Mot So Pp Giai Nhanh Bai Tap Trac Ltdh

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<p>MT S PHNG PHP GII NHANH BI TP TRC NGHIM (LTH)PHNG PHAP 1: PHNG PHAP GIAI TOAN AP DUNG NH LUAT BAO TOAN KHOI LNG 1. Kien thc can ghi nh - Neu co PTHH tong quat: A + B C + D Th theo nh luat bao toan khoi lng ta co: mA + mB = mC + mD Nh vay trong phan ng co n chat neu biet khoi lng cua n 1 chat th tnh c khoi lng chat con lai. - Trong phan ng kh oxit kim loai bang CO, H2. Ta co: nO trong oxit = n CO2 = n H2O Vay: moxit = mO trong oxit + m kim loai - Trong phn ng gia kim loi vi dd axit giai phong kh hyro th n axit = 2 .nH2 2. Bai Tap Bai 1: Cho 24.4g hon hp Na2CO3 va K2CO3 tac dung va u vi dung dch BaCl2. sau phan ng thu c 39.4g ket tua. Loc tach ket tua co can dung dch thu c m gam muoi clorua. M co gia tr la a. 13.3g b. 15g c. 26.6g d. 63.8g ap an: c Giai: Phng trnh tong quat M2CO3 + BaCl2 BaCO3 + 2MCl n BaCO3 = 0,2mol theo phng trnh n BaCl2 = n BaCO3 = 0,2mol m BaCl2 = 0,2 . 208 = 41,6g theo nh luat bao toan khoi lng m M2CO3 + m BaCl2 = m BaCO3 + m MCl =&gt; m MCl = 24,4 + 41,6 39,4 = 26,6g Bai 2: Hoa tan hoan toan 20 gam hon hp Mg va Fe vao dung dch HCl d thay co 11.2 lt kh thoat ra (ktc) va dung dch X. co can dung dch X th khoi lng muoi khan thu c la a. 19g b. 19,5g c. 55,5g d. 37,25g ap an: c Giai: M + 2HCl MCl2 + H2 n H2 = 0.5 mol =&gt; m H2 = 1g n HCl = 1 mol =&gt; m HCl = 36,5g m M + m HCl = m X + m H2 =&gt; m X = 20 + 36,5 1 = 55,5 g Bai 3: Kh hoan toan 40.1g hon hp A gom ZnO va Fe 2O3 bang kh H2 thay tao ra 12.6g H2O khoi lng hon hp kim loai thu c la a. 28,9g b. 51,3g c. 27,5g d. 52,7g ap an a</p> <p>Giai: n H2O = 0.7 mol to ZnO + H2 Zn + H2O to Fe2O3 + 3H2 2Fe + 3H2O T phng trnh hoa hoc ta co: n oxi trong oxit = n H2O = 0,7mol mO = 0,7 .16 = 11,2g mA = m kim loai + moxi trong oxit =&gt; mkim loai = 40,1 11,2 = 28,9 g Bai 4: Kh hoan toan 32g hon hp A gom CuO va Fe2O3 bang kh H2 thay tao ra 9g H2O. khoi lng hon hp kim loai thu c la: a. 12g b. 24g c. 23g d. 41g ap an : b Giai: n H2O = 0.5 mol to CuO + H2 Cu + H2O to Fe2O3 + 3H2 2Fe + 3H2O nO = n H2O = 0,5 mol -&gt; mo = 0,5.16 = 8g mA = m kim loai + mO mkim loai = 32 8 = 24g Bai 5: Hoa tan hoan toan 18.4g hon hp 2 kim loai hoa tr II va hoa tr III trong dung dch HCl ngi ta thu c dung dch A va kh B. ot chay hoan toan lng kh B thu c 9g nc. Co can dung dch A thu c khoi lng hon hp muoi khan la a. 9,4 g b. 53,9g c. 55,9g d. 27,4g ap an b Giai: nH2O = 0,5mol X + 2HCl XCl2 + H2 (1) 2Y + 6HCl 2YCl3 + 3H2 (2) to 2H2 + O2 2H2O (3) Theo (3) =&gt; n H2 = 0,5mol =&gt; m H2 = 0,5.2 = 1g theo (1) va(2) ta