20 mL of methane is completely burnt using 50 mL of oxygen. The volume of the gas left after cooling to room temperature is? A) 80 B)40 C) 60 D) 30

Start by writing a balanced chemical equation for your reaction

#"CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((l))#

Notice that I added water as a liquid, #(l)#, because the problem tells you that after the reaction is complete, the resulting gaseous mixture is cooled down to room temperature.

Now, notice that the reaction consumes #2# moles of oxygen gas for every #1# mole of methane that takes part in the reaction and produces #1# mole of carbon dioxide.

When your reaction involves gases kept under the same conditions for pressure and temperature, you can treat the mole ratios that exist between them in the balanced chemical reaction as volume ratios.

In your case, you can say that the reaction consumes #"2 mL"# of oxygen gas for every #"1 mL"# of methane that takes part in the reaction and produces #"1 mL"# of carbon dioxide.

This means that in order for the reaction to consume all the methane present in the sample, you need

#20 color(red)(cancel(color(black)("mL CH"_4))) * "2 mL O"_2/(1color(red)(cancel(color(black)("mL CH"_4)))) = "40 mL O"_2#

As you can see, you have more oxygen gas than you need to ensure that all the methane reacts #-># oxygen gas is in excess, which is equivalent to saying that methane is a limiting reagent.

So, the reaction will consume #"20 mL"# of methane and #"40 mL"# of oxygen gas and produce

#20 color(red)(cancel(color(black)("mL CH"_4))) * "1 mL CO"_2/(1color(red)(cancel(color(black)("mL CH"_4)))) = "20 mL CO"_2#

After the reaction is complete, you will be left with

#"50 mL O"_2 - "40 mL O"_2 = "10 mL O"_2#

that are not consumed by the reaction and with #"20 mL"# of carbon dioxide. Therefore, you can say that after the reaction is complete, your mixture will contain

#"10 mL O"_2 + "20 mL CO"_2 = "30 mL gas"#