10 PHUONG PHAP GIAI NHANH BAI TAP TRAC NGHIEM HOA HOC

www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 oc 01 PHNG PHP BO TON KHI LNG D Ta iL ie moxit mCO,C,H2 mran mCO2 ,H2O uO nT hi Xột phn ng A + B C + D ta cú ; mA + mB = mC + mD Mt s dng thng gp : - Hn hp oxit tỏc dng vi axit to mui: MO + HCl, H2SO4 loóng moxit + maxit = mmui + mnc Trong ú s mol nc c tớnh theo axit - Hn hp oxit tỏc dng CO, C hoc H2 : H Nguyờn tc ro up s/ Trong ú s mol CO2, H2O c tớnh theo CO, C v H2 - Hn hp kim loi tỏc dng vi axit gii phúng H2 mkim loi + maxit = mmui + mH2 s mol H2 = 2HCl = H2SO4 - Hn hp mui cacbonat tỏc dng vi axit mmui (1) + maxit = mmui + mH2O + mCO2 om /g Cỏc vớ d minh ho w w w fa ce bo ok c Vớ d 1: (2007 - Khi A) Ho tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO 500 ml axit H2SO4 0,1M (va ) Sau phn ng, hn hp mui sunfat khan thu c cụ cn dung dch cú lng l A 6,81 gam B 4,81 gam C 3,81 gam D 5,81 gam Hng dn Fe2(SO4)3 + 3H2O (1) Fe2O3 + 3H2SO4 MgSO4 + H2O (2) MgO + H2SO4 ZnSO4 + H2O (3) ZnO + H2SO4 Theo cỏc pt hoỏ hc (1, 2, 3): n H2O = n H2SO4 = 0,5 0,1 = 0,05 (mol) p dng nh lut bo ton lng: m hh mui khan = 2,81 + 98 0,05 18 0,05 = 6,81 (g) Vớ d 2: Hn hp X gm Fe, FeO v Fe2O3 Cho mt lung khớ CO i qua ng s ng m gam hn hp X nung núng Sau kt thỳc thớ nghim thu c 64 gam cht A ng s v 11,2 lớt khớ B (ktc) cú t so vi H2 l 20,4 Giỏ tr ca m l A 105,6 B 35,2 C 52,8 D 70,4 Hng dn Cỏc phng trỡnh hoỏ hc ca phn ng kh oxit st cú th cú: t 2Fe3O4 + CO2 (1) 3Fe2O3 + CO Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 t Fe3O4 + CO 3FeO + CO2 (2) D nT 44x + 28(0,5 x) = 0,5 20,4 = 20,4 x = 0,4 (mol) Do ú n CO phn ng = 0,4 (mol) p dng nh lut bo ton lng: m = mA + mCO2 mCO H 11, = 0,5 (mol) 22, hi nB = oc t FeO + CO Fe + CO2 (3) Nhn xột: Cht rn A cú th gm cht Fe, FeO, Fe3O4 hoc ớt hn, iu quan trng l s mol CO phn ng bao gi cng bng s mol CO2 to thnh Gi x l s mol CO2 to thnh up s/ Ta iL ie uO = 64 + 44 0,4 28 0,4 = 70,4 (g) Vớ d 3: Ho tan hon ton gam hn hp kim loi bng dung dch HCl thu c dung dch A v khớ B Cụ cn dung dch A thỡ c 5,71 gam mui khan Tớnh th tớch khớ B (o ktc) Hng dn Gi kim loi ó cho l X v Y 2X + 2m HCl 2XClm + m H2 (1) ro 2Y + 2n HCl 2YCln + n H2 (2) Theo (1, 2): n HCl = n H2 om /g p dng nh lut bo ton lng: + 36,5 n H2 = 5,71 + n H2 n H2 = 0,01 (mol) Vy VH2 (ktc) = 0,01 22,4 = 0,224 (l) bo ok c Vớ d 4: (2009 - Khi A)Cho 3,68 gam hn hp gm Al v Zn tỏc dng vi mt lng va dung dch H2SO4 10%, thu c 2,24 lớt khớ H2 ( ktc) Khi lng dung dch thu c sau phn ng l A 101,68 gam B 88,20 gam C 101,48 gam D 97,80 gam w w fa ce Hng dn Al2(SO4)3 + 3H2 (1) 2Al + 3H2SO4 ZnSO4 + H2 (2) Zn + H2SO4 T (1, 2): n H2SO4 = n H2 = w mdd H2SO4 = 01 2, 24 = 0,1 (mol) 22, 98 0,1 100 = 98 (g) 10 p dng nh lut bo ton lng: mdd sau phn ng = mhh + mdd H2SO4 mH = 3,68 + 98 0,1 = 101,48 (g) Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 oc Li gii : Khớ i sau phn ng gm CO2 v CO d cho i qua dung dch Ca(OH)2 d : H CO2 + Ca(OH)2 CaCO3 + H2O D 40 0,4 (mol) 100 0,4 (mol) CO Fe + CO2 Ta iL ie 28.0,4 + m = 64 + 44.0,4 m = 70,4g uO + nT hi S phn ng: FeO Fe2O3 Fe3O4 up s/ Vớ d Ngi ta cho t t lung khớ H2 i qua mt ng s ng 5,44 gam hn hp gm FeO, Fe3O4, Fe2O3, CuO nung núng, kt thỳc phn ng thu c m gam hn hp cht rn A v 1,62 gam H2O Vy m cú giỏ tr l A 4g B 5g C 4,5g D 3,4g Li gii : nH2 nH2O 0, 09 (mol) ok c om /g ro S phn ng: FeO H2 + Fe2O3 A + H2O Fe3O4 CuO 0,09.2 + 5,44 = m + 1,62 m = 4g fa ce bo Vớ d Cho 35g hn hp Na2CO3, K2CO3 tỏc dng va vi dung dch BaCl2 Sau phn ng thu c 59,1g kt ta Lc tỏch kt ta, cụ cn dd thu c m(g) mui clorua Vy m cú giỏ tr l A 38,3g B 22,6g C 26,6g D 6,26g w w w Li gii : S phn ng: Na CO3 K CO3 01 Vớ d Cho t t mt lung khớ CO i qua ng s ng m gam hn hp gm Fe, FeO, Fe3O4, Fe2O3 nung núng, kt thỳc phn ng thu c 64g st, khớ i gm CO v CO2 cho sc qua dung dch Ca(OH)2 d c 40g kt ta Vy m cú giỏ tr l A 70,4g B 74g C 47g D 104g BaCO3 + + BaCl2 NaCl KCl nBaCl2 nBaCO3 0,3 (mol) p dng nh lut bo ton lng: mhh mBaCl2 m m dd m = 35 + 0,3.208 59,1 = 38,3 (g) Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 Li gii : S phn ng: Na2SO4 K2SO4 + Ba(NO3)2 BaSO4 + (NH4)2SO4 01 Vớ d Cho 4,48g hn hp Na2SO4, K2SO4, (NH4)2SO4 tỏc dng va vi 300 ml dung dch Ba(NO3)2 0,1M Kt thỳc phn ng thu c kt ta A v dung dch B Lc tỏch kt ta, cụ cn dung dch thu c m(g) mui nitrat Vy m cú giỏ tr l A 5,32g B 5,23g C 5,26g D 6,25g H oc NaNO3 KNO3 NH4NO3 D nBa(NO3 )2 nBaSO4 0, 03 (mol) ; 4, 48 7,83 6,99 mB mB 5,32 (g) uO nT hi Vớ d Ho tan 2,57g hp kim Cu, Mg, Al bng mt lng va dung dch H 2SO4 loóng thu c 1,456 lớt khớ X (ktc), 1,28g cht rn Y v dung dch Z Cụ cn Z thu c m gam mui khan, m cú giỏ tr l A 7,53g B 3,25g C 5,79g D 5,58g Ta iL ie Li gii: S phn ng : up s/ Cu MgSO4 + Cu + H2 Mg + H2SO4 Al (SO ) Al m m (Al Mg) m SO2 (2,57 1,28) 0, 065.96 7,53 (g) om /g ro Vớ d Ho tan hon ton 3,72g hn hp kim loi A, B dung dch HCl d thy to 1,344 lớt khớ H2 (ktc) Cụ cn dung dch sau phn ng thu c mui khan cú lng l A 7,12g B 7,98g C 3,42g D 6,12g H n ok n c Li gii : Theo phng trỡnh in li : Cl 1,344 0,12(mol) 22, bo mmui = mKL + mCl = 3,72 + 0,12.35,5 = 7,98 (g) w w w fa ce Vớ d 10 Nung m gam hn hp A gm mui MgCO3 v CaCO3 cho n khụng cũn khớ thoỏt thu c 3,52g cht rn B v khớ C Cho ton b khớ C hp th ht bi lớt dung dch Ba(OH)2 thu c 7,88g kt ta un núng dung dch li thy to thnh thờm 3,94g kt ta na Nu cỏc phn ng xy hon ton thỡ m cú giỏ tr l A 7,44g B 7,40g C 7,04g D 4,74g Li gii: m = mB + mCO2 BaCO3 + H2O CO2 + Ba(OH)2 Ba(HCO3)2 2CO2 + Ba(OH)2 m = 3,52 + ( 7,88 3,94 ).44 7, 04 (g) 197 197 Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 Nguyờn tc H oc Tng s mol nguyờn t ca mt nguyờn t trc v sau phn ng luụn bng Tớnh s mol nguyờn t ca mt nguyờn t : nnguyờn t A = x.nX = (s nguyờn t A X).s mol X vớ d : nO = 4.nH2SO4 01 PHNG PHP BO TON MOL NGUYấN T D Cỏc vớ d minh ho uO nT hi Vớ d 1: (2007 - Khi A) Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo axit HNO3 (va ), thu c dung dch X (ch cha hai mui sunfat) v khớ nht NO Giỏ tr ca a l A 0,04 B 0,075 C 0,12 D 0,06 Hng dn Ta iL ie Fe2 (SO4 )3 FeS2 HNO3 NO H 2O Cu 2S CuSO4 S phn ng: ce bo ok c om /g ro up s/ 2FeS2 Fe2(SO4)3 0,12 0,06 Cu2S 2CuSO4 a 2a p dng nh lut bo ton nguyờn t i vi S, ta cú: 0,12 + a = 0,06 + 2a a = 0,06 (mol) Vớ d 2: Dn t t V lớt khớ CO ( ktc) i qua mt ng s ng lng d hn hp rn gm CuO, Fe2O3 ( nhit cao) Sau cỏc phn ng xy hon ton, thu c khớ X Dn ton b khớ X trờn vo lng d dung dch Ca(OH)2 thỡ to thnh gam kt ta Giỏ tr ca V l A 1,120 B 0,896 C 0,448 D 0,224 (Trớch thi TSC nm 2008 - Khi A, B) Hng dn CuO t0 + CO CO2 (X) + hn hp rn Fe O3 w w w fa S phn ng: CaCO3 + H2O CO2 + Ca(OH)2 d p dng s bo ton s mol i vi nguyờn t C: n C CO = n C CO2 = n C CaCO3 n C CO = n CaCO = = 0,04 (mol) 100 Vy V = 0,04 22,4 = 0,896 (l) Vớ d 3: Ho tan hn hp X gm 0,2 mol Fe v 0,1 mol Fe2O3 vo dung dch HCl d c dung dch Y Cho dung dch Y tỏc dng vi dung dch NaOH d thu c kt Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 ta Lc kt ta, sch em nung khụng khớ n lng khụng i thu c m gam cht rn Z Giỏ tr ca m l A 16,0 B 24,0 C 28,8 D 32,0 Hng dn S cỏc phn ng: oc Fe(OH) t Z (Fe2O3) Fe(OH)3 p dng s bo ton s mol i vi nguyờn t Fe: n Fe Z = n Fe X = n Fe 2n Fe2O3 = 0,2 + 2.0,1 = 0,4 (mol) D n Fe Z = 0,2 (mol) Vy m = 0,2 160 = 32,0 (g) hi n Fe2O3 Z = H FeCl2 NaOH Fe HCl X Y Fe2O3 FeCl3 up s/ Ta iL ie uO nT Vớ d 4: Cho 2,13 gam hn hp X gm ba kim loi Mg, Cu v Al dng bt tỏc dng hon ton vi oxi thu c hn hp Y gm cỏc oxit cú lng 3,33 gam Th tớch dung dch HCl 2M va phn ng ht vi Y l A 57 ml B 50 ml C 75 ml D 90 ml (Trớch thi TSH nm 2008 - Khi A) Hng dn MgO Mg O2 X Cu Y CuO Al O Al om /g ro MgCl2 + H2O (1) MgO + 2HCl CuCl2 + H2O (2) CuO + 2HCl 2AlCl3 + 3H2O (3) Al2O3 + 6HCl Ta cú mO / hhY = 3,33 2,13 = 1,2 (g) hay n O / hhY = 1, = 0,075 (mol) 16 c Theo (1, 2, 3): n HCl = n O / hhY = 0,075 = 0,15 (mol) Vy Vdd HCl = 0,15 = 0,075 w w w fa ce bo ok (l) = 75 (ml) Vớ d 5: Cho mt mu Na lõu khụng khớ, b chuyn hoỏ thnh hn hp rn X gm Na, Na2O, NaOH, Na2CO3 Ho tan hon ton hn hp X bng H2SO4 loóng, sau phn ng thu c dung dch Y Lm bay hi nc t t thu c 8,05 gam tinh th Na2SO4.10H2O Khi lng mu Na l A 0,575 gam B 1,15 gam C 2,3 gam D 1,725 gam Hng dn Na O NaOH