co nHCl = 2 n H2 = 1mol m HCl = 36,5g theo nh luat bao toan khoi lng m(X+Y) + m(HCl) = mA + mB =&gt; mA = 18,4 + 36,5 -1 = 53,9g PHNG PHAP 2: PHNG PHAP NG CHEO 1. Kien thc can ghi nh a. Cac chat cung nong o phan tram m1 C%1 C%2 C% C%1 2 =&gt; m = C % C % 2 1</p> <p>m</p> <p>C% C%</p> <p>m2 C%2 C% C%1 m1 la khoi lng cua dung dch co nong o C%1 m2 la khoi lng cua dung dch co nong o C%2 C% la nong o phan tram dung dch thu c sau khi tron lan b. Cac chat cung nong o mol</p> <p>V1 V2</p> <p>CM1 CM</p> <p>CM2 CMV C</p> <p>1 M2 M =&gt; V = C C 2 M M1</p> <p>C</p> <p>CM2 CM CM1 V1 la the tch cua dung dch co nong o CM1 V2 la the tch cua dung dch co nong o CM2 CM la nong o mol dung dch thu c sau khi tron lan c. Cac chat kh khong tac dung c vi nhau V1 M1 M2 M M V2 M2 M M11 2 =&gt; V = M M 2 1</p> <p>V</p> <p>M M</p> <p>M la khoi lng mol trung bnh thu c khi tron lan cac kh M1 &lt; M &lt; M2</p> <p>V1 la the tch chat kh co phan t khoi la M1 V2 la the tch chat kh co phan t khoi la M2 2. BAI TAP Bai 1: Dung dch HCl co nong o 45% va dung dch HCl khac co nong o 15% e co dung dch HCl co nong o 20% th phai pha che ve khoi lng gia 2 dung dch HCl 45% va HCl 15% theo t le la a. 1:3 b. 1:5 c. 3:1 d. 5:1 ap an: b Giai: Ap dung phng phap ng cheo ta co m1 45% 5% 20%1 =&gt; m = 25% = 5 2</p> <p>m</p> <p>5%</p> <p>1</p> <p>m2 15% 25% Bai 2: The tch nc va dung dch MgSO4 2M can e pha c 100ml dung dch MgSO4 0.4M lan lt la a. 50ml va 50ml b. 40ml va 60ml c. 80ml va 20ml d. 20ml va 80ml ap an: b Giai: Goi V la the tch nc =&gt; the tch dd MgSO4 = 100 - V V 0 1.6 0.4 =&gt; 100 - V 2 0.4 Vay VH2O = 80ml va V MgSO4 = 20ml Bai 3: ieu che hon hp 26 lit kh hiro va kh cacbonoxit co t khoi hi oi vi kh metan la 1.5 th the tch kh hiro va cacbonoxit can lay la: a. 4 lit va 22 lit b. 8 lit va 44 lit c. 22 lit va 4 lit d. 44 lit va 8 lit ap an: a Giai: Mhon hp = 1,5.16 = 24g Ap dung phng phap ng cheo ta coV 1.6 = = 4 =&gt; V = 80ml 100 V 0.4</p> <p>VH 2</p> <p>2 24</p> <p>4 =&gt;VH 2 VCO = 4 22</p> <p>28 22 Bai 4: The tch H2O va dung dch NaCl 0.2M can e pha c 50ml dung dch NaCl 0.1M lan lt la a. 45ml va 5ml b. 10ml va 40ml c. 25ml va 25ml d. 5ml va 45ml ap an: c Giai: Goi V la the tch nc =&gt; the tch dd NaCl = 50 - V V 0 0.1VCO</p> <p>0.1</p> <p>=&gt;</p> <p>50 - V 0.2 0.1 Vay VH2O = 25ml va V NaCl = 25ml Bai 5: Khoi lng H2O va khoi lng dung dch ng 15% can e pha che c 50g dung dch ng 5% la: a. 2.5g va 47.5g b. 16.7g va 33.3g c. 47.5g va 2.5g d. 33.3g va 16.7g ap an: d Giai: Goi khoi lng nc la m th khoi lng ng 15% la 50 - m m 0% 10% 5% 50-m 15% Vay m H2O = 33,3g =&gt; 5% va