Chemical calculation based on mass volume relationship

Chemical calculation based on mass volume relationship

Example:

Calculate the volume of carbon dioxide formed at STP in 'ml' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16)

Solution:

As per the equation,

40 + 12 + (16 x 3) 1 mole

100g 22.4 litres

100 g 24400 ml

Volume of carbon dioxide formed from 100 g of CaCO₃ = 22400 ml

Volume of carbon dioxide from 3.125 g of CaCO₃ = ?

CaCO₃ : CO₂

100 g : 22400 ml

3.125 g : x

Volume of carbon dioxide formed = 700 ml.

Example:

Calculate the volume of ammonia formed at STP in 'ml' by treating 2.675 g of ammonium chloride with excess of calcium hydroxide (Relative Atomic Mass of N=14, O=16, H=1, Cl=35.5, Ca=40).

Solution:

As per the equation,

2 (14 + 4 + 35.5) 2 vols.(of NH₃)

107 2 x 22.4 litres

107 g 44.8 litres or 44800 ml

NH₄Cl : NH₃

107 g : 44800 ml

2.675 g : x

Volume of ammonia formed at STP = 1120 ml.

Example: 31Calculate the volume of sulphur dioxide formed at STP in 'ml', by treating 4.8 g of copper with excess of hot concentrated sulphuric acid. (Relative atomic mass of Cu=64, H=1, S=32, O=16).

Solution:

As per the equation,

64 g 1 volume (of SO₂)

64 g 22400 ml

Volume of SO₂ formed using 64 g of copper = 22400 ml

Volume of SO₂ formed using 4.8 g of copper = ?

Cu : SO₂

64 g : 22400 ml

4.8 g : x

Volume of sulphur dioxide formed at STP = 1680 ml.

Example:

Calculate the volume of water gas obtained by passing steam over 96 kg of white hot coke (Relative atomic mass of C=12, H=1, O=16).

Solution:

As per the equation,

12 g 1 vol. + 1 vol.

12 g (22.4 + 22.4) litres

12 g 44.8 litres

Volume of water gas formed from 12 g of coke = 44.8 litres

Volume of water gas formed from 96 kg of coke =?

C : Water gas

12 g : 44.8 litres

96000 g : x litres

Volume of water gas formed = 358400 litres.