Tài liệu Thần tốc luyện đề thpt quốc gia môn hóa học

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THAN TQC LUYEN DE THPT QUOC GIA 2016 M6N 111HOC ÿ Sr » vr t i 1 LUC TRAN Chii bien CHU THI HANH THAN TDC LUYEN DE THPT QUOC GIA 2016 ides inhoc u 20 DE THEN CHOT DEDAT DIEM CAO LUYEN TAP • • « * Bam sat de thi dai hoc 2016, c£u true ra de cua Bo Gi£o DucBaoTao ÿ De dang on tap thong qua Ibi gicii chi tiet dude nhan xet va binh luan. ÿ Nang cao t a duy vdi nhieu cong thite, meo thuc tien thong qua Ibi giai chi tiet. NHA XUAT BAN DAI HOC QUOC GIA HA NOl THAY LCfl NOB DAU MEGABOOK MUON CAC EM HIEU DlTtfC GlA TRI CUA VIEC TIT HOC HOC dAnh THITC tiem nAng trong ban TIT • ÿ » Chao cac em hoc sinh than men. Megabook ra ddi bo sach nhOng bo sach co tinh til hpc, til on tap cao, nham muc dfch giup cdc em nang cao kha nang til hoc vh dac biet phat trien tit duy cua minh ve mon hoc do. Megabook hieu diidc viec phat trien til duy, tri tue con ngu'di de tao nen sil thanh cong nhit Bill Gates, Steve Job hay Mark Zuckerberg... la nhd 80% dila vao viec til hoc, til nghien cdu den say me chil khong phai la ngoi tren ghe nha trildng, nghe giao huan. Viec til hoc khong han thong qua sach vd, ma thong qua sil quan sat cuoc song xung quanh, qua internet, hay dcin gian la hoc hoi kinh nghiem cua ngtfcri di trifdc. Viec til hoc se giup cac em phdt huy tiem nang cua ban than, nhan thay nhflng kha nang, sd triidng cua chinh minh c6n dang an giau dau do trong tiem thiic ma cac em chiia nhan ra. Viec til hoc giup cac em tang kha nang tif duy, xif ly cac van de nhanh nhay, thich nghi va dap ilng tot hdn vdi si1thay doi cua moi tritdng va xa hoi. Viec til hoc xay dilng bÿn nang sinh t6n, phÿn xa t6t hon cho moi con ngitdi. Sinh ra d tren ddi moi dila tre da biet til Jiqc hoi nhu' vi£c quan sat, nhin moi vat xung quanh, nghe nhieu va roi biet n6i. Viec til hoc that ra rat til nhien, d&i trildng la mot phufcrng phap giup kich thich si! til hoc. Va thay co chi c6 the hUdng dan va tao cam hilng chil khong the day chung ta moi thil. Tom lai vifc til hoc se giup moi ngildi dot pha trong sil nghiep va cuoc song. M<)t ky sit biet til hoc se dot pha cho nhflng cong trinh vi dai, mot bac sy say me nghien cilu se dot pha trd thkrih bac sy tai nang cilu chila bao nhieu ngitdi, mpt giao vien til nang cao chuyen mon moi ngay se bien nhilng gid hoc nham chan thanh dSy cÿm hilng va thii vi. Bdi vay vifie til hoc se giup bat ky ai thanh cong hdn va hanh phuc hdn trong cuoc song. DanDan Xu Huting Sdch Luyen 7W Biet til hoc => Nang cao kha nang tiiduy, x\i ly van de nhanh Bi£t tii hoc => Tang kha nang thich nghi, phan xa nhanh vdi moi triidng Biet tii hoc => Tao ra nhilng thien tai giup dat nildc va nhan loai Biet tii hoc => Giup moi ngiicfi thanh cong trong cuoc song, dot pha trong sli nghiep Biet tii hoc => Tao xa hoi v<5i nhflng cong dan liu tii. de sir dung cuon sAch nAy hieu quA nhat Biiofc 1: Lap ke hoach thdi gian lam de. Moi tuan 2 de la hop ly em nhe. (it nhiing ma chat) Biidc 2: Bam thtii gian lam de, lam that can than, chac chan, chinh xac khong can nhanh. Btfdc 3: Xem dap an, doc ldi giai can than. Trong ldi giai co nhac lai kien thdc, cau true, til vtfng vi the cac em on tap lai diiOc luon. Bildc 4: Luu laihanh trinh luyen thi Thanh Cong d sau moi de, tiic la ghi lai minh du'Oc bao nhieu diem, sai cau nao, kien thiic can nhd trong tarn. Bildc 5: Sau khi lam de tii tin hay thildng xuyen thi thil tren trang Vtest.vn de ren luyenky nang til duy, lam bai that nhanh em nhe. GIQ" HAY BAT BAU LUYEN DE NHE CAC EM! LET'S GO! CHUYEN OE LYTHUYeT VA BAI TAP BIEN HiNH VE PEPTIT - Lien ket cua nh6m CO v6i nhom NH gifla hai dcfn vi a aminoaxit diiac goi la lien ket peptit. - Peptit \k nhting hap