd do: dung dich Ca(OH)2 du', thu du'dc 40 gam ket tiia Hoa tan cha't ran Y •; H ^ + O^- - > H2O 1,6 => V^f,, dung dich HCl du' thay c6 4,48 lit H2 bay (dktc) Gia tri cila m la: 0,8 1,6 = , lit Dap an B CuO nung nong thu du'dc chS't ran Y, khoi ong du'dc dan vao blnh diTng A 24 B 16 C 32 D 12 Hi/