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VL-B0BV 7-7 D8;,;<*-;<!;<;,%)'&;/_ ‡ 7-7‡_ B #'zp 5 B %5 B C0 Gi=S*_'s%8"Ej8_0Gi8-=S*5 B %5 B C"?-8R "3.:A $ _c:V_ a B5C c aJ:aBD a V5 B aND B C J8B88 B5C c a6:aJD a V5 B CaZD B C 6J [0_VNB-}$ 0VJ8a6}}$ 8aVN-6YBB-7VJ-LZ}}}$ B68a77$YJ-LZV 7-7}d$ W}$-}}$-}}}$-}d$A 8VJ-7Z VJ- 0_VNB- 0VN-O 1x"8_VO0KzV -%2_VBL0d.NI/+I 0W8A [...]... cỏc ion sau K+ :0,15 mol, Mg2+ : 0,1 mol,NH4+:0,25 mol,H+ :0,2 mol, Cl- :0,1 mol SO4 2- :0,075 mol NO 3- :0,25 mol,NO 3- :0,25 mol v CO3 2- :0,15 mol Mt trong 2 dung dch trờn cha A: K+,Mg2+,SO4 2- v Cl-; B : K+,NH4+,CO3 2- v ClC: NH4+,H+,NO 3-, v SO42D : Mg2+,H+,SO4 2- v Cl2+ 2+ Cõu 3: Dung dch Y cha Ca 0,1 mol ,Mg 0,3 mol,Cl 0,4 mol,HCO 3- y mol Khi cụ cn dung dch Y thỡ c mui khan thu c l : A: 37,4 gam B 49,8... Cht rn D l Cu: Cu + 1/2O2 = CuO (x + y + z) = 12/80 = 0,15(mol) (2) - Khi cho 1/2 dung dch B + NaOH s xy ra cỏc phn ng: +OH +OH 2Zn 2+ Zn(OH) 2 ZnO 2 (tan) - o +O 2 ,+H 2 O +OH t 2Fe 2+ 2Fe(OH) 2 2Fe(OH)3 Fe2 O3 0,5y +OHto Cu 2+ Cu(OH) 2 CuO 0,5(0,1 7- x - y) 0,5(0,1 7- x - y) Ta cú phng trỡnh: (0,25y 160) + 0,5(0,17 - x - y) 80 = 5,2 (3) 0,25y Gii h phng trỡnh (1), (2), (3) ta c: x = 0,04... + 3a - 0,085 2 = 0,05 (mol) nC : nH : nO = 0,08 : 0,12 : 0,05 = 8 : 12 : 5 Vy cụng thc ca cht hu c A l C8H12O5 cú MA < 203 gam ỏp ỏn D Vớ d 7: Cho 0,1 mol este to bi 2 ln axit v ancol mt ln ancol tỏc dng hon ton vi NaOH thu c 6,4 gam ancol v mt lng mui cú khi lng nhiu hn lng este l 13,56% (so vi lng este) CTCT ca este l: A CH3 - COO - CH3 B CH3OCO - COO - CH3 C CH3COO - COOCH3 D CH3COO - CH2 - COOCH3... nHCl = 2nH2 =0,3 mol S mol HCl hũa tan cỏc oxit =0, 7- 0,3 = 0,4 mol Theo nh lut bo ton din tớch ta cú nO 2-( oxit) =1/2 nCl- = 0,2 mol nFe (trong X) =moxit - moxi /56 =(2 0-0 ,2 x 16)/56 = 0,3 mol Cú th coi : 2Fe (trong X ) Fe2O3 nFe2O3 =1,5 mol mFe2O3 = 24 gam ỏp ỏn C III BI TP T LUYN Cõu 1: Dung dch X cú cha a mol Na+ ,b mol Mg2+ ,C mol Cl- v d mol SO4 2- Biu thc liờn h gia a,b,c,d l A: a+2b = c+2d B:... thy: + Bi toỏn cú 3 i lng: m, m1 v ne nhn (2) (hoc V khớ (2)) Khi bit 2 trong 3 i lng trờn ta tớnh c ngay i lng cũn li + giai on (2) bi cú th cho s mol, th tớch hoc khi lng ca 1 khớ hay nhiu khớ; giai on (1) cú th cho s lng cht rn c th l cỏc oxit hoc hn hp gm kim loi d v cỏc oxit 2 Phm vi ỏp dng v mt s chỳ ý + Ch dựng khớ HNO3 (hoc (H2SO4 c, núng) ly d hoc va + Cụng thc kinh nghim trờn ch ỏp dng... lớt B 0,25 lớt C 0,25 lớt D 0,52 lớt Hng dn : Dung dch