!" # $%& ' & ( # &) * +, # ,! - " . /! * )0 ! # 0 ' 1 # ' &23 ! " # " $ %&%'%(' ) " *+, !-, . / ! *)/%/ 01 20, !30 " !% 4 4 &20 # # &3. " 5 0, " 56 % )1 0 " 5 0 " 5- # " % )1 7 ! " " 1 - " 4 2 8 9 0 " 4 : 1 ;295 0 " :2< 5 ) " &49(< = 1 - 3 3 ; :2< 9 44 < × + × 8 4 &20 # # 5 1 5 > 50 " ? )% !# " / ? ? ) 50 " ? )5 8 + → 5 5 7 ! " " :2@) 1 -%1 A24@ 5 ? -, !0 " 5 . " % )1 %/ 1 ) " 8;2@&@2@@(42@ 7 - % 35 B5 C5 3 5B % 5 3 A24@ < 3:2@3:28;@) Yahoo: tat_trung151 5 5 C5 8 ↓ 5 5 8 35 3:28;@) 5 8 3:28;@::38;2@ 7 ! " " 1 -)// - ! " # 1 2). 5 %0 " 29 5 ! " # " $ %&%'%(' 53 29 < 3:;>:2@+, ! ) " % 7 ! " " 1 -)// - ! " # 1 24). 5 %0 " @2 5 ) " 9 0 " 8 < & 8 < 0 " 4 : 4 : 0 " @ ( @ 0 " 9 4 D!), 53 @2 < 324)E5 3) 5>5 ⇒ , !%F ) " ? !/ . " + 8 + 5 C 5 ( ) + 5 6 24 = + C 32@C 7 ! " " 1 - " %0 " %'? -, !), " )1 . " G 5 @ 0 " . " H5# 2, ! . " #4242. " #929D )% 1 -) " :2:9&:2:A:2:8(:2:4@ Yahoo: tat_trung151 9 8 < 53 424 < 3:28E5 3 929 44 3:24 % 3 5I5 3:28I:243:2:A) 7 ! 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" / . 1 ? ) 1 2 0 " " N 6G61 ./ *01 *5"*'N' - 1 ' OI5 5 : 8 2) ' → OI55 6'-, ! )'' 1C)N Yahoo: tat_trung151 [...]... dd Na 2CO3 thi thu c V lit CO2 (ktc) va dd muụ i.Cụ ca n dd thi thu c 28,96g muụ i Gia tri cua V la: A 4,84 lit B 4,48 lit C 2,24 lit D 2,42 lit E Kờ t qua khac Suy luõn: Go i cụng thc trung binh cua 2 axit la: R COOH Ptpu: 2 R COOH + Na 2CO3 2 R COONa + CO2 + H2O Theo pt: 2 mol 2 mol 1 mol m = 2.(23 - 11) = 44g Theo ờ bai: Khụ i l ng tng 28,96 20,15 = 8,81g 8,81 = 0,2mol Thờ tich CO2 : V... dd Na 2CO3 thi thu c V lit CO 2 (ktc) va dd muụi.Cụ can dd thi thu c 28,96g muụi Gia tri cua V la: A 4,84 lit B 4,48 lit C 2,24 lit D 2,42 lit Suy luõn: Goi cụng thc trung binh cua 2 axit la: R COOH Ptpu: 2 R COOH + Na 2CO3 2 R COONa + CO2 + H2O Theo pt: 2 mol 2 mol m = 2.(23 - 11) = 44g Theo ờ bai: Khụi lng tng 28,96 20,15 = 8,81g Yahoo: tat_trung151 1 mol Nguyn Tt Trungvocalcords... axit Vớ d 1: Cho t t mt lung khớ CO qua ng s ng m gam hn hp Fe v cỏc Yahoo: tat_trung151 Nguyn Tt Trungvocalcords T:05002461803 oxit ca Fe un núng thu c 64 gam Fe, khi i ra sau PU to 40 gam kt ta vi dung dch Ca(OH)2 d Tớnh m A 7,04 g B 74,2 g C 70,4 g D 74 g Gii Ta cú: nCO2 = nCaCO3 = 40/100 = 0,4 mol mCO + m = mFe + mCO2 m nCO pu = nCO2 = 0,4 nờn: m = mFe + mCO2 - mCO = 64 + 0,4.44 - 0,4.28 = 70,4... hợp Na 2CO3 0,1 M và (NH4) 2CO3 0,25 M thu đợc 39,7 gam kết tủa X Tính phần trăm khối lợng các chất ttrong X A 42,62% và 53,38% B 40,70% và 50,30% C 60% và 40% D 70,80% và 20,20% Gii: CO3 2- + Ba2+ -> BaCO3 CO3 2- + Ca2+ -> CaCO3 Khi chuyển 1 mol muối BaCl2 hay CaCl2 thành BaCO3 hay CaCO3 khối lợng bị giảm đi : 71-6o = 