# 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?

Aug 18, 2017

${\text{H"_2:"O}}_{2} = 4 : 1$

#### 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 $x$ $\text{g}$ to be the mass of hydrogen gas present in the mixture, you can say that the mass of oxygen gas will be equal to $\left(4 \cdot x\right)$ $\text{g}$ $\to$ the $1 : 4$ mass ratio comes into play here.

The molar ratio of the two gases will be equal to

$\left(x \textcolor{w h i t e}{.} {\text{g H"_2)/(4xcolor(white)(.)"g O"_2) = (x color(white)(.)color(red)(cancel(color(black)("g H"_2))) * "1 mole H"_2/(2color(red)(cancel(color(black)("g H"_2)))))/(4xcolor(red)(cancel(color(black)("g O"_2))) * "1 mole O"_2/(32color(red)(cancel(color(black)("g O"_2))))) = ((x/2) color(white)(.)"moles H"_2)/( ((4x)/32)color(white)(.)"moles O}}_{2}\right)$

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 $4 : 1$ mole ratio.