Originally Posted by z987k
hmmm, anyone want to figure out how much H2O2 yields enough O2 for properly oxygenated wort?
OK. I'll bite.
Assumptions: 2H2O2 --> 2H2O + O2
Neglecting everything in the wort that isn't water, 5 gallons is roughly 19 liters.
8 PPM target value for oxygenation.
All oxygen released by H2O2 goes into solution (this probably isn't the case).
19 l water * 56 (mol/l) = 1064 moles of water.
1064 mol water * ( 8 mol O2 / 1,000,000 mol water) = 8.5E-3 moles O2 needed.
Since 2 mol H2O2 yields 1 mol O2, we'd need 17.02E-3 moles of H2O2
With a molecular weight of 34, you'd need 34 * 71.02E-3 = 0.578 grams, or about a half a gram of H2O2. Using a 3% solution from the drugstore would mean you'd need around 19.3 grams dilute H2O2. Doesn't seem like much, but that assumes 100% efficiency at getting the released O2 into solution.
3% H2O2 has a specific gravity similar to water, so we're talking about 20ml of dilute hydrogen peroxide.
Anybody want to give it a shot?