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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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Stoichiometry of Chemical Reaction involving Mass and Mole Part one

   


   Stoichiometry is the study of the relationships which exist between reactants and products in a chemical reaction. This relationship can easily be seen in a balanced chemical equation. For example,

  K2SO4 ------ 2K+ + SO42-

      1             2           1

In the ionization reaction above, notice that one-mole potassium tetraoxosulphate (VI) gives two moles of potassium ions and one mole of tetraoxosulphate (VI) ion in solution. This ionization equation also shows the mole ratio of the reactants and the products, which is known as the stoichiometry of the chemical reaction. It is necessary to note that the mole of a substance is not affected if the reactants or products are ions, atoms or molecules etc. in a chemical reaction.

CaCO3(s) + 2HCl(aq)  ------ CaCl2(s)  + H2O(l) + CO(g) 

1mole          2mole             1mole      1 mole     1mole

Consider the chemical reaction above, and note the relationship between the reactants and the products in moles. The reacting masses of the above reaction can be calculated, since, the molar mass of a substance is the mass of one mole of the substance expressed in grams.

Mole = reacting mass/molar mass

Then reacting mass = mole x molar mass

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Example question 1

Calculate the mass of lead chloride that can be obtained from 30g of lead trioxonitrate (V) in the presence of excess hydrochloric acid. (Pb=207, Cl =35.5,N= 14, O=16, H=1)

  Pb(NO3)2(s) +  2HCl(aq)  -------- PbCl2(s)  + HNO3(aq)

Solution

Pb(NO3)2(s) +  2HCl(aq)  -------- PbCl2(s)  + HNO3(aq)

 1mole                                      1mole

 331g/mol                               278g/mol

From the equation

The reacting mass of lead trioxonitrate (V) = 1mole x 331g/mol = 331g

The mass of lead chloride that would produce = 1mole x 278g/mol = 278g

331g of Pb(NO3)2  = 278g of PbCl2

30g of Pb(NO3)2  = 278÷331 x30  = 25.2g of PbCl2

Or

By considering the number of moles involved in the reaction

  1mole of Pb(NO3)2   =   1mole of PbCl2 

Then find the number of moles in 30g of Pb(NO3)2

Mole = reacting mass/molar mass

Reacting mass given = 30g

Molar mass = 331g/mol

Mole = 30/331 = 0.09063mol

Since 1mole of Pb(NO3)2   =  1mole of PbCl2

then

0.09063mole of Pb(NO3)2    = 0.09063mole of PbCl2

Mass in 0.09063mole of PbCl2  = 0.09063  x 278  = 25.2g

Example question 2

What mass of silver trioxonitrate (V) would be required to yield 14g of silver chloride in addition of excess sodium chloride? (Na=23, Cl = 35.5,Ag=108, N=14,O=16)

 NaCl(aq) + AgNO3(aq) --------- NaNO3(aq)  + AgCl(s)

Solution

NaCl(aq) + AgNO3(aq) --------- NaNO3(aq)  + AgCl(s)

1mole                                          1mole

170g/mol                                    143.5g/mol

From the equation

Reacting mass of AgNO3 = 1mole x 170g/mol = 170g

Mass of AgCl produced = 1mole x 143.5g/mol = 143.5g

143.5g of AgCl required = 170g of AgNO3

14g of AgCl will require = 170/143.5 x 14/1 = 16.6g

Or

By considering the number of moles involved in the reaction

NaCl(aq) + AgNO3(aq) --------- NaNO3(aq)  + AgCl(s)

                  1mole                                           1mole

From the equation

1mole of AgCl    =  1 mole of AgNO3 

Number of moles in 14g of AgCl  = 

Reacting mass given = 14g

Molar mass     = 143.5g/mol

Mole = 14/143.5  =0.0975mole

Since

1mole of AgCl = 1mole of AgNO3 

then

0.0975mole of AgCl = 0.0975mole of AgNO3

So, mass of AgNO3   in 0.0975 = 0.0975 x 170 =16.6g

Example question 3

Calculate the mass of magnesium chloride that will be formed, when 20g of magnesium is added to excess dilute hydrochloric acid. ( Mg = 24, Cl=35.5, H=1)

 Mg(s) + 2HCl(aq) ----------- MgCl2(aq) + H2(g)

Solution

 Mg(s) + 2HCl(aq) ----------- MgCl2(aq) + H2(g)

  1mole                              1mole

  24g/mol                           95g/mol

From the equation

Reacting mass of Mg = 24g/mol x 1mole =24g

Mass of MgCl2 produced = 95g/mol x 1mole= 95g

If 24g of Mg = 95g of MgCl2

Then

20g of Mg = 95/24 x 20/1 = 79.2 g of MgCl2 

Or

By considering the number of moles involved in the reaction

1mole of Mg  = 1mole of MgCl2

Reacting mass given = 20g

Molar mass of Mg = 24g/mol

Number of moles in 20g of Mg = 20/24  = 0.8333mole

Since

1mole of Mg = 1mole of MgCl2

Then

0.8333mole of MgCl2  =  0.8333 mole of Mg

Mass of MgCl2 in 0.8333mole  =   0.8333 x 95  =79.2g of MgCl2

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Application of Gay-Lussac’s Law of Combining Volumes

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