2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Question

In the equation, if 14 L of ethane (C2H6) and 14 L of oxygen (O2) combined and burned to completion, which gas will be leftover after the reaction, and what is the volume of that gas remaining?

Answer 1

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

In all gas systems volume may be used as mols and that cuts out a step or two.
14 L ethane x (4 mols CO2/2 mol C2H6) = about 28 L CO2 possibly formed
14 L O2 x (4 mols CO2/7 mols O2) = 8 L CO2 possibly formed.
Therefore, 8 L CO2 will be formed, all of the O2 will be used and some of the ethane will remain. How much will remain? That's 14 L O2 x (2 mols C2H6/7 mols O2) = 4 mols ethane used to react with the 14 L O2 which leaves 10 L ethane remaining.

Answer 2

ethane, 10L