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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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Titration of acid and base (calculation)

 Titration is the method employed in volumetric analysis. In this method, an unknown concentration of a liquid substance is titrated against another of known concentration. We are going to learn how to calculate the concentration of the substance using the question below.

A is a dilute tetraoxosulphate (VI) acid. B contains 2g of sodium hydroxide per 250 cm3 of solution. 25cm3 of B requires 23cm3 of A for complete neutralization. Calculate

i.                     Concentration of solution B in mol/dm3

ii.                   Concentration of A in mol/dm3

iii.                  The number of hydrogen ions in 1dm3 of solution A

(Na=23, H=1, S=32, O=16)

Equation of reaction

H2SO4(aq)  +  2NaOH(aq) → Na2SO4(aq)  + 2H2O(l)

Solution

i.                     Concentration of solution B in mol/dm3

250cm3   = 2g

1000cm3 = 2/250 x1000 = 8g

Molar mass of NaOH = 40g/mol

Concentration in mol/dm3   = 8/40   = 0.2mol/dm3

Or

C = n/V

Where C = concentration in mol/dm3

             .n = amount in mole

              V = volume in dm3

First convert 250cm3 to dm3 

250 /1000 = 0.25dm3

Amount of NaOH  = 2/40  =  0.05mole

C = 0.05/0.25 = 0.2mol/dm3   

ii.                   Concentration of A in mol/dm3 

 

CaVa/CbVb  = a/b

 

Ca  = ??

Va  = 23cm3

Cb  = 0.2mol/dm3

Vb= 25cm3

.a  =  1

.b = 2

 

Ca x 23/0.2 x25 = 1/2.

 

Ca = 0.2 x 25/46 = 0.109mol/dm3

 

iii.                  Number of hydrogen ions in 1 dm3  of solution A

 

H2SO4   → 2H+   + SO42-

  1          :     2

0.109    = 2x 0.109 = 0.22mole

 

 

 

 

 

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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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