# 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.

Related Biology and Chemistry Homework Help News on Phys.org
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!