m ng 15% = 16,7gm 10% 2 = = =&gt; m = 33.3 g 50 m 5% 1</p> <p>V 0.1 = = 1 =&gt; V = 25ml 50 V 0.1</p> <p>PHNG PHAP 3: PHNG PHAP TANG GIAM KHOI LNG 1. Kien thc can ghi nh Da vao s tang giam khoi lng khi chuyen t chat nay sang chat khac e xac nh khoi lng hon hp hay mot chat + Da vao phng trnh hoa hoc tm s thay oi ve khoi lng cua 1 mol chat trong phan ng. + Tnh so mol cac chat tham gia phan ng va ngc lai. + o tang Khoi lng kim loai = m Kim loai sinh ra m kim loai tan. o giam Khoi lng kim loai = m kim loai tan m kim loai sinh ra. 2. Bai tap Bai 1: Nhung 1 thanh nhom nang 45g vao 400ml dung dch CuSO4 0.5M sau 1 thi gian lay thanh nhom ra can nang 46.38g. Khoi lng ong thoat ra la: a. 0.64g b. 1.92g c. 1.28g d. 2.56g ap an b giai: 2Al + 3 CuSO4 Al2(SO4)3 + 3CuTheo phng trnh c 2 mol Al 3 mol Cu khoi lng tang 3 . 64 2 . 27 = 138g Theo e bai n mol Cu khoi lng tang 46,38 45 = 1,38g</p> <p>Vay n Cu = 1,38 .</p> <p>3 = 0.03 mol =&gt; m Cu = 0,03 . 64 = 1,92g 138</p> <p>Bai 2: Nhung thanh sat co khoi lng 56g vao 100ml dd CuSO4 0.5M en phan ng hoan toan. Coi toan bo lng ong sinh ra eu bam vao thanh sat. Khoi lng thanh sat sau phan ng a. 59,2g b. 56,4g c. 53,2g d. 57,2g ap an b Giai : Fe + CuSO4 FeSO4 +Cu n CuSO4 = 0,05 mol m kim loai tang = 64 . 0,05 56 . 0,05 = 0,4g m sat sau phan ng = m sat trc phan ng + m kim loai tang = 56 + 0,4 = 56,4 g Bai 3: Cho 2.52g mot kim loai cha ro hoa tr tac dung vi dd H2SO4 loang d thu c 6.84 g muoi sunfat. Kim loai a dung la a. Fe b.Zn c.Al d.Mg ap an: a Giai: C 1 mol kim loai tac dung -&gt; muoi sunfat khoi lng tang 96g n mol kim loai theo e bai khoi lng tang 6,84 2,52 = 4,32g n kim loai = 4,32 .2,52 1 = 0,045 mol =&gt; M kim loai = 0, 045 = 56g 96</p> <p>vay kim loai o la Fe Bai 4: Hoa tan 39,4g muoi cacbonat cua kim loai hoa tr II bang H2SO4 loang d thu c 46.6g muoi sunfat ket tua. Kim loai o la a Fe b. Ca c. Ba d. Pb ap an: c Giai: RCO3 + H2SO4 RSO4 + CO2 + H2OC 1 mol muoi cacbonat tac dung -&gt; muoi sufat khoi lng tang 96 60 = 36g n mol muoi cacbonat Khoi lng tang 46,6 39,4 = 7,2g</p> <p>n muoi cacbonat = 7,2 .39, 4</p> <p>1 = 0.2 mol 36</p> <p>M muoi cacbonat = 0, 2 = 197g MR = 197 60 = 137 =&gt; R la Ba Bai 5: Cho 50g kim loai ch vao 100 ml dung dch cha 2 muoi Cu(NO3)2 0.5M va AgNO3 2M. sau phan ng lay Pb ra khoi dung dch lam kho th khoi lng ch la a. 43,75g b. 56,25g c. 42,85g d. 50,9g ap an: a Giai: n Cu(NO3)2 = 0,05mol n AgNO3 = 0,2mol Pb + Cu(NO3)2 Pb(NO3)2 + Cu (1)</p> <p>Pb + 2AgNO3 Pb(NO3)2 + 2Ag (2) Theo (1) m Pb giam = 207 . 0,05 64 . 0,05 = 7,15g Theo (2) m Pb tang = 0,2 . 108 0,1 . 