chat chila tii2 den 50 goc a - aminoaxit lien ket vdi nhau blng cac lien ket peptit. - Oligopeptit gom cac peptit co tit 2 d£n 10 goc a - aminoaxit. Vi du neu co hai goc thi gQi la dipeptit, ba goc thi goi la tripeptit (cac goc co the giong hoac khac nhau). - Polipeptit gom cac peptit co tif 11 den 50 g6c a - aminoaxit. Polipeptit la ca sd tao nen protein. - Su' thay d6i vi tri cdc goc a - aminoaxit tao nen cac peptit khac nhau. Phan til co n goc a aminoaxit khac nhau se c6 n! dong phan. - Aminoaxit dau N la aminoaxit ma nhom amin d vi tri a chtfa tao lien ket peptit c6n aminoaxit dau C la aminoaxit ma nh6m -COOH chita tao li£n ket peptit. - T£n peptit = goc axyl cua cac a-aminoaxit bat dau tit dau chiia N, a-aminoaxit cuoi cung gitf nguyen ten goi. Vi du: Ala - Gly - Lys thi ten goi la Alanyl Glyxyl Lysin. l.PHANtfNGMAUBIURE Peptit va protein tac dung vdi Cu(OH)2 tao dung dich co mau tfm dac tritng. Dipeptit khong co ph&n ling nay. 2. PHAN CfNG THUY PHAN HOAN TOAN TAO CAC a - AMINOAXIT Khi thuy phan hoan toan tiiy theo m6i trd&ng ma san pham cua phan ling khac nhau: - Trong moi tritdng trung tinh: n-peptit + (n - 1)H?0 aminoaxit. - Trong moi tritdng axit HC1: n-peptit + (n - 1)H20 + (n + x)HCl -» muoi amoniclorua cua aminoaxit. Iin Trong do x la so mat xich Lysin trong n - peptit - Trong m6i trifting bazo NaOH: n-peptit + (n + y) NaOH -> muoi natri cua aminoaxit + (y + 1) H20 Vdi y \k so mat xich Glutamic trong n-peptit. Tnidng hop thuy phan khong hoan toan peptit thi chung ta thu diidc h6n hop cac aminoaxit va cac oligopeptit. Khi gap bai toan dang nay chung ta c6 thd' sti dung bao tokn so mdt xich cua mot loÿi aminoaxit n&o do ket hgfp vdi bko to&n khoi ltfdng. CAC DANG BAI TAP CO BAN ÿ DANG 1. PHAN 0NG THUY PHAN CUA PEPTIT: Trong cdc loÿi aminoaxit thi chi co loai a-amino axit mtii la don phan cau tao nen peptit va protein. Trong phan t£f peptit hay protein thi lien ket peptit 1& moi lien ket y£u nhat, de bi dtit dkn d£n tinh chat co b&n nhat cua peptit vk protein la phan ting thuy phan trong moi trtfting axit va bazd. DS* Giai: nhanh dti(?c cac bai tap thuy phan peptit vk protein can thtfc hien ckc biitic sau: Biidc 1: Dat cong thtic tong quit cho peptit: Gia sti don phan cau tao nen peptit chtia mot nhtim NH2 vk mot nhtim COOH co cong thtic \k: NH2-R-COOH thi cong thtic t6ng qudt cua peptit la [NH2-R-C00H]n(l-n)H20. Neu a-amino axit la no, mach hti chtia mot nhom NH2 va mot nhom COOH thi co cong thtic t6ng quat la: [CaH2a+102N] n(l-n)H20 (Vtii n la so goc a-amino axit cau tao nen peptit). Ki hieu peptit tao btii n don vi amino axit la Xn. Vf dy dipeptit la X2, tripeptit la X3... Biidc 2: Viet phiidng trinh phin tingthuyphan Phiidng trinh phan ting thuy phan hoan toan: Trong m6i triidng axit va bazo nhifng khong dtfa m6i triidng vao phiidng trinh phan ting: NH2-R-COOH]n(l-n)H20 + (n-l)H20 H*/OH" ÿ> nNH2-R-COOH Hay: Xn + (n-l)H20 ÿ» nXr Khi dun nong trong moi triidng axit nhii HC1: [NH2-R-C00H]n(l-n)H20 + (n-l)H20 +nHCl ., nNH3Cl-R-COOH Khi dun nting trong moi triiting bazo nhii NaOH: [NH2-R-C00H]n(l-n)H20+ nNaOH ., nNH2-R-COONa +HzO Phiidng trinh thuy phan khdng hokn toan: mXn + (n - m)H20 ., nXm Bitdc3:Dita vio phiictng trinh thuy phan, du ki$n baicho va cic djnh lu&t xic dinh d& kien bai . yeu cau V Dtfa vao dinh luat bao toan khoi liiyng: mpeptil + mH0Q = maminoaxi! mMOM + mH2o+mHa = mmu6i mpeptit + mNaOH V 8 mmuoi + mH0O Diia vao dinh luat bao toan moi goc a-amino axit: i o Chii y: Diia vao phiicfng trinh thuy phan de tlm moi quan he so mol cua cac chat trong mot phutfng trinh phan ilng de xac dinh so mol hoac loai peptit. S Khoi litpng mol cua n-peptit = a-amino axit x n-18.