S GIO DC V O TO GIA LAI K THI CHN HC SINH GII LP Mụn thi: Húa hc Thi gian: 150 phỳt ( khụng k thi gian phỏt ) CHNH THC BI : ( thi cú 06 cõu) Cõu 1: ( im) Cú th tn ti ng thi hn hp cỏc cht sau õy c khụng ? Vỡ ? a) Na2CO3 (r) , Ca(OH)2 (r) , NaCl (r) , Ca(HSO4)2 (r) ; b) SO2 (k) ; H2S (k) ; Cl2 (k) c) NaHSO4(dd) , KOH (dd) ; Na2SO4 (dd) ; d) (NH4)2CO3 (dd) ; NaHSO4 (dd) Cõu 2: ( im) Trong CN sn xut NaOH ngi ta in phõn dung dch NaCl bóo ho, cú mng ngn xp a) Vit phng trỡnh phn ng xy b) Sn phm thu c cú ln NaCl, lm th no cú c NaOH tinh khit ( Bit S NaOH > SNaCl ) c) Cho bit SNaCl 250C l 36 gam Hóy tớnh lng ca dung dch bóo ho NaCl cn dựng sn xut c tn dung dch NaOH 40%, bit hiu sut phn ng in phõn l 90% Cõu 3: ( im) Mt loi phõn bún phc hp NPK cú ghi trờn nhón l: 15 11 12 a) Thụng tin trờn cú ý ngha gỡ ? b) Tớnh lng mi mui : KCl, NH 4NO3, Ca(H2PO4)2 cn dựng pha trn thnh 50kg phõn bún núi trờn Cõu 4: ( im) t chỏy hon ton 15,6 gam hp cht hu c (A) thu c 26,88 lớt CO ( ktc)v 10,8 gam H2O 1) Xỏc nh CTG ca (A) 2) Xỏc nh CTCT v vit PTHH xy ca (A) trng hp sau õy: a Bit : 1mol (A) + mol H2 ( xỳc tỏc Ni, t0C); mol (A) + 1mol Br2/CCl4 v d(A)/He =26 b Bit : 0,1 mol (A) phn ng vi AgNO d/ NH3 thỡ thu c 15,9 gam kt ta (A) cú dng mch h Cõu 5: ( im) Hn hp X gm Al v mt kim loi M húa tr II tan hon ton dung dch H 2SO4 c núng thỡ cho dung dch Y v khớ SO2, hp th khớ vo dung dch NaOH d, to 50,4 gam mui Khi thờm vo X mt lng kim loi M gp ụi lng kim loi M cú sn hn hp X ( gi nguyờn lng Al) thỡ lng mui thu c tng 32 gam Cũn nu gim ẵ lng Al trong X (gi nguyờn lng M), thỡ thu c 5,6 dm3 khớ B ( o ktc) a) Xỏc nh tờn kim loi M b) Tớnh thnh phn phn trm v lng ca cỏc kim loi hn hp X Cõu 6: ( im) t chỏy hon ton x gam hn hp ru CnH2n + OH v CmH2m + OH thu c a gam CO2 v b gam H2O a) Lp biu thc tớnh x theo a v b b) Chng minh rng nu m n = k thỡ : 9(1 + k)a 22kb 9a SNaCl nờn lm gim nhit ca dung dch hn hp, thỡ NaCl s kt tinh v tỏch dung dch phng phỏp kt tinh phõn on ( Hoc cụ cn t t dung dch thỡ NaCl s kt tinh trc v tỏch dung dch ) c) tn dung dch NaOH 40% cú m NaOH = 0,4 tn 0, 4.106 s mol NaOH = = 0, 01.106 ( mol) 40 Theo phng trỡnh húa hc : s mol NaCl = 0,01.106 ( mol) Khi lng dung dch NaCl bóo hũa ( 250C) cn dựng l: 0, 01.106.58,5.136.100 = 2, 455 (taỏn) 106.36.90 a) Cho bit t l lng ca N, P2O5, K2O ln lt l 15:11:12 b) Chon N = 15 g ta cú 11 gam P2O5 