207 = 0,9g Vay m Pb giam = 7,15 0,9 = 6.25g m Pb = 50 6,25 = 43,75g</p> <p>PHNG PHAP 4: BO TON MOL ELECTRONTrc ht cn nhn mnh y khng phi l phng php cn bng phn ng oxi ha - kh, mc d phng php thng bng electron dng cn bng phn ng oxi ha - kh cng da trn s bo ton electron. Nguyn tc ca phng php nh sau: khi c nhiu cht oxi ha, cht kh trong mt hn hp phn ng (nhiu phn ng hoc phn ng qua nhiu giai on) th tng s electron ca cc cht kh cho phi bng tng s electron m cc cht oxi ha nhn. Ta ch cn nhn nh ng trng thi u v trng thi cui ca cc cht oxi ha hoc cht kh, thm ch khng cn quan tm n vic cn bng cc phng trnh phn ng. Phng php ny c bit l th i vi cc bi ton cn phi bin lun nhiu trng hp c th xy ra. Sau y l mt s v d in hnh.V d 1: Oxi ha hon ton 0,728 gam bt Fe ta thu c 1,016 gam hn hp hai oxit st (hn hp A).</p> <p>1. Ha tan hn hp A bng dung dch axit nitric long d. Tnh th tch kh NO duy nht bay ra ( ktc). A. 2,24 ml. B. 22,4 ml. C. 33,6 ml. D. 44,8 ml. 2. Cng hn hp A trn trn vi 5,4 gam bt Al ri tin hnh phn ng nhit nhm (hiu sut 100%). Ha tan hn hp thu c sau phn ng bng dung dch HCl d. Tnh th tch bay ra ( ktc). A. 6,608 lt. B. 0,6608 lt. C. 3,304 lt. D. 33,04. ltHng dn gii</p> <p>1. Cc phn ng c th c: t 2Fe + O2 2FeO (1) t 2Fe + 1,5O2 Fe2O3 (2) t 3Fe + 2O2 Fe3O4 (3) Cc phn ng ha tan c th c: 3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O (4) Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O (5) 3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + 14H2O (6) 0 +3 +5 Ta nhn thy tt c Fe t Fe b oxi ha thnh Fe , cn N b kh thnh N+2, O20 b kh thnh 2O 2 nn phng trnh bo ton electron l:o o o</p> <p>3n + 0,009 4 =</p> <p>0,728 3 = 0,039 mol. 56</p> <p>trong ,</p> <p>l s mol NO thot ra. Ta d dng rt ra n = 0,001 mol; VNO = 0,001 22,4 = 0,0224 lt = 22,4 ml. (p n B) 2. Cc phn ng c th c: t 2Al + 3FeO 3Fe + Al2O3 t 2Al + Fe2O3 2Fe + Al2O3 t 8Al + 3Fe3O4 9Fe + 4Al2O3 Fe + 2HCl FeCl2 + H2 2Al + 6HCl 2AlCl3 + 3H2no o o</p> <p>(7) (8) (9) (10) (11)</p> <p>Xt cc phn ng (1, 2, 3, 7, 8, 9, 10, 11) ta thy Fe0 cui cng thnh Fe+2, Al0 thnh Al+3, O20 thnh 2O v 2H+ thnh H2 nn ta c phng trnh bo ton electron nh sau: 2</p> <p>0,013 2 +</p> <p>5,4 3 = 0,009 4 + n 2 27</p> <p>Fe0 Fe+2 Al0 Al+3 O20 2O 2 2H+ H2 n = 0,295 mol VH = 0,295 22,4 = 6,608 lt. (p n A) Nhn xt: Trong bi ton trn cc bn khng cn phi bn khon l to thnh hai oxit st (hn hp A) gm nhng oxit no v cng khng cn phi cn bng 11 phng trnh nh trn m ch cn quan tm ti trng thi u v trng thi cui ca cc cht oxi ha v cht kh ri p dng lut bo ton electron tnh lc bt c cc giai on trung gian ta s tnh nhm nhanh c bi ton.2</p> <p>V d 2: Trn 0,81 gam bt nhm vi bt Fe2O3 v CuO ri t nng tin hnh phn ng nhit nhm thu c hn hp A. Ho tan hon ton A trong dung dch HNO 3 un nng thu c V lt kh NO (sn phm kh duy nht) ktc. Gi tr ca V l</p> <p>A. 0,224 lt.Hng dn gii</p> <p>B. 0,672 lt. C. 2,24 lt.