# Question 61ae6

Jul 5, 2017

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

#### Explanation:

The idea here is that you need to use the molar masses of the two elements to convert the mass ratio to a mole ratio.

You know that you have

M_ ("M H"_2) ~~ "2 g mol"^(-1)

M_ ("M O"_2) ~~ "32 g mol"^(-1)

Right from the start, you can say that in a mixture that contains equal masses of hydrogen gas and oxygen gas, the two gases will be in a $16 : 1$ mole ratio, i.e. you will have $16$ times more moles of hydrogen gas than of oxygen gas.

Now, you know that your mixture contains hydrogen gas and oxygen gas in a $1 : 4$ mass ratio.

This means that you have $4$ times as many grams of oxygen gas than of hydrogen gas, which implies that you have $4$ times as many moles of oxygen gas than you would have if the two gases had equal masses in the mixture.

This implies that in this case, the $16 : 1$ mole ratio will be equal to $16 : 4 = 4 : 1$.

Therefore, you can say that the mole ratio that exists between the two gases is equal to

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

To double-check the logic, take $m$ to be the mass of hydrogen gas present in the sample $\to$ the mass of oxygen gas will be $4 \cdot m$.

The number of moles of hydrogen gas will be

m color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2color(red)(cancel(color(black)("g")))) = (m/2) ${\text{moles H}}_{2}$

The number of moles of oxygen gas will be

4m color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32color(red)(cancel(color(black)("g")))) = (m/8) ${\text{moles O}}_{2}$

The mole ratio that exists between the two gases will be equal to

"H"_2/"O"_2 = (color(red)(cancel(color(black)("m")))/2 color(red)(cancel(color(black)("moles"))))/(color(red)(cancel(color(black)("m")))/8 color(red)(cancel(color(black)("moles")))) = 1/2 * 8/1 = 4/1#