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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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Avogadro’s number and the mole concept

 


   Matter has different types of particles, such as atoms, ions, molecules, etc. An element is made up of atoms.

A mole is used to represent the number of atoms in an element just as a crate is used to represent 30 eggs. Avogadro’s number is the number of units in a mole of any substance and it is equal to 6.02 x 1023. Therefore, one mole of any substance is represented by 6.02x1023. A diatomic element like chlorine is 6.02x1023 molecules of chlorine. One mole of tetraoxosulphate (VI) oxide and one mole of methane each represents 6.02x1023 molecules of these compounds. Similarly, sodium and calcium elements each also contain 6.02x1023 atoms.

Example Question 1

 How many atoms are there in 10g of sodium? (Na = 23, Avogadro’s number = 6.02x1023)

Since one mole is equal to 6.02x1023 and one mole is equal to relative atomic mass expressed in grams.

    1mole = 6.02x1023

    1mole = 23g

23g   = 6.02x1023

10g = 6.02x1023/23 x 10/1 = 2.62x1023

Example Question 2

Calculate the mass in grams of 3.4x1021 atoms of copper

(Cu = 63.5, Avogadro’s number = 6.02x1023)

6.02 x 1023  = 63.5g

3.4x1021 =   63.5/6.02x1023 x 3.4x1021 = 0.35g of Cu

Avogadro’s law and Molar Volume of gases

Avogadro’s law states that the equal volume of all gases at standard temperate and pressure contain the same molecules. According to this law, one mole of any gas at s.t.p. has a volume of 22.4dm23 or 22400cm3.

So, one mole of oxygen gas has 22.4dm23 at s.t.p.

One mole of hydrogen at s.t.p has 22.4dm23

Example Question 1

Calculate the volume of carbon (IV) oxide that would be liberated if 20g of sodium trioxocarbonate (IV) dissolved in dilute hydrochloric acid at S.T.P

(Na =23, Cl = 35.5, C=12, O=16, H= 1, Molar volume =22.4dm3)

Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

 

From the equation,

 106g of sodium trioxocarbonate IV = 22.4dm3 of Carbon IV oxide

20g of sodium trioxocarbonate IV   = 22.4/106 x 20  = 4.23dm3

 

Example Question 2

Calculate the volume of HCl gas that can be obtained at S.T.P from 12g of sodium chloride. (H=1,Na=23, Cl =35.5,, molar volume at S.T.P = 22.4dm3)

2NaCl(s) + H2SO4 → Na2SO4(aq) + 2HCl(g) 

Molar mass of NaCl = 58.5g/mol

From the equation,

2x 58.5 of NaCl→ 2x22.4 dm3 of HCl gas

117 of NaCl   = 44.8 dm3

12g  = 44.8/117 x 12/1   = 4.595dm3

 

 

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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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