[Stevee1967]

What would the ratios of H20, I2 and RP be to end up with an aqueous 57% solution of HI?
I know that HI is formed at each step during the creation of (i)H2P02, (ii)H2PO3 and (iii)H2PO4, I think its 2 HI's during the formation of (i), another 1 when (i) changes to (ii) and another 1 when (ii) changes to (iii).
trying to work it out is doing my head in!

[lost and found]

2 P + 3 I2 + 6 H2O --> 6 HI + 2 H3PO3

Why do you want HI? It's used to make some illegal drugs.

MB

[Stevee1967]

Really? Like what?
I guess it's useful in organic chem due to it's strong reducing properties?
Actually I'm interested in it purely for the fact that it is a very corrosive substance.
So does anyone know the actual reaction and molar ratios required?
Phosphorous chemistry always does my head in.
It's an interesting reaction given that hypophosphorous acid, phosphorous acid and phosphoric acid are all made during the generation of HI.  I am under the impression that the first stage is where  4 waters, 2P atoms and an I2 combine to create two HI molecules and two hypophosphorous acid molecules
4H20 + 2P + I2  -->  2(H3PO2) + 2(HI),
Second stage involves each of the hypophosphorous acid molecules taking on an oxygen, (from two waters)and the remaining H2's from each combining with I2's to make two HI's from each of the hypophosphorous acids which due to the extra oxygens have now become phosphorous acid
2(H3PO2) + 2(H20) +2(I2)  -->  2(H3PO3) + 4(HI)
In the last step, the previous process is repeated, turning each of the phosphorous acids into phosphoric acids, the extra O coming from water again which leaves 2H to combine with I2 for each of the phosphoric acids.
2(H3PO4) + 2(H2O) + 2(I2)  -->  2(H3PO4) + 4(HI)

The total reaction from start to finish would then be
8(H2O) + 2P + 5(I2)  -->  2(H3PO4) + 10(HI)
So 8 moles of water and 2 moles of phosphouros and 5 moles of I2 would give 10 moles of HI
8 moles H20 = 8*18= 144g
2 moles phosphorous = 2*31=62g
5 moles I2= 1270g
this should give 1280g HI
Now say I wanted
57% HI by mass, then with 1280g of HI I'd need 965g water.
If I just added that to the 144 grams required for the reaction from the start, then I'd start off with 1109 grams H20. That should leave me with 57% HI by mass after the reaction is complete right?  (disregard the presence of phosphoric acid)

Is this correct or am I way off?

[AWK]

I do not believe reaction H3PO3 + I2 + H2O = H3PO4 + .
can proceed

For stoichiometry just use this reaction only
P + I2  + H2O = H3PO3 + HI (unbalanced)

[Sev]

Really? Like what?

methamphetamine.