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

1 Answer
Sep 10, 2016

Answer:

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

Explanation:

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

The balanced equation for the combustion 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#