Page 133 - Chemistry ICSE Class X

Mole Concept and Stoichiometry 119
So,
To produce 2 vol of steam, one requires 4 vol of N O.
2
To produce 150 mL of steam, one requires 4 vol × 150 mL of N O
2 vol 2
= 300 mL of N O
2

EXAMPLE 5.5. 200 mL of ethylene (C H ) is burnt in just sufficient air (containing
2
4
20% oxygen) to form carbon dioxide gas and steam. If all measurements are made
at constant pressure and 100°C, find the composition of resulting mixture.
Solution: The reaction is
100°C
C H (g) + 3O (g) o 2CO (g) + 2H O(g)
2
2
4
2
2
1 vol 3 vol 2 vol 2 vol
200 mL
From the Gay-Lussac’s law,
Volume of oxygen required = 3 vol × 200 mL = 600 mL
1 vol
Air contains 20% oxygen by volume, so
Volume of nitrogen in the mixture = 80 × 600 mL = 2400 mL
20
From the reaction stoichiometry,
Volume of carbon dioxide produced = 2 vol × 200 mL = 400 mL
1 vol
Volume of steam produced = 2 vol × 200 mL = 400 mL
1 vol
Therefore, the composition of the gaseous mixture is
Ethylene = 200 mL – 200 mL = 0 mL
Oxygen = 600 mL – 600 mL = 0 mL
Nitrogen = 2400 mL
Carbon dioxide = 400 mL
Steam = 400 mL
3
EXAMPLE 5.6. A mixture of hydrogen and chlorine occupying 36 cm of volume
3
was exploded. On shaking it with water, 4 cm of hydrogen was left behind. Find
the composition of the mixture.
Solution: The balanced chemical equation for the reaction between hydrogen and
chlorine is
H (g) + Cl (g) o 2HCl(g)
2
2
1 vol 1 vol 2 vol
Total volume of hydrogen and chlorine = 36 cm 3
Volume of hydrogen left behind = 4 cm 3
3
So, Volume of the reaction mixture reacted = 36 cm – 4 cm 3
3
= 32 cm
Hydrogen and chlorine react in the ratio of 1 : 1. So,
Volume of hydrogen used up in the reaction = 32 cm 3 = 16 cm 3
2

Volume of chlorine used up in the reaction = 32 cm 3 = 16 cm 3
2
Therefore,
Total volume of hydrogen present initially in the mixture
3
3
= 16 cm + 4 cm = 20 cm 3
Volume of chlorine present initially in the mixture
= 16 cm 3