</p> <p>D. 6,72 lt.</p> <p>Tm tt theo s : Fe2O3 t o ha tan hon ton 0,81 gam Al + hn h p A VNO = ? dung d HNO3 ch CuO</p> <p>Thc cht trong bi ton ny ch c qu trnh cho v nhn electron ca nguyn t Al v N. Al Al+3 + 3e0,81 27</p> <p>0,09 mol</p> <p>N+5 + 3e N+2 0,09 mol 0,03 mol VNO = 0,03 22,4 = 0,672 lt. (p n D) Nhn xt: Phn ng nhit nhm cha bit l hon ton hay khng hon ton do hn hp A khng xc nh c chnh xc gm nhng cht no nn vic vit phng trnh ha hc v cn bng phng trnh phc tp. Khi ha tan hon ton hn hp A trong axit HNO 3 th Al0 to thnh Al+3, nguyn t Fe v Cu c bo ton ha tr. C bn s thc mc lng kh NO cn c to bi kim loi Fe v Cu trong hn hp A. Thc cht lng Al phn ng b li lng Fe v Cu to thnh. vV d 3: Cho 8,3 gam hn hp X gm Al, Fe (nAl = nFe) vo 100 ml dung dch Y gm Cu(NO3)2 v AgNO3. Sau khi phn ng kt thc thu c cht rn A gm 3 kim loi. Ha tan hon ton cht rn A vo dung dch HCl d thy c 1,12 lt kh thot ra (ktc) v cn li 28 gam cht rn khng tan B. Nng C M ca Cu(NO3)2 v ca AgNO3 ln lt l</p> <p>A. 2M v 1M. C. 0,2M v 0,1M.Tm tt s : 8,3 gam hn h p X (nAl =nFe )</p> <p>B. 1M v 2M. D. kt qu khc.</p> <p>: x mol AgNO3 Al + 100 ml dung dch Y Fe Cu(NO3 )2 :y mol 1 l H 2 ,12 t Cht r A n + HCl d Z ] (3 kim loi) 2,8 gamcht r khng tan B n Hng dn gii 8,3 = 0,1mol. Ta c: nAl = nFe = 83 t n AgNO3 = x mol v n Cu( NO3 )2 = y mol</p> <p> X + Y Cht rn A gm 3 kim loi. Al ht, Fe cha phn ng hoc cn d. Hn hp hai mui ht. Qu trnh oxi ha: Al Al3+ + 3e Fe Fe2+ + 2e 0,1 0,3 0,1 0,2 Tng s mol e nhng bng 0,5 mol. Qu trnh kh: Ag+ + 1e Ag Cu2+ + 2e Cu 2H+ + 2e H2 x x x y 2y y 0,1 0,05 Tng s e mol nhn bng (x + 2y + 0,1). Theo nh lut bo ton electron, ta c phng trnh: x + 2y + 0,1 = 0,5 hay x + 2y = 0,4 (1) Mt khc, cht rn B khng tan l: Ag: x mol ; Cu: y mol. 108x + 64y = 28 (2) Gii h (1), (2) ta c: x = 0,2 mol ; y = 0,1 mol.0,2 0,1 = 2M; C M Cu( NO3 )2 = 0,1 = 1M. (p n B) 0,1 V d 4: Ha tan 15 gam hn hp X gm hai kim loi Mg v Al vo dung dch Y gm HNO3 v H2SO4 c thu c 0,1 mol mi kh SO2, NO, NO2, N2O. Phn trm khi lng ca Al v Mg trong X ln lt l</p> <p>C M AgNO3 =</p> <p>A. 63% v 37%. C. 50% v 50%.Hng dn gii</p> <p>B. 36% v 64%. D. 46% v 54%.</p> <p>t nMg = x mol ; nAl = y mol. Ta c: 24x + 27y = 15. (1) Qu trnh oxi ha: Mg Mg2+ + 2e Al Al3+ + 3e x 2x y 3y Tng s mol e nhng bng (2x + 3y). Qu trnh kh: N+5 + 3e N+2 2N+5 + 2 4e 2N+1 0,3 0,1 0,8 0,2 +5 +4 +6 +4 N + 1e N S + 2e S 0,1 0,1 0,2 0,1 Tng s mol e nhn bng 1,4 mol. Theo nh lut bo ton electron: 2x + 3y = 1,4(2) Gii h (1), (2) ta c: x = 0,4 mol ; y = 0,2 mol. %Al = 27 0,2 100% = 36%. 