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Subject: Hydriodic Acid FAQ 1.0 by Iyemtetum
Date: Sun, 05 Oct 1997 19:39:28 GMT

The Hydriodic Acid FAQ 1.0
By: Iyemtetum

This FAQ covers only one of many methods for the production of
Hydriodic acid. No information regarding its use is included.

This procedure requires the use of iodine, red phosphorus, and
distilled water.

Iodine has the chemical symbol I. It occurs in its stable form as a
diatomic molecule, I2. Iodine has a formula weight of 253.809. 1 mol
of iodine has a mass of 253.809 grams.

Red phosphorus has the chemical symbol P. In its stable form,
phosphorus occurs as four molecules bound together, P4. It has a
formula weight of 30.973. 1 mol of phosphorus has a mass of 30.973
grams.

Water has a chemical formula of H2O. Its formula weight is 18.015. 1
mol of water has a mass of 18.015 grams, or 18.015 milliliters.

Hydrogen iodide gas has a chemical formula of HI. Its formula weight
is 127.912. 1 mol of HI gas has a mass of 127.912 grams.

Phosphoric acid has a chemical formula of H3PO4 its formula weight is
99.995. 1 mol of phosphoric acid has a mass of 99.995 grams.

Hydrogen iodide gas can be formed through the reaction between iodine,
water, and red phosphorus. This is shown by the chemical equation
below.

16(H20) + 10(I2) + 2(P4) ---> 20(HI) + 4(H3PO4) +P4

16 mol of water added to a mixture of 10 mol iodine and 2 mol red
phosphorus yields 20 mol hydrogen iodide gas, 4 mol phosphoric acid,
and 1 mol red phosphorus. The extra mol of red phosphorus in the
first half of the equation is to ensure that the reaction will be
carried out. As you can see, the additional mol remains unchanged,
and can be filtered out for later use. You can NOT add 16 grams of
water to 10 grams of iodine and 2 grams of phosphorus to produce 20
grams of hydrogen iodide gas, 4 grams of phosphoric acid, and one
grams of phosphorus. In order to find the quantities of reagents
necessary to carry out this reaction, the mol’s must be converted to
their mass equivalent. For example:

16 mol H2O 18.015 grams
-------------------- * ---------------- = 288.82 milliliters H2O
1 1 mol

The mol units cancel out leaving grams.


16 mol H2O = 288.82 milliliters of water
10 mol I2 = 2538.09 grams
2 mol P4 = 61.95 grams
20 mol HI = 2558.24 grams
4 mol H3PO4 = 399.98 milliliters
1 mol P4 = 30.97 grams

This give you:
288.82 milliliters of water plus 2538.09 grams of iodine plus 61.95
grams of red phosphorus will produce 2558.24 grams of hydrogen iodide
gas plus 399.98 milliliters of phosphoric acid plus 30.97 grams of red
phosphorus.

A 57% solution of hydriodic acid can be produced by bubbling 57 grams
of HI gas through 100 milliliters of water. To ensure the required
concentration is obtained, 60 grams of HI gas for every 100
milliliters of water will provide a sufficient safety net.

The above quantities of reagents will produce enough HI gas to make
approximately 4 Liters of hydriodic acid.

How To Make 300 ml of Hydriodic Acid

Materials:

198.42 grams Iodine
4.84 grams red phosphorus
250ml Erlenmeyer flask
500ml Erlenmeyer flask filled with 300 ml distilled water
one hole stopper
two hole stopper
addition funnel filled with 22.52 ml distilled water
long stem medicine dropper
inch section of glass tubing
6 inches of rubber tubing
hot plate

Procedure:

Step 1.

In a 250 ml flask, mix 198.42 grams of iodine and 4.84 grams
of red phosphorus.

Step 2.

Prepare the two hole stopper by inserting the addition funnel
through one hole, and the 2 inch section of glass tubing in the other.
This stopper will go into the 250 ml flask containing the red
phosphorus and iodine. Make sure that both the stem of the addition
funnel and a portion of the glass tubing are exposed on the side of
the stopper that will be inside the flask.


Step 3.

To the exposed end of the glass tubing, attach the 6 inch
section of rubber tubing. At the other end of the rubber tubing insert
the medicine dropper.

Step 4.

Insert the medicine dropper into the one hole stopper and
LOOSELY insert the stopper into the 500 ml flask filled with 300 ml
water.

Step 5.

Put some goggles on and slowly open the addition funnel valve
to allow water into the 250 ml flask. A significant amount of heat
will evolve. HI gas will begin bubbling through the rubber tubing
into the water in the 500 ml flask. If you are adventuresome, VERY
VERY gently heat the 500 ml flask with the hot plate. Take
precautions to ensure that the pressure inside the flask does not
become too great. The increased pressure will increase the amount of
HI gas that can be absorbed by the water, however this does not
benefit very much because as the pressure is equalized when you remove
the stopper, any HI gas that was absorbed over 57 grams will become
gaseous once again.

Congratulations, you have succeeded in producing approximately 300 ml
of 57% hydriodic acid.

If the quantities of the chemicals are changed at all, do it
proportionally!! If they are not changed proportionally, another
reaction could occur involving the production of a phosphoric acid,
and much less HI than the method detailed. For those of you
interested in the phosphoric acid.... Maybe later.

Have fun, be careful, Don't take any of this as law, research, enjoy.

Special thanks to Phiber Optik for all of his help!
Also, Scott, Thanks for for the constructive criticism. I hope you
agree with all this!

