Question

Question #61ae6

Answer 1

`"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

`"H"_2/"O"_2 =4/1`

To double-check the logic, take `m` to be the mass of hydrogen gas present in the sample `->` the mass of oxygen gas will be `4 * 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)` `"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)` `"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`