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 />
- Nung D ®Õn khèi l−îng kh«ng ®æi:<br />
2Fe(OH) 3 (r) Fe 2 O 3 (r) + 3H 2 O (h) (6)<br />
t<br />
4Fe(OH) 2 (r) + O 2 (k)<br />
o<br />
2Fe 2 O 3 (r) + 4H 2 O (h)<br />
→E lµ Fe 2 O 3<br />
- Cho dßng khÝ CO qua E:<br />
3CO (k) + Fe 2 O 3 (r)<br />
t o<br />
2Fe (r) + 3CO 2 (k)<br />
→ G lµ Fe. KhÝ X gåm: CO 2 , CO d−.<br />
- Sôc khÝ X vµo dd Ba(OH) 2 th× thu ®−îc kÕt tña Y vµ dung dÞch Z. Läc bá<br />
Y, ®un nãng dung dÞch Z l¹i t¹o kÕt tña Y → phn øng gi÷a X vµ dd<br />
Ba(OH) 2 t¹o hai muèi:<br />
CO 2 (k) + Ba(OH) 2 (dd) BaCO 3 (r) + H 2 O(l)<br />
2CO 2 (k) + Ba(OH) 2 (dd) Ba(HCO 3 ) 2 (dd)<br />
t o<br />
Ba(HCO 3 ) 2 (dd) BaCO 3 (r) + CO 2 (k) + H 2 O(l)<br />
- Cho CaCO 3 d− vµo hçn hîp ban ®Çu, råi ch−ng cÊt ®Ó thu lÊy r−îu:<br />
2CH 3 COOH (dd) + CaCO 3 (r) (CH 3 COO) 2 Ca(dd) + CO 2 (k) + H 2 O (l)<br />
- Thu r−îu råi lµm khan ®−îc r−îu etylic tinh khiÕt.<br />
Cho dung dÞch H 2 SO 4 vµo dung dÞch cßn l¹i sau phn øng trªn råi<br />
ch−ng cÊt ®Ó thu CH 3 COOH<br />
(CH 3 COO) 2 Ca(dd) + H 2 SO 4 2CH 3 COOH (dd) + CaSO 4 (r)<br />
0,<strong>25</strong><br />
0,<strong>25</strong><br />
0,5<br />
0,5<br />
C©u ý §¸p ¸n §iÓm<br />
4 2,0<br />
-PTHH ACO 3 (r) + H 2 SO 4 (dd) ASO 4 (dd) + CO 2 (k) + H 2 O (l) (1)<br />
BCO 3 (r) + H 2 SO 4 (dd) BSO 4 (dd) + CO 2 (k) + H 2 O (l) (2) 0,<strong>25</strong><br />
1<br />
muèi thu ®−îc trong dd X lµ ASO 4 , BSO 4 ; nCO<br />
= 0, 05(mol)<br />
2<br />
* TÝnh tæng khèi l−îng c¸c muèi t¹o thµnh trong dung dÞch X:<br />
- Theo (1), (2): nH2SO = n<br />
4 H2O = nCO2<br />
= 0, 05(mol)<br />
0,<strong>25</strong><br />
- Theo §LBTKL: m muèi = 4,68 + 0,05.98 - 0,05.44 - 0,05.18 = 6,48 (g)<br />
2<br />
* T×m c¸c kim lo¹i A, B vµ tÝnh % khèi l−îng cña mçi muèi ban ®Çu:<br />
- §Æt: nACO<br />
= 2x (mol) → n<br />
3 BCO<br />
= 3x (mol) (v× n<br />
3<br />
ACO<br />
: n<br />
3 BCO<br />
= 2:3)<br />
3<br />
M A = 3a (g) → M B = 5a (g) (v× M A : M B = 3 : 5)<br />
- Theo (1), (2): nCO = n<br />
2<br />
ACO<br />
+ n<br />
3<br />
BCO<br />
= 5x = 0,05(mol) → x = 0,01 (mol)<br />
3<br />
n = 0,02 (mol) → n = 0,03 (mol)<br />
→<br />
ACO 3 BCO 3<br />
0,<strong>25</strong><br />
→ 0,02(3a + 60) + 0,03(5a + 60) = 4,68 (g) → a = 8<br />
→ M A = 24 g, M B = 40 g → A lµ Mg. B lµ Ca. 0,<strong>25</strong><br />
0,02.84<br />
%m = .100% = 35,9% ; %m<br />
MgCO<br />
= 64,1%<br />
0,<strong>25</strong><br />
3<br />
46,8<br />
→ MgCO3<br />
3<br />
* TÝnh nång ®é mol cña dung dÞch Ba(OH) 2<br />
- Theo bµi ra: hÊp thô hÕt l−îng khÝ CO 2 ë trªn vµo dd Ba(OH) 2 ®−îc<br />
kÕt tña → kÕt tña lµ BaCO 3 → nBaCO<br />
= 1,97 :197 = 0,01 (mol)<br />
3<br />
- Gi sö phn øng chØ t¹o muèi trung hoµ:<br />
CO 2 (k) + Ba(OH) 2 (dd) BaCO 3 (r) + H 2 O(l) (4)<br />
- Theo (4): nCO<br />
= n<br />
2 BaCO nh−ng thùc tÕ n<br />
3<br />
CO<br />
> n<br />
2 BaCO → ®iÒu g/s sai.<br />
3<br />
0.<strong>25</strong><br />
0,<strong>25</strong><br />
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