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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

How to calculate Limiting and Excess Reagents



The calculation of limiting reagents and excess reagents is a real-life application of the stoichiometric relationship between reactants and products.

Limiting Reagent: this is the reagent which gets consumed entirely in the chemical reaction. It determines how long a chemical reaction will last or the amount of products to be produced. In other words, the limiting reagent or reactant stops the chemical reaction.

 Excess Reagent: this is the reactant that could continue to react if the other reactant is still available for the reaction.

 Let us look at this example, and assume that all the requirements of producing a tricycle are three wheels and a wheel steering and a tricycle company has 600 wheels and 250 wheel-steerings for the production of tricycles.

Which of the materials will finish first and how many of the other materials will be left over?

      3 wheels + 1 steering = 1tricycle

The first thing to do is to find the number of tricycles that can be produced by each of the materials.

 3wheels = 1tricycle

600 wheels = 1/3 X 600 = 200 tricycles.

1 steering = 1 tricycle

250 steering = 250 tricycles.

The material that will finish first is the wheel because it produced the least amount of tricycles, so, the wheel is the limiting material.

Since 200 tricycles will be produced, the material in excess is steering.

The amount of excess wheel steering is 250 -200 = 50 steering.

Now, let us apply the same principle to chemical reactions.

Question One:

In the Haber process of ammonia, 10cm3 of nitrogen combines with 20cm3 of hydrogen to produce ammonia gas. Calculate the:

1)      Limiting reagent or reactant

2)      Volume of excess reagent or reactant

The equation for the reaction is

           N2(g) + 3H2(g) 2NH3(g)   

 

Calculating the Limiting Reactant

To find the limiting reagent, we need to calculate the volume of ammonia produced by each of the reactants.

From the equation,

1cm3 of nitrogen   = 2cm3 of ammonia

10 cm3 of nitrogen = 2/1 * 10 = 20 cm3 of ammonia

Similarly,

3 cm3 of hydrogen = 2cm3 of ammonia

20cm3 of hydrogen = 2/3 * 20 = 13.3 cm3 of ammonia

Since hydrogen produced less volume of ammonia, it is the limiting reactant or reagent.

Calculating the volume of excess reagent

To find the volume of excess reagent, we need to calculate the volume of nitrogen required to combine completely with the limiting reagent.

From the equation,

3cm3 of hydrogen = 1cm3 of nitrogen

20cm3 of hydrogen = 1/3 *20 = 6.7cm3 of nitrogen

Volume of excess nitrogen = 10 – 6.7 = 3.3cm3 

To be continued in the next class.

 

 

  

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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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