H2SO4 kk dd Y (dd Na2SO4) Na2SO4.10H2O Na X Na CO3 Na 8,05 Ta cú n Na 2SO4 10H2O = = 0,025 (mol) n Na ca mu Na = n Na 322 n Na 2SO4 10H2O = 0,05 (mol) Vy lng mu Na l: 0,05 23 = 1,15 (g) Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 01 Header Page of 16 tinh th = www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Vớ d 6: Cho hn hp A gm ba kim loi X, Y, Z cú hoỏ tr ln lt l 3, 2, v t l s mol ln lt l : : 3, ú s mol ca X bng x mol Ho tan hon ton A bng dung dch cú cha y mol HNO3 Sau phn ng thu c dung dch B khụng cha NH4NO3 v V lớt hn hp khớ E ( ktc) gm NO2 v NO Biu thc tớnh y theo x v V l V 22, B 6x + V 22, V 22, C 5x + D 10x + V 22, V (mol) 22, D Ta cú n hh E = H Hng dn oc A 8x + Ta iL ie uO nT hi Theo bi n X : n Y : n Z = : : M n X = x (mol) nờn n Y = 2x (mol), n Z = 3x (mol) S phn ng: X , Y , Z + HNO3 X(NO3)3, Y(NO3)2, ZNO3 + NO2 , NO + H2O x 2x 3x x 2x 3x S mol mi mui bng s mol mi kim loi tng ng Da vo s bo ton s mol i vi nguyờn t N: n HNO3 = n N = n X(NO3 )3 + n Y(NO3 )2 + n ZNO3 + n NO2 + n NO V V = 10x + (mol) 22, 22, up s/ = x + 2x + 3x + om /g ro Vớ d Cho hn hp A gm 0,1 mol Cu, 0,2 mol Ag tan va ht V lớt dung dch HNO3 1M thu c dung dch X v hn hp Y gm khớ NO, NO2 ( n NO n NO2 0,1 mol ) V cú giỏ tr l A lớt B 0,6 lớt C 1,5 lớt D lớt + HNO3 ok Ag c Li gii S phn ng: Cu Cu(NO3 )2 AgNO3 + NO NO2 + H2 O bo p dng s bo ton nguyờn t Cu, Ag ta cú : nCu = nCu(NO3 )2 = 0,1 mol v nAg = ce n AgNO3 = 0,2 mol fa p dng cho nguyờn t N : n N (HNO3 ) = n N(Cu(NO3 )2 AgNO3 NONO2 ) w w w n(HNO3 ) = 2nCu(NO3 )2 nAgNO3 n NO n NO2 n(HNO3 ) = 2.0,1 + 0,2 + 0,1 + 0,1 = 0,6 mol VHNO3 0,6 0,6(lit) Vớ d 8: Cho hn hp gm : FeO (0,01 mol), Fe2O3 (0,02 mol), Fe3O4 (0,03 mol) tan va ht dung dch HNO3 thu c mt mui nht v 0,448 lớt khớ N2O4 (ktc) Khi lng mui v s mol HNO3 tham gia phn ng l Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 01 Header Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 A 32,8 g ; 0,4 mol C 33,88 g ; 0,06 mol B 33,88 g ; 0,46 D 33,28 g ; 0,46 mol mol Li gii S phn ng : H oc 01 FeO Fe2O3 + HNO3 Fe(NO3)3 + NO2 + H2O Fe3O4 p dng s bo ton nguyờn t Fe : nFe [Fe(NO3 )3 ] nFe [FeO,Fe2O3 ,Fe3O4 ] hi D nFe [Fe(NO3 )3 ] = nFeO 2nFe2O3 3nFe3O4 = 0, 01 2.0, 02 3.0, 03 0,14 (mol) nT mFe(NO )3 0,14.242 33,88 (g) uO p dng s bo ton nguyờn t N : n N [HNO3 ] n N [Fe(NO3 )3 N 2O4 ] Ta iL ie n HNO3 3n Fe(NO3 )3 n N2O4 3.0,14 2.0, 02 0, 46 (mol) Li gii : S phn ng : Al + HCl H SO hn hp mui ( X+T ) + H2 c Fe om /g ro up s/ Vớ d 9: Cho 1,1 gam hn hp Fe, Al phn ng vi dung dch HCl thu c dung dch X, cht rn Y v khớ Z, ho tan ht Y cn s mol H2SO4 (loóng) bng 1/2 s mol HCl trờn thu c dung dch T v khớ Z Tng th tớch khớ Z (ktc) l 0,896 lớt Tng lng mui sinh hai trng hp trờn l A 2,54 gam B 2,77 gam C 3,36 gam D 1,06 gam ok t x= nHCl ; n H2SO4 = y (mol) w w w fa ce bo p dng s bo ton nguyờn t H: n H [HClH2SO4 ] n H [H2 ] n HCl 2n H2SO4 2n H2 0, 04 (mol) x + 2y = 0,04 x x = 0,02 ; y = 0,01 mmui = m(Al,Fe) + mCl mSO = 1,1 + 0,02.35,5 + 0,01.96 = 2,77 (gam) y= Vớ d 10 Cho 1,48 g hn hp kim loi Fe, Al, Zn tan hon ton dung dch H2SO4 loóng, ta thu c 0,784 lớt khớ H2 (ktc) Khi cụ cn dung dch lng mui khan thu c l A 4,84 g B 5,65 g C 5,56 g D 4,56 g Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page of 16 Li gii : S phn ng : Fe FeSO4 0, 784 = 0,035 (mol) 22, oc n H2SO4 = n H = 01 Al + H2SO4 Al2(SO4)3 + H2 Zn ZnSO4 p dng s bo ton nguyờn t H: 4 w w w fa ce bo ok c om /g ro up s/ Ta iL ie uO nT mmui = m(Fe, Al, Zn) + mgc axit = 1,48 + 0,055.96 = 4,84 (gam) D hi H p dng s bo ton nguyờn t S: = 0,035 (mol) n n SO [H SO ] SO [muối] Footer Page of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 10 of 16 PHNG PHP TNG HOC GIM KHI LNG oc 01 H Nguyờn tc om /g ro up s/ Ta iL ie uO nT hi D Khi chuyn t cht X (thng tớnh cho mol) thnh cht Y (khụng nht thit trc tip, cú th b qua nhiu giai on trung gian), lng tng hay gim bao nhiờu gam Da vo lng thay i ú ta tớnh c s mol cỏc cht cn thit hoc ngc li Ghi nh: Trng hp kim loi A y kim loi B dung dch mui thnh kim loi B t Ta cú: Khi lng A tng = mB bỏm vo mA tan Khi lng A gim = mA