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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 />

150×<br />

15<br />

xg<br />

g<br />

100<br />

Theo phương trình (2)<br />

150× 15×<br />

56<br />

x = = 17, 26 gFe<br />

100×<br />

73<br />

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

y<br />

2FeO + C 2Fe + CO 2<br />

144g<br />

112g<br />

yg 17,26g<br />

17, 26×<br />

144<br />

= = 22,19g<br />

FeO trong hỗn hợp đầu<br />

112<br />

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

Theo PTHH: (3) CO 2 + Ca(OH) 2 CaCO 3 + H 2 O<br />

44g<br />

100g<br />

zg<br />

20g<br />

20×<br />

44<br />

z = = 8,8g<br />

100<br />

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

Theo PTHH (2)<br />

2FeO + C 2Fe + CO 2<br />

144g<br />

44g<br />

22,19g<br />

44 × 22,19<br />

= 6, 78 g<br />

144<br />

Lượng CO 2 do phản ứng CuO với C: 8,8 - 6,78 = 2,02g<br />

2CuO + C CO 2 + 2Cu<br />

160g<br />

44g<br />

? 2,02g<br />

Khối lượng<br />

2,02×<br />

160<br />

CuO = g = 7,35g<br />

44<br />

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

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

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

0,5<br />

Theå tích 8,8g CO 2<br />

44gCO 2 coù theå tích 22,4l ôû ÑKTC<br />

8,8gCO 2<br />

Theå tích CO 2 laø 4,48l<br />

2 2 , 4 × 8, 8<br />

l<br />

4 4<br />

49

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