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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 />

b. Tõ B viÕt s¬ ®å phn øng ®iÒu chÕ CH 3 COOCH 3 vµ CH 3 COO –CH -- CH 3<br />

CH 3<br />

(Cho: O=16, H=1, C=12, Ca=40, Ba=137, Na=23, S=32, Cl=35,5 )<br />

--------------------------------------------- HÕt -----------------------------------------------<br />

L−u ý: Häc <strong>sinh</strong> ®−îc sö dông m¸y tÝnh th«ng th−êng, kh«ng ®−îc sö dông bÊt k× tµi<br />

liÖu g× (kÓ c bng tuÇn hoµn c¸c nguyªn tè hãa häc).<br />

H−íng dÉn chÊm bµi Thi<br />

§¸p ¸n<br />

M«n : Ho¸ häc<br />

C©u 1: 6,5®<br />

1. 1,5<br />

C¸c ph−¬ng tr×nh hãa häc:<br />

2Al + 2NaOH + 2H 2 O → 2NaAlO 2 + 3H 2 ↑<br />

.....................................................................<br />

NaOH + NH 4 Cl → NaCl + NH 3 ↑+ H 2 O<br />

NaAlO 2 + NH 4 Cl + H 2 O → Al(OH) 3 ↓ +NH 3 + NaCl -----------------------------<br />

--------<br />

=> Dung dÞch X 1 chøa NaOH d− vµ NaAlO 2<br />

- KhÝ A 2 lµ H 2 .<br />

- KÕt tña A 3 lµ Al(OH) 3<br />

- KhÝ A 4 lµ NH 3 .<br />

................................................................<br />

2. 1,5<br />

C¸c ph−¬ng tr×nh hãa häc:<br />

t<br />

MgCO 3 ⎯⎯→<br />

0<br />

MgO + CO 2<br />

CO 2 + NaOH → NaHCO 3<br />

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

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

.........................................................................<br />

Na 2 CO 3 + HCl → NaHCO 3 + NaCl<br />

NaHCO 3 + Ca(OH) 2 → CaCO 3 + NaOH + H 2 O<br />

Na 2 CO 3 + CaCl 2 → CaCO 3 + 2NaCl<br />

........................................................................<br />

=> B lµ CO 2 , A lµ muèi cacbonnat dÔ bÞ nhiÖt ph©n nh− MgCO 3 , BaCO 3 ..., C<br />

lµ NaHCO 3 , D lµ Na 2 CO 3 , E lµ Ca(OH) 2 , F lµ muèi tan cña canxi nh− CaCl 2 ,<br />

Ca(NO 3 ) 2 ..., H lµ CaCO 3 . ..............................................................................<br />

3. 2,0<br />

a. 0,5<br />

Cho hçn hîp qua dd NaOH d−, cßn l¹i O 2 :<br />

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

SO 3 + 2NaOH → Na 2 SO 4 + H 2 O<br />

dung dÞch thu ®−îc t¸c dông víi H 2 SO 4 lo·ng:<br />

Thang<br />

®iÓm<br />

0,5<br />

0,5<br />

0,5<br />

0,5<br />

0,5<br />

0,5<br />

0,<strong>25</strong><br />

23

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