tan mB bỏm vo Mt s dng thng gp : HCl + mol kim loi mui Cl thỡ lng tng 35,5n gam (n l s oxi húa ca kl) + mol mui CO32- mol Cl lng tng 35,5.2 - 60 = 11 gam 2+ mol O (trong oxit) mol SO4 (trong mui) thỡ lng tng 96 - 16 = 80 gam + mol O (trong oxit) mol Cl (trong mui) thỡ lng tng 35,5.2 - 16 = 55 gam .c Cỏc vớ d minh ho w w w fa ce bo ok Vớ d 1: Ho tan hon ton gam hn hp kim loi bng dung dch HCl ta thu c dung dch A v khớ B Cụ cn dung dch A thỡ c 5,71 gam mui khan Th tớch khớ B (o ktc) l A 0,224 lớt B 0,448 lớt C 0,112 lớt D 0,336 lớt Hng dn Gi cụng thc chung ca kim loi l M v cú hoỏ tr l n n MCln + H2 M + n HCl M g (M + 35,5n) g Theo pt hoỏ hc, c mol kim loi to thnh mol mui thỡ lng tng 35,5n gam v cú n mol H2 bay Theo bi, lng tng 5,71 = 0,71 gam thỡ s mol H2 bay l: n = 0,01 (mol); Vy V = 22,4 0,01 = 0,224 (l) H2 35,5n 0,71 Footer Page 10 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 45 of 16 Hng dn Cỏc phng trỡnh hoỏ hc dng ion: Phn 1: Fe3 + OH Fe(OH)3 (1) 0,01 0,01 NH3 + H2O (2) NH + OH n Fe(OH) = 1,07 = 0,01 (mol) n BaSO = 4,66 = 0,02 (mol) oc hi nT D 107 0,672 = 0,03 (mol) n NH = 22, 233 Ta iL ie Vỡ dung dch X luụn trung ho v in tớch nờn: 0,01 + 0,03 = 0,02 + n Cl uO Ta cú (3) H Phn 2: Ba + SO24 BaSO4 0,02 0,02 01 0,03 0,03 n Cl = 0,02 (mol) Vy tng lng ca cỏc cht tan dung dch Z l (56.0,01 + 18.0,03 + 96.0,02 + 35,5.0,02) = 7,46 (g) Li gii om /g ro up s/ Bi Cho dung dch A cha ng thi axit H2SO4 1,5M v HCl 2M vo 200ml dung dch B cha NaOH 1,8M v KOH 1,2M Khi mụi trng dung dch trung tớnh thỡ th tớch dung dch A cn l A 100 ml B 120 ml C 200 ml D 250 ml nH V.(2.1,5 2) 5V (mol) nOH 0, 2.(1,8 1, 2) 0, (mol) c bo ok Khi mụi trng trung tớnh : 5V = 0,6 V= 120 ml Li gii 60.500 n H 0, 2(1 2) 100.1000 0, 9(mol) nOH 0,3(0,8 x)(mol) w w w fa ce Bi Cho 200ml dung dch A cha ng thi axit HCl 1M v HNO3 2M vo 300ml dung dch B cha NaOH 0,8M v KOH x(M) thu c dung dch C Bit rng trung ho 100ml dung dch C cn 60ml dung dch HCl 1M x cú giỏ tr l A 2,2M B 3,2M C 2,3M D 2M Mụi trng trung tớnh: 0,9 = 0,3(0,8+x) x = 2,2M Footer Page 45 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 46 of 16 Bi 10 Ho tan hon ton 23,8g hn hp gm mt mui cacbonat ca kim loi hoỏ tr I v mt mui cacbonat ca kim loi hoỏ tr II vo dung dch HCl thu c 0,2 mol khớ CO2 Khi lng mui mi to dung dch l A 16g B 26g C 36g D 46g + HCl + H2 O uO nT Phng trỡnh ion thu gn : 2H+ + CO32 CO2 + H2O 0,4 0,2 mol Theo phng phỏp tng gim lng, ta cú: mmui = 23,8 + 0,2.11 = 26 (g) D RCl2 hi RCO3 + CO2 oc MCl H M2 CO3 01 Li gii t cụng thc ca cỏc mui l M2CO3 v RCO3 Ta iL ie Bi 11 Cho 100ml dung dch A cha NaCl 1,5M v HCl 3M vo 100ml dung dch B cha AgNO3 1M v Pb(NO3)2 1M thu c dung dch C v kt ta D Khi lng kt ta D l A 56,72 g B 49,13 g C 34,48 g D 50,10 g up s/ Li gii Cl n Ag 0,1.(1,5 3) 0, 45 (mol) n ( ) om /g n ro AgCl Ag+ + Cl Pb2+ + 2Cl PbCl2 0,1 mol ; n Pb2 0,1 mol n ( ) 0,1 0,1.2 0,3 (mol) Ion Cl d: mmui = 108.0,1 + 0,1.207 + 0,3.62 = 50,10 (gam) Li gii w fa ce bo ok c Bi 12 Dung dch A cha axit HCl a M v HNO3 b M trung ho 100 ml dung dch A cn dựng 200 ml dung dch hn hp NaOH 0,05M v Ba(OH) 0,15M Mt khỏc kt ta hon ton ion Cl cú 50ml dung dch A cn 100ml dung dch AgNO3 0,1M Cỏc giỏ tr a, b ln lt l A 0,2M; 0,1M B 0,2M; 0,2M C 0,2M; 0,3M D 0,1M; 0,2M n H 0,1.(a b) mol a b 0, 04 nOH 0, 2.(0, 05 0,15) mol w w Ag Cl AgCl a 0, 2M 0, 05a 0, 01 b 0,1M Footer Page 46 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 47 of 16 Bi 13 Cho 46,4g hn hp A gm FeO, Fe2O3, Fe3O4 ( n FeO n Fe2O3 ) ho tan hon 01 ton V lớt dung dch H2SO4 0,2M v HCl 0,6M (ó ly d 20% so vi lng cn thit) V cú giỏ tr l A 1,80 B 1,92 C 1,94 D 1,56 n FeO : n Fe2O3 1:1 , m FeO.Fe2O3 = Fe3O4, vy A xem nh cht Fe3O4 D 0,16 0,192 1, 92 (lớt) 0,1 uO Do ly d 20% 0,16 + 0,16.0,2 = 0,1V V = hi 0,02 H 4, 64 0, 02 (mol) 232 2+ 3+ Fe3O4 + 8H+ Fe + 2Fe + 4H2O nT n A n Fe3O4 oc Li gii up s/ Ta iL ie Bi 14 Cho kim loi Fe, Mg tỏc dng vi 200ml dung dch A gm HCl 0,1M, H2SO4 0,2M thu c dung dch B v khớ C Cho t t dung dch D gm NaOH 0,3M, KOH 0,1M vo B tỏc dng va vi cỏc cht B thỡ th tớch dung dch D l A 0,15 lớt B 