v 12 gam K2O - Lp t l s mol : 15 11 12 nN : nP : nK = : ì2 : ì2 = 1, 07 : 0,155 : 0, 255 6,9 :1:1, 65 14 142 94 Gi x l s mol ca nguyờn t P, ta cú: x Ca(H2PO4)2 = KCl = 1,65x x NH4NO3 = 6,9 ì x x Vy 80 ì6,9 ì + 234 ì + 1,65x ì74,5=50.10 2 Suy : 0,097.103 mol x Do ú : NH4NO3 = 6,9 ì = 26,772 kg x Ca(HCO3)2 = = 11,35 kg KCl = 1,65x=11,92kg Nhn xột : Phõn bún N,P,K khụng phi l hn hp ch cú mui trờn Nu ly tng mui trờn bng 100% thỡ khụng chớnh xỏc C th : Nu th li t l % ca N phõn ny khụng bng 15% (Sai vi kin thc SGK Húa tr.39) im 4(5,0) 26,88 ì12 = 14, gam 22, 10,8 mH = ì2 = 1, gam 18 Vỡ mC + mH = mA , suy cht A khụng cha Oxi Do ú cụng thc n gin ca A l CxHy vi x : y = 1,2 : 1,2 = 1:1 A = (CH)n 2a dA/He = 26 MA = 104 (CH)n = 104, suy n = A = C8H8 , cú bt bóo hũa a = ( Vỡ 2ì8 + a = ) - 1mol A + molBr2 , suy A cú liờn kt ụi C = C - 1molA + mol H2, suy cú liờn kt pi, ú cú liờn kt pi khụng phn ng vi Br2 Kt hp vi a = nờn hp cht A phi cú vũng benzen Cụng thc cu to ca A l : m C = CH=CH2 ( Hc sinh phi vit cỏc phng trỡnh phn ng xy ) 2b 2CnHn + xAg2O 2CnHn xAgx + xH2O 0,1mol 0,1 mol 15,9 = 159 suy : 13n + 107x = 159 MKT = 0,1 Chn x = v n = Cụng thc phõn t C4H4, cụng thc cu to ca A: HCC-CH=CH2 ( Hc sinh phi vit phng trỡnh phn ng) 5(3,0) a) t x,y ln lt l s mol ca Al v M hn hp X 2Al + 6H2SO4 Al2(SO4)3 + 6H2O + 3SO2 x 3x 0,5x 1,5x M + 2H2SO4 MSO4 + 2H2O + SO2 y 2y y y SO2 + 2NaOH Na2SO3 + H2O (1,5x + y) (1,5x + y) mol Theo ta cú : 1,5x + y = 50,4 : 126 = 0,4 (1) Khi thờm 2y (mol) M thỡ lng mui tng thờm 2y (mol) MSO4 32g l lng ca 2y (mol) l MSO4 16 Ta cú phng trỡnh : 2y(M+96) = 32 M + 96 = (2) y Nu lng M khụng i, gim ẵ lng Al thỡ s mol SO2 l: 5, n SO = =, 025 0,75x + y = 0,25 ( 3) 22, 1,5x + y = 0, T (1) v(3) ta cú h phng trỡnh : gii : x =0,2 ; y = 0,1 0, 75x + y = 0, 25 16 = 160 M =64 ( Cu) Th y = 0,1 vo (2) ta cú : M+96 = 0,1 b) Phn trm lng ca mi nguyờn t hn hp X 0, 2.27.100 = 45, 76% ; %Cu = 54,24% %Al = 0, 2.27 + 0,1.64 6(3,0) a) t cụng thc trung bỡnh ca ru l : C n H 2n +1OH 3n C n H 2n +1OH + O2 n CO2 + ( n +1) H2O X a b (mol) 14n + 18 44 18 a 3n 3a = S mol O2 = 44 2n 88 p dng nh lut BTKL ta cú: 3a 96a 11b a 32 = a + b = a x = a + b 88 88 11 b Ta cú : n < n < m a(n + 1) bn = Theo phng trỡnh phn ng ta cú: 44 18 9a Bin i c : n = 22b 9a M : m n =k m = n + k 9a Nờn ta cú : n < ... : m n =k m = n + k 9a Nờn ta cú : n <