(n - 1) S chay peptit tao ra tii a-amino axit no, mach hd chda mot nhom COOH theo phuto'ng trinh tong quat sau: S Dot NH2 va mpt nhom 3an 1, 5n 2an — n + 2TT/~. ÿ x t -Hÿ0 + ÿ )Mr2a + -- °ÿa" C02 + 2 2 2 •/ Doi vrti 2 Peptit khi thuy phan co ti le so mol bang nhau, thi ta xem 2 Peptit do la mpt Peptit va ghi phan ilng ta nen ghi gop. Khoi litpng mol cua Peptit chinh la tong kh6i ldpng mol cua 2 Peptit do. ÿ -a r ÿ /-vx-n tt 2«, 2 s-ÿr\ /•» Vi du: Tripeptit H[NHCH2CO]3OH va Tetrapeptit H[NHCH2CO]4OH (co so mol bang nhau) thi ta xem 2 Peptit do la Heptapeptit: H(NHCH2CO]7OH v& M = 435g/mol (TOT) CUBE'* X la mot Tetrapeptit cau tao tit Amino axit A, trong phan tit A co 1 nhom(-NH2), 1 nhom (-COOH), no, mach hd. Trong A Oxi chiem 42,67% khoi litpng. Thuy phan m gam X trong moi triidng acid thi thu diipc 28,35(g) tripeptit; 79.2(g) dipeptit vk 101.25(g) A. Gia tri cua m la? B. 258,3. A. 184,5. C. 405,9. D. 202,95. Giai: Tit % khoi litpng Oxi trong A ta xac dinh difpc A la Gli ( H2NCH2COOH) vdi M = 75 Cong thiic cua Tetrapeptit la H[NHCH2CO]„OH vdi M = 75.4 - 3.18 = 246g/mol Tinh so mol: Tripeptit la: Dipeptit la: —'— 132 ÿ8.35 189 = 0,6 (mol) = 0,15(mol) Glyxin(A) la: 101'25 = l,35(mol). 75 Giai: gon nhit sau: Bat mat xich NHCH2CO = X Ghi scr do ph&n dng: (X)4 —-*(X)3 + X 0,15 0,15 0,15 mol (X)4 -- 2 (X)2 *ÿ 0,3 0,6 mol (X)4 --** 4X 0,3 1,2 mol Tit sd do tren ta tinh diipc: So mol X phan ilng la: (0,15 + 0,3 + 0,3) = 0,75mol m = 0,75.246 = 184,5 gam Chon Dap an A. > Thuy phan hoan toan 143,45 gam hon hpp A gom hai tetrapeptit thu ditpc 159,74 gam hon hop X gom cac aminoacid (C&c aminoacid chi chiia lnhom -COOH vk 1 nh6m -NH2). Cho toan bo X tac dung vdi dung dich HC1dit, sau do co can dung dich thi nhan ditpc m (gam) muoi khan. Tinh khoi litpng nitdc ph&n dng va gia tri cua m lan lifpt bang? IH DmiOau Xu Hitting SachlMyenlhi A. 8,145(g) va 203,78(g). B. 32,58(g) v* 10,15(g). C. 16,2(g) va 203,78(g) D. 16,29(g) va 203,78(g). Giai: Dat cong thtic chung cho h6n hop A la H[NHRCO]4OH Ta co phan ting: H[NHRCO]4OH + 3H20 >- 4 H2NRCOOH Hay: X4 + 3H20 >4X, - Ap dung DLBTKL => nHÿ0 = TO phan ting => 4 - 0,905(mol ) => mH20 = 16,29 gam. 4 = — .0,905 mol nx=-nHo 3 o ÿ Phan i3ng cua X tac dung vcfi HC1:X + HC1> X.HC1 Ap dung BTKL =>rnmu. = mx + mHQ = 159,74 + -.0,905.36,5 = 203.78(g) 3 Chon Dap an D. : /Ny y 1111111 Tripeptit M va Tetrapeptit Q dope tao ra tti mot amino axit X mach hd, trong phan tti chi chtia 1nhom amino. Phan tram khoi lOo'ng nitd trong X bang 18,67%. Thuy phan khong hoan toan 8,389 gam hon hop K gom hai peptit M, Q trong dug dich HC1 thu dope 0,945 gam tripeptit M; 4,62 gam dipeptit va 3,75 gam X. Ti Id ve s6 mol tripeptit M v& tetrapeptit Q trong hon hop K \k: C. 1:1 A. 1:2 B. 3 : 2 D.2:l Giai: Dap an C. = 75 => cong thiic cua X: H2NCH2COOH (C2H502N) Mx = Cong thiic M: C6H„04N3, Q: C8HK05N, Sau phin ting thuy phan: nM = 0,945 - 0,005 mol 189 = 0'035 m0l> = °'05 mDl "dipeptit = => Neu phan ting thuy phan hoan toan thi £ nx = 3nM + 2ndjpeptjt + nx= 0,135 mol => 2 % = 3nMbandgu + 4nQband.u = 0,135 mol ma mK = 189nMban(Hu+ 246nQbanflju = 8,389 gam d&u /% Uqbandau = Mband3u * -1•1 -> X, Y, Z, T la cac peptit deu dope tao bdi cac anpha-amino axit no chtia mpt nhom -NH2 va 1 nhom -COOH va co tong so nguyen tti oxi la 12. Dot chay 13,98 gam hon hop E chtia X, Y, Z, T can dung 14,112 lit 02 (dktc) thu dope C02, va N2. Mat khac dun n6ng 0,135 mol hon hop E bang dung dich NaOH (lay do 20% so vPi phan ting), co can dung dich sau pMn ting thu dope lOpng ran khan la. H20 A. 31,5 gam B. 24,51 gam C. 36,05 gam D. 25,84 gam Giai: Dap an A. Tong so nguyen tti O cua X, Y, Z, T la 12 => Tong so don vi a - amino axit tao cac peptit =12-4 = 8 => X, Y, Z, T deu la dipeptit. Quy doi hon hpp peptit X, Y, Z, T tu'dng ting vPi 1 peptit tao bPi 2 don vi CnH2n + 3N02, peptit c6 CTTQ = 2CnH2n+1NQ2 - H2Q = C2 H4nN203 I c & tip ÿ 7Juin toe Luyende THPT Quoc Gia 2016 ntonHoa hoc mm D6t chay peptiL C2nH4nN2°3 + (3n ~ ~)02 ~ÿ 2nC°2 + 2nH20 + N2 3 ~ => n00, = (3n —)npcpiM 22 li( ÿ 3 13 98 14 1 12 -= (3n —ÿ).