0,25 lớt C 0,35 lớt D 0,45 lớt Fe2 Fe H2 H ro H OH H2O Mg2 om /g Mg Fe(OH)2 nh lut bo ton in tớch : n() B = nH A = nOH Mg(OH)2 D .c Dung dch trung tớnh : nOH 0,2(0,1 0,2.2) V(0,3 0,1) V 0,25 (lớt) ok nH w fa ce bo Bi 15 tỏc dng va vi 0,96g hiroxit ca kim loi kim hai chu kỡ liờn tip bng tun hon, phi dựng 20ml dung dch HCl 0,4M v H2SO4 0,3M Tờn kim loi ú l A Na, K B Li, Na C K, Rb D Na, Rb w w ROH n H H R H2 O 0, 02(0, 0,3.2) 0, 02(mol) M ROH 0, 96 48 (g / mol) M R 17 48 M R 31 Na, K 0, 02 Footer Page 47 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 (1) Ta iL ie uO nT hi D H Ta cú: FeO + Fe2O3 Fe3O4 0,1 0,1 0,1 Hn hp X coi nh gm: 0,2 mol Fe3O4; 0,1 mol Fe + dung dch Y: Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O 0,2 0,2 0,4 + 2+ Fe + 2H Fe + H2 (2) 0,05 0,05 0,05 2+ 3+ Dung dch Z cha Fe (0,35 mol), Fe (0,35 mol), H+ d, Cl, SO 24 Nh t t dung dch Cu(NO3)2 vo dung dch Z : Fe + Fe3+ 2Fe2+ 0,05 0,05 0,1 oc Bi 16 Hn hp cht rn X gm Fe, Fe2O3, Fe3O4 v FeO (cú s mol bng l 0,1 mol) Hũa tan ht X vo dung dch Y gm HCl v H2SO4 loóng (d), thu c dung dch Z v 1,12 lớt khớ H2 (ktc) Nh t t dung dch Cu(NO3)2 1M vo dung dch Z cho ti ngng khớ NO thoỏt thỡ dng li Th tớch dung dch Cu(NO3)2 ó dựng l A 58,3 ml B 40,5 ml C 50,2 ml D 60,5 ml 3Fe2+ + NO (3) up s/ 0,35 3Fe3+ + NO + 2H2O + 4H+ 0,35 om /g ro 0,35 (3) n NO n Fe3 (mol) n Cu(NO3 )2 n NO 0, 05 mol 3 3 VCu(NO3 )2 50 ml n OH = nNaOH = 0,3375 mol 8Al 3NO3 5OH 18H 2O 8[Al(OH) ] 3NH3 w w w fa ce bo ok c Bi 17 Cho 12,15 gam bt Al vo 112,5 ml dung dch hn hp NaNO3 1M v NaOH 3M, khuy u cho n ngng khớ thoỏt thỡ dng li Th tớch khớ thoỏt ktc l A 2,24 lớt B 7,56 lớt C 15,92 lớt D 10,08 lớt nAl = 0,45 (mol) ; n NO n NaNO3 0,11205 mol ; Ban đầu : 0, 45 0,1125 Phn ứng : 0,3 0,1125 Dư : 0,15 0,3375 0,1875 0,15 0,3 0,1125 Footer Page 48 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 (1) 01 Header Page 48 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 49 of 16 2Al 2OH 6H 2O 2[Al(OH) ] 3H 0,15 0,15 0, 225 (1) v (2) nKhớ = 0,1125 + 0,225 = 0,3375 (mol) Vkhớ = 0,3375.22,4 = oc 7,56 (lớt) 01 Ban đầu : 0,15 Phn ứng : 0,15 Dư : (2) nT hi D H Bi 18 Cho 6,4 gam Cu tỏc dng vi 120 ml dung dch hn hp gm HNO31M v H2SO4 0,5 M loóng, thu c V lớt khớ NO nht (ktc), phn ng xy hon ton Giỏ tr ca V l A 0,672 lớt B 0,896 lớt C 1,344 lớt D 2,24 lớt nCu = 0,1 mol; n H n NO n HNO3 0,12 mol ; uO n H 2n H2SO4 0,12.0,5 0, 06 (mol) n H 0,12 2.0, 06 0, 24 (mol) 2+ + 2NO + 8H+ + 2NO + 4H2O 3Cu 0,24 0,24 0,06 up s/ Ban u : 0,1 0,12 Phn ng : 0,09 0,06 D : 0,01 0,06 VNO = 0,06.22,4 = 1,344 (lớt) Ta iL ie 3Cu om /g ro Bi 19 Hũa tan hon ton 9,65 gam hn hp kim loi Fe, Al dung dch hn hp HCl v H2SO4 loóng, kt thỳc phn ng thu c 7,28 lớt H2 (ktc) Phn trm lng ca nhụm hn hp ban u l A 40,32% B 35,53% C 41,19 % D 56,48% ok c Li gii n H = 0,325 mol ; t x, y ln lt l s mol ca Al, Fe cha hn hp Ta w w w fa ce bo cú: 27x + 56y = 9,65 (1) Phng trỡnh ion rỳt gn ca cỏc phn ng 2Al + 6H+ 2Al3+ + 3H2 (2) x 1,5x + 2+ Fe + 2H Fe + H2 (3) y y T (2) v (3) VH2 = 1,5 x + y = 0,325 y = 0,325 1,5x (4) T (4) v (1) ta tớnh c : x = 0,15 mol %mAl = 0,15.27 100% 41,19% 9, 65 Footer Page 49 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 50 of 16 PHNG PHP T CHN LNG CHT Ta iL ie uO nT hi D H oc Cỏch 1: Chn mt mol nguyờn t, phõn t hoc mt mol hn hp cỏc cht phn ng Cỏch 2: Chn ỳng t l lng cht u bi ó cho Cỏch 3: Chn cho thụng s mt giỏ tr phự hp chuyn phõn s phc v s n gin tớnh toỏn Sau õy l mt s vớ d in hỡnh: Cỏch 1: CHN MOL CHT HOC HN HP CHT PHN NG Vớ d 1: Ho tan mt mui cacbonat kim loi M húa tr n bng mt lng va dung dch H2SO4 9,8% ta thu c dung dch mui sunfat 14,18% M l kim loi gỡ? A Cu B Fe C Al D Zn Hng dn gii Chn mol mui M2(CO3)n M2(CO3)n + nH2SO4 M2(SO4)n + nCO2 + nH2O C (2M + 60n) gam 98n gam (2M + 96n) gam 01 10 98n 100 1000n gam 9,8 mdd H2SO4 mdd muối mM2 (CO3 )n mdd H2SO4 mCO2 ro up s/ = 2M + 60n + 1000.n 44.n = (2M + 1016.n) gam 2M 96 100 14,18 om /g C%dd muối 2M 1016 n w w w fa ce bo ok c M = 28.n n = ; M = 56 l phự hp vy M l Fe (ỏp ỏn B) Vớ d 2: (Cõu - Mó 231 - Khi A - TSC 2007) Khi hũa tan