Have a chemistry question:

It has been noted that mixing a solution of Hydrogen peroxide %3 and Vinegar %5 can be used as a cleaning solution for silencers. It removes lead build up. Apparently the lead is converted to lead acetate which is a nasty poisonous & toxic compound requiring proper disposal. Anyway, the solution indeed works but in the case of heavy leading, it seems to play out quickly. Case in point. As an experiment I put a 40gr .22 projectile in approx 50ml of the solution for 24 hours. Roughly half of the projectile was disolved. An additional 50 ml was required to complete the job.

I know there has to be some relationship to the amount of lead that can be disolved by a certain amount of solution. What I'm trying to determine is if I have a silencer with for example 2oz of lead build up, how much solution will be requred to remoe it. What if the concentration was increased etc.

Any brilliant scientific minds care to weigh in? No pun intended.

2187.5 ml

It looks like you have all of the information you need right at your fingertips.

You know it took 50mls 24 hours to dissolve 20 grains of lead.

2 ounces of lead is (I believe) 875 grains.  Or 43.75 times as much as what dissolved from your
bullet.

So you would need 2187.5 mls of your solution.  

(as out-of-ammo has already stated)

I was kinda looking for something more exact as in molar formulas. But I guess a SWAG will work too.

Its been a long time (and I'm into a few beers right now) since I've had a chemistry class, but I'd say that you need to look at the actual chemical equation of how lead reacts with hydrogen peroxide (HO) and vinegar (C2H4O2). From the equation, determine the limiting reagent. From that you can determine how much of a certian molar solution is needed to react with a certian amount of lead. Google "emprical formula," "stoichiometric calculation," and "molar weight".

Here's a start in the right direction.

ETA: From your anecdote, it looks like hydrogen peroxide or vinegar is the limiting reagent. I'd say that, yes, a higher concentration of peroxide, or vinegar, or both would be more effective. Use the right balanced chemical equation, and you could indeed find out how much solution would be needed to remove 2oz of lead.

I'm into recreational chemistry.

Jeez, just stick to Hoppes Number 9, will ya. It's tried and true. Everyone has to think they have "the edge". Junior chemists at work!