# Volume of Oxygen Liberated from Hydrogen Peroxide

## Homework Statement

Hydrogen peroxide breaks down easily to give water and oxygen as follows:
2H2O2(aq)----> 2H2O2(l) + O2(g)
Bottles of hydrogen peroxide are sometimes labelled as 10 volume as well as 3%. This means the volume of oxygen that can be liberated is 10 times the volume of the solution. Remembering that 1 mol at STP of any gas occupies 22.4L, calculate the volume of oxygen at STP that could be produced from 1L of a 3% solution. Is 10L a good approximation to your answer?

## Homework Equations

2H2O2(aq)----> 2H2O2(l) + O2(g)
3% refers to 3g/100mL
Mole equations and density

## The Attempt at a Solution

Knowing that I have a 1L solution, I make 3g/100ml into 30g/L.
Working from there, I divide 30g H2O2 by the molar mass, 34g/mol, to find moles of H2O2.
The equation states there is a 2:1 ratio between hydrogen peroxide and oxygen, so I divide the moles by 2.
I then multiply the moles of Oxygen by 22.4 to find the volume, which I find is 9.88L (in line with the question's estimate of 10L as well).

I quite literally just thought of this possible solution when I was posting this for help. I'm uncertain as to whether or not I went about this properly, particularly in finding the grams of H2O2 in the 1L solution.

As always, appreciate any help on the matter.

Borek
Mentor
2H2O2(aq)----> 2H2O2(l) + O2(g)
Beware - you are not producing H2O2 but H2O (but I suppose it just a typo).

Knowing that I have a 1L solution, I make 3g/100ml into 30g/L.
Working from there, I divide 30g H2O2 by the molar mass, 34g/mol, to find moles of H2O2.
The equation states there is a 2:1 ratio between hydrogen peroxide and oxygen, so I divide the moles by 2.
I then multiply the moles of Oxygen by 22.4 to find the volume, which I find is 9.88L (in line with the question's estimate of 10L as well).
Looks OK.

Yes. I assume I copied it down wrong, but I left it like that just in case.
Thanks for the help!