Answer:- 10 L of ethane.
Solution:- The given balanced equation is:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)[/tex]
From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.
Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:
[tex]14LO_2(\frac{2LC_2H_6}{7LO_2})[/tex]
= [tex]4LC_2H_6[/tex]
From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L
Hence, 10 L of ethane will be remaining.
