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Tuyển tập 25 đề thi học sinh giỏi Hóa học lớp 9 (kèm đáp án) (by Dameva)

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TUYỂN TẬP 50 ĐỀ THI HỌC SINH GIỎI MÔN HÓA HỌC LỚP 9 (<strong>kèm</strong> <strong>đáp</strong> <strong>án</strong>)<br />

C©u Néi dung §iÓm<br />

1)<br />

0,5<br />

2FeCl 3(dd) +3Na 2 CO 3(dd) +3H 2 O (l) →2Fe(OH) 3(r) +3CO 2(k) +6NaCl (dd)<br />

(n©u ®á)<br />

0,5<br />

1<br />

(4,0 ® )<br />

t<br />

2Fe(OH) 3(r) ⎯⎯→<br />

o<br />

Fe 2 O 3(r) + 3H 2 O (h)<br />

t<br />

2) CuO + H 2 ⎯⎯→<br />

o<br />

Cu + H 2 O<br />

t<br />

Fe 2 O 3 + 3H 2 ⎯⎯→<br />

o<br />

2Fe + 3H 2 O<br />

Na 2 O + H 2 O ⎯⎯→<br />

t o<br />

2NaOH<br />

Sn phÈm trong mçi ènglµ CaO, Cu, Al 2 O 3 , Fe, NaOH<br />

- Cho t¸c dông víi CO 2<br />

CaO + CO 2 → CaCO 3<br />

2NaOH + CO 2 → Na 2 CO 3 + H 2 O<br />

- Cho t¸c dông víi dung dÞch HCl<br />

CaO + 2HCl → CaCl 2 + H 2 O<br />

Al 2 O 3 + 6HCl → 2AlCl 3 + 3 H 2 O<br />

Fe + 2HCl → FeCl 2 + H 2<br />

- Cho t¸c dông víi dung dÞch AgNO 3<br />

Fe + 2AgNO 3 → Fe(NO 3 ) 2 + 2Ag↓<br />

Cu + 2AgNO 3 → Cu(NO 3 ) 2 + 2Ag↓<br />

NÕu AgNO 3 d− th×:<br />

Fe(NO 3 ) 2 + AgNO 3 → Fe(NO 3 ) 3 + Ag ↓<br />

CaO + H 2 O → Ca(OH) 2<br />

Ca(OH) 2 + 2AgNO 3 → 2AgOH↓ + Ca(NO 3 ) 2<br />

NaOH + AgNO 3 → AgOH↓ + NaNO 3<br />

2AgOH ↓<br />

⎯⎯→<br />

o<br />

t<br />

Ag 2 O (r) + H 2 O<br />

(®en)<br />

0,5<br />

0,5<br />

1,0<br />

0,5<br />

0,5<br />

2<br />

(3,5 ® )<br />

+ Hoµ tan hçn hîp vµo dung dÞch NaOH (d−), sau khi phn øng<br />

xy ra hoµn toµn, läc bá phÇn chÊt r¾n, sôc khÝ CO 2 d− vµo dung<br />

dÞch.<br />

2Al + 2NaOH + 2H 2 O → 2NaAlO 2(dd) + 3H 2(K)<br />

NaAlO 2 + CO 2 + 2H 2 O → Al(OH) 3(r) + NaHCO 3<br />

Nung kÕt tña ®Õn khi khèi l−îng kh«ng ®æi<br />

t<br />

2Al(OH) 3(r) ⎯⎯→<br />

o<br />

Al 2 O 3 + 3H 2 O<br />

§iÖn ph©n nãng chy chÊt r¾n thu ®−îc víi xóc t¸c lµ Criolit, ta<br />

thu ®−îc Al<br />

§iÖn ph©n nãng chy<br />

2Al 2 O 3 4 Al + 3O<br />

Criolit<br />

2<br />

+ Hoµ tan chÊt r¾n cßn l¹i vµo dung dÞch HCl d−<br />

Fe + 2HCl → FeCl 2 + H 2<br />

Läc chÊt r¾n cßn l¹i, cho dung dÞch t¸c dông víi dung dÞch<br />

NaOH d−.<br />

FeCl 2 + NaOH → Fe(OH) 2(r) + 2NaCl<br />

Nung chÊt r¾n vµ cho dßng khÝ H 2 ®i qua ®Õn khi khèi l−îng<br />

kh«ng ®æi ta thu ®−îc s¾t.<br />

t<br />

4Fe(OH) 2 + O 2 ⎯⎯→<br />

o<br />

2Fe 2 O 3 + 4H 2 O<br />

1,0<br />

1,0<br />

7

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