Response: - After the reaction, the gas volume in the flask totals 156.0 L.
Solution: The equation representing the combustion of ethane is:

The balanced equation indicates that ethane and oxygen react in a 2:7 mol ratio, or 2:7 vol ratio assuming ideal conditions.
To determine the limiting reactant, we can first calculate how much oxygen is necessary for 36.0 L of ethane:

= 126 L 
To completely react with 36.0 L of ethane, 126 L of oxygen is needed, but only 105.0 L is available—therefore, oxygen is the limiting reactant.
Next, we compute the volumes of products formed from 105.0 L of oxygen:

= 60.0 L 
Now, let's calculate the volume of water vapor produced:

= 90.0 L 
Since there is excess ethane, some will still be present in the flask.
First, we need to find out how many liters reacted with 105.0 L of oxygen and subtract that from the initial volume of ethane:

= 30.0 L 
The remaining ethane volume = 36.0 L - 30.0 L = 6.0 L
Thus, the total gas volume in the flask post-reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Therefore, the final answer is 156.0 L.