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| Covering Chapters |
1. |
The Periodic Table - Periodic Properties |
8. |
Electrolysis |
2. |
Chemical Bonding |
9. |
Metallurgy |
3. |
Acids, Bases And Salts |
10. |
Hydrogen Chloride |
4. |
Practical Chemistry |
11. |
Ammonia |
5. |
Gases, Gas Laws and Mole concept |
12. |
Nitric Acid |
6. |
Empirical and Molecular Formula |
13. |
Study of Compounds: Sulphur Dioxide,Hydrogen Sulphide, Sulphuric Acid |
7. |
Percentage Composition |
14. |
Hydrocarbons |
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| Questions |
| Questions 1 (g) (i) |
[2] |
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Find the weight of ammonia required for the preparation of 66 g of ammonium sulphate, according to the following equation: |
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[Atomic weights: N = 14, H = 1, S = 32, O = 16] |
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| Questions 1 (g) (ii) |
[2] |
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4g of a metallic chloride contains 1.89 g of the metal 'X'. Calculate the empirical formula of the metallic chloride. |
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[Atomic weights: X = 64, Cl=35.5]4 g of a metallic chlo |
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| Questions 1 (g) (iii) |
[1] |
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How many moles of nitrogen dioxide are present in |
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| Questions 1 (b) (ii) |
[3] |
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From the equation: |
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| Calculate : |
(i) |
the mass of copper needed to react with 63 g of nitric acid. |
(ii) |
the volume of nitric oxide collected at the same time. |
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[Cu = 64, H = 1, O = 16, N = 14] |
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| Solutions Tips and Guidelines |
| Solution (c) |
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| 132 g of ammonium sulphate requires ammonia = 34 g |
| 66 g of ammonium sulphate requires ammonia = (34x6/132)g = 17g |
| Tips |
(i) |
First calculate the molecular weight of ammonia and ammonium sulphate |
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(ii) |
From the equation, 1 mole of ammonium sulphate is produced by 2 moles of ammonia. |
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(iii) |
Now place the values and calculate how much ammonia is required for producing 66 g of ammonium sulphate. |
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| Solution 1 (g) (ii) |
Element |
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Atomic Weight |
Atomic Weight |
Simplest Ratio |
x |
1.89 |
64 |
1.89/64=0.029 |
0.029/0.029=1 |
cl |
2.11 |
35.5 |
2.11/35.5=0.059 |
0.059/0.029=2 |
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