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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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2022 NECO Exam Chemistry Practical Qualitative Analysis Review

 


   In this qualitative analysis, we are going to review the possible qualitative analysis practical exam questions you are likely to expect in the coming NECO examination. You are expected to analyse the ions in the following compounds

·         Zinc trioxocarbonate IV (labelled Xn ) watch video 1

·         Aluminium chloride  ( labelled Yn ) watch video 2

The reagents for the qualitative analysis of these compounds include

·         Dilute sodium hydroxide solution

·         Dilute aqueous ammonia

·         Dilute hydrochloric acid

·         Distilled water

·         Red and blue litmus paper

·         Barium chloride

·         Phenolphthalein

·         Calcium hydroxide

·         Methyl orange

·         One boiling tube

·         Filtration apparatus

·         Source of  heat

s/n

Test

Observation

Inference

A(I)

X + dilute HCl, divide into two portions

Effervescence, odourless, colourless gas, turns blue litmus red, turns Ca(OH)2 milky

CO2 gas confirmed, CO32-or HCO3-  present

II

To the first portion from a(I) above  add dilute NaOH in drops then excess

White precipitate, soluble in excess

Pb2+, Al3+, Zn2+  likely present

III

To the second portion a(I) from above, add aqueous ammonia

White gelatinous precipitate, soluble in excess , gives a colourless solution

Zn2+ confirmed

B(I)

Y + distilled water, divide into four portions

Soluble, colourless solution

Transition ions absent, soluble ions like Na+, NH+, K+ likely present

II

To the first portion from b(I) above add dilute NaOH solution in drops then excess

White precipitate, soluble in excess

Zn2+, Pb2+ , Al3+ likely present

III

To the second portion from b(I) above add aqueous ammonia

White precipitate, insoluble in excess

 Pb2+, Al3+ likely present

IV

To the third from b(I) above add dilute HCl or dilute BaCl2 solution

No visible reaction

Al3+ confirmed

 

Note:

·          All trioxocarbonate IV liberate carbon IV oxide when they react with dilute acid and they also liberate carbon IV oxide when they are heated strongly.

·         Lead ions, aluminium ions and Zinc ions show similar reactions with dilute

Sodium hydroxide solution.

·         Among the three ions, Zinc is the only soluble in excess of aqueous ammonia.

·         Lead ions give a white precipitate with dilute hydrochloric acid or any chloride while aluminium ions do not. To watch the Youtube video click here, you can watch part 2.


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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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