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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3

The qualitative Analysis of Copper (II) chloride

  Copper (II) chloride is one of the compounds of copper which can dissolve in water. It is a good electrolyte and it is green in colour. The ions to be analyzed in this compound are copper ions and chloride ions. The methods of identifying copper ions have been discussed in our previous analysis that is in Qualitative analysis of unknown salt(copper (II) tetraoxosulphate VI). In this topic, we will explain how to identify chloride gas and chloride ions.

Chlorine is a member of group seven elements in the Periodic Table and has seven valence electrons.

How to Test for Chloride ion and Chlorine gas

To an unknown solution of the salt sample, add manganese (VI)oxide and concentrated tetraoxosulphate (VI) acid, if the gas evolved has greenish-yellow colour, with a pungent smell, it means that chlorine gas is likely to be the possible gas. To identify the gas carry out the following tests:

·       To the unknown gas given off from the unknown salt, bring a damp blue litmus paper closer to the gas, if the litmus paper turns red and bleaches, chlorine gas is likely to be present.

·       The second test is to bring a damp starch-iodide paper closer to the gas, if the paper turns dark-blue, then the gas is chlorine.

·       Another test is the reaction of chlorine gas with potassium iodide solution. Bubble the gas into the solution of potassium iodide, if the gas is chlorine the solution will turn brown.

Test for Chloride ion

To the solution of the unknown substance, add silver trioxonitrate V solution then add dilute trioxonitrate V acid, if the unknown substance contains chloride ion, it will form a white precipitate which remains insoluble when dilute trioxonitrate V acid is added, the white precipitate darken on exposure to sunlight and soluble to aqueous ammonia solution.

Note: Silver has two white insoluble salts: silver trioxocarbonate IV and silver chloride, but silver trioxocarbonate IV is soluble in dilute acid.

Example Question

X is an inorganic compound Carry out the following tests and complete the table below.

 

Test

Observation

inference

A (i)

X plus 10ml of distilled water and shake divide into four portions

Soluble to give a green-coloured solution


ii

To the first portion from (i) above, add dilute NaOH in drops and then excess.


Cu2+ likely present

iii

To the second portion add aqueous ammonia solution in drops and then in excess

Pale blue gelatinous precipitate but soluble in excess ammonia solution to give a deep blue solution.


B (i)

To the third portion add AgNO3 then dilute HNO3


Chloride ion present.

ii

To the mixture above, add aqueous ammonia solution in excess

Precipitate dissolves to form a clear solution


C (i)

To the fourth portion add MnO2 then a concentrated H2SO4 and heat if necessary


 Chlorine gas is likely present

ii

Bring damp starch-iodide paper to the gas given off

The paper  turns black-blue


 

Answer 

 

Test

Observation

inference

A (i)

X plus 10ml of distilled water and shake divide into four portions

Soluble to give a green-coloured solution

Soluble ions of transition elements are likely present (Fe2+, Cu2+, Fe3+).

ii

To the first portion from (i) above, add dilute NaOH in drops and then excess.

Pale blue precipitate and insoluble in excess NaOH solution

Cu2+ likely present

iii

To the second portion add aqueous ammonia solution in drops and then in excess

Pale blue gelatinous precipitate but soluble in excess ammonia solution.

Cu2+ confirmed

B (i)

To the third portion add AgNO3 then dilute HNO3

A white precipitate formed which darkens on exposure to sunlight and is insoluble to dilute HNO3

Chloride ion present.

ii

To the mixture above, add aqueous ammonia solution in excess

Precipitate dissolves to form a clear solution

Chloride ion confirmed

C (i)

To the fourth portion add MnO2 then a concentrated H2SO4 and heat if necessary

Greenish-yellow gas evolves with pungent which turns damp blue litmus paper red and bleaches it

 Chlorine gas is likely present

ii

Bring damp starch-iodide paper to the gas given off

The paper  turns black-blue

Chlorine gas confirmed

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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3

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