<=> n = 2,833 => M — 2gn + 76 22>4 . =155,33 >*pm E + NaOH -> muoi + nifcfc. nNaOH = 2npeptit • 120% = 0. 324 mol, nHj0 = nE = 0, 135 mol => mchs,rjn khan = mE + mNaOH -mHi0 = 0,135 .15 5,33 + 40.0,324 - 18.0, 135 = 3 1,5 gam Nhan x£t: Day la bai kha hay va kho tuy cac bifdc ti'nh toan khong dai. Van de mau chot de Giai: quyet diiac bai toan la lam sao dat diidc cong thiic chung cho h6n hop peptit. Nhieu ban se boi roi khi cho dfl kien tong so nguyen til O la 12. Con so 12 n£y co y nghia quan trong, neu la mpt gia trj kliac thi se kho dat cong thiic chung cho hon hop va bai toan trd nen phiic tap hdn rat nhifiu. Thuy phan m gam h6n hop X gom mot tetrapeptit A va mQt pentapeptit B b&ng dung dich NaOH vila du roi co can thu duoc ( m + 23,7) gam hon hop muoi cua Gly va A1A. Dot chay toan b() lifting muoi sinh ra bang mpt lifting oxi via du thu dti(?c Na2C03 vk h6n hop htii Y gom C02, H20 vk N2. Dan to&n bo h6n hop htii Y di rat cham qua binh diing dung dich NaOH dac dif thay khoi lifting binh tang 84,06 gam va co 7,392 lit mot khi duy nhat (dktc) thoat ra khoi binh. Thanh phan phan tram khoi lifting cua A trong h6n hop X la: A. 53.06%. B. 35,37%. C. 30,95%. D. 55,92%. 7 392 Ik N-,2. => n*. = -= 0, 33 mol Giai: Khi thoat ra khoi binh difng NaOH dac • n2 22,4 Goi a, b Ik so mol cua tetrapeptit A va pentapeptit B — H[NHRCO]4OH; H[NHRCO]sOH Bko to&n nguyen to nitti 4a + 5b = 2nNÿ = 0,33. 2 = 0,66 mol H[NHRCO]4OH +4 NaOH -> 4 NH2RCOONa + H20 a (1) 4a a -WTV H[NHRCO]5OH + 5 NaOH -> 4 NH2RCOONa + H20 b b 5b © Ap dung bdo toan khoi lifting : + mNa0H = mÿ + mHÿ0 => mNa0H - mH20 = 23,7 gam => 40(4a + 5b) - 18(a + b) = 23,7 gam /ÿy> mx (2) T£f 1va 2 t> a = 0,09 va b = 0,06 • Goi x, y Ik so mol cua muoi Gly va Ala H2NCH2COONa:x; H2NCH(CH3)COONa:y (3) 75 x + 89 y = m + 0,09.18.3 + 0,06.18.4 bm = 75x + 89y- 9,18 97 x + 111y = m + 23,7 (2) bx + y = 0,66 (4) + mbinh T = mco2 + mH2o = 84, 06 gam => 44 [2x +3y- (4fl 56) ] + 18(x +3y) = 84,06 l> 2 x+ 3y = 1,59 (5) 2 Tif (4) va ( 5) P x = 0,39 mol va y = 0,27 mol thay vko (3)1> m = 44,1 (g) Tif do CT cua A : Gly-Gly-Gly-Ala B : Ala-Ala-Ala-Gly-Gly Mega book DrnBau mmmSdckLuyfn Thi 100% = 53, 06% => Chpn Dap an A. ÿ DANG 2: PHAN CNG CHAY CUA PEPTIT: Tripeptit mach hd X va Tetrapeptit mach hd Y dupe tao tti mQt amino axit no, hd trong phan tti co 1 nhom (-NH2) va 1 nhom (-COOH). Dot chay X va Y. Vay lam the nao de dÿt CTPT cho X,Y? Ta lam nhti sau: Tti CTPT cua Amino axit no 3CnH2n+102N - 2H20 th&nh C3nH6n_ ,04N3 (day la cong thtic Tripeptit) va 4CH2rvfl02N - 3H20 thanh CÿHÿ 205N4 (day Ik cong thtic Tetrapeptit)... Neu d6t ch&y lien quan den lupng ntidc va cacbonic thi ta chi can can bang C, H d£ tinh toan cho nhanh. + p02 -3nC02 + (3n-0,5)H2O + N2 C4nH8n-205N4 + P02 ** 4nC02 + (4ll-l)H20 + N, Chu y: Trong phan ting d6t chay mpt peptit hay hon hop peptit, ta co the quy d6i ltipng san pham tUdng dtidng vdi skn pham d6t ch&y amino axit tao peptit, trong do lupng H20 thu dupe tang them trong trUdng hop dot amino axit chinh banglupng H20 can de thuy ph&n peptit hoan toan thanh ckc amino axit do. Dieu nay dtipc ting dung rat nhieiu trong qua trinh Giai: cac bai toan dot chay peptit. >ÿ ( ÿ 1du: ) > Cho hai chat hflu cd X, Y lan ltipt \k tripeptit vk hexapeptit dupe tao thanh tti ciing mpt amino axit no, mach hd, co mot nhom cacboxyl va mot nhom amino. Dot chay hoan toan 0,1 mol X bang 02 vtia du thu dupe san pham chay co tong khoi ltipng 40,5 gam. Neu cho 0,15 mol Y cho tac dung hoan tokn vdi NaOH (lay dti 20% so vdi ltipng can thiet), sau ph&n ting co can dung dich thi khoi ltipng chat ran thu dtipc \k: A. 87,3 gam B. 9,99 gam Giai: Gik sti X va Y deu dtipc tao bdi amino axit la C. 107,1 gam D. 94,5 gam CnH2n+1N02 X = 3CnH2n+1N03 - 2H20 = CA-M Dot chay X: = mC0; + mHj0 + mNj = 44. 