Hi All,
I'm a little confused by a paper I'm trying to replicate. They say they use "enough sodium percarbonate to produce .025% H202 in water". I assume they mean weight/volume. The confusing part to me is when buying sodium percarbonate, it's specified as "avail. H2O2 20-30 %". I'm not sure how that comes into play or where that number comes from.
This is how I proceeded, if I'm off in my methods, could you please let me know?
I am making 100 mL and want .025% H202.
The mass I need is thus .00025 g/mL * 100 mL = .025 g H202.
Converting that to moles, .025g/34.0147 g/mol = 1.041 x10
-3 mol H202
The reaction for producing H202 from sodium percarbonate is:
2Na
2CO
3ยท3H
2O
2 2Na
2CO
3 + 3H
2O
2Therefore I need 2/3 *1.041 * x 10
-3 mol sodium percarbonate or 6.94 x 10
-4 mol sodium percarbonate
That's .095 g of sodium percarbonate, or 95 mg.
I was going to say since worst case scenario is that only 20% of the sodium percarbonate is available for production of H202, that I then just divide by 0.2, yielding 475 mg. Does this seem right or am I completely missing something?
Thanks