Page 132 - Chemistry ICSE Class X

118 ICSE Chemistry – 10
Thus from the Gay-Lussac’s law, 30 mL of methane will require 60 mL
of oxygen for complete combustion. So,
Volume of the unreacted oxygen left = 80 mL – 60 mL
= 20 mL
EXAMPLE 5.2. What volume of oxygen would be required to burn 400 mL
of ethyne (acetylene, C H ) completely? Also calculate the volume of carbon
2
2
dioxide formed.
Solution: The burning of C H in oxygen can be described by the balanced
2
2
chemical equation.
2C H (g) + 5O (g) o 4CO 2 + 2H O
2
2
2
2
2 vol 5 vol 4 vol
400 mL ? ?
From the stoichiometry, 2 units of acetylene requires 5 units of oxygen
for complete combustion and 4 units of CO are produced. So
2
Volume of oxygen required = 5 × 400 mL = 1000 mL
2
and Volume of carbon dioxide produced = 4 × 400 mL = 800 mL
2
EXAMPLE 5.3. 20 mL of each of oxygen and hydrogen and 10 mL of carbon
monoxide are exploded in an enclosure. What will be the volume and
composition of the mixture of gases when it is cooled to room temperature?

2H (g) + O (g) o 2H O(l)
2
2
2
2CO(g) + O (g) o 2CO (g)
2
2
Solution: The chemical equations describing the reactions and the stoichiometric
volume ratios are:
2H (g) + O (g) o 2H O(l)
2
2
2
2 vol 1 vol 0 vol
Initial volume : 20 mL 20 mL 0
Volume consumed: 20 mL 10 mL 0
Volume left : 0 (20 – 10) mL = 10 mL 0
2CO(g) + O (g) o 2CO (g)
2
2
Initial volume: 2 vol 1 vol 2 vol
Volume consumed
produced: 10 mL 5 mL 10 mL
So,
Total oxygen consumed in the reaction = 10 mL + 5 mL = 15 mL
Oxygen left unreacted = 20 mL – 15 mL = 5 mL
So, Carbon dioxide produced in the reaction = 10 mL
Therefore, the reaction mixture when cooled to room temperature will contain,
V (oxygen) = 5 mL
V (carbon dioxide) = 10 mL

EXAMPLE 5.4. The reaction,
4N O(g) + CH (g) o CO (g) + 2H O(g) + 4N (g)
2
2
2
4
2
takes place in the gaseous state. If all the volumes are measured at the same
temperature and pressure, calculate the volume of dinitrogen oxide (N O)
2
required to produce 150 mL of steam. (N = 14, O = 16, C = 12 and H = 1).
Solution: The stoichiometry of the reaction gives
4N O(g) + CH (g) o CO (g) + 2H O(g) + 4N (g)
2
2
4
2
2
Volume ratio: 4 vol 1 vol 1 vol 2 vol 4 vol
Volume
required/produced: ? — — 150 mL —