Rhodium, that was the weight of the hcl form.
WizardX's writeup gave me the idea to use less hypo. Actually what i wanted to do was to dissolve all the pseudo in dh20 so that i would know that there was enough water in the rxn to start with. Even with all that water if i didn't swirl the flask as i added the hypo it would immediately begin to react. However, since i make my hypo from sodium hypophosphite, i have no definite way of knowing how strong it is. I estimate it at 50% based on the density which is 1.27g per ml. I also wanted to see what the rxn would bee like if the hypo was added to the iodine and water because i was reading some old chemistry literature which says to add 1 part phosphorous to 10 parts iodine and 4 parts water to make hydriodic acid. Here is what that same literature says about the square crystals:
A number of complicated changes take place during the preparation of this gas, from the reaction of the different substances mixed together and part of the newly formed products.
Small cubical crystals may frequently be seen in the neck of the flask or retort employed; they consist of hydriodic acid and Phosphureted hydrogen, and are rapidly decomposed by water with effervescence, this fluid combining with the hydriodic acid. The hydriodic acid gas is produced by the iodine combining with the hydrogen of a portion of water which is decomposed, the oxygen uniting with the phosphorus.
Heres part of WizardX's writeup about having an excess of hypo:
HYPOPHOSPHOROUS TO HI CALCULATIONS
H3PO2 + H2O + I2 ==>> H3PO3 + 2HI
Hypophosphorous 50% w/w. F.W = 66 g/mol. Density = 1.274 g/ml.
100mls (0.1Lt) of Hypophosphorous 50% w/w contains :
(1.274 / 50)/100 = 0.637 g/ml H3PO2 = 0.00965 mol/ml H3PO2.
0.00965 mol/ml H3PO2 x 100 = 0.965 mol/100ml H3PO2.
OR
(0.637/66) x 1000 = 9.65 moles H3PO2 per 1000 mls. (mol/Lt)
Since we use 100mls (0.1Lt), then 9.65 x 0.1 = 0.965 mol/100ml H3PO2.
Now since the ratio of Ephedrine : HI is 1:3 = (3/1), we require 0.75 moles of HI for every 0.25 moles of ephedrine hydrochloride.
Since we have 0.965 mol of H3PO2 and 0.39 moles of I2 (99/253.8 = 0.39), then the ratio of I2 : HI is 1: 2 = (2/1); so 0.39 moles of I2 reacts with the Hypophosphorous acid to form 0.39 x 2 = 0.78 moles of HI.
Finally, the excess Hypophosphorous acid, H3PO2 is 0.965-0.39 = 0.575 moles of H3PO2 is excess. The ratio for H3PO2 : I2 is 1:1, so only 0.39 moles of H3PO2 is needed to react with 0.39 moles of I2 to form 0.78 moles of HI.
Not only do we have enough HI; 0.78 moles to reduce 0.25 moles of ephedrine hydrochloride, but a large excess of 0.575 moles of H3PO2.