How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Question

28 days ago

6

n (under constant pressure and temperature) to form carbon dioxide gas and water vapor? Assume ideal gas behavior.

1 answer:

castortr0y [2.9K] · Answer 1 · 2026-03-11T09:39:46+03:00

Response: - After the reaction, the gas volume in the flask totals 156.0 L.

Solution: The equation representing the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

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:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

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:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

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

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

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:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

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.