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
There are two samples of
compounds to be examined in this coming examination: zinc oxide and glucose. Glucose
is tagged Cn and zinc
oxide is tagged Dn.
Glucose is an example of a monosaccharide. It is a strong reducing agent
because of the presence of the –CHO group. It is dehydrated to carbon by
concentrated tetraoxosulphate (VI) acid like all other carbohydrates.
Zinc oxide is a white insoluble powder which turns yellow when it is hot.
As an amphoteric oxide, it can dissolve in both acids and alkalis.
Analysis of Glucose
The reducing property of glucose
can be identified by the following experiments:
·
Its reaction with acidified potassium tetraoxomanganate
(VII) solution. Glucose decolorizes acidified potassium tetraoxomanganate (VII)
solution from purple to colourless.
·
Its reaction with acidified potassium
heptaoxodichromate (VI) solution. Glucose turns acidified potassium heptaoxodichromate
(VI) solution from orange to green.
·
It also reduces ammoniacal silver trioxonitrate (V)
(Tollen’s reagent) to metallic silver. This is one of the confirmatory tests
for glucose
·
Glucose forms brick red precipitate with Fehling’s
solution which is another confirmatory test for glucose
·
With Benedict’s solution, glucose forms a yellow
precipitate. This test can also be used to confirm the presence of glucose
Analysis of zinc oxide
The common reagents for the test
are sodium hydroxide and aqueous ammonia
Test with sodium hydroxide
Add a few drops of dilute sodium
hydroxide to a solution of the test sample. The presence of zinc ions can be
identified by the formation of a white gelatinous precipitate which is soluble
in excess sodium hydroxide. However, lead ions and aluminium ions give similar results.
Test with Aqueous ammonia
Add some drops of aqueous ammonia
to a solution of the unknown sample. The formation of a white gelatinous precipitate
which is soluble in excess aqueous ammonia confirms the presence of Zinc ions. Lead
ions and aluminium ions form a precipitate which is insoluble in excess.
Example Question
C is a sample of an organic compound and D is an inorganic compound. Carry out the tests below, and record
your observation and inference.
|
s/n |
Test |
Observation |
Inference |
|
1 (a) |
C +
distilled water. Divide into two portions |
Soluble and gives a clear solution |
|
|
(b) |
To the first portion from 1(a) above, add
acidified K2Cr2O7 solution. |
|
C is a
reducing agent. Reducing sugar (glucose) is likely present |
|
(c) |
A brick |
A brick-red precipitate is formed |
|
|
2 (I) |
D +
distilled water |
|
Insoluble ions of Ca, Al, Pb, and Zn are likely to present |
|
(II) |
D + dilute HCl, and divide into two
portions |
Soluble |
|
|
(III) |
To the first portion from 2(II) above, add dilute
NaOH solution in drops then in excess |
|
Ions like Pb, Al or Zn are likely to present. |
|
(IV) |
To the second portion from 2(II) above, add aqueous
ammonia |
The white gelatinous precipitate, soluble in
excess |
|
Answers
|
s/n |
Test |
Observation |
Inference |
|
1 (a) |
C +
distilled water. Divide into two portions |
Soluble and gives a clear solution |
Soluble organic compounds like reducing sugar
(glucose), organic acids likely present |
|
(b) |
To the first portion from 1(a) above, add
acidified K2Cr2O7 solution. |
It turns from orange to green |
C is a
reducing agent. Reducing sugar (glucose) is likely present |
|
(c) |
To the second portion from 1(a) above, add Fehling’s
solution |
A brick-red precipitate is formed |
Reducing sugar (glucose) confirmed |
|
2 (I) |
D +
distilled water |
White insoluble compound |
Insoluble ions of Ca, Al, Pb, and Zn are likely to
present |
|
(II) |
D + dilute HCl, and divide into two
portions |
soluble |
Soluble ions of Ca, Al, and Zn are likely to present |
|
(II) |
To the first portion from 2(II) above, add dilute
NaOH solution in drops then in excess |
The white gelatinous precipitate, soluble in
excess NaOH |
Ions like Pb, Al or Zn are likely to present. |
|
(IV) |
To the second portion from 2(II) above, add
aqueous ammonia in drops then excess |
The white gelatinous precipitate, soluble in
excess aqueous ammonia |
Zinc ion confirmed |
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