Hasil (Melayu) 1:

Kumpulan Berita: alt.drugs.chemistry,alt.drugs,rec.drugs,rec.drugs.chemistrySubjek: asid Hydriodic Soalan Lazim 1.0 oleh IyemtetumDate: Sun, 05 Okt 1997 19:39:28 GMTAsid Hydriodic Soalan Lazim 1.0Oleh: IyemtetumFAQ ini meliputi hanya salah satu daripada banyak kaedah untuk pengeluaranAsid hydriodic. Tiada maklumat mengenai penggunaannya juga disediakan.Prosedur ini memerlukan penggunaan fosforus iodin, merah, danair suling.Iodin mempunyai simbol kimia saya. Ia berlaku dalam bentuknya yang stabil sebagai sebuahmolekul diatomic, I2. Iodin mempunyai berat formula 253.809. 1 moliodin mempunyai jisim 253.809 gram. Fosforus merah mempunyai simbol kimia P. Dalam bentuknya yang stabil,fosforus berlaku kerana molekul empat terikat bersama-sama, P4. Ia mempunyai sebuahFormula berat 30.973. 1 mol fosforus mempunyai jisim 30.973gram.Air mempunyai formula kimia H2O. Beratnya formula adalah 18.015. 1mol air mempunyai jisim gram 18.015, atau 18.015 milliliters.Gas hidrogen iodide mempunyai formula kimia HI. Beratnya formulaadalah 127.912. 1 mol gas HI mempunyai jisim 127.912 gram.Anodizing asid mempunyai formula kimia H3PO4 yang beratnya formula adalah99.995. 1 mol asid Anodizing mempunyai jisim 99.995 gram.Gas hidrogen iodide boleh dibentuk melalui tindak balas antara iodin,air, dan fosforus merah. Hal ini ditunjukkan oleh persamaan kimiadi bawah.16(H20) + 10(I2) + 2(P4)---> 20(HI) + 4(H3PO4) + P416 mol air ditambah kepada campuran 10 mol iodin dan 2 mol merahphosphorus yields 20 mol hydrogen iodide gas, 4 mol phosphoric acid,and 1 mol red phosphorus. The extra mol of red phosphorus in thefirst half of the equation is to ensure that the reaction will becarried out. As you can see, the additional mol remains unchanged,and can be filtered out for later use. You can NOT add 16 grams ofwater to 10 grams of iodine and 2 grams of phosphorus to produce 20grams of hydrogen iodide gas, 4 grams of phosphoric acid, and onegrams of phosphorus. In order to find the quantities of reagentsnecessary to carry out this reaction, the mol’s must be converted totheir mass equivalent. For example:16 mol H2O 18.015 grams-------------------- * ---------------- = 288.82 milliliters H2O 1 1 molThe mol units cancel out leaving grams.16 mol H2O = 288.82 milliliters of water10 mol I2 = 2538.09 grams2 mol P4 = 61.95 grams20 mol HI = 2558.24 grams4 mol H3PO4 = 399.98 milliliters1 mol P4 = 30.97 gramsThis give you:288.82 milliliters of water plus 2538.09 grams of iodine plus 61.95grams of red phosphorus will produce 2558.24 grams of hydrogen iodidegas plus 399.98 milliliters of phosphoric acid plus 30.97 grams of redphosphorus.A 57% solution of hydriodic acid can be produced by bubbling 57 gramsof HI gas through 100 milliliters of water. To ensure the requiredconcentration is obtained, 60 grams of HI gas for every 100milliliters of water will provide a sufficient safety net.The above quantities of reagents will produce enough HI gas to makeapproximately 4 Liters of hydriodic acid.How To Make 300 ml of Hydriodic AcidMaterials:198.42 grams Iodine4.84 grams red phosphorus250ml Erlenmeyer flask500ml Erlenmeyer flask filled with 300 ml distilled waterone hole stoppertwo hole stopperaddition funnel filled with 22.52 ml distilled waterlong stem medicine dropperinch section of glass tubing6 inches of rubber tubinghot plateProcedure:Step 1. In a 250 ml flask, mix 198.42 grams of iodine and 4.84 gramsof red phosphorus. Step 2. Prepare the two hole stopper by inserting the addition funnelthrough one hole, and the 2 inch section of glass tubing in the other.This stopper will go into the 250 ml flask containing the redphosphorus and iodine. Make sure that both the stem of the additionfunnel and a portion of the glass tubing are exposed on the side ofthe stopper that will be inside the flask.Step 3. To the exposed end of the glass tubing, attach the 6 inchsection of rubber tubing. At the other end of the rubber tubing insertthe medicine dropper. Step 4. Insert the medicine dropper into the one hole stopper andLOOSELY insert the stopper into the 500 ml flask filled with 300 mlwater. Step 5. Put some goggles on and slowly open the addition funnel valveto allow water into the 250 ml flask. A significant amount of heatwill evolve. HI gas will begin bubbling through the rubber tubinginto the water in the 500 ml flask. If you are adventuresome, VERY
VERY gently heat the 500 ml flask with the hot plate. Take
precautions to ensure that the pressure inside the flask does not
become too great. The increased pressure will increase the amount of
HI gas that can be absorbed by the water, however this does not
benefit very much because as the pressure is equalized when you remove
the stopper, any HI gas that was absorbed over 57 grams will become
gaseous once again.

Congratulations, you have succeeded in producing approximately 300 ml
of 57% hydriodic acid.

If the quantities of the chemicals are changed at all, do it
proportionally!! If they are not changed proportionally, another
reaction could occur involving the production of a phosphoric acid,
and much less HI than the method detailed. For those of you
interested in the phosphoric acid.... Maybe later.

Have fun, be careful, Don't take any of this as law, research, enjoy.

Special thanks to Phiber Optik for all of his help!
Also, Scott, Thanks for for the constructive criticism. I hope you
agree with all this!