How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP?

Question

Ellie Andrews · Accepted Answer

The volume of oxygen required is #"0.50 dm"^3#.

We can apply Gay-Lussac's Law of Combining Volumes to this problem: The ratio of the volumes of the reactant gases to the products can be expressed as simple whole numbers if the temperature and pressure are both constant. The combustion's balanced equation is #color(white)(l)"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O"# #"1 dm"^3color(white)(ll)"2 dm"^3# According to Gay-Lussac, #"1 dm"^3color(white)(l) "of CH"_4# requires #"2 dm"^3color(white)(l) "of O"_2#. ∴ #"Volume of O"_2 = 0.25 color(red)(cancel(color(black)("dm"^3 color(white)(l)"CH"_4))) × ("2 dm"^3color(white)(l) "O"_2)/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "CH"_4)))) = "0.50 dm"^3color(white)(l)"O"_2#

Olivia Anton · Answer

undefined To calculate the volume of oxygen required for the complete combustion of methane at STP, you use the stoichiometry of the balanced chemical equation for the combustion reaction of methane. The balanced equation is:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)

According to the stoichiometry of this equation, 1 mole of methane reacts with 2 moles of oxygen gas. Since the volume of gases at STP is directly proportional to the number of moles, you can use the volume ratio from the balanced equation to determine the volume of oxygen required.

Given that the volume of methane is 0.25 dm^3, you can set up a proportion:

0.25 dm^3 CH4 / 1 * (2 dm^3 O2 / 1 dm^3 CH4) = 0.5 dm^3 O2

So, 0.5 dm^3 of oxygen gas is required for the complete combustion of 0.25 dm^3 of methane at STP.