2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
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?

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  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.

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  2. IDK

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  3. ethane, 10L

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