11 gam Nh vậy tổng số mol 2 muối cacbonat = = 0,3 mol 43-39,7 11 Còn số mol của CO2 2-... P1: em ụt chay hoan toan thu c 1,08g H2O P2: tac dung vi H2 d (Ni, t0) thi thu hụn hp A em A ụt chay hoan toan thi thờ tich CO2 (ktc) thu c la: A 1,434 lit B 1,443 lit C 1,344 lit D 1,444 lit Suy luõn: Vi anehit no n chc nờn sụ mol CO2 = sụ mol H2O = 0,06 mol nCO2 ( P 2) = nC ( P 2) = 0,06mol Theo BTNT va BTKL ta co: nC ( P 2) = nC ( A ) = 0,06mol nCO2 ( A ) = 0,06mol VCO2 = 22,4.0,06 = 1,344... hụn h p A em A ụ t chay hoan toan thi thờ tich CO2 (ktc) thu c la: A 1,434 lit B 1,443 lit C 1,344 lit D 1,444 lit Suy luõn: Vi anehit no n chc nờn sụ mol CO2 = sụ mol H2O = 0,06 mol nCO2 ( P 2) = nC ( P 2) = 0,06mol Theo BTNT va BTKL ta co: nC ( P 2) = nC ( A ) = 0,06mol nCO2 ( A ) = 0,06mol Yahoo: tat_trung151 Nguyn Tt Trungvocalcords T:05002461803 VCO2 = 22,4.0,06 = 1,344 lit Thi du 4: Tach nc... tat_trung151 Nguyn Tt Trungvocalcords T:05002461803 5 ụt chay ankin: nCO2 > nH2O va nankin (chay) = nCO2 nH2O Thi du 1: ụt chay hoan toan hụn hp 2 hidrocacbon liờn tiờp trong day ụng ng thu c 22,4 lit CO2 (ktc) va 25,2g H2O Hai hidrocacbon o la: A C2H6 va C3H8 B C 3H8 va C4H10 C C4H10 va C5H12 D C 5H12 va C6H14 Suy luõn: nH2O = nH2O > nCO2 25, 2 = 1,4 mol ; nCO2 = 1mol 18 2 chõt thu c day ankan Goi n... Ta co: 3n + 1 O2 2 n CO2 n 1 = n = 2,5 n + 1 1, 4 + ( n + 1) H2O C2H6 C3H8 Yahoo: tat_trung151 Nguyn Tt Trungvocalcords T:05002461803 Thi du 2: ụt chay hoan toan V lit (ktc) 1 ankin thu c 10,8g H2O Nờu cho tõt ca san phõm chay hõp thu hờt vao binh ng nc vụi trong thi khụi lng binh tng 50,4g V co gia tri la: A 3,36 lit B 2,24 lit C 6,72 lit D 4,48 lit Suy luõn: Nc vụi trong hõp thu ca CO2 ... thu ca CO2 va H2O mCO2 + mH2O = 50,4g ; mCO2 = 50,4 10,8 = 39,6g nCO2 = 39,6 = 0,9 mol 44 nankin = nCO2 nH2O = 0,9 10,8 = 0,3 mol 4418 II- Một số phơng pháp giải nhanh trắc nghiệm hoá vô cơ 1 Bo ton khi lng:-Nguyờn tc: +Trong PUHH thỡ tng khi lng cỏc sn phm bng tng khi lng cỏc cht tham gia PU +Khi cụ cn dung dch thỡ khi lng hn hp mui thu c bng tng khi lng cỏc cation kim loi v anion gc axit Vớ d 1:... axit m = 45 29 = 16g Võy nờu ờ cho m anehit, maxit nanehit, nAg CTPT anehit ụi vi axit: Xet phan ng vi kiờm R(COOH) x + xNaOH R(COONa) x + xH2O Hoc RCOOH + NaOH RCOONa + H2O Yahoo: tat_trung151 Nguyn Tt Trungvocalcords T:05002461803 1 mol m = 1 mol 22g ụi vi este: xet phan ng xa phong hoa RCOOR + NaOH 1 mol RCOONa + ROH m = 1 mol 23 MR ụi vi aminoaxit: xet phan ng vi HCl HOOC-R-NH2 . ⇒ ∑ ∑ '' 3 2 :2 9 = /! * 6 $ !" $ # $ (& ' +5 0 # &!" * ),'- . :2A:8&:28A:2A4(:2A8 4 !" $ # $ # +)! 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