Question

When a certain amount of ethylene was combusted, 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ. The volume of $O_2$ (at NTP) that entered into the reaction is :

• A. 296.3 mL
• B. 296.3 L
• C. $6226\times 22.4L$
• D. 22.4 L

Answer 1

The correct option is B 296.3 L

Reaction:

$C_2$$H_4$ + $3O_2$ $\to$ $3C{O}_2$ + $2H_{2}O$

From the heat of combustion data, one mole of ethylene releases 1411 KJ of heat. If 6226 KJ of heat had been released, the amount of ethylene taken would be, 4.41 moles.

From the equation, it is clear that, 1 mole of ethylene requires 3 moles of oxygen gas. Then 4.41 moles of ethylene would requires 13.23 moles of oxygen gas.

At NTP, the volume required would be, 13.23 x 22.4 = 296.3 L