hiroxit kim loi M(OH)2 bng mt lng va dung dch H2SO4 20% thu c dung dch mui trung ho cú nng 27,21% Kim loi M l A Cu B Zn C Fe D Mg Xột mol M(OH)2 tham gia phn ng M(OH)2 + H2SO4 MSO4 + 2H2O C (M + 34) gam 98 gam (M + 96) gam 98 100 490 gam 20 M 96 100 M 34 490 27, 21 mdd H2SO4 mdd MSO4 M = 64 M l Cu (ỏp ỏn A) Footer Page 50 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc Cỏch 2: Chn theo t l lng cht ó cho Vớ d 1: Hn hp X gm N2 v cú H2 cú t hi so vi H2 bng 3,6 Sau tin hnh phn ng tng hp c hn hp Y cú t hi so vi H2 bng Hiu sut phn ng tng hp l A 10% B 15% C 20% D 25% Hng dn Theo pp ng chộo, N2 : H2 bng : Chn s mol N2 v H2 l mol v mol o xt, t 2NH3 3H2 p H + hi D Ban u: Phn ng: x 3x 2x Sau phn ng: (1 x) (4 3x) 2x nY = (5 2x) mol p dng nh lut bo ton lng ta cú mX = mY uO n1 M2 0,25 n 0,25 H 100 25% n2 M1 2n 7,2 nT N2 Ta iL ie Vớ d 2: Phúng in qua O2 c hn hp khớ O2, O3 cú M 33 gam Hiu sut phn ng l A 7,09% B 9,09% C 11,09% D.13,09% Hng dn bo ok c om /g ro up s/ TLĐ 3O2 2O3 Theo ng chộo, t l mol O2 : O3 = 15 : Chn s mol O2 v O3 ln lt l 15 v mol Tớnh theo phn ng c s mol O2 phn ng l 1,5; ban u l 16,5 H = 9,09% Cỏch 3: CHN GI TR CHO HN HP Vớ d 1: X l hp kim gm (Fe, C, Fe3C), ú hm lng tng cng ca Fe l 96%, hm lng C n cht l 3,1%, hm lng Fe3C l a% Giỏ tr a l A 10,5 B 13,5 C 14,5 D 16 Hng dn gii Xột 100 gam hn hp X ta cú mC = 3,1 gam, mFe3C = a gam v s gam Fe tng cng l ce 96 gam w w w fa mC Fe3C 100 96 3,1 01 Header Page 51 of 16 12a 180 a = 13,5 (ỏp ỏn B) Vớ d 2: Nung m gam ỏ X cha 80% lng gam CaCO3 (phn cũn li l cht tr) mt thi gian thu c cht rn Y cha 45,65 % CaO Tớnh hiu sut phõn hy CaCO3 A 50% B 75% C 80% D 70% Hng dn gii Chn mX = 100 gam mCaCO 80 gam v lng cht bng 20 gam Footer Page 51 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 52 of 16 o t CaCO3 + CO2 CaO Phng trỡnh: 100 gam 56 gam 80 gam Phn ng: 56.80 h 100 (hiu sut = h) 44 gam 44.80 h 100 Khi lng cht rn cũn li sau nung l 01 44.80.h 100 56 80 45, 65 44 80 h h 100 100 100 100 h = 0,75 hiu sut phn ng bng 75% (ỏp ỏn B) hi D H oc mX mCO2 100 67m lng hn hp ban u Thnh phn % v lng cht rn 100 uO lng bng nT Vớ d Mt hn hp gm CaCO3, Al2O3 v Fe2O3 ú Al2O3 chim 10,2%, Fe2O3 chim 9,8% Nung m gam hn hp ny nhit cao thu c cht rn cú Ta iL ie CaO, Al2O3, Fe2O3, CaCO3 (d) th t l A 62,6%; 1,22%; 12,62%; 17,4% B 62,6%; 1,22%; 14,62%; 7,4% C 66,6%; 1,82%; 14,62%; 7,4% D 52,6%; 1,22%; 14,62%; 7,4% om /g ro up s/ Li gii Vi mi m ta u tỡm c thnh phn % hn hp, n gin ta chn m = 100g gam thỡ Al2O3 = 10,2 g; Fe2O3 = 9,8 g v CaCO3 = 80 g Khi nung hn hp ch cú : CaCO3 CaO + CO2 gim lng = 100 67 = 33 g chớnh l lng CO2 = 0,75 mol bay i nCaCO3 0,75 (mol) mCaCO3 75 (g) , CaCO3 d l gam ok c Hn hp sau nung cú : 10,2 g Al2O3 = 15,22% 9,8 g Fe2O3 = 14,62% g CaCO3 d = 7,4% v 62,6% CaO w w w fa ce bo Vớ d Cho m gam hn hp Na v Fe tỏc dng ht vi dung dch HCl thu c dung dch X Cho X tỏc dng vi dung dch KOH d, lc ly kt ta em nung khụng khớ n lng khụng i thu c cht rn nng m gam % v lng Na, Fe ban u l A 30% ; 70% B 40% ; 60% C 35% ; 65% D 50% ; 50% Li gii S phn ng : Na H2 + + HCl Fe NaCl O2 Fe2O3 + H2O Fe(OH)2 + KOH to FeCl2 Footer Page 52 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 53 of 16 Vi mi m ta u tỡm c thnh phn % hn hp, n gin chn m = 160g n Fe 2n Fe2O3 160 (mol) 160 2.56 70% 160 30% Vớ d Cho m gam hn hp hai kim loi Mg, Al tỏc dng vi mt lng d dung 4m gam 1, 02 H dch HCl thu c dung dch A v khớ B Lng khớ B ny kh va ht oc %m Na 01 %m Fe nT hi D CuO thnh Cu kim loi Thnh phn % v lng cỏc kim loi Mg, Al l A 47,06%; 52,94% B 50,00%; 50,00% C 40,00%; 60,00% D 47,25%; 52,75% up s/ Ta iL ie uO Li gii S phn ng : Mg MgCl2 + HCl + H2 + CuO Cu + H2O Al AlCl3 Vi mi m u tỡm c thnh phn hn hp, n gin chn m = 1,02g Gi n Mg a (mol);nAl b (mol) Ta cú: 24a + 27b = 1,02 (1) Tng s mol e m Mg, Al cho l : 2a + 3b ro = 0,05mol) nhn theo na phn ng sau : 80 Cu2+ + om /g S mol e ny c Cu2+ (CuO : 2e Cu 0,05 0,1 ok c 24a + 27b = 1,02 a 0, 02 %m Mg 47, 06% 2a+3b=0,1 b 0, 02 %mAl 52, 94% Thu c h w w w fa ce bo Vớ d Hn hp A gm cỏc kim loi Mg, Al, Cu Oxi hoỏ hon ton m gam A bng oxi khụng khớ thỡ thu c 1,72m gam hn hp oxit Mt khỏc ho tan m gam A bng dung dch HCl d thu c 0,952m lớt khớ H2 (ktc) % Khi lng Mg, Al, Cu A l A 30%; 50%; 20% B 25%; 55%; 20% C 30%; 54%; 16% D 25%; 50%; 25% Li gii n Mg x(mol) ; n Al y(mol) ; nCu z(mol) S phn ng : Vi mi m ta u tỡm c thnh phn hn hp, n gin chn m=100 g p dng s bo ton electron, bo ton lng ta cú : Footer Page 53 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 54 of 16 ne cho 2x 3y 2z Trong (1) : 172 100 4,5 (mol) n e nhận 16 ne cho 2x 3y Trong (2) : Cht nhn H2 : 0, 952.100 4, 25 (mol) n e nhận 8,5 22, 01 Cht nhn O: H oc 24x 2y 64z 100 x 1, 25 %m Mg 30% y %m Al 54% 2x 3y 2z 2x 3y 8,5 z 0, 25 %m 16% Cu nT hi D Vớ d Cho m gam oxit kim loi tan V lớt dung dch HCl 7,3% va thu c dung dch mui cú nng 12,5% CTPT ca oxit kim loi l A Fe2O3 B Al2O3 C CuO D MgO Ta iL ie uO Hng dn Vi mi m, V ta luụn tỡm c CT oxit Chn m, V lng dung dch HCl l 100g CuO M2On + 2nHCl 2MCln + n H2O 0, 02 2n 0, 02 n up s/ 0,02 om /g ro 0, 02 (M M 71n) 12,5 n M M 64 CuO 0, 02 100 (2M M 16n) 100 2n Vớ d 4: Hn hp X gm N2 v H2 cú MX 12, Dn X i qua bỡnh ng bt Fe ri nung núng bit rng hiu sut tng hp NH3 t 40% thỡ thu c hn hp Y M Y cú giỏ tr l o N2 + xt, t 2NH3 (vi hiu sut 40%) 3H2 p w w w fa ce bo ok c A 15,12 B 18,23 C 14,76 D 13,48 Hng dn gii Xột mol hn hp X mX = 12,4 gam gm a mol N2 v (1 a) mol H2 28a + 2(1 a) = 12,4 a = 0,4 mol n H2 0,6 mol Ban u: 0,4 0,6 Phn ng: 0,08 0,60,4 0,16 mol Sau phn ng: 0,32 0,36 Tng: nY = 0,32 + 0,36 + 0,16 = 0,84 mol; Theo nh lut bo ton lng ta cú: mX = mY MY 0,16 mol 12, 14, 76 gam (ỏp ỏn C) 0,84 Footer Page 54 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 2R 96n gam muối 2R 96n 5R R= H C R (gam) oc Vớ d 6: Ho tan hon ton mt lng kim loi R húa tr n bng dung dch H2SO4 loóng ri cụ cn dung dch sau phn ng thu c mt lng mui khan cú lng gp ln lng kim loi R ban u em ho tan Kim loi R ú l A Al B Ba C Zn D Mg Hng dn gii Xột mol kim loi ng vi R (gam) tham gia phn ng 2R + nH2SO4 R2(SO4)n + nH2 Ta iL ie uO nT hi D 12n tha vi n = Vy: R = 24 (Mg) (ỏp ỏn D) Cỏc vớ d phn hu c : Vớ d 1: Cho dung dch axit axetic cú nng x% tỏc dng va vi dung dch NaOH 10% thỡ thu c dung dch mui cú nng 10,25% Vy x cú giỏ tr no sau õy? A 20% B 16% C 15% D.13% Hng dn gii Xột mol CH3COOH: CH3COOH + NaOH CH3COONa + H2O 60 gam 40 gam 82 gam 60 100 gam x 40 100 mddNaOH 400 gam 10 60 100 82 100 gam mdd muối 400 x 10, 25 om /g ro up s/ mdd CH3COOH ce bo ok c x = 15% (ỏp ỏn C) Vớ d 2: Hn hp A gm mt Anken v hiro cú t so vi H2 bng 6,4 Cho A i qua niken nung núng c hn hp B cú t so vi H2 bng (gi thit hiu sut phn ng xy l 100%) Cụng thc phõn t ca anken l A C2H4 B C3H6 C C4H8 D C5H10 Xột mol hn hp A gm (a mol CnH2n v (1a) mol H2) Ta cú: 14.n.a + 2(1 a) = 12,8 (1) Hn hp B cú M 16 14n (vi n 2) hn hp B cú H2 d w w w fa o Ni, t CnH2n+2 CnH2n + H2 Ban u: a mol (1a) mol Phn ng: a a a mol Sau phn ng hn hp B gm (1 2a) mol H2 d v a mol CnH2n+2 tng nB = 2a p dng nh lut bo ton lng ta cú mA = mB nB mB MB 2a 12,8 16 a = 0,2 mol Footer Page 55 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 01 Header Page 55 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc Thay a = 0,2 vo (1) ta cú 140,2n + 2(1 0,2) = 12,8 n = anken l C4H8 (ỏp ỏn C) Vớ d 3: Oxi húa C2H5OH bng CuO nung núng, thu c hn hp cht lng gm CH3CHO, C2H5OH d v H2O cú M = 40 vC Hiu sut phn ng oxi húa l A 25% B 35% C 45% D 55% Xột mol C2H5OH t a mol C2H5OH b oxi húa Vy a l hiu sut ca phn ng oxi húa ru o D nT hi 46(1 a) 44a 18a 40 a H t C2H5OH + CuO CH3CHO + H2O + Cu Ban u: mol Oxi húa: a mol a mol a mol Sau phn ng: (1 a) mol C2H5OH d a mol a mol M y y up s/ Ta iL ie uO a = 0,25 hay hiu sut l 25% (ỏp ỏn A) Vớ d 4: (Cõu 48 - Mó 182 - A - TSH 2007) Hn hp gm hirocacbon X v oxi cú t l s mol tng ng l 1:10 t chỏy hon ton hn hp trờn thu c hn hp khớ Y Cho Y qua dung dch H 2SO4 c, thu c hn hp khớ Z cú t i vi hiro bng 19 Cụng thc phõn t ca X l A C3H8 B C3H6 C C4H8 D C3H4 Hng dn gii t hn hp gm hirocacbon X gm CxHy (1 mol) v O2 (10 mol ) y ro CxHy + x O2 xCO2 + H2O y mol om /g mol x mol x mol y Hn hp khớ Z gm x mol CO2 v 10 x mol O2 d c ce bo ok MZ 19 38 w w w fa Vy: (n CO2 ) 44 38 (n O2 ) 32 x 10 x y n co2 n o2 1 8x = 40 y x = 4, y = tho ỏp ỏn C Vớ d 5: A l hn hp gm mt s hirocacbon th khớ, B l khụng khớ Trn A vi B cựng nhit ỏp sut theo t l th tớch (1:15) c hn hp khớ D Cho D vo bỡnh kớn dung tớch khụng i V Nhit v ỏp sut bỡnh l toC v p atm Sau t chỏy A bỡnh ch cú N2, CO2 v hi nc vi VCO2 : VH2O : a bỡnh v toC p sut bỡnh sau t l p1 cú giỏ tr l Footer Page 56 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 01 Header Page 56 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 57 of 16 A p1 47 p 48 y CxHy + x O2 t A: 16 p 17 y xCO2 + H 2O B p1 = p C p1 D p1 p Vỡ phn ng ch cú N2, H2O, CO2 cỏc hirocacbon b chỏy ht v O2 va 01 y x x : y : x= H n N2 4n O2 12 mol ; y= 3 uO nT p1 12 47 47 p1 p (ỏp ỏn A) 48 p 15 48 hi Vỡ nhit v th tớch khụng i nờn ỏp sut t l vi s mol khớ, ta cú: D y 15 mol oc Chn n Cx H y nB = 15 mol n O2 p.ứ x Ta iL ie Vớ d 6: t chỏy hon ton a gam hn hp X hai hirocacbon A, B thu c 132.a 41 45a gam H 2O Nu thờm vo hn hp X mt na lng A cú hn 41 165a 60, 75a hp X ri t chỏy hon ton thỡ thu c gam CO2 v gam H 2O Bit A, 41 41 gam CO2 v bo ok c om /g ro up s/ B khụng lm mt mu nc Br2 a) Cụng thc phõn t ca A l A C2H2 B C2H6 C C6H12 D C6H14 b) Cụng thc phõn t ca B l A C2H2 B C6H6 C C4H4 D C8H8 c) Phn trm s mol ca A, B hn hp X l A 60%; 40% B 25%; 75% C 50%; 50% D 30%; 70% Hng dn gii a) Chn a = 41 gam n CO2 t A thu c (3,75 3) = 0,75 mol CO2 v (3,375 2,5) = 0,875 mol H2O t chỏy A thu c n CO2 1,5 mol v n H2O 1,75 mol w w fa t X A w 132 45 mol v n H2O 2,5 mol 44 18 165 60, 75 3, 75 mol v n H2O 3,375 mol 44 18 n CO2 ce t X vỡ n H2O n CO2 A thuc loi ankan, ú: Cn H2n 3n O2 nCO2 n H2O Footer Page 57 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 58 of 16 n CO2 n H 2O n 1,5 n 1, 75 n = A l C6H14 (ỏp ỏn D) b) t B thu c (3 1,5) = 1,5 mol CO2 v (2,5 1,75) = 0,75 mol H2O nC 1,5 n H 0, 75 cụng thc tng quỏt ca B l (CH)n vỡ X khụng lm hi D H oc mt mu nc Brom nờn B thuc aren B l C6H6 (ỏp ỏn B) c) Vỡ A, B cú cựng s nguyờn t C (6C) m lng CO2 A, B to bng (1,5 mol) nA = nB %nA = %nB = 50% (ỏp ỏn C) Vớ d 7: Trn a gam hn hp X gm hirocacbon C6H14 v C6H6 theo t l s mol (1:1) vi m gam mt hirocacbon D ri t chỏy hon ton thỡ thu c 01 Nh vy nT 275a 94,5a gam H2O gam CO2 v 82 82 up s/ Ta iL ie uO a) D thuc loi hirocacbon no A CnH2n+2 B CmH2m2 C CnH2n D CnHn b) Giỏ tr m l A 2,75 gam B 3,75 gam C gam D 3,5 gam Hng dn gii a) Chn a = 82 gam t X v m gam D (CxHy) ta cú: bo t D: ok c om /g ro 275 n CO2 44 6, 25 mol 94,5 n 5, 25 mol H 2O 18 19 C6H14 + O2 6CO2 + 7H2O 15 C6H6 + O2 6CO2 + 3H2O y y Cx H y x O2 xCO2 H2O t n C6H14 n C6H6 b mol ta cú: w fa ce 86b + 78b = 82 b = 0,5 mol t 82 gam hn hp X thu c: n H2O 0,5 mol w w n CO2 0,5 mol t chỏy m gam D thu c: n CO2 6, 25 0, 25 mol n H2O 5, 25 0, 25 mol Footer Page 58 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Header Page 59 of 16 Do n CO2 n H2O D thuc CnH2n (ỏp ỏn C) mD = mC + mH = 0,25(12 + 2) = 3,5 gam (ỏp ỏn D) Vớ d t chỏy hon ton m gam cht hu c A cha C, H, O thu c iu kin) CTPT ca A l A C2H6O2 B C3H6O3 C C3H4O3 D C2H4O2 hi D Li gii Vi mi m ta u xỏc nh c CTPT ca A, chn m = 15 gam n C(A) n CO2 0,5 mol nT 15 0,5.12 (mol) ; n O 0,5 (mol) 18 16 3, n C : n H : n O : :1 (CH 2O) n 90 n 0, 04 CTPT A l C 3H O3 w w w fa ce bo ok c om /g ro up s/ Ta iL ie uO n H(A) 2n H2O Footer Page 59 of 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc 3m g H2O Bit rng 3,6g hi A cú th tớch bng th tớch ca 1,76g CO2 ( cựng H v 22m g CO2 15 01 b) ... + mX = 0,26 g 2a .108 – MX.a = 0,26  MX.a = 2a .108 – 0,26 (2) up s/  Chất X Cd 64.2,5 .10 3  0,12 2,5 .10 3  112 (g/mol) ro  x = 2,5 .10 3 mol  MX = (1) bo ok c om /g Ví dụ 10 Cho dung dịch... tức chiếm 50% Do hỗn hợp đầu có 10 lít SO2 10 lít O2 Gọi V số lít O2 cần thêm vào Như hỗn hợp sau có 10 lít SO2 (10 + V) lít O2 bo Ta có M sau = 64  10  32  (10  V) = 2,5  16 = 40 20  V... – 20 H O: 300 600 x    x  40% 300 x  20 Lời giải m 100 20 50 30 50  m 20   m  40 g 100 50 up s/ 100 Ta iL ie uO nT Ví dụ 10 Từ 100 g dung dịch KOH 30% để có dung dịch 50% cần thêm vào