Question 1
State and explain the observation made when excess ammonia gas reacts with chlorine gas.
Answer
Observation – white fumes
Ammonia is ignited in chlorine to form hydrogen chloride.
The hydrogen chloride reacts with excess ammonia to form ammonium chloride.
Ammonia is ignited in chlorine to form hydrogen chloride.
The hydrogen chloride reacts with excess ammonia to form ammonium chloride.
Question 2
Hydrogen gas reacts with ethene to form ethane. Calculate the volume of hydrogen required to convert 14g of ethene to ethane at S.T.P.
C2H2(g) → C2H6
(C=12, H=1, molar gas volume at S.T.P is 22.4 litres)
Answer
Question 3
The table below shows the relative molecular masses and boiling points of propan-l-ol and Ethanoic acid.
| Relative Molecule Mass | Boiling Point (oC) | |
| Propan -1-ol | 60 | 36 |
| Ethanolacid | 60 | 118 |
Explain why the boiling point of Ethanoic acid is higher than that of propan –l-ol and yet they have same molecular mass.
Answer
Ethanoic acid has ability to form hydrogen bonds which are stronger than those of Propan-1-ol. Hence requires more energy to break hydrogen bond in ethanoic acid
Question 4
In an experiment an equal amount of iron fillings and sulphur powder was heated in a test tube. The mixture was left to cool then dilute hydrochloric acid added to it.
a) State the observations that were made;
- In the test tube.
- Dilute hydrochloric acid was added to the mixture after cooling.
b) Write an equation for the reaction which occurred in a) (ii) above.
Answer
a)
i) - Red glow
- Black/ grey solid formed
ii) - Gas with a small of a rotten egg
- Effervescence
b) FeS(s) + 2 HCL(ag) → FeCL2(ag) + H2S(s)
- Black/ grey solid formed
ii) - Gas with a small of a rotten egg
- Effervescence
b) FeS(s) + 2 HCL(ag) → FeCL2(ag) + H2S(s)
Question 5
a) What is meant by double decomposition?
b) Starting with 1M sodium sulphate solution, describe how you would prepare dry lead II sulphate.
Answer
a) It is a reaction between two soluble salts to give one soluble and one insoluble (precipitate).
b) - Add solution of 1M sodium sulphate to an equal volume of 1M to lead (II) nitrate solution
- Filter to obtain Lead sulphate as a residue and sodium nitrote as a filtrate
- Dry the residue with filter papers to obtain the solid lead (II) sulphate.
b) - Add solution of 1M sodium sulphate to an equal volume of 1M to lead (II) nitrate solution
- Filter to obtain Lead sulphate as a residue and sodium nitrote as a filtrate
- Dry the residue with filter papers to obtain the solid lead (II) sulphate.