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 />
2<br />
(3 ® )<br />
B 1 : NaAlO 2 ; NaOH d−<br />
B 2 : Na 2 SO 4 ; Al 2 (SO 4 ) 3 ; H 2 SO 4 d−<br />
B 3 : Fe 2 (SO 4 ) 3<br />
B 4 :FeSO 4 ; Al 2 (SO 4 ) 3<br />
C 1 : H 2 ; C 2 : SO 2 ; C 3 : NO<br />
B 5 : Fe 2 (SO 4 ) 3 ; Fe(NO 3 ) 3 ; Al 2 (SO 4 ) 3<br />
C 4 : NO 2<br />
b) C¸c PTP¦:<br />
1) 2Al + 2NaOH + 2H 2 O → 2NaAlO 2 + 3H 2 ↑<br />
2) Al 2 O 3 + 2NaOH → 2NaAlO 2 + H 2 O<br />
t<br />
3) Fe 3 O 4 + 4H 2 ⎯⎯→<br />
o<br />
3Fe + 4H 2 O<br />
t<br />
4) 8Al + 3Fe 3 O 4 ⎯⎯→<br />
o<br />
4Al 2 O 3 + 9Fe<br />
5) 2NaOH + H 2 SO 4 → Na 2 SO 4 +2H 2 O<br />
6) 2NaAlO 2 + 4H 2 SO 4 → Na 2 SO 4 + Al 2 (SO 4 ) 3 + 4H 2 O<br />
t<br />
7) 2Fe + 6H 2 SO 4 (®Æc) ⎯⎯→<br />
o<br />
Fe 2 (SO 4 ) 3 + 3SO 2 ↑ + 6H 2 O<br />
t<br />
8) 2Al + 6H 2 SO 4 (®Æc) ⎯⎯→<br />
o<br />
Al 2 (SO 4 ) 3 + 3SO 2 ↑ + 6H 2 O<br />
9) Al 2 O 3 + 3H 2 SO 4 → Al 2 (SO 4 ) 3 + 3H 2 O<br />
10) Fe + Fe 2 (SO 4 ) 3 → 3FeSO 4<br />
11) 3FeSO 4 + 4HNO 3 (lo·ng) → Fe 2 (SO 4 ) 3 + Fe(NO 3 ) 3 + NO↑<br />
12) 2NO + O 2 → 2NO 2<br />
a) Cho lÇn l−ît tõng dung dÞch t¸c dông víi nhau, ta thu ®−îc kÕt<br />
qu:<br />
KOH HCl FeCl 3 Cu(NO 3 ) 2 Al(NO 3 ) 3 NH 4 Cl<br />
KOH ↓ ↓ ↓ ↑<br />
HCl<br />
FeCl 3 ↓<br />
Cu(NO 3 ) 2 ↓<br />
Al(NO 3 ) 3 ↓<br />
NH 4 Cl ↑<br />
- Dung dÞch nµo khi cho vµo c¸c dung dÞch cßn l¹i: cho 1 kÕt<br />
tña n©u ®á, 1 kÕt tña xanh l¬, 1 kÕt tña tr¾ng (nÕu cho d− ↓<br />
tan), 1 khÝ mïi khai bay lªn ⇒ dung dÞch KOH.<br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,375<br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
1,0<br />
0,5<br />
1) 3KOH + FeCl 3 → Fe(OH) 3 ↓ + 3KCl<br />
(n©u ®á)<br />
2) 2KOH + Cu(NO 3 ) 2 → Cu(OH) 2 + 2KNO 3<br />
(xanh l¬)<br />
3) 3KOH + Al(NO 3 ) 3 → Al(OH) 3 + 3KNO 3<br />
(tr¾ng)<br />
4) KOH (d−) + Al(OH) 3 → KalO 2 (dd) + 2H 2 O<br />
5) KOH + NH 4 Cl → NH 3 + KCl + H 2 O<br />
(mïi khai)<br />
0,<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
0,1<strong>25</strong><br />
107
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