Chemical calculation based on volume – volume relationship

Chemical calculation based on volume – volume relationship

Example:

Calculate the approximate volume of air needed at STP, to oxidize 4480 litres of ammonia to nitric oxide in the Ostwald's process.

Solution:

As per the equation,

4 vols. : 5 vols.

4 x 22.4 litres : 5 x 22.4 litres

89.6 litres : 112 litres

Volume of oxygen required to oxidize 89.6 litres of ammonia = 122 litres

Volume of oxygen required to oxidize 4480 litres of ammonia =?

NH₃ : O₂

89.6 : 112

4480 : x

But oxygen forms only ^th the volume of air.

Hence volume of air needed = 5600 x 5 = 28000 litres

Volume of air = 28000 litres.

Example:

About 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. Calculate the volume of carbon dioxide formed, and the volume of oxygen left behind in excess.

Solution:

As per the equation,

2 vols. : 1 vol. 2 vol.

2 x 22400 ml : 22400 ml 2 x 22400 ml

44800 ml : 22400 ml 44800 ml

Volume of CO₂ formed by burning 2 volumes of CO = 2 volumes

Therefore, by igniting 560 ml of CO the volume of CO₂ formed = 560 ml

Volume of O₂ needed to burn 2 volumes of CO = 1 vol.

Volume of O₂ needed to burn 560 ml of CO = ?

2CO : O₂

2 vol. : 1 vol

560 ml : x

Hence volume of excess oxygen = 500 - 280 = 220 ml

Volume of carbon dioxide formed = 560 ml

Volume of excess oxygen = 220 ml.

Example:

Calculate the volume of gaseous products obtained by burning 4.8 litres of methane in requisite quantity of oxygen.

Solution:

As per the equation,

1 vol. 1 vol. + 2 vols.

1 vol. 3 vols.

Volume of gaseous products obtained by burning 1 volume of CH₄ = 3vols.

Volume of gaseous products obtained by burning 4.8 litres of CH₄ =?

Methane : (carbon dioxide + water)

1 vol. : 3 vols.

4.8 : x

Volumes of gaseous products formed = 14.4 litres.

Example:

Chlorine combines with excess ammonia to form ammonium chloride and nitrogen. Calculate the minimum volume of ammonia required to combine completely with 21 litres of chlorine.

Solution:

As per the equation,

8 vols. : 3 vols.

Volume of NH₃ required to combine with 3 volumes of Cl₂ = 8 vols

Volume of NH₃ required to combine with 21 litres of Cl₂ =?

NH₃ : Cl₂

8 : 3

x : 21

Minimum volume of ammonia required = 56 litres.