# A mixture of gases contains #"H"_2# and #"O"_2# gases in the ratio of #"1:4 w/w"#. What is the molar ratio of two gases in the mixture?

##### 1 Answer

#### Explanation:

All you have to do here is to convert the *mass ratio* to a **mole ratio** by using the conversion factor that takes you from grams to moles or vice versa, i.e. the **molar mass**.

You know that you have

#M_ ("M H"_ 2) ~~ "2 g mol"^(-1)#

#M_ ("M O"_ 2) ~~ "32 g mol"^(-1)#

Now, if you take **mass ratio** comes into play here.

The molar ratio of the two gases will be equal to

#(x color(white)(.)"g H"_2)/(4xcolor(white)(.)"g O"_2) = (x color(white)(.)color(red)(cancel(color(black)("g H"_2))) * "1 mole H"_2/(2color(red)(cancel(color(black)("g H"_2)))))/(4xcolor(red)(cancel(color(black)("g O"_2))) * "1 mole O"_2/(32color(red)(cancel(color(black)("g O"_2))))) = ((x/2) color(white)(.)"moles H"_2)/( ((4x)/32)color(white)(.)"moles O"_2) #

This can be simplified to

#((color(red)(cancel(color(black)(x)))/2) color(white)(.)"moles H"_2)/( ((4color(red)(cancel(color(black)(x))))/32)color(white)(.)"moles O"_2) = (1/2 * 32/4)"moles H"_2/"moles O"_2 = "4 moles H"_2/"1 mole O"_2#

Therefore, the mixture contains hydrogen gas and oxygen gas in a **mole ratio**.