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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 />

t<br />

4) 2M(OH) X ⎯⎯→<br />

o<br />

M 2 O X + xH 2 O<br />

36,5<br />

⇒ C% dd HCl = ⋅ 100%<br />

≈ 16%<br />

228<br />

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

Tõ (3) vµ (4) ta cã: 2MCl x → M 2 O x<br />

38 −16<br />

0,4<br />

⇒ nM<br />

2 O<br />

= =<br />

x<br />

71x<br />

−16x<br />

x<br />

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

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

0,4<br />

(2M + 16 x). = 16 ⇔ M = 12x<br />

x<br />

x 1 12 3 ...<br />

M 12 24 36<br />

VËy M = 24 tho m·n ⇒ M lµ Magie (Mg)<br />

0,5<br />

0,4<br />

⇒ nM 0,2 ( mol)<br />

2O<br />

= =<br />

â<br />

2<br />

Ta cã: m dd HCl = m dd E + mH<br />

+ mCO<br />

− mdd<br />

NaHCO<br />

− m<br />

2<br />

2<br />

3 M<br />

Theo (1), (3) vµ (4) ⇒ nH = 2n<br />

2 M 2Oâ<br />

= 0,4( mol)<br />

0,1<strong>25</strong><br />

⇒ n M = 0,4 (mol)<br />

⇒ m dd HCl = 468 + 0,4 . 2 + 0,2 . 44 – 240 – 0,4 . 24 = 228(g)<br />

0,1<strong>25</strong><br />

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

Theo (1) vµ (2) ⇒ ∑n HCl = 2n M + n<br />

NaHCO3<br />

= 0,8 + 0,2 = 1 (mol)<br />

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

⇒ m HCl = 36,5(g)<br />

5<br />

(3,75 ® )<br />

Ni, t<br />

1) C 2 H 2 + H 2 ⎯⎯ O<br />

→ C 2 H 4 (1)<br />

Ni, t<br />

C 2 H 4 + H 2 ⎯⎯ O<br />

→ C 2 H 6<br />

KhÝ duy nhÊt cßn l¹i lµ C 2 H 6<br />

⇒ c¸c p/− (1) vµ (2) xy ra hoµn toµn, H 2 , C 2 H 2 ®Òu hÕt.<br />

V hçn hîp A gim lµ do VH<br />

2 p/− = 8- 5 = 3 (lit)<br />

1<br />

⇒ V<br />

C 2 H<br />

ban ®Çu = V<br />

2<br />

H<br />

= 1, 5(lit)<br />

2<br />

2<br />

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

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

0,1<strong>25</strong><br />

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

0,1<strong>25</strong><br />

110

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