3n.0,l + 9.(6n- 1).0,1+ 14.3.0,1 = 40, 5gam => n = 2 => Amino axit la H2NCH2COOH Y + NaOH -» H2NCH2COONa => mchÿtrSn = mmu5i + niNaOHda - 97.0,15.6 + 40.20%.6.0,15 = 94,5 gam Chpn Dap an D. X, Y, Z la 3 peptit dtipc tao bdi tti cac anpha-amino axit no chtia 1 nhom NH2 va 1nhom -COOH. Dun nong 0,1 mol hon hop E chtia X, Y, Z b&ng dung dich NaOH (vtia du). Co can dung djch sau ph&n ting thu dope m gam muoi khan. Dot chay toan bo ltipng muoi nky thu dupe 0,2 mol Na2C03 va hon hpp gom C02, H20, N2 trong do tong khoi lupng cua C02 va H20 la 65,6 gam. M£t khac dot chay 1,51m gam hon hpp E can dung a mol 02, thu dupe C02, H20 vk N2. Gia tri cua a 1k. A. 3,33 B. 2,98 C. 1,89 D. 2,76 Gi&i: 0,1 mol E phan ting vtia du vtii 2.0,2 = 0,4 mol NaOH => E cd so don vi amino axit trung binhla 4.=> Dat cong thtic chung cho cac axit amin tao E la CnH2n + ,N02=> Cong thtic muoi thu dtipc 1k CnH2nN02Na, nmu6J = 4nE = 0,4 mol i CTTQ cua E la C4nH8n_2N405 Dot chay muoi thu ditoc nCOi = 0, 4n — nNazC03 = 0, 4n - 0, 2 mol,nH20 = 0, 4n mol => 44.(0,4n - 0,2) + 18.0,4n = 65,6 gam => n = 3 => m = 111.0,4 = 44,4 gam 44 4 1 51 => 1,51m gam E titong ting v6i —L-12— = 0, 222 mol E. 302 nC02 =0,222.12 = 2, 664mol Dot chay 1,51m gam E thu dtf So — 3, 33 mol C>2 * dung = —2 Chon Dap an A. 2,664.2 2,442-0,222.5 Tripeptit mach hd X va Tetrapeptit machhd Y deu dtfoc tao ra til mot aminoacid no,mach hd co 1 nhom -COOH va 1 nhom -NH2 .Dot chay hoan toan 0,1 mol X thu difOc san pham gom H20, C02 vk N2 trong d6 t6ng khoi liiong C02 va H20 bang 36,3(g) .Neu dot chay hoan toan 0,2 mol Y thi so mol 02 can phan ling la? C. l,875(mol). D. 3,375 (mol) A. 2,8 (mol). B. l,8(mol). Giai: Ro rang X,Y deu sinh ra do amino axit c6 CT CH2n+102N. Do vay ta co CT cua X,Y tifcing ting la: C3nH6n_ ,04N3(X) ,C4nH8n_205N4(Y). Phan ting chay X: C3nH6n-i°4N3 + p03 -> 3nCO? + (3n - 0,5)H2O N2 0,3(3n - 0,5) mol 0,3n 0,1 + Ta co phtiOng trinh tong khoi ItiOng H20 va C02 : 0,3. [44n + 18.(3n - 0,5)] = 36,3 => n = 2 . Phan ting chay Y: C4 Hen_205N4 + p02 -4nC02 + (4n 1)H20 + N2 . 0,2 mol 0,2.p 0,8n - (0,8n - 0,2) Ap dung bao toÿn nguyen t6 O: 0,2.5 + 0,2.2p = 0,8.2.2 + (0,8.2 - 0,2) => p = 9 => n02 = 9.0,2 = l,8(mol) Chon dap an D Ihuy phan hoan toan m gam hon hop gom peptit X va peptit Y bangdung dich NaOH thu dtiOc 151,2 gam hon hop gom cac muoi natri cua Gly, Ala va Val. Mat khac, de dot chay hoan toan m gam h6n hop X, Y d tren can 107,52 lit khi 02 (dktc) va thu dtiOc 64,8 gam H20. Gia tri cua m la A. 102,4. B. 97,0. C. 92,5. D. 107,8. Giai: Gly: H2NCH2COOH Ala: CH3CH(NH2)COOH Val: (CH3)2CHCH(NH2)COOH Cac peptit deu tao bdfi cac amino axit no, dcfn chtic CnH2n+]N02. Quy doi h6n hop peptit X, Y ttiong dtiong vdi peptit tao bdi a dcfn vi CnH2n+1N02. 111IMl@gJ£8 book Dan Dau Ku Hu&ng Sack Luyen TM => CTPT peptit = aCDH2n+1N02 - (a - 1)H20 = CÿN.O.ÿ (x mol) ® TNI: CanH2an+2NaOa+1 + aNaOH -> muoi + H20 Bao toan khoi luong: m + 40ax = 151,2+ 18x • TN2: CanH2an+2NaOa+1 02 (1) >anC02 + (an + l)H20 + -ÿN2 Ap dung bao toan nguyen to O co: nCOi = (a + 1)x + 2A8-3,6 _ (a + l)x-6 Ap dung bao toan khoi lilting c6: m + 32.4,8 = 44. + —- + 64,8+ 14.ax =>36ax + 22x = m + 91,8 Thay lan liiot cac dap an vao ( 1) va (2) giai he ( tim ax va x) thay dap an A duy nhat cho nghiem dilOng => Chpn dap an A. Bai 1:Thuy phln hoan toan 60(g) h6n hop hai dipeptit thu dooc 63,6(g) h6n hop X gom cac Aminoacid no mach ho' (H2NRCOOOH). Neu lay 1/10 hon