Trng THCS ng Tin

Trng THCS ng Tin Yn Phong Bc Ninh KIM TRA HC KY II

MN HO HC LP 8

s 7 (Thi gian lm bi: 45 pht)

I - Trc nghim khch quan (3 im) Khoanh trn vo ch A hoc B, C, D trc phng n chn ng.Cu 1. tan ca 1 cht trong nc nhit xc nh lA. s gam cht tan trong 100 g nc.

B. s gam cht tan trong 100 g dung dch.C. s ml cht tan trong 100 ml dung dch.

D. s gam cht tan trong 100 g nc to thnh dung dch bo ho.Cu 2. C th dng thuc th no sau y nhn bit 3 cht rn sau : CaO, P2O5, Al2O3A. Qu tm ; B. Nc ;C. Qu tm v nc; D. Dung dch HCl.Cu 3. Dy cht no sau y ch gm CTHH ca baz ?A. KOH, CuCl2, H2S ; B. NaOH, Cu(OH)2, Mg(OH)2 ;C. Na2S, H2SO4, MgCl2 ; D. NaOH, HCl, Cu(OH)2.Cu 4. Ho tan 20g ng vo 180 g nc. Nng phn trm ca dung dch ng l A. 10%. B. 15%. C. 12%. D. 20%.Cu 5. Cho cc cht sau : (1) km, (2) ng, (3) st, (4) HCl, (5) H2SO4 long, (6) NaOH.Nhng cht no c th dng iu ch H2 trong phng th nghim ?A. (1), (2), (4), (5) ; B. (2), (3), (5), (6) ;C. (1), (3), (4), (5) ; D. (1), (2), (4), (6).Cu 6. Cho cc phng trnh ho hc ca cc phn ng sau :

Zn + CuCl2 ZnCl2 + Cu (1) Fe + H2SO4 FeSO4 + H2 (2)HCl + NaOH NaCl + H2O (3) Fe2O3 + 3CO 2Fe + 3CO2 (4)Phn ng no l phn ng th ? A. (1), (3) ; B. (1), (2) ; C. (2), (3) ; D. (2), (4).II - T lun (7 im)

Cu 7. (2 im) Vit cc phng trnh phn ng theo dy chuyn ho sau : (1) (2) (1) (2) a) Ca CaO Ca(OH)2 b) S SO2 H2SO3Cu 8. (1,5 im) Vit cng thc ho hc cc mui c tn gi sau : a) St(III) sunfat. b) Km clorua. c) Natri cacbonat.Cu 9. (3,5 im) Dng 500 ml dung dch H2SO4 1,2M ho tan ht lng kim loi st.a) Vit phng trnh ho hc ca phn ng.b) Tnh khi lng mui st(II) sunfat thu c.c) Tnh th tch kh H2 thot ra ( ktc) ?(Bit H = 1, S = 32, O = 16, Fe = 56)p n v biu imI - Trc nghim khch quan (3 im)- Mi cu ng : 0,5 imcu 1 : D cu 2 : C

cu 3 : B cu 4 : A

cu 5 : C cu 6 : BII. T lun (7 im)Cu 7. (2 im) - Vit v cn bng ng, mi phng trnh cho 0,5 im. - Thiu cn bng hoc sai tr 0,25 imb) (1) S + O2 SO2a) (1) 2Ca + O2 2CaO(2) CaO + H2O Ca(OH)2(2) SO2 + H2O H2SO3Cu 8. ( 1,5 im) Mi cng thc ng 0,5 ima) Fe2(SO4)3b) ZnCl2c) Na2CO3Cu 9. (3,5 im)n H 2SO4 = CM. V = 1,2.0,5 = 0,6 mol- Vit ng phng trnh ho hc : 1 im Fe + H2SO4 FeSO4 + H2 1 mol 1 mol 1 mol 0,6 (mol) x(mol) y(mol)Do x = y = 0,6 (mol)b) Tnh m FeSO4 : (1 im)m FeSO4 = 0,6.152 = 91,2 gc) Tnh v H 2 : 1 imVH 2 = 0,6.22,4 = 13,44 (1it)* Nu hc sinh tnh gp : -Theo PTHH ta c n FeSO4 = n H 2 = n H 2SO4 = 0,6 mol th vn cho 1 im.* Nu tnh ton sai kt qu th tr 0,25 im