15</p> <p>V d 5: Trn 60 gam bt Fe vi 30 gam bt lu hunh ri un nng (khng c khng kh) thu c cht rn A. Ho tan A bng dung dch axit HCl d c dung dch B v kh C. t chy C cn V lt O 2 (ktc). Bit cc phn ng xy ra hon ton. V c gi tr l</p> <p>%Mg = 100% 36% = 64%. (p n B)</p> <p>A. 11,2 lt.Hng dn gii</p> <p>B. 21 lt.</p> <p>C. 33 lt.</p> <p>D. 49 lt.</p> <p>V n Fe &gt; n S =</p> <p>30 nn Fe d v S ht. 32</p> <p>Kh C l hn hp H2S v H2. t C thu c SO2 v H2O. Kt qu cui cng ca qu trnh phn ng l Fe v S nhng e, cn O2 thu e. Nhng e: Fe Fe2+ + 2e60 mol 56 2 60 mol 56</p> <p>S</p> <p>30 mol 32</p> <p> S+4 +</p> <p>4e4 30 mol 32</p> <p>Thu e: Gi s mol O2 l x mol. O2 + 4e 2O-2 x mol 4x Ta c: 4x = </p> <p>60 30 2 + 4 gii ra x = 1,4732 mol. 56 32 VO2 = 22,4 1,4732 = 33 lt. (p n C)</p> <p>V d 6: Hn hp A gm 2 kim loi R1, R2 c ho tr x, y khng i (R1, R2 khng tc dng vi nc v ng trc Cu trong dy hot ng ha hc ca kim loi). Cho hn hp A phn ng hon ton vi dung dch HNO3 d thu c 1,12 lt kh NO duy nht ktc. Nu cho lng hn hp A trn phn ng hon ton vi dung dch HNO3 th thu c bao nhiu lt N2. Cc th tch kh o ktc.</p> <p>A. 0,224 lt.Hng dn gii</p> <p>B. 0,336 lt. C. 0,448 lt.</p> <p>D. 0,672 lt.</p> <p>Trong bi ton ny c 2 th nghim: +5 +2 TN1: R1 v R2 nhng e cho Cu2+ chuyn thnh Cu sau Cu li nhng e cho N thnh N (NO). S mol e do R1 v R2 nhng ra l +5 +2 N + 3e N 0,15 1,12 = 0,05 22 ,4+5</p> <p>TN2: R1 v R2 trc tip nhng e cho N to ra N2. Gi x l s mol N2, th s mol e thu vo l +5 2 N + 10e N 0 2 10x x mol Ta c: 10x = 0,15 x = 0,015 VN = 22,4.0,015 = 0,336 lt. (p n B) 2</p> <p>V d 7: Cho 1,35 gam hn hp gm Cu, Mg, Al tc dng ht vi dung dch HNO3 thu c hn hp kh gm 0,01 mol NO v 0,04 mol NO2. Tnh khi lng mui to ra trong dung dch.</p> <p>A. 10,08 gam. B. 6,59 gam. C. 5,69 gam. D. 5,96 gam.Hng dn gii</p> <p>Cch 1: t x, y, z ln lt l s mol Cu, Mg, Al. 2+ 2+ Nhng e: Cu = Cu + 2e Mg = Mg + 2e</p> <p>Al = Al + 3e x x 2x y y 2y z z 3z +5 +2 +5 +4 Thu e: N + 3e = N (NO) N + 1e = N (NO2) 0,03 0,01 0,04 0,04 Ta c: 2x + 2y + 3z = 0,03 + 0,04 = 0,07 v 0,07 cng chnh l s mol NO3 Khi lng mui nitrat l: 1,35 + 62 0,07 = 5,69 gam. (p n C) Cch 2:</p> <p>3+</p> <p>Nhn nh mi: Khi cho kim loi hoc hn hp kim loi tc dng vi dung dch axit HNO3 to hn hp 2 kh NO v NO2 thn HNO3 = 2n NO2 + 4n NO n HNO3 = 2 0,04 + 4 0,01 = 0,12 mol nH O = 0,06mol p dng nh lut bo ton khi lng:2</p> <p>mKL + mHNO3 = mmui + m + m 2 + m 2O NO NO H</p> <p>V d 8: (Cu 19 - M 182 - Khi A - TSH - 2007) Ha tan hon ton...</p>