hop X tac dung vdi dung dich HC1 do thu dufo'c m(g) muoi. Gia tri cua m la? A, 7,82. B. 8,72. />/o"V C. 7,09. D.16,3. Bai 2: (De thi tuyen sinh daihoc Khoi B- 2010): Dipeptit mach hd X va mach hd Y deu dilOc tao ra tit mot loai amino axit no, mach hd co mpt nhom NH2 va mot nhom COOH. Dot chay hoan toan 0, 1 mol Y thu difoc san pham gom C02, H20, va N2 trong do t6ng khoi litong C02 va H?0 b&ng 54,9 gam. Neu dot chay hoan toan 0,2 mol X, san pham thu dii(?c cho l<)i qua dung dich niidc voi trong dii thi thu dooc m gam ket tua. Gia tri cua m la: A. 45. B,12a C30' D-60' Bai 3: X la mpt Hexapeptit dtu tao t is mot amino axit H2N-CnH2n-COOH(Y). Y co tong % khoi lilting Oxi va Nito la 61,33%. Thuy phan het m(g) X trong moi triidng acid thu diitic 30,3 (g) pentapeptit, 19,8(g) dipeptit va 37,5(g) Y. Gia tri cua m la? C. 100 gam. D.78 gam. B. 84 gam. A. 69 gam. Bai 4: X la mot tetrapeptit cau tao tit mot amino axit (A) no, mach hd co 1 nhom -COOH ; 1 nhom -NH2. Trong A %N = 15,73% (ve khoi lilting). Thuy phan m gam X trong moi triidng axit thu dilpc 41 ,58 gam tripeptit ; 25,6 gam dipeptit va 92,56 gam A. Gia tri cua m la : A. 149 gam. B. 161 gam. C. 143,45 gam. D. 159 gam. Bai 5: X la tetrapeptit Ala-Gly-Val-Ala, Y la tripeptit Val-Gly-Val. Dun nong m gam h6n hop X va Y co ti le so mol nX : nY = 1 : 3 vdi 780 ml dung dich NaOH 1M (vita du), sau khi phan ilng ket thuc thu diipc dung dich Z. Co can dung dich thu diitic 94,98 gam muoi. m co gia tri la C. 77,04 gam. A. 68,1 gam. B. 64,86 gam. D. 65,13 gam. ftsBSa Softs Than toe luyen de ThPT Quoc Gia 2016 mon Hoa hpc Bai 6: Dipeptit mach hd X va Tripeptit mach hd Y deu ditpc tao ra tit mot amino axit no,mach hd co 1 nhdm -COOH va 1 nhdm -NH2 .Dot chay hoan toan 0,1 mol Y thu ditpc sdn pham gom H20,C02 vk N2 trong do tdng khdi litpng C02 va H20 bang 54,9(g) .Neu dot chay hoan toan 0,2 mol X,san phSm thu ditpc cho lpi qua dung djch nitdc voi trong dit thi ditpc m(g) ket tua .Gia tri cua m la? C.30. D.60. A. 45. B. 120. Bai 7: X va Y lan litpt \k tripeptit va hexapeptit deu mach hd va deu ditpc tao thanh tit cimg mpt amoni axit no mach hd (chiia mpt nhom -COOH vk mot nhdm -NH2). Dot chay hoan toan 0,1 mol X bang 02 vita du thu ditpc san pham gom C02, H20 vk N2 co tong khoi litpng Ik 40,5 gam. Neu cho 0,15 mol Y tac dung hoan toan vdi NaOH (lay dit 20% so vdi litpng phan ilng), sau phan ling co can dung dich thi thu ditpc ch&\ ran khan co khoi litpng la C. 107,1 gam. B. 9,99 gam. D. 94,5 gam. A. 87,3 gam. Bai 8: Khi thuy phan khong hoan toan mot peptit X (Mx = 293) thu ditpc hai peptit Y va Z. Biet 0,472 gam Y phan ling vita du vdi 18 ml dung dich HC1 0,222 M dun nong vk 0,666 gam Z phan ling vita du vdi 14,7 ml dung dich NaOH 1,6% (d = 1,022 gam/ml) dun nong. Bift rSng khi thuy phan hoan toan X thu ditpc hon hpp 3 amino axit la glyxin, alanin va phenyl alanin. Cdng thiic cau tao cua X la C. Ala-Phe-Gly-AlA. D. Gly- Ala-Phe A. Ala-Phe-Glv. B. Gly-Phe-Ala-Gly. Bai 9: Thuy phan hoan tokn 27,52 gam hon hop dipeptit thi thu ditpc 31,12 gam hon hpp X gom cac aminoaxit (cac amino axit chi cd mot nhom amino va mot nhdm cacboxyl trong phan tit). Neu cho litpng hon hop X nky tac dung vdi dung dung dich HC1 dit, co can c£n than dung dich, thi litpng muoi khan thu ditpc 1k ? D. 42,12 gam. C. 31,12 gam. B. 58,64 gam. A. 45,72 gam. Bai 10:Thuy phan hoan toan 60 gam hon hop hai dipetit thu ditpc 63,6 gam hon hop X gom cac amino axit (cac amino axit chi co mpt nhdm amino vk mpt nhdm cacboxyl trong phan tit). Neu cho 1/10 hon hop X tac dung vdi dung dich HC1 (dit), c6 can can than dung dich, thi litpng muoi khan thu ditpc la: ÿ C. 8,15 gam B. 16,30 gam D. 7,82 gam. A. 7,09 gam. 15 IKS CHUYEN BE BAlTAP BIEN HlNH VE HN03 VA ION NO3TRONG MOlTRl/dNG H+ ÿ DANG 1: BAl TOAN KIM LOAI TAC DUNG V('Jl DUNG DICH HN03 HN03 la axit co tinh oxi hoa manh vi so oxi hoa toi da cua N la +5 vii phan tCf kern ben, the - hien ca khi loang. - HN03 dtf t&c dung vdi kim loai ( trit Au, Pt) cho muoi nitrat kim loai co so oxi hoa toi da: 'NO2 NO + H20 + HNOj—» M(N03)n + n2o n2 M NH3(NH4N03) Chu y: © Al, Fe, Cr bi thu dong hoa trong dung djch HN03 dac ngudi ® Tinh khoi litong muoi thu dii NaN02 + NaN03 + H20 2N02 ÿN204 NO la chal khi khong mau, bi hoa nau ngo&i khong khi NO + -O,2 2 — NO,2 Dung djch muoi sau phan uing thu diipc nho dung dich chda OH- thay co khi thoat ra dieu do chiing to dung dich thu diioc co NH4N03 0 NH4 + OH" 16 i! -» NH3 1 + H20 Cho 29 gam hon hap gom Al, Cu va Ag tac dung vifa du v6i 950 ml dung dich HN03 1,5M, * thu diiac dung dich chiia m gam muoi va 5,6 lit hon hnNH4NO, = 0,0125 mol mmu6l mkimlo5i + ÿnetrao«l6i+ mNH4N03 = 29 + 62.(3.0,2 + 8.0,05 + 8.0,0125) + 80.0,0125 = 98,2 gam ÿ Dap an C het 6 gam h6n hap X gom Mg, Al, Zn trong dung djch HNOs vifa du, Hoa klii phan * ling hodn toin thu du'ac dung dich Y va hon hap khi gom 0,02 mol NO va 0,02 mol N?0. Lam tan sail bay hai dung dich Y thu diiac 25,4 gam muoi khan. Tong so mol ion nitrat b| khii trong cdc phdn ling tren Id: A. 0,07 mol B. 0,05 mol D. 0,09 mol C. 0,06 mol Giai: Khi cho hon h Trong dung dich Y co the ton tai NH4 N0* ÿ Dat mol NH4N03 la a mol ÿtong so mol electron nhan: ne(+)=3.nN0+8.nN20 + 8.n nh4no3 — 0,22 + 8a Ta CO: nNQ. mutf(k|mk>ai _rlenh$n ~ nenhi/dng ~~ + =>mmudl = mklmloal + ÿO; mu<5iklmloai + mNH a =0,01(mol) => Tong so mol ion nitrat bi Idtiif la : nN* ÿ= 0,02 +0,02.2 + 0,01= 0,07 (mol) Dap an A ÿ Cho 5,04 gam hon hap Mg vd Al co ti le mol tiidng ing la 3:2 tac dung vdi dung dich HN03 loang, dif thu dif ac dung djch X va 0,896 lit (dktc) hon hap hai khi khong mau, khong h6a nau trong khong khi co ti khoi hai so vdi H2 bang 18. So mol HN03 bi khif trong qud trinh tren la A. 0,095 mol. Giai: Co B. 0,11mol. 24nMg + 27nA1 = 5, 04 gam nMg : n/u =3:2 C. 0,1 mol. fnMg = 0,12mol = 0,08 mol Hai klii thu difcfc khong hoa nau trong khong khi Ma Mx = 1 8.2 = 36 suy ra 2 khi la N20 va N2 D. 0,08 mol. book Ml nN2o + nN2 = Dan BailXu Huting Stick Luyen fnNjO=0,02mol = 0,04 mol 44nN.20 + 28nN2 -0,04.36 = 1,44gam \aÿ2 = 0>02mo1 Co 2.0,12 + 3.0,08 = 0,48 mol > 8.0,02 + 10.0,02 = 0,36 mol => chting to san pham khti c6n co => So mol HN03 bi khti = NH4N03 : 4N03 2nNÿQ 4- 2nNa + nNH4NQ3 = g = 0,095 mol = 0,015 mol => Chon Dap an A. Hon hop X gom Mg (0,10 mol), Al (0,04 mol) va Zn (0,15 mol). Cho X tdc dung vdi dung dich HN03 loang (do), sau phan ting khoi lOOng dung dich tang 13,23 gam. So mol HN03 tham gia phan ting l& B. 1,2400 mol. C. 0,6975 mol. A. 0,6200 mol. D. 0,7750 mol. Giai: Co mdungdjchang = - mN0 = 24.0,1 + 27.0,04 + 65.0,15 - mN0 = 13,23 gam mx => mN0 = 0 => Chting to san pham khti tao thanh la NH4N03 Ap dyng bao toan electron co: = 2.0,1 + 3.0,04 + 2.0,15 = 0,62 mol ÿ nNH4N03 0775 mol ÿnÿQÿy =8nNH4NQ3 + 2x1ÿÿ =0,775 mol Dap an D. Hon hop X gom: Mg (0,15mol), Al (0,lmol), Zn (0,12mol). Cho X tac dyng vdi dung djch HN03 (loang, do). Sau phan ting hoÿin toan, thay khoi lilting dung dich tang 14,10 gam so vdi dung dich HN03 ban dau. So mol HN03 da tham gia ph&n ting gan nhat vdi gia tri nao dtfdi day? A. 0,98 B. 1,08 Giai; Co mdungd|cht4ng = mx - C. 1,17 D. 0,92 14,1 gam - 14,1 = 0 => Chting to san pham khti tao thanh NH4N03. => mkh( = 24.0,15 + 27.0,1 + 65.0,12 Ap dyng bao toan electron cd 8nnh4nc)3 =2.0,15 + 3.0,1 + 2.0,12 => Hnh4No3 =0,105mol => pu = 2.0, 15 + 3 .0,1+ 2.0,12 + 2.0,105 = 1,05 mol gan vdi gia trj 1,08 nhat. => Ch?n Dap an D Hoa tan hoan toan 11,6 gam hon hop A gom Fe, Cu vao 700 ml HN03 1M. Sau klii kim loÿi tan het thu diicfc dung dich B va m gam h6n h mF > = 0,5.85 = 42,5 gam > 41,05gam => Chting to KOH phan ting dil, D nung trong khong khi d£n kh6i lilting khong doi ditdc Fe203 vaCuO + 64b = ll)6gam J a = 0,15 J56a j b = 0, 05 [80a + 80b = 16 gam Dat x, y lan lilcrt l Co ltf Acid phan ling het. =>%2o =ÿnHN03 =0,35mol Ap dung bao toan khoi lifpng co : mA + h1hno3 = mmu6j + mkhiC + mn2o => m = 11,6 + 63.0,7 - (11,6 + 0,45.62) - 18.0,35 = 9,9 gam gan vdi gia tri 10 nhat. Dap an B. Dot m gam h6n hop X gom Fe, Cu, Mg, Zn trong oxi thu ddpc 29,7 gam hon hop chat ran Y. Hoa tan het Y bang dung dich HN03 dac nong dif thu ddpc 17,92 lit khi N02 (dktc). Cho m gam hon hop X tac dung vdi dung dich H?S04 dac, nong, dd thu dope dung djch Z chda 84,1 gam muoi va khi S02. Biet rang N02 va S02 la cac san pham khd duy nhat cua HN03 va H2S04. Gia tri cua m la D.23,3 C. 26,5. B. 20,1. A. 20,9. Giai: Xet thi nghiem khi cho m gam hon hop X tac dung vPi H2S04 dac nong dii. _ n 84,1-m n ' enhi/cfng ' 96. — *4enhi/6ng 'kimloai + 2 Fe ÿ Cu Xet thi nghiem *:. X- Mg+ O2 -> 29, 7 gamY 4|IW0> > NO2 ( 0, 8 mol) Zn S6 mol e nhan : nenh$n = 4.n0z +nNOz Theo dinh luat bao toan: =4.ÿÿ-ÿ+0,8 2.ÿ-—— =4.ÿÿ—— + 0,8 =>m=26,5(gam) 32 96 Dap an C. Dung dich X chda HC1 4M va HN03 aM. Cho tii tii Mg vao 100ml dung dich X cho to'i khi khi ngiing thoat ra thay ton het b gam Mg, thu ditpc dung dich B chi chda cac muoi cua Mg va thoat ra 8,96 lit hon hpp khi Y gom 3 khi (trong do co mot khi khong mau, de h6a nau trong khong khi). Cho Y qua dung dich NaOH dii thay con lai 5,6 lit hon hpp khi Z thoat ra. H khoi hoi cua Z so vPi hidro la 3,8. Cac phan dng xay ra hoan toan. The tich cac khi do d dktc. Gia tri cua a va b lan liipt la? A. 2,5 va 4,2 Giai: nY Y = 22,4 B.8,4va5,0 = 0,4mol,n7z = 22,4 Vi di qua dung dich NaOH chi c6 D. 5,0 va 8,4 0,15mol = o, 25 mol => nNO N°2 = 0,4-0,25 = . N02 hap thu nen Z phai chda H va khi A (Mz =7,6) 1 °'4 o o - nH2 = -nHCI = — = 0,2mol t Taco C.4,2va2,5 ÿPhan tfl khdi khi A = 7,6"°'25 20,2 = 30 =» KhiA la NO 0,25-0,2 Ap dung bao toan electron co: 2nMg =2nH2 + nN(>2 + 3nN0 = 2.0, 2 + 0,15 + 3.0, 05 = 0, 7 mol => nMg = 0,35 mol => b = 24.0,35 = 8,4 gam Co nHNo3 =2nMg(N03)2 +nN02 + °no = 2.(0, 35 - 0,2) + 0,15 + 0,05 = 0,5mol =>a= — = 5M 0,1 => Chon Ddp an D. ÿ DANG 2: HOP CHAT TAG DUNG V0I DUNG DICH HN03 p Hoa tan hoan toan hon hop gom FeS2 va Fe304 bang 100 gam dung dich HN03 a% (vita du) thu difoc 15,344 lit hon hop khi gom NO va N02 co khoi litOng 31,35 gam va dung djch chi chiia 30,15 gam hon hop muoi. Gia tri cua a gan nhat vdi? A. 63 B. 57 C. 43 <0ÿ D. 46 Giai: Dap an D Dat a, b lan loot la so mol cua NO va N02 15 344 a + b = —--= 0,658 mol 22,4 30a + 46b = 31, 35 gam fa = 0,01 <=> < [b = 0,675 Dat x, y lan loot la so mol cua FeS2 va Fe304 Ap dung bao toan so mol =>15x + y = 3a + b = 0,705 mol Ap dung bao toan nguyen to co = 2x=> nN?. ÿ muoj (1) = 3x + 3.3y - 2.2x = 9y - x => mmu6j = 56.(x + 3y) + 96.2x + 62.(9y ~ x) = 30,15 gam Til (1) va (2) (2) fx =0,045 I y = 0,03 nHNO, =9y-x+a+b= 0,91mol => a% = 0,91.63 100 .100% = 57,33% Vay a gan vdi gia tri 57 nhat. Hoa tan het hon hop gom 0,01 mol Cu2S; 0,04 mol FeC03 va x mol FeS2 bang dung dich HNOj vila du. Sau khi cac phan ting xay ra hoan toan, thu difdc V lit (dktc) hon hop hai khi, trong do co mot khi mau nau do va dung dich chi chiia mu6i cua Cu2+, Fe3+ vdi mot anion. Gia tri cua V la A. 51,072. B. 46,592. C. 47,488. D. 50,176. Giai: Sau phan ting chi thu dilpc muoi cua Cu2+, Fe3i vdi mot anion + 2 khi (co 1 khi mau nau do). => Anion la SO4 ,2 khi la N02 va C02 => NO" chuyen het thanh N02. 2 FeS2 + 30 HNO3 -> Fe2(S04)3 + 30 N02 + H2S04 + 14 H/) x -> 15x 0,5xmol
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