Question

When a certain amount of ethylene was combusted,6226 KJ heat was evolved. If the heat of combustion of ethylene is 1411KJ, the volume of ${O_2}$(at NTP) that entered into the reaction is:
A. 296.5 ml
B. 296.5 litre
C. 6226×22.4 litres
D. 22.4 litres

Answer 1

Hint: Combustion is defined as the chemical process in which a substance rapidly reacts with oxygen and gives off heat. The substance which undergoes combustion is known as fuel and the source of oxygen is known as an oxidizer.

Complete step by step answer:
On the combustion of ethylene, there is the formation of carbon dioxide and water. The balanced chemical reaction for the combustion of ethylene is given as:
${C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O$
Since given that the heat of combustion of ethylene is 1411 KJ or we can say that:
1411 KJ heat evolved when there is the combustion of one mole of ethylene then the number of moles when the 6226 KJ of heat is evolved, given as:
Number of moles =$\dfrac{{6226}}{{1411}} = 4.41moles$
Since for combustion, we require three moles of oxygen for one mole of ethylene, then 4.41 mole of ethylene for combustion requires
$3 \times 4.41 = 13.23$moles of oxygen.
As we know the volume of one mole of any gas at NTP is 22.4 litre. Similarly, the volume of one mole of oxygen gas at NTP is 22.4 litre. Then the volume of 13.23 mole of oxygen gas is given as: $22.4 \times 13.23 = 296.35$litre.

So, the correct answer is Option B.

Note: The combustion of hydrocarbon refers to that chemical reaction where the hydrocarbon reacts with oxygen to give carbon dioxide, water, and heat. The general equation for the combustion of hydrocarbon is given as: