Answer:

Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ

2. Mass of O₂

The volume of a sample of ammonia gas : 5.152 L
<h3>Further explanation</h3>
Given
0.23 moles of ammonia
Required
The volume of a sample
Solution
Assumed on STP
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 0.23 moles :
= 0.23 x 22.4 L
= 5.152 L