X cha cỏc ion Na+ ; AlO 2- ; OH- d (cú th) p dng nh lut Bo ton in tớch: n AlO 2- + n OH- = n Na+ = 0,5 Khi cho HCl vaof dung dch X: H+ + OH H2O (1) + H + AlO2 + H2O Al(OH)3 (2) 3H+ + Al(OH)3 Al3+ + 3H2O (3) kt ta l ln nht, suy ra khụng xy ra (3) v n H+ = n AlO 2- + n OH- = 0,5 Suy ra th tớch HCl = 0,5/2 = 0,25 (lớt) ỏp ỏn B Dng 5 : Bi toỏn tng... mol Fe + O2 Cỏc oxi st + HNO3 Fe(NO3)3 + NO + NO2 + H2O Ta cú cỏc quỏ trỡnh cho nhn electron: Cho 3+ Fe - 3e = Fe x 3x Nhn O2 + 4e = 2O2(5,04 - 56x)/32 4(5,04 - 56x)/32 NO 3- + 3e + 4H+ = NO + 2H2O 3a a NO 3- + e + 2H+ = NO2 + H2O b b Suy ra: a + b = 0,035 (30a + 46b) / (a + b) = 19 2 = 38 4(5,04 - 56x)/32 + 3a + 3b = 3x a = 0,0175; b = 0,0175; x = 0,07 Bi 7: m gam phụi bo st (A) ngoi khụng khớ, sau... NO + 14H2O b Cn c vo s phn ng: a mol Fe + O2 Fe, FeO, Fe2O3, Fe3O4 + HNO3 Fe(NO3)3 + NO + H2O Ta cú cỏc quỏ trỡnh cho, nhn electron: Cho Fe - 3e = Fe3+ a 3a Nhn O2 + 4e = 2O2(12 - 56a)/32 4(12 - 56a)/32 NO 3- + 3e + 4H+ = NO + 2H2O 0,03 mol 0,1 mol 3a = 4(12 - 56a)/32 + 0,3 a = 0,18 m = 56a = 10,08g Tuyt chiờu s 6 (Bo ton khi lng) P DNG PHNG PHP BO TON KHI LNG Nguyờn tc ca phng phỏp ny khỏ n gin,... NO duy nht, Giỏ tr ca x l A 0,045 B 0,09 C 0,135 D 0,18 Hng dn: - p dng bo ton nguyờn t: Fe3+ : x mol ; Cu2+ : 0,09 ; SO4 2- : ( x + 0,045) mol - p dng nh lut bo ton in tớch (trong dung dch ch cha mui sunfat) Ta cú : 3x + 2.0,09 = 2(x + 0,045) x = 0,09 ỏp ỏn B Vớ D 5: Dng dch X cú cha 5 ion : Mg2+ , Ba2+ , Ca2+ , 0,1 mol Cl- v 0,2 mol NO 3- Thờm dn V lớt dung dch K2CO3 1M vo X n khi c lng kt ta ln nht... = 83,68 -3 2 x 0,78 = 58,72 (gam) Cho cht rn B tỏc dng vi 0,18 mol K2CO3 CaCl +K CO CaCO +2KCl (4) 2 2 3 3 0,36 mol hn hp D Hn hp B 0,18 ơ 0,18 KCl (B) 0 0 0 0 0 m KCl(B) = m B - m CaCl2 (B) = 58,72 - 0,18 x 111 = 38,74 (gam) m KCl(D) = m KCl(B) + m KCl(pt4) = 38,74 +0,36 x 74,5 = 65,56 (gam) 3 3 m KCl(D) = x 65,56 = 8,94 (gam) 22 22 m KCl (pt1) = m KCl (B) - m KCl (A) = 38,74 - 8,94 = . :: ="8A BL8a0_VB-7Z 8aVJ-BZ 8aBV6aJa6: aV - BYBB-7VJ-JZ 77aB6$Ya$V 6-6 0BV L-N :VJ-776YBB-7VJ-JB [8VJ- VJ-Z VJ-J VJ-B :VJ-JB N854ONI/P ./5S*9:#'>Z-J7H@*=40D8H "D5C . 1KCC58a1|CD→ -7 78*-J68*Z-ZN8* ‰v•IWo - 8 * 58CD V-J6aZ-ZNc -7 7VZ-B8*$ 58CD VZ-BY7JVJ- *$ 1KCC58 1|CD 1KCC1| -J6 _ V V6Z-B 8*$ 1V6-B J- Z-ZN _. c a6:aJD a V5 B CaZD B C 6J [0_VNB-}$ 0VJ8a6}}$ 8aVN-6YBB-7VJ-LZ}}}$ B68a77$YJ-LZV 7-7 }d$ W} $- }} $- }}} $- }d$A 8VJ-7Z VJ- 0_VNB- 0